The Poynting vector for the matter-wave

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

Original post:

In my various posts on the wavefunction – which I summarized in my e-book – I wrote at the length on the structural similarities between the matter-wave and the electromagnetic wave. Look at the following images once more:

Animation 5d_euler_f

Both are the same, and then they are not. The illustration on the right-hand side is a regular quantum-mechanical wavefunction, i.e. an amplitude wavefunction: the x-axis represents time, so we are looking at the wavefunction at some particular point in space. [Of course, we could just switch the dimensions and it would all look the same.] The illustration on the left-hand side looks similar, but it is not an amplitude wavefunction. The animation shows how the electric field vector (E) of an electromagnetic wave travels through space. Its shape is the same. So it is the same function. Is it also the same reality?

Yes and no. The two energy propagation mechanisms are structurally similar. The key difference is that, in electromagnetics, we get two waves for the price of one. Indeed, the animation above does not show the accompanying magnetic field vector (B), which is equally essential. But, for the rest, Schrödinger’s equation and Maxwell’s equation model a similar energy propagation mechanism, as shown below.

They have to, as the force laws are similar too:

Coulomb Law

gravitation law

The only difference is that mass comes in one color only, so to speak: it’s always positive. In contrast, electric charge comes in two colors: positive and negative. You can now guess what comes next: quantum chromodynamics, but I won’t write about that here, because I haven’t studied that yet. I won’t repeat what I wrote elsewhere, but I want to make good on one promise, and that is to develop the idea of the Poynting vector for the matter-wave. So let’s do that now. Let me first remind you of the basic ideas, however.

Basics

The animation below shows the two components of the archetypal wavefunction, i.e. the sine and cosine:

circle_cos_sin

Think of the two oscillations as (each) packing half of the total energy of a particle (like an electron or a photon, for example). Look at how the sine and cosine mutually feed into each other: the sine reaches zero as the cosine reaches plus or minus one, and vice versa. Look at how the moving dot accelerates as it goes to the center point of the axis, and how it decelerates when reaching the end points, so as to switch direction. The two functions are exactly the same function, but for a phase difference of 90 degrees, i.e. a right angle. Now, I love engines, and so it makes me think of a V-2 engine with the pistons at a 90-degree angle. Look at the illustration below. If there is no friction, we have a perpetual motion machine: it would store energy in its moving parts, while not requiring any external energy to keep it going.

two-timer-576-px-photo-369911-s-original

If it is easier for you, you can replace each piston by a physical spring, as I did below. However, I should learn how to make animations myself, because the image below does not capture the phase difference. Hence, it does not show how the real and imaginary part of the wavefunction mutually feed into each other, which is (one of the reasons) why I like the V-2 image much better. 🙂

summary 2

The point to note is: all of the illustrations above are true representations – whatever that means – of (idealized) stationary particles, and both for matter (fermions) as well as for force-carrying particles (bosons). Let me give you an example. The (rest) energy of an electron is tiny: about 8.2×10⁻¹⁴ joule. Note the minus 14 exponent: that’s an unimaginably small amount. It sounds better when using the more commonly used electronvolt scale for the energy of elementary particles: 0.511 MeV. Despite its tiny mass (or energy, I should say, but then mass and energy are directly proportional to each other: the proportionality coefficient is given by the E = m·c² formula), the frequency of the matter-wave of the electron is of the order of 1×10²⁰ = 100,000,000,000,000,000,000 cycles per second. That’s an unimaginably large number and – as I will show when we get there – that’s not because the second is a huge unit at the atomic or sub-atomic scale.

We may refer to this as the natural frequency of the electron. Higher rest masses increase the frequency and, hence, give the wavefunction an even higher density in spacetime. Let me summarize things in a very simple way:

The (total) energy that is stored in an oscillating spring is the sum of the kinetic and potential energy (T and U) and is given by the following formula: E = T + U = a₀²·m·ω₀²/2. The a₀factor is the maximum amplitude – which depends on the initial conditions, i.e. the initial pull or push. The ω₀in the formula is the natural frequency of our spring, which is a function of the stiffness of the spring (k) and the mass on the spring (m): ω₀² = k/m.
Hence, the total energy that’s stored in two springs is equal to a₀²·m·ω₀².
The similarity between the E = a₀²·m·ω₀² and the E = m·c² formula is much more than just striking. It is fundamental: the two oscillating components of the wavefunction each store half of the total energy of our particle.
To emphasize the point: ω₀ = √(k/m) is, obviously, a characteristic of the system. Likewise, c = √(E/m) is just the same: a property of spacetime.

Of course, the key question is: what is that is oscillating here? In our V-2 engine, we have the moving parts. Now what exactly is moving when it comes to the wavefunction? The easy answer is: it’s the same thing. The V-2 engine, or our springs, store energy because of the moving parts. Hence, energy is equivalent only to mass that moves, and the frequency of the oscillation obviously matters, as evidenced by the E = a₀²·m·ω₀²/2 formula for the energy in a oscillating spring. Mass. Energy is moving mass. To be precise, it’s oscillating mass. Think of it: mass and energy are equivalent, but they are not the same. That’s why the dimension of the c² factor in Einstein’s famous E = m·c² formula matters. The equivalent energy of a 1 kg object is approximately 9×10¹⁶ joule. To be precise, it is the following monstrous number:

89,875,517,873,681,764 kg·m²/s²

Note its dimension: the joule is the product of the mass unit and the square of the velocity unit. So that, then, is, perhaps, the true meaning of Einstein’s famous formula: energy is not just equivalent to mass. It’s equivalent to mass that’s moving. In this case, an oscillating mass. But we should explore the question much more rigorously, which is what I do in the next section. Let me warn you: it is not an easy matter and, even if you are able to work your way through all of the other material below in order to understand the answer, I cannot promise you that the answer will satisfy you entirely. However, it will surely help you to phrase the question.

The Poynting vector for the matter-wave

For the photon, we have the electric and magnetic field vectors E and B. The boldface highlights the fact that these are vectors indeed: they have a direction as well as a magnitude. Their magnitude has a physical dimension. The dimension of E is straightforward: the electric field strength (E) is a quantity expressed in newton per coulomb (N/C), i.e. force per unit charge. This follows straight from the F = q·E force relation.

The dimension of B is much less obvious: the magnetic field strength (B) is measured in (N/C)/(m/s) = (N/C)·(s/m). That’s what comes out of the F = q·v×B force relation. Just to make sure you understand: v×B is a vector cross product, and yields another vector, which is given by the following formula:

a×b = |a×b|·n = |a|·|b|·sinφ·n

The φ in this formula is the angle between a and b (in the plane containing them) and, hence, is always some angle between 0 and π. The n is the unit vector that is perpendicular to the plane containing a and b in the direction given by the right-hand rule. The animation below shows it works for some rather special angles:

We may also need the vector dot product, so let me quickly give you that formula too. The vector dot product yields a scalar given by the following formula:

a•b = |a|·|b|·cosφ

Let’s get back to the F = q·v×B relation. A dimensional analysis shows that the dimension of B must involve the reciprocal of the velocity dimension in order to ensure the dimensions come out alright:

[F]= [q·v×B] = [q]·[v]·[B] = C·(m/s)·(N/C)·(s/m) = N

We can derive the same result in a different way. First, note that the magnitude of B will always be equal to E/c (except when none of the charges is moving, so B is zero), which implies the same:

[B] = [E/c] = [E]/[c] = (N/C)/(m/s) = (N/C)·(s/m)

Finally, the Maxwell equation we used to derive the wavefunction of the photon was ∂E/∂t = c²∇×B, which also tells us the physical dimension of B must involve that s/m factor. Otherwise, the dimensional analysis would not work out:

[∂E/∂t] = (N/C)/s = N/(C·s)
[c²∇×B] = [c²]·[∇×B] = (m²/s²)·[(N/C)·(s/m)]/m= N/(C·s)

This analysis involves the curl operator ∇×, which is a rather special vector operator. It gives us the (infinitesimal) rotation of a three-dimensional vector field. You should look it up so you understand what we’re doing here.

Now, when deriving the wavefunction for the photon, we gave you a purely geometric formula for B:

B = e_x×E = i·E

Now I am going to ask you to be extremely flexible: wouldn’t you agree that the B = E/c and the B = e_x×E = i·E formulas, jointly, only make sense if we’d assign the s/m dimension to e_x and/or to i? I know you’ll think that’s nonsense because you’ve learned to think of the e_x× and/or i· operation as a rotation only. What I am saying here is that it also transforms the physical dimension of the vector on which we do the operation: it multiplies it with the reciprocal of the velocity dimension. Don’t think too much about it, because I’ll do yet another hat trick. We can think of the real and imaginary part of the wavefunction as being geometrically equivalent to the E and B vector. Just compare the illustrations below:

e-and-b Rising_circular

Of course, you are smart, and you’ll note the phase difference between the sine and the cosine (illustrated below). So what should we do with that? Not sure. Let’s hold our breath for the moment.

Let’s first think about what dimension we could possible assign to the real part of the wavefunction. We said this oscillation stores half of the energy of the elementary particle that is being described by the wavefunction. How does that storage work for the E vector? As I explained in my post on the topic, the Poynting vector describes the energy flow in a varying electromagnetic field. It’s a bit of a convoluted story (which I won’t repeat here), but the upshot is that the energy density is given by the following formula:

Its shape should not surprise you. The formula is quite intuitive really, even if its derivation is not. The formula represents the one thing that everyone knows about a wave, electromagnetic or not: the energy in it is proportional to the square of its amplitude, and so that’s E•E = E² and B•B = B². You should also note he c²factor that comes with the B•B product. It does two things here:

As a physical constant, with some dimension of its own, it ensures that the dimensions on both sides of the equation come out alright.
The magnitude of B is 1/c of that of E, so cB = E, and so that explains the extra c² factor in the second term: we do get two waves for the price of one here and, therefore, twice the energy.

Speaking of dimensions, let’s quickly do the dimensional analysis:

E is measured in newton per coulomb, so [E•E] = [E²] = N²/C².
B is measured in (N/C)/(m/s), so we get [B•B] = [B²] = (N²/C²)·(s²/m²). However, the dimension of our c² factor is (m²/s²) and so we’re left with N²/C². That’s nice, because we need to add stuff that’s expressed in the same units.
The ε₀ is that ubiquitous physical constant in electromagnetic theory: the electric constant, aka as the vacuum permittivity. Besides ensuring proportionality, it also ‘fixes’ our units, and so we should trust it to do the same thing here, and it does: [ε₀] = C²/(N·m²), so if we multiply that with N²/C², we find that u is expressed in N/m².

Why is N/m² an energy density? The correct answer to that question involves a rather complicated analysis, but there is an easier way to think about it: just multiply N/m²with m/m, and then its dimension becomes N·m/m³= J/m³, so that’s joule per cubic meter. That looks more like an energy density dimension, doesn’t it? But it’s actually the same thing. In any case, I need to move on.

We talked about the Poynting vector, and said it represents an energy flow. So how does that work? It is also quite intuitive, as its formula really speaks for itself. Let me write it down:

Just look at it: u is the energy density, so that’s the amount of energy per unit volume at a given point, and so whatever flows out of that point must represent its time rate of change. As for the –∇•S expression… Well… The ∇• operator is the divergence, and so it give us the magnitude of a (vector) field’s source or sink at a given point. If C is a vector field (any vector field, really), then ∇•C is a scalar, and if it’s positive in a region, then that region is a source. Conversely, if it’s negative, then it’s a sink. To be precise, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. So, in this case, it gives us the volume density of the flux of S. If you’re somewhat familiar with electromagnetic theory, then you will immediately note that the formula has exactly the same shape as the ∇•j = −∂ρ/∂t formula, which represents a flow of electric charge.

But I need to get on with my own story here. In order to not create confusion, I will denote the total energy by U, rather than E, because we will continue to use E for the magnitude of the electric field. We said the real and the imaginary component of the wavefunction were like the E and B vector, but what’s their dimension? It must involve force, but it should obviously not involve any electric charge. So what are our options here? You know the electric force law (i.e. Coulomb’s Law) and the gravitational force law are structurally similar:

Coulomb Law

gravitation law

So what if we would just guess that the dimension of the real and imaginary component of our wavefunction should involve a newton per kg factor (N/kg), so that’s force per mass unit rather than force per unit charge? But… Hey! Wait a minute! Newton’s force law defines the newton in terms of mass and acceleration, so we can do a substitution here: 1 N = 1 kg·m/s² ⇔ 1 kg = 1 N·s²/m. Hence, our N/kg dimension becomes:

N/kg = N/(N·s²/m)= m/s²

What is this: m/s²? Is that the dimension of the a·cosθ term in the a·e^−i·θ= a·cosθ − i·a·sinθ wavefunction? I hear you. This is getting quite crazy, but let’s see where it leads us. To calculate the equivalent energy density, we’d then need an equivalent for the ε₀ factor, which – replacing the C by kg in the [ε₀] = C²/(N·m²) expression – would be equal to kg²/(N·m²). Because we know what we want (energy is defined using the force unit, not the mass unit), we’ll want to substitute the kg unit once again, so – temporarily using the μ₀ symbol for the equivalent of that ε₀ constant – we get:

[μ₀] = [N·s²/m]²/(N·m²) = N·s⁴/m⁴

Hence, the dimension of the equivalent of that ε₀·E² term becomes:

[(μ₀/2)]·[cosθ]² = (N·s⁴/m⁴)·m²/s⁴= N/m²

Bingo! How does it work for the other component? The other component has the imaginary unit (i) in front. If we continue to pursue our comparison with the E and B vectors, we should assign an extra s/m dimension because of the e_x and/or i factor, so the physical dimension of the i·sinθ term would be (m/s²)·(s/m) = s. What? Just the second? Relax. That second term in the energy density formula has the c² factor, so it all works out:

[(μ₀/2)]·[c²]·[i·sinθ]² = [(μ₀/2)]·[c²]·[i]²·[sinθ]² (N·s⁴/m⁴)·(m²/s²)·(s²/m²)·m²/s⁴= N/m²

As weird as it is, it all works out. We can calculate u and, hence, we can now also calculate the equivalent Poynting vector (S). However, I will let you think about that as an exercise. 🙂 Just note the grand conclusions:

The physical dimension of the argument of the wavefunction is physical action (newton·meter·second) and Planck’s quantum of action is the scaling factor.
The physical dimension of both the real and imaginary component of the elementary wavefunction is newton per kg (N/kg). This allows us to analyze the wavefunction as an energy propagation mechanism that is structurally similar to Maxwell’s equations, which represent the energy propagation mechanism when electromagnetic energy is involved.

As such, all we presented so far was a deep exploration of the mathematical equivalence between the gravitational and electromagnetic force laws:

Coulomb Law

gravitation law

Despite our grand conclusions, you should note we have not answered the most fundamental question of all. What is mass? What is electric charge? We have all these relations and equations, but are we any wiser, really? The answer to that question probably lies in general relativity: mass is that what curves spacetime. Likewise, we may look at electric charge as causing a very special type of spacetime curvature. However, even such answer – which would involve a much more complicated mathematical analysis – may not satisfy you. In any case, I will let you digest this post. I hope you enjoyed it as much as I enjoyed writing it. 🙂

Post scriptum: Of all of the weird stuff I presented here, I think the dimensional analyses were the most interesting. Think of the N/kg = N/(N·s²/m)= m/s²identity, for example. The m/s² dimension is the dimension of physical acceleration (or deceleration): the rate of change of the velocity of an object. The identity comes straight out of Newton’s force law:

F = m·a ⇔ F/m = a

Now look, once again, at the animation, and remember the formula for the argument of the wavefunction: θ = E₀∙t’. The energy of the particle that is being described is the (angular) frequency of the real and imaginary components of the wavefunction.

circle_cos_sin

The relation between (1) the (angular) frequency of a harmonic oscillator (which is what the sine and cosine represent here) and (2) the acceleration along the axis is given by the following equation:

a(x) = −ω₀²·x

I’ll let you think about what that means. I know you will struggle with it – because I did – and, hence, let me give you the following hint:

The energy of an ordinary string wave, like a guitar string oscillating in one dimension only, will be proportional to the square of the frequency.
However, for two-dimensional waves – such as an electromagnetic wave – we find that the energy is directly proportional to the frequency. Think of Einstein’s E = h·f = ħ·ω relation, for example. There is no squaring here!

It is a strange observation. Those two-dimensional waves – the matter-wave, or the electromagnetic wave – give us two waves for the price of one, each carrying half of the total energy but, as a result, we no longer have that square function. Think about it. Solving the mystery will make you feel like you’ve squared the circle, which – as you know – is impossible. 🙂

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Quantum Mechanics: The Other Introduction

About three weeks ago, I brought my most substantial posts together in one document: it’s the Deep Blue page of this site. I also published it on Amazon/Kindle. It’s nice. It crowns many years of self-study, and many nights of short and bad sleep – as I was mulling over yet another paradox haunting me in my dreams. It’s been an extraordinary climb but, frankly, the view from the top is magnificent. 🙂

The offer is there: anyone who is willing to go through it and offer constructive and/or substantial comments will be included in the book’s acknowledgements section when I go for a second edition (which it needs, I think). First person to be acknowledged here is my wife though, Maria Elena Barron, as she has given me the spacetime and, more importantly, the freedom to take this bull by its horns.Below I just copy the foreword, just to give you a taste of it. 🙂

Foreword

Another introduction to quantum mechanics? Yep. I am not hoping to sell many copies, but I do hope my unusual background—I graduated as an economist, not as a physicist—will encourage you to take on the challenge and grind through this.

I’ve always wanted to thoroughly understand, rather than just vaguely know, those quintessential equations: the Lorentz transformations, the wavefunction and, above all, Schrödinger’s wave equation. In my bookcase, I’ve always had what is probably the most famous physics course in the history of physics: Richard Feynman’s Lectures on Physics, which have been used for decades, not only at Caltech but at many of the best universities in the world. Plus a few dozen other books. Popular books—which I now regret I ever read, because they were an utter waste of time: the language of physics is math and, hence, one should read physics in math—not in any other language.

But Feynman’s Lectures on Physics—three volumes of about fifty chapters each—are not easy to read. However, the experimental verification of the existence of the Higgs particle in CERN’s LHC accelerator a couple of years ago, and the award of the Nobel prize to the scientists who had predicted its existence (including Peter Higgs and François Englert), convinced me it was about time I take the bull by its horns. While, I consider myself to be of average intelligence only, I do feel there’s value in the ideal of the ‘Renaissance man’ and, hence, I think stuff like this is something we all should try to understand—somehow. So I started to read, and I also started a blog (www.readingfeynman.org) to externalize my frustration as I tried to cope with the difficulties involved. The site attracted hundreds of visitors every week and, hence, it encouraged me to publish this booklet.

So what is it about? What makes it special? In essence, it is a common-sense introduction to the key concepts in quantum physics. However, while common-sense, it does not shy away from the math, which is complicated, but not impossible. So this little book is surely not a Guide to the Universe for Dummies. I do hope it will guide some Not-So-Dummies. It basically recycles what I consider to be my more interesting posts, but combines them in a comprehensive structure.

It is a bit of a philosophical analysis of quantum mechanics as well, as I will – hopefully – do a better job than others in distinguishing the mathematical concepts from what they are supposed to describe, i.e. physical reality.

Last but not least, it does offer some new didactic perspectives. For those who know the subject already, let me briefly point these out:

I. Few, if any, of the popular writers seems to have noted that the argument of the wavefunction (θ = E·t – p·t) – using natural units (hence, the numerical value of ħ and c is one), and for an object moving at constant velocity (hence, x = v·t) – can be written as the product of the proper time of the object and its rest mass:

θ = E·t – p·x = E·t − p·x = m_v·t − m_v·v·x = m_v·(t − v·x)

⇔ θ = m₀·(t − v·x)/√(1 – v²) = m₀·t’

Hence, the argument of the wavefunction is just the proper time of the object with the rest mass acting as a scaling factor for the time: the internal clock of the object ticks much faster if it’s heavier. This symmetry between the argument of the wavefunction of the object as measured in its own (inertial) reference frame, and its argument as measured by us, in our own reference frame, is remarkable, and allows to understand the nature of the wavefunction in a more intuitive way.

While this approach reflects Feynman’s idea of the photon stopwatch, the presentation in this booklet generalizes the concept for all wavefunctions, first and foremost the wavefunction of the matter-particles that we’re used to (e.g. electrons).

II. Few, if any, have thought of looking at Schrödinger’s wave equation as an energy propagation mechanism. In fact, when helping my daughter out as she was trying to understand non-linear regression (logit and Poisson regressions), it suddenly realized we can analyze the wavefunction as a link function that connects two physical spaces: the physical space of our moving object, and a physical energy space.

Re-inserting Planck’s quantum of action in the argument of the wavefunction – so we write θ as θ = (E/ħ)·t – (p/ħ)·x = [E·t – p·x]/ħ – we may assign a physical dimension to it: when interpreting ħ as a scaling factor only (and, hence, when we only consider its numerical value, not its physical dimension), θ becomes a quantity expressed in newton·meter·second, i.e. the (physical) dimension of action. It is only natural, then, that we would associate the real and imaginary part of the wavefunction with some physical dimension too, and a dimensional analysis of Schrödinger’s equation tells us this dimension must be energy.

This perspective allows us to look at the wavefunction as an energy propagation mechanism, with the real and imaginary part of the probability amplitude interacting in very much the same way as the electric and magnetic field vectors E and B. This leads me to the next point, which I make rather emphatically in this booklet: the propagation mechanism for electromagnetic energy – as described by Maxwell’s equations – is mathematically equivalent to the propagation mechanism that’s implicit in the Schrödinger equation.

I am, therefore, able to present the Schrödinger equation in a much more coherent way, describing not only how this famous equation works for electrons, or matter-particles in general (i.e. fermions or spin-1/2 particles), which is probably the only use of the Schrödinger equation you are familiar with, but also how it works for bosons, including the photon, of course, but also the theoretical zero-spin boson!

In fact, I am personally rather proud of this. Not because I am doing something that hasn’t been done before (I am sure many have come to the same conclusions before me), but because one always has to trust one’s intuition. So let me say something about that third innovation: the photon wavefunction.

III. Let me tell you the little story behind my photon wavefunction. One of my acquaintances is a retired nuclear scientist. While he knew I was delving into it all, I knew he had little time to answer any of my queries. However, when I asked him about the wavefunction for photons, he bluntly told me photons didn’t have a wavefunction. I should just study Maxwell’s equations and that’s it: there’s no wavefunction for photons: just this traveling electric and a magnetic field vector. Look at Feynman’s Lectures, or any textbook, he said. None of them talk about photon wavefunctions. That’s true, but I knew he had to be wrong. I mulled over it for several months, and then just sat down and started doing to fiddle with Maxwell’s equations, assuming the oscillations of the E and B vector could be described by regular sinusoids. And – Lo and behold! – I derived a wavefunction for the photon. It’s fully equivalent to the classical description, but the new expression solves the Schrödinger equation, if we modify it in a rather logical way: we have to double the diffusion constant, which makes sense, because E and B give you two waves for the price of one!

[…]

In any case, I am getting ahead of myself here, and so I should wrap up this rather long introduction. Let me just say that, through my rather long journey in search of understanding – rather than knowledge alone – I have learned there are so many wrong answers out there: wrong answers that hamper rather than promote a better understanding. Moreover, I was most shocked to find out that such wrong answers are not the preserve of amateurs alone! This emboldened me to write what I write here, and to publish it. Quantum mechanics is a logical and coherent framework, and it is not all that difficult to understand. One just needs good pointers, and that’s what I want to provide here.

As of now, it focuses on the mechanics in particular, i.e. the concept of the wavefunction and wave equation (better known as Schrödinger’s equation). The other aspect of quantum mechanics – i.e. the idea of uncertainty as implied by the quantum idea – will receive more attention in a later version of this document. I should also say I will limit myself to quantum electrodynamics (QED) only, so I won’t discuss quarks (i.e. quantum chromodynamics, which is an entirely different realm), nor will I delve into any of the other more recent advances of physics.

In the end, you’ll still be left with lots of unanswered questions. However, that’s quite OK, as Richard Feynman himself was of the opinion that he himself did not understand the topic the way he would like to understand it. But then that’s exactly what draws all of us to quantum physics: a common search for a deep and full understanding of reality, rather than just some superficial description of it, i.e. knowledge alone.

So let’s get on with it. I am not saying this is going to be easy reading. In fact, I blogged about much easier stuff than this in my blog—treating only aspects of the whole theory. This is the whole thing, and it’s not easy to swallow. In fact, it may well too big to swallow as a whole. But please do give it a try. I wanted this to be an intuitive but formally correct introduction to quantum math. However, when everything is said and done, you are the only who can judge if I reached that goal.

Of course, I should not forget the acknowledgements but… Well… It was a rather lonely venture, so I am only going to acknowledge my wife here, Maria, who gave me all of the spacetime and all of the freedom I needed, as I would get up early, or work late after coming home from my regular job. I sacrificed weekends, which we could have spent together, and – when mulling over yet another paradox – the nights were often short and bad. Frankly, it’s been an extraordinary climb, but the view from the top is magnificent.

I just need to insert one caution, my site (www.readingfeynman.org) includes animations, which make it much easier to grasp some of the mathematical concepts that I will be explaining. Hence, I warmly recommend you also have a look at that site, and its Deep Blue page in particular – as that page has the same contents, more or less, but the animations make it a much easier read.

Have fun with it!

Jean Louis Van Belle, BA, MA, BPhil, Drs.

The Imaginary Energy Space

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In addition, I see the dark force has amused himself by removing some material even here!

Original post:

Intriguing title, isn’t it? You’ll think this is going to be highly speculative and you’re right. In fact, I could also have written: the imaginary action space, or the imaginary momentum space. Whatever. It all works ! It’s an imaginary space – but a very real one, because it holds energy, or momentum, or a combination of both, i.e. action. 🙂

So the title is either going to deter you or, else, encourage you to read on. I hope it’s the latter. 🙂

In my post on Richard Feynman’s exposé on how Schrödinger got his famous wave equation, I noted an ambiguity in how he deals with the energy concept. I wrote that piece in February, and we are now May. In-between, I looked at Schrödinger’s equation from various perspectives, as evidenced from the many posts that followed that February post, which I summarized on my Deep Blue page, where I note the following:

The argument of the wavefunction (i.e. θ = ωt – kx = [E·t – p·x]/ħ) is just the proper time of the object that’s being represented by the wavefunction (which, in most cases, is an elementary particle—an electron, for example).
The 1/2 factor in Schrödinger’s equation (∂ψ/∂t = i·(ħ/2m)·∇²ψ) doesn’t make all that much sense, so we should just drop it. Writing ∂ψ/∂t = i·(m/ħ)∇²ψ (i.e. Schrödinger’s equation without the 1/2 factor) does away with the mentioned ambiguities and, more importantly, avoids obvious contradictions.

Both remarks are rather unusual—especially the second one. In fact, if you’re not shocked by what I wrote above (Schrödinger got something wrong!), then stop reading—because then you’re likely not to understand a thing of what follows. 🙂 In any case, I thought it would be good to follow up by devoting a separate post to this matter.

The argument of the wavefunction as the proper time

Frankly, it took me quite a while to see that the argument of the wavefunction is nothing but the t’ = (t − v∙x)/√(1−v²)] formula that we know from the Lorentz transformation of spacetime. Let me quickly give you the formulas (just substitute the u for v):

relativity

In fact, let me be precise: the argument of the wavefunction also has the particle’s rest mass m₀ in it. That mass factor (m₀) appears in it as a general scaling factor, so it determines the density of the wavefunction both in time as well as in space. Let me jot it down:

ψ(x, t) = a·e^{−i·(m_v·t − p∙x)}= a·e^{−i·[(m₀/√(1−v²))·t − (m₀·v/√(1−v²))∙x]}= a·e^{−i·m₀·(t − v∙x)/√(1−v²)}

Huh? Yes. Let me show you how we get from θ = ωt – kx = [E·t – p·x]/ħ to θ = m_v·t − p∙x. It’s really easy. We first need to choose our units such that the speed of light and Planck’s constant are numerically equal to one, so we write: c = 1 and ħ = 1. So now the 1/ħ factor no longer appears.

[Let me note something here: using natural units does not do away with the dimensions: the dimensions of whatever is there remain what they are. For example, energy remains what it is, and so that’s force over distance: 1 joule = 1 newton·meter (1 J = 1 N·m. Likewise, momentum remains what it is: force times time (or mass times velocity). Finally, the dimension of the quantum of action doesn’t disappear either: it remains the product of force, distance and time (N·m·s). So you should distinguish between the numerical value of our variables and their dimension. Always! That’s where physics is different from algebra: the equations actually mean something!]

Now, because we’re working in natural units, the numerical value of both c and c²will be equal to 1. It’s obvious, then, that Einstein’s mass-energy equivalence relation reduces from E = m_vc² to E = m_v. You can work out the rest yourself – noting that p = m_v·v and m_v= m₀/√(1−v²). Done! For a more intuitive explanation, I refer you to the above-mentioned page.

So that’s for the wavefunction. Let’s now look at Schrödinger’s wave equation, i.e. that differential equation of which our wavefunction is a solution. In my introduction, I bluntly said there was something wrong with it: that 1/2 factor shouldn’t be there. Why not?

What’s wrong with Schrödinger’s equation?

When deriving his famous equation, Schrödinger uses the mass concept as it appears in the classical kinetic energy formula: K.E. = m·v²/2, and that’s why – after all the complicated turns – that 1/2 factor is there. There are many reasons why that factor doesn’t make sense. Let me sum up a few.

[I] The most important reason is that de Broglie made it quite clear that the energy concept in his equations for the temporal and spatial frequency for the wavefunction – i.e. the ω = E/ħ and k = p/ħ relations – is the total energy, including rest energy (m₀), kinetic energy (m·v²/2) and any potential energy (V). In fact, if we just multiply the two de Broglie (aka as matter-wave equations) and use the old-fashioned v = f·λ relation (so we write E as E = ω·ħ = (2π·f)·(h/2π) = f·h, and p as p = k·ħ = (2π/λ)·(h/2π) = h/λ and, therefore, we have f = E/h and p = h/p), we find that the energy concept that’s implicit in the two matter-wave equations is equal to E = m∙v², as shown below:

f·λ = (E/h)·(h/p) = E/p
v = f·λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v) ⇒ E = m·v²

Huh? E = m∙v²? Yes. Not E = m∙c² or m·v²/2 or whatever else you might be thinking of. In fact, this E = m∙v² formula makes a lot of sense in light of the two following points.

Skeptical note: You may – and actually should – wonder whether we can use that v = f·λ relation for a wave like this, i.e. a wave with both a real (cos(-θ)) as well as an imaginary component (i·sin(-θ). It’s a deep question, and I’ll come back to it later. But… Yes. It’s the right question to ask. 😦

[II] Newton told us that force is mass time acceleration. Newton’s law is still valid in Einstein’s world. The only difference between Newton’s and Einstein’s world is that, since Einstein, we should treat the mass factor as a variable as well. We write: F = m_v·a = m_v·a = [m₀/√(1−v²)]·a. This formula gives us the definition of the newton as a force unit: 1 N = 1 kg·(m/s)/s = 1 kg·m/s². [Note that the 1/√(1−v²) factor – i.e. the Lorentz factor (γ) – has no dimension, because v is measured as a relative velocity here, i.e. as a fraction between 0 and 1.]

Now, you’ll agree the definition of energy as a force over some distance is valid in Einstein’s world as well. Hence, if 1 joule is 1 N·m, then 1 J is also equal to 1 (kg·m/s²)·m = 1 kg·(m²/s²), so this also reflects the E = m∙v² concept. [I can hear you mutter: that kg factor refers to the rest mass, no? No. It doesn’t. The kg is just a measure of inertia: as a unit, it applies to both m₀as well as m_v. Full stop.]

Very skeptical note: You will say this doesn’t prove anything – because this argument just shows the dimensional analysis for both equations (i.e. E = m∙v² and E = m∙c²) is OK. Hmm… Yes. You’re right. 🙂 But the next point will surely convince you! 🙂

[III] The third argument is the most intricate and the most beautiful at the same time—not because it’s simple (like the arguments above) but because it gives us an interpretation of what’s going on here. It’s fairly easy to verify that Schrödinger’s equation, ∂ψ/∂t = i·(ħ/2m)·∇²ψ equation (including the 1/2 factor to which I object), is equivalent to the following set of two equations:

Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ)

[In case you don’t see it immediately, note that two complex numbers a + i·b and c + i·d are equal if, and only if, their real and imaginary parts are the same. However, here we have something like this: a + i·b = i·(c + i·d) = i·c + i²·d = − d + i·c (remember i²= −1).]

Now, before we proceed (i.e. before I show you what’s wrong here with that 1/2 factor), let us look at the dimensions first. For that, we’d better analyze the complete Schrödinger equation so as to make sure we’re not doing anything stupid here by looking at one aspect of the equation only. The complete equation, in its original form, is:

Notice that, to simplify the analysis above, I had moved the i and the ħ on the left-hand side to the right-hand side (note that 1/i = −i, so −(ħ²/2m)/(i·ħ) = ħ/2m). Now, the ħ²factor on the right-hand side is expressed in J²·s². Now that doesn’t make much sense, but then that mass factor in the denominator makes everything come out alright. Indeed, we can use the mass-equivalence relation to express m in J/(m/s)² units. So our ħ²/2m coefficient is expressed in (J²·s²)/[J/(m/s)²] = J·m². Now we multiply that by that Laplacian operating on some scalar, which yields some quantity per square meter. So the whole right-hand side becomes some amount expressed in joule, i.e. the unit of energy! Interesting, isn’t it?

On the left-hand side, we have i and ħ. We shouldn’t worry about the imaginary unit because we can treat that as just another number, albeit a very special number (because its square is minus 1). However, in this equation, it’s like a mathematical constant and you can think of it as something like π or e. [Think of the magical formula: e^iπ= i² = −1.] In contrast, ħ is a physical constant, and so that constant comes with some dimension and, therefore, we cannot just do what we want. [I’ll show, later, that even moving it to the other side of the equation comes with interpretation problems, so be careful with physical constants, as they really mean something!] In this case, its dimension is the action dimension: J·s = N·m·s, so that’s force times distance times time. So we multiply that with a time derivative and we get joule once again (N·m·s/s = N·m = J), so that’s the unit of energy. So it works out: we have joule units both left and right in Schrödinger’s equation. Nice! Yes. But what does it mean? 🙂

Well… You know that we can – and should – think of Schrödinger’s equation as a diffusion equation – just like a heat diffusion equation, for example – but then one describing the diffusion of a probability amplitude. [In case you are not familiar with this interpretation, please do check my post on it, or my Deep Blue page.] But then we didn’t describe the mechanism in very much detail, so let me try to do that now and, in the process, finally explain the problem with the 1/2 factor.

The missing energy

There are various ways to explain the problem. One of them involves calculating group and phase velocities of the elementary wavefunction satisfying Schrödinger’s equation but that’s a more complicated approach and I’ve done that elsewhere, so just click the reference if you prefer the more complicated stuff. I find it easier to just use those two equations above:

Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ)

The argument is the following: if our elementary wavefunction is equal to eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt), then it’s easy to proof that this pair of conditions is fulfilled if, and only if, ω = k²·(ħ/2m). [Note that I am omitting the normalization coefficient in front of the wavefunction: you can put it back in if you want. The argument here is valid, with or without normalization coefficients.] Easy? Yes. Check it out. The time derivative on the left-hand side is equal to:

∂ψ/∂t = −iω·ieⁱ^{(kx − ωt)}= ω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·cos(kx − ωt) + iω·sin(kx − ωt)

And the second-order derivative on the right-hand side is equal to:

∇²ψ= ∂²ψ/∂x²= i·k²·eⁱ^{(kx − ωt)}= k²·cos(kx − ωt) + i·k²·sin(kx − ωt)

So the two equations above are equivalent to writing:

Re(∂ψ_B/∂t) = −(ħ/2m)·Im(∇²ψ_B) ⇔ ω·cos(kx − ωt) = k²·(ħ/2m)·cos(kx − ωt)
Im(∂ψ_B/∂t) = (ħ/2m)·Re(∇²ψ_B) ⇔ ω·sin(kx − ωt) = k²·(ħ/2m)·sin(kx − ωt)

So both conditions are fulfilled if, and only if, ω = k²·(ħ/2m). You’ll say: so what? Well… We have a contradiction here—something that doesn’t make sense. Indeed, the second of the two de Broglie equations (always look at them as a pair) tells us that k = p/ħ, so we can re-write the ω = k²·(ħ/2m) condition as:

ω/k = v_p = k²·(ħ/2m)/k = k·ħ/(2m) = (p/ħ)·(ħ/2m) = p/2m ⇔ p = 2m

You’ll say: so what? Well… Stop reading, I’d say. That p = 2m doesn’t make sense—at all! Nope! In fact, if you thought that the E = m·v² is weird—which, I hope, is no longer the case by now—then… Well… This p = 2m equation is much weirder. In fact, it’s plain nonsense: this condition makes no sense whatsoever. The only way out is to remove the 1/2 factor, and to re-write the Schrödinger equation as I wrote it, i.e. with an ħ/m coefficient only, rather than an (1/2)·(ħ/m) coefficient.

Huh? Yes.

As mentioned above, I could do those group and phase velocity calculations to show you what rubbish that 1/2 factor leads to – and I’ll do that eventually – but let me first find yet another way to present the same paradox. Let’s simplify our life by choosing our units such that c = ħ = 1, so we’re using so-called natural units rather than our SI units. [Again, note that switching to natural units doesn’t do anything to the physical dimensions: a force remains a force, a distance remains a distance, and so on.] Our mass-energy equivalence then becomes: E = m·c²= m·1²= m. [Again, note that switching to natural units doesn’t do anything to the physical dimensions: a force remains a force, a distance remains a distance, and so on. So we’d still measure energy and mass in different but equivalent units. Hence, the equality sign should not make you think mass and energy are actually the same: energy is energy (i.e. force times distance), while mass is mass (i.e. a measure of inertia). I am saying this because it’s important, and because it took me a while to make these rather subtle distinctions.]

Let’s now go one step further and imagine a hypothetical particle with zero rest mass, so m₀ = 0. Hence, all its energy is kinetic and so we write: K.E. = m_v·v/2. Now, because this particle has zero rest mass, the slightest acceleration will make it travel at the speed of light. In fact, we would expect it to travel at the speed, so m_v = m_c and, according to the mass-energy equivalence relation, its total energy is, effectively, E = m_v = m_c. However, we just said its total energy is kinetic energy only. Hence, its total energy must be equal to E = K.E. = m_c·c/2 = m_c/2. So we’ve got only half the energy we need. Where’s the other half? Where’s the missing energy? Quid est veritas? Is its energy E = m_c or E = m_c/2?

It’s just a paradox, of course, but one we have to solve. Of course, we may just say we trust Einstein’s E = m·c²formula more than the kinetic energy formula, but that answer is not very scientific. 🙂 We’ve got a problem here and, in order to solve it, I’ve come to the following conclusion: just because of its sheer existence, our zero-mass particle must have some hidden energy, and that hidden energy is also equal to E = m·c²/2. Hence, the kinetic and the hidden energy add up to E = m·c² and all is alright.

Huh? Hidden energy? I must be joking, right?

Well… No. Let me explain. Oh. And just in case you wonder why I bother to try to imagine zero-mass particles. Let me tell you: it’s the first step towards finding a wavefunction for a photon and, secondly, you’ll see it just amounts to modeling the propagation mechanism of energy itself. 🙂

The hidden energy as imaginary energy

I am tempted to refer to the missing energy as imaginary energy, because it’s linked to the imaginary part of the wavefunction. However, it’s anything but imaginary: it’s as real as the imaginary part of the wavefunction. [I know that sounds a bit nonsensical, but… Well… Think about it. And read on!]

Back to that factor 1/2. As mentioned above, it also pops up when calculating the group and the phase velocity of the wavefunction. In fact, let me show you that calculation now. [Sorry. Just hang in there.] It goes like this.

The de Broglie relations tell us that the k and the ω in the eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction (i.e. the spatial and temporal frequency respectively) are equal to k = p/ħ, and ω = E/ħ. Let’s now think of that zero-mass particle once more, so we assume all of its energy is kinetic: no rest energy, no potential! So… If we now use the kinetic energy formula E = m·v²/2 – which we can also write as E = m·v·v/2 = p·v/2 = p·p/2m = p²/2m, with v = p/m the classical velocity of the elementary particle that Louis de Broglie was thinking of – then we can calculate the group velocity of our eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction as:

v_g = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂[p²/2m]/∂p = 2p/2m = p/m = v

[Don’t tell me I can’t treat m as a constant when calculating ∂ω/∂k: I can. Think about it.]

Fine. Now the phase velocity. For the phase velocity of our eⁱ^{(kx − ωt)} wavefunction, we find:

v_p = ω/k = (E/ħ)/(p/ħ) = E/p = (p²/2m)/p = p/2m = v/2

So that’s only half of v: it’s the 1/2 factor once more! Strange, isn’t it? Why would we get a different value for the phase velocity here? It’s not like we have two different frequencies here, do we? Well… No. You may also note that the phase velocity turns out to be smaller than the group velocity (as mentioned, it’s only half of the group velocity), which is quite exceptional as well! So… Well… What’s the matter here? We’ve got a problem!

What’s going on here? We have only one wave here—one frequency and, hence, only one k and ω. However, on the other hand, it’s also true that the eⁱ^{(kx − ωt)} wavefunction gives us two functions for the price of one—one real and one imaginary: eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt). So the question here is: are we adding waves, or are we not? It’s a deep question. If we’re adding waves, we may get different group and phase velocities, but if we’re not, then… Well… Then the group and phase velocity of our wave should be the same, right? The answer is: we are and we aren’t. It all depends on what you mean by ‘adding’ waves. I know you don’t like that answer, but that’s the way it is, really. 🙂

Let me make a small digression here that will make you feel even more confused. You know – or you should know – that the sine and the cosine function are the same except for a phase difference of 90 degrees: sinθ = cos(θ + π/2). Now, at the same time, multiplying something with i amounts to a rotation by 90 degrees, as shown below.

Hence, in order to sort of visualize what our eⁱ^{(kx − ωt)} function really looks like, we may want to super-impose the two graphs and think of something like this:

You’ll have to admit that, when you see this, our formulas for the group or phase velocity, or our v = f·λ relation, do no longer make much sense, do they? 🙂

Having said that, that 1/2 factor is and remains puzzling, and there must be some logical reason for it. For example, it also pops up in the Uncertainty Relations:

Δx·Δp ≥ ħ/2 and ΔE·Δt ≥ ħ/2

So we have ħ/2 in both, not ħ. Why do we need to divide the quantum of action here? How do we solve all these paradoxes? It’s easy to see how: the apparent contradiction (i.e. the different group and phase velocity) gets solved if we’d use the E = m∙v² formula rather than the kinetic energy E = m∙v²/2. But then… What energy formula is the correct one: E = m∙v² or m∙c²? Einstein’s formula is always right, isn’t it? It must be, so let me postpone the discussion a bit by looking at a limit situation. If v = c, then we don’t need to make a choice, obviously. 🙂 So let’s look at that limit situation first. So we’re discussing our zero-mass particle once again, assuming it travels at the speed of light. What do we get?

Well… Measuring time and distance in natural units, so c = 1, we have:

E = m∙c² = m and p = m∙c= m, so we get: E = m = p

Waw ! E = m = p ! What a weird combination, isn’t it? Well… Yes. But it’s fully OK. [You tell me why it wouldn’t be OK. It’s true we’re glossing over the dimensions here, but natural units are natural units and, hence, the numerical value of c and c²is 1. Just figure it out for yourself.] The point to note is that the E = m = p equality yields extremely simple but also very sensible results. For the group velocity of our eⁱ^{(kx − ωt)} wavefunction, we get:

v_g = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂p/∂p = 1

So that’s the velocity of our zero-mass particle (remember: the 1 stands for c here, i.e. the speed of light) expressed in natural units once more—just like what we found before. For the phase velocity, we get:

v_p = ω/k = (E/ħ)/(p/ħ) = E/p = p/p = 1

Same result! No factor 1/2 here! Isn’t that great? My ‘hidden energy theory’ makes a lot of sense.

However, if there’s hidden energy, we still need to show where it’s hidden. 🙂 Now that question is linked to the propagation mechanism that’s described by those two equations, which now – leaving the 1/2 factor out, simplify to:

Re(∂ψ/∂t) = −(ħ/m)·Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/m)·Re(∇²ψ)

Propagation mechanism? Yes. That’s what we’re talking about here: the propagation mechanism of energy. Huh? Yes. Let me explain in another separate section, so as to improve readability. Before I do, however, let me add another note—for the skeptics among you. 🙂

Indeed, the skeptics among you may wonder whether our zero-mass particle wavefunction makes any sense at all, and they should do so for the following reason: if x = 0 at t = 0, and it’s traveling at the speed of light, then x(t) = t. Always. So if E = m = p, the argument of our wavefunction becomes E·t – p·x = E·t – E·t = 0! So what’s that? The proper time of our zero-mass particle is zero—always and everywhere!?

Well… Yes. That’s why our zero-mass particle – as a point-like object – does not really exist. What we’re talking about is energy itself, and its propagation mechanism. 🙂

While I am sure that, by now, you’re very tired of my rambling, I beg you to read on. Frankly, if you got as far as you have, then you should really be able to work yourself through the rest of this post. 🙂 And I am sure that – if anything – you’ll find it stimulating! 🙂

The imaginary energy space

Look at the propagation mechanism for the electromagnetic wave in free space, which (for c = 1) is represented by the following two equations:

∂B/∂t = –∇×E
∂E/∂t = ∇×B

[In case you wonder, these are Maxwell’s equations for free space, so we have no stationary nor moving charges around.] See how similar this is to the two equations above? In fact, in my Deep Blue page, I use these two equations to derive the quantum-mechanical wavefunction for the photon (which is not the same as that hypothetical zero-mass particle I introduced above), but I won’t bother you with that here. Just note the so-called curl operator in the two equations above (∇×) can be related to the Laplacian we’ve used so far (∇²). It’s not the same thing, though: for starters, the curl operator operates on a vector quantity, while the Laplacian operates on a scalar (including complex scalars). But don’t get distracted now. Let’s look at the revised Schrödinger’s equation, i.e. the one without the 1/2 factor:

∂ψ/∂t = i·(ħ/m)·∇²ψ

On the left-hand side, we have a time derivative, so that’s a flow per second. On the right-hand side we have the Laplacian and the i·ħ/m factor. Now, written like this, Schrödinger’s equation really looks exactly the same as the general diffusion equation, which is written as: ∂φ/∂t = D·∇²φ, except for the imaginary unit, which makes it clear we’re getting two equations for the price of one here, rather than one only! 🙂 The point is: we may now look at that ħ/m factor as a diffusion constant, because it does exactly the same thing as the diffusion constant D in the diffusion equation ∂φ/∂t = D·∇²φ, i.e:

As a constant of proportionality, it quantifies the relationship between both derivatives.
As a physical constant, it ensures the dimensions on both sides of the equation are compatible.

So the diffusion constant for Schrödinger’s equation is ħ/m. What is its dimension? That’s easy: (N·m·s)/(N·s²/m) = m²/s. [Remember: 1 N = 1 kg·m/s².] But then we multiply it with the Laplacian, so that’s something expressed per square meter, so we get something per second on both sides.

Of course, you wonder: what per second? Not sure. That’s hard to say. Let’s continue with our analogy with the heat diffusion equation so as to try to get a better understanding of what’s being written here. Let me give you that heat diffusion equation here. Assuming the heat per unit volume (q) is proportional to the temperature (T) – which is the case when expressing T in degrees Kelvin (K), so we can write q as q = k·T – we can write it as:

So that’s structurally similar to Schrödinger’s equation, and to the two equivalent equations we jotted down above. So we’ve got T (temperature) in the role of ψ here—or, to be precise, in the role of ψ ‘s real and imaginary part respectively. So what’s temperature? From the kinetic theory of gases, we know that temperature is not just a scalar: temperature measures the mean (kinetic) energy of the molecules in the gas. That’s why we can confidently state that the heat diffusion equation models an energy flow, both in space as well as in time.

Let me make the point by doing the dimensional analysis for that heat diffusion equation. The time derivative on the left-hand side (∂T/∂t) is expressed in K/s (Kelvin per second). Weird, isn’t it? What’s a Kelvin per second? Well… Think of a Kelvin as some very small amount of energy in some equally small amount of space—think of the space that one molecule needs, and its (mean) energy—and then it all makes sense, doesn’t it?

However, in case you find that a bit difficult, just work out the dimensions of all the other constants and variables. The constant in front (k) makes sense of it. That coefficient (k) is the (volume) heat capacity of the substance, which is expressed in J/(m³·K). So the dimension of the whole thing on the left-hand side (k·∂T/∂t) is J/(m³·s), so that’s energy (J) per cubic meter (m³) and per second (s). Nice, isn’t it? What about the right-hand side? On the right-hand side we have the Laplacian operator – i.e. ∇²= ∇·∇, with ∇ = (∂/∂x, ∂/∂y, ∂/∂z) – operating on T. The Laplacian operator, when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m²). In this case, it’s operating on T, so the dimension of ∇²T is K/m². Again, that doesn’t tell us very much (what’s the meaning of a Kelvin per square meter?) but we multiply it by the thermal conductivity (κ), whose dimension is W/(m·K) = J/(m·s·K). Hence, the dimension of the product is the same as the left-hand side: J/(m³·s). So that’s OK again, as energy (J) per cubic meter (m³) and per second (s) is definitely something we can associate with an energy flow.

In fact, we can play with this. We can bring k from the left- to the right-hand side of the equation, for example. The dimension of κ/k is m²/s (check it!), and multiplying that by K/m²(i.e. the dimension of ∇²T) gives us some quantity expressed in Kelvin per second, and so that’s the same dimension as that of ∂T/∂t. Done!

In fact, we’ve got two different ways of writing Schrödinger’s diffusion equation. We can write it as ∂ψ/∂t = i·(ħ/m)·∇²ψ or, else, we can write it as ħ·∂ψ/∂t = i·(ħ²/m)·∇²ψ. Does it matter? I don’t think it does. The dimensions come out OK in both cases. However, interestingly, if we do a dimensional analysis of the ħ·∂ψ/∂t = i·(ħ²/m)·∇²ψ equation, we get joule on both sides. Interesting, isn’t it? The key question, of course, is: what is it that is flowing here?

I don’t have a very convincing answer to that, but the answer I have is interesting—I think. 🙂 Think of the following: we can multiply Schrödinger’s equation with whatever we want, and then we get all kinds of flows. For example, if we multiply both sides with 1/(m²·s) or 1/(m³·s), we get a equation expressing the energy conservation law, indeed! [And you may want to think about the minus sign of the right-hand side of Schrödinger’s equation now, because it makes much more sense now!]

We could also multiply both sides with s, so then we get J·s on both sides, i.e. the dimension of physical action (J·s = N·m·s). So then the equation expresses the conservation of action! Huh? Yes. Let me re-phrase that: then it expresses the conservation of angular momentum—as you’ll surely remember that the dimension of action and angular momentum are the same. 🙂

And then we can divide both sides by m, so then we get N·s on both sides, so that’s momentum. So then Schrödinger’s equation embodies the momentum conservation law.

Isn’t it just wonderful? Schrödinger’s equation packs all of the conservation laws! The only catch is that it flows back and forth from the real to the imaginary space, using that propagation mechanism as described in those two equations.

Now that is really interesting, because it does provide an explanation – as fuzzy as it may seem – for all those weird concepts one encounters when studying physics, such as the tunneling effect, which amounts to energy flowing from the imaginary space to the real space and, then, inevitably, flowing back. It also allows for borrowing time from the imaginary space. Hmm… Interesting! [I know I still need to make these points much more formally, but… Well… You kinda get what I mean, don’t you?]

To conclude, let me re-baptize my real and imaginary ‘space’ by referring to them to what they really are: a real and imaginary energy space respectively. Although… Now that I think of it: it could also be real and imaginary momentum space, or a real and imaginary action space. Hmm… The latter term may be the best. 🙂

Isn’t this all great? I mean… I could go on and on—but I’ll stop here, so you can freewheel around yourself. For example, you may wonder how similar that energy propagation mechanism actually is as compared to the propagation mechanism of the electromagnetic wave? The answer is: very similar. You can check how similar in one of my posts on the photon wavefunction or, if you’d want a more general argument, check my Deep Blue page. Have fun exploring! 🙂

So… Well… That’s it, folks. I hope you enjoyed this post—if only because I really enjoyed writing it. 🙂

[…]

OK. You’re right. I still haven’t answered the fundamental question.

So what about the 1/2 factor?

What about that 1/2 factor? Did Schrödinger miss it? Well… Think about it for yourself. First, I’d encourage you to further explore that weird graph with the real and imaginary part of the wavefunction. I copied it below, but with an added 45º line—yes, the green diagonal. To make it somewhat more real, imagine you’re the zero-mass point-like particle moving along that line, and we observe you from our inertial frame of reference, using equivalent time and distance units.

So we’ve got that cosine (cosθ) varying as you travel, and we’ve also got the i·sinθ part of the wavefunction going while you’re zipping through spacetime. Now, THINK of it: the phase velocity of the cosine bit (i.e. the red graph) contributes as much to your lightning speed as the i·sinθ bit, doesn’t it? Should we apply Pythagoras’ basic r² = x² + y²Theorem here? Yes: the velocity vector along the green diagonal is going to be the sum of the velocity vectors along the horizontal and vertical axes. So… That’s great.

Yes. It is. However, we still have a problem here: it’s the velocity vectors that add up—not their magnitudes. Indeed, if we denote the velocity vector along the green diagonal as u, then we can calculate its magnitude as:

u = √u² = √[(v/2)² + (v/2)²] = √[2·(v²/4) = √[v²/2] = v/√2 ≈ 0.7·v

So, as mentioned, we’re adding the vectors, but not their magnitudes. We’re somewhat better off than we were in terms of showing that the phase velocity of those sine and cosine velocities add up—somehow, that is—but… Well… We’re not quite there.

Fortunately, Einstein saves us once again. Remember we’re actually transforming our reference frame when working with the wavefunction? Well… Look at the diagram below (for which I thank the author)

special relativity

In fact, let me insert an animated illustration, which shows what happens when the velocity u goes up and down from (close to) −c to +c and back again. It’s beautiful, and I must credit the author here too. It sort of speaks for itself, but please do click the link as the accompanying text is quite illuminating. 🙂

Animated_Lorentz_Transformation

The point is: for our zero-mass particle, the x’ and t’ axis will rotate into the diagonal itself which, as I mentioned a couple of times already, represents the speed of light and, therefore, our zero-mass particle traveling at c. It’s obvious that we’re now adding two vectors that point in the same direction and, hence, their magnitudes just add without any square root factor. So, instead of u = √[(v/2)² + (v/2)²], we just have v/2 + v/2 = v! Done! We solved the phase velocity paradox! 🙂

So… I still haven’t answered that question. Should that 1/2 factor in Schrödinger’s equation be there or not? The answer is, obviously: yes. It should be there. And as for Schrödinger using the mass concept as it appears in the classical kinetic energy formula: K.E. = m·v²/2… Well… What other mass concept would he use? I probably got a bit confused with Feynman’s exposé – especially this notion of ‘choosing the zero point for the energy’ – but then I should probably just re-visit the thing and adjust the language here and there. But the formula is correct.

Thinking it all through, the ħ/2m constant in Schrödinger’s equation should be thought of as the reciprocal of m/(ħ/2). So what we’re doing basically is measuring the mass of our object in units of ħ/2, rather than units of ħ. That makes perfect sense, if only because it’s ħ/2, rather than ħthe factor that appears in the Uncertainty Relations Δx·Δp ≥ ħ/2 and ΔE·Δt ≥ ħ/2. In fact, in my post on the wavefunction of the zero-mass particle, I noted its elementary wavefunction should use the m = E = p = ħ/2 values, so it becomes ψ(x, t) = a·e^{−i∙[(ħ/2)∙t − (ħ/2)∙x]/ħ}= a·e^{−i∙[t − x]/2}.

Isn’t that just nice? 🙂 I need to stop here, however, because it looks like this post is becoming a book. Oh—and note that nothing what I wrote above discredits my ‘hidden energy’ theory. On the contrary, it confirms it. In fact, the nice thing about those illustrations above is that it associates the imaginary component of our wavefunction with travel in time, while the real component is associated with travel in space. That makes our theory quite complete: the ‘hidden’ energy is the energy that moves time forward. The only thing I need to do is to connect it to that idea of action expressing itself in time or in space, cf. what I wrote on my Deep Blue page: we can look at the dimension of Planck’s constant, or at the concept of action in general, in two very different ways—from two different perspectives, so to speak:

[Planck’s constant] = [action] = N∙m∙s = (N∙m)∙s = [energy]∙[time]
[Planck’s constant] = [action] = N∙m∙s = (N∙s)∙m = [momentum]∙[distance]

Hmm… I need to combine that with the idea of the quantum vacuum, i.e. the mathematical space that’s associated with time and distance becoming countable variables…. In any case. Next time. 🙂

Before I sign off, however, let’s quickly check if our a·e^{−i∙[t − x]/2}wavefunction solves the Schrödinger equation:

∂ψ/∂t = −a·e^{−i∙[t − x]/2}·(i/2)
∇²ψ = ∂²[a·e^{−i∙[t − x]/2}]/∂x²= ∂[a·e^{−i∙[t − x]/2}·(i/2)]/∂x = −a·e^{−i∙[t − x]/2}·(1/4)

So the ∂ψ/∂t = i·(ħ/2m)·∇²ψ equation becomes:

−a·e^{−i∙[t − x]/2}·(i/2) = −i·(ħ/[2·(ħ/2)])·a·e^{−i∙[t − x]/2}·(1/4)

⇔ 1/2 = 1/4 !?

The damn 1/2 factor. Schrödinger wants it in his wave equation, but not in the wavefunction—apparently! So what if we take the m = E = p = ħ solution? We get:

∂ψ/∂t = −a·i·e^{−i∙[t − x]}
∇²ψ = ∂²[a·e^{−i∙[t − x]}]/∂x²= ∂[a·i·e^{−i∙[t − x]}]/∂x = −a·e^{−i∙[t − x]}

So the ∂ψ/∂t = i·(ħ/2m)·∇²ψ equation now becomes:

−a·i·e^{−i∙[t − x]} = −i·(ħ/[2·ħ])·a·e^{−i∙[t − x]}

⇔ 1 = 1/2 !?

We’re still in trouble! So… Was Schrödinger wrong after all? There’s no difficulty whatsoever with the ∂ψ/∂t = i·(ħ/m)·∇²ψ equation:

−a·e^{−i∙[t − x]/2}·(i/2) = −i·[ħ/(ħ/2)]·a·e^{−i∙[t − x]/2}·(1/4) ⇔ 1 = 1
−a·i·e^{−i∙[t − x]} = −i·(ħ/ħ)·a·e^{−i∙[t − x]}⇔ 1 = 1

What these equations might tell us is that we should measure mass, energy and momentum in terms of ħ (and not in terms of ħ/2) but that the fundamental uncertainty is ± ħ/2. That solves it all. So the magnitude of the uncertainty is ħ but it separates not 0 and ± 1, but −ħ/2 and −ħ/2. Or, more generally, the following series:

…, −7ħ/2, −5ħ/2, −3ħ/2, −ħ/2, +ħ/2, +3ħ/2,+5ħ/2, +7ħ/2,…

Why are we not surprised? The series represent the energy values that a spin one-half particle can possibly have, and ordinary matter – i.e. all fermions – is composed of spin one-half particles.

To conclude this post, let’s see if we can get any indication on the energy concepts that Schrödinger’s revised wave equation implies. We’ll do so by just calculating the derivatives in the ∂ψ/∂t = i·(ħ/m)·∇²ψ equation (i.e. the equation without the 1/2 factor). Let’s also not assume we’re measuring stuff in natural units, so our wavefunction is just what it is: a·e^{−i·[E·t − p∙x]/ħ}. The derivatives now become:

∂ψ/∂t = −a·i·(E/ħ)·e^{−i∙[E·t − p∙x]/ħ}
∇²ψ = ∂²[a·e^{−i∙[E·t − p∙x]/ħ}]/∂x²= ∂[a·i·(p/ħ)·e^{−i∙[E·t − p∙x]/ħ}]/∂x = −a·(p²/ħ²)·e^{−i∙[E·t − p∙x]/ħ}

So the ∂ψ/∂t = i·(ħ/m)·∇²ψ = i·(1/m)·∇²ψ equation now becomes:

−a·i·(E/ħ)·e^{−i∙[E·t − p∙x]/ħ} = −i·(ħ/m)·a·(p²/ħ²)·e^{−i∙[E·t − p∙x]/ħ}⇔ E = p²/m = m·v²

It all works like a charm. Note that we do not assume stuff like E = m = p here. It’s all quite general. Also note that the E = p²/m closely resembles the kinetic energy formula one often sees: K.E. = m·v²/2 = m·m·v²/(2m) = p²/(2m). We just don’t have the 1/2 factor in our E = p²/m formula, which is great—because we don’t want it! Of course, if you’d add the 1/2 factor in Schrödinger’s equation again, you’d get it back in your energy formula, which would just be that old kinetic energy formula which gave us all these contradictions and ambiguities. 😦

Finally, and just to make sure: let me add that, when we wrote that E = m = p – like we did above – we mean their numerical values are the same. Their dimensions remain what they are, of course. Just to make sure you get that subtle point, we’ll do a quick dimensional analysis of that E = p²/m formula:

[E] = [p²/m] ⇔ N·m = N²·s²/kg = N²·s²/[N·m/s²] = N·m = joule (J)

So… Well… It’s all perfect. 🙂

Post scriptum: I revised my Deep Blue page after writing this post, and I think that a number of the ideas that I express above are presented more consistently and coherently there. In any case, the missing energy theory makes sense. Think of it: any oscillator involves both kinetic as well as potential energy, and they both add up to twice the average kinetic (or potential) energy. So why not here? When everything is said and done, our elementary wavefunction does describe an oscillator. 🙂

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https://wordpress.com/support/copyright-and-the-dmca/

Freewheeling…

In my previous post, I copied a simple animation from Wikipedia to show how one can move from Cartesian to polar coordinates. It’s really neat. Just watch it a few times to appreciate what’s going on here.

Cartesian_to_polar First, the function is being inverted, so we go from y = f(x) to x = g(y) with g = f⁻¹. In this case, we know that if y = sin(6x) + 2 (that’s the function above), then x = (1/6)·arcsin(y – 2). [Note the troublesome convention to denote the inverse function by the -1 superscript: it’s troublesome because that superscript is also used for a reciprocal—and f⁻¹has, obviously, nothing to do with 1/f. In any case, let’s move on.] So we swap the x-axis for the y-axis, and vice versa. In fact, to be precise, we reflect them about the diagonal. In fact, w’re reflecting the whole space here, including the graph of the function. Note that, in three-dimensional space, this reflection can also be looked at as a rotation – again, of all space, including the graph and the axes – by 180 degrees. The axis of rotation is, obviously, the same diagonal. [I like how the animation visualizes this. Neat! It made me think!]

Of course, if we swap the axes, then the domain and the range of the function get swapped too. Let’s see how that works here: x goes from −π to +π, so that’s one cycle (but one that starts from −π and goes to +π, rather than from 0 to 2π), and, hence, y ranges between 1 and 3. [Whatever its argument, the sine function always yields a value between −1 and +1, but we add 2 to every value it takes, so we get the [1, 3] interval now.] After swapping the x- and y-axis, the angle, i.e. the interval between −π and +π, is now on the vertical axis. That’s clear enough. So far so good. 🙂 The operation that follows, however, is a much more complicated transformation of space and, therefore, much more interesting.

The transformation bends the graph around the origin so its head and tail meet. That’s easy to see. What’s a bit more difficult to understand is how the coordinate axes transform. I had to look at the animation several times – so please do the same. Note how this transformation wraps all of the vertical lines around a circle, and how the radius of those circles depends on the distance of those lines from the origin (as measured along the horizontal axis). What about the vertical axis? The animation is somewhat misleading here, as it gives the impression we’re first making another circle out of it, which we then sort of shrink—all the way down to a circle with zero radius! So the vertical axis becomes the origin of our new space. However, there’s no shrinking really. What happens is that we also wrap it around a circle—but one with zero radius indeed!

It’s a very weird operation because we’re dealing with a non-linear transformation here (unlike rotation or reflection) and, therefore, we’re not familiar with it. Even weirder is what happens to the horizontal axis: somehow, this axis becomes an infinite disc, so the distance out is now measured from the center outwards. I should figure out the math here, but that’s for later. The point is: the r = sin(6θ) + 2 function in the final graph (i.e. the curve that looks like a petaled flower) is the same as that y = sin(6x) + 2 curve, so y = r and x = θ, and so we can write what’s written above: r(θ) = sin(6·θ) + 2.

You’ll say: nice, but so what? Well… When I saw this animation, my first reaction was: what if the x and y would be time and space respectively? You’ll say: what space? Well… Just space: three-dimensional space. So think of one of the axes as packing three dimensions really, or three directions—like what’s depicted below. Now think of some point-like object traveling through spacetime, as shown below. It doesn’t need to be point-like, of course—just small enough so we can represent its trajectory by a line. You can also think of the movement of its center-of-mass if you don’t like point-like assumptions. 🙂

trajectory

Of course, you’ll immediately say the trajectory above is not kosher, as our object travels back in time in three sections of this ‘itinerary’.

You’re right. Let’s correct that. It’s easy to see how we should correct it. We just need to ensure the itinerary is a well-defined function, which isn’t the case with the function above: for one value of t, we have only one value of x everywhere—except where we allow our particle to travel back in time. So… Well… We shouldn’t allow that. The concept of a well-defined function implies we need to choose one direction in time. 🙂 That’s neat, because this gives us an explanation for the unique direction of time without having to invoke entropy or other macro-concepts. So let’s replace that thing above by something more kosher traveling in spacetime, like the thing below.

trajectory 2 Now think of wrapping that around some circle. We’d get something like below. [Don’t worry about the precise shape of the graph, as I made up a new one. Note the remark on the need to have a well-behaved function applies here too!]

trajectory 4 Neat, you’ll say, but so what? All we’ve done so far is show that we can represent some itinerary in spacetime in two different ways. In the first representation, we measure time along some linear axis, while, in the second representation, time becomes some angle—an angle that increases, counter-clockwise. To put it differently: time becomes an angular velocity.

Likewise, the spatial dimension was a linear feature in the first representation, while in the second we think of it as some distance measured from some zero point. Well… In fact… No. That’s not correct. The r above has got nothing to do with the distance traveled: the distance traveled would need to be measured along the curve.

Hmm… What’s the deal here?

Frankly, I am not sure. Now that I look at it once more, I note that the exercise with our graph above involved one cycle of a periodic function only—so it’s really not like some object traveling in spacetime, because that’s not a periodic thing. But… Well… Does that matter all that much? It’s easy to imagine how our new representation would just involve some thing that keeps going around and around, as illustrated below.

trajectory 5

So, in this representation, any movement in spacetime – regular or irregular – does become something periodic. But what is periodic here? My first answer is the simplest and, hence, probably the correct one: it’s just time. Time is the periodic thing here.

Having said that, I immediately thought of something else that’s periodic: the wavefunction that’s associated with this object—any object traveling in spacetime, really—is periodic too. So my guts instinct tells me there’s something here that we might want to explore further. 🙂 Could we replace the function for the trajectory with the wavefunction?

Huh? Yes. The wavefunction also associates each x and t, although the association is a bit more complex—literally, because we’ll associate it with two periodic functions: the real part and the imaginary part of the (complex-valued) wavefunction. But for the rest, no problem, I’d say. Remember our wavefunction, when squared, represents the probability of our object being there. [I should say “absolute-squared” rather than squared, but that sounds so weird.]

But… Yes? Well… Don’t we get in trouble here because the same complex number (i.e. r·e^θ = x + i·y) may be related to two points in spacetime—as shown in the example above? My answer is the same: I don’t think so. It’s the same thing: our new representation implies stuff keeps going around and around in it. In fact, that just captures the periodicity of the wavefunction. So… Well… It’s fine. 🙂

The more important question is: what can we do with this new representation? Here I do not have any good answer. Nothing much for the moment. I just wanted to jot it down, because it triggers some deep thoughts—things I don’t quite understand myself, as yet.

First, I like the connection between a regular trajectory in spacetime – as represented by a well-defined function – and the unique direction in time it implies. It’s a simple thing: we know something can travel in any direction in space – forward, backwards, sideways, whatever – but time has one direction only. At least we can see why now: both in Cartesian as well as polar coordinates, we’d want to see a well-behaved function. 🙂 Otherwise we couldn’t work with it.

Another thought is the following. We associate the momentum of a particle with a linear trajectory in spacetime. But what’s linear in curved spacetime? Remember how we struggle to represent – or imagine, I would say – curved spacetime, as evidenced by the fact that most illustrations of curved spacetime represent a two-dimensional space in three-dimensional Cartesian space? Think of the typical illustration, like that rubber surface with the ball deforming it.

That’s why this transformation of a Cartesian coordinate space into a polar coordinate space is such an interesting exercise. We now measure distance along the circle. [Note that we suddenly need to keep track of the number of rotations, which we can do by keeping track of time, as time units become some angle, and linear speed becomes angular speed.] The whole thing underscores, in my view, that’s it’s only our mind that separates time and space: the reality of the object is just its movement or velocity – and that’s one movement.

My guts instinct tells me that this is what the periodicity of the wavefunction (or its component waves, I should say) captures, somehow. If the movement is linear, it’s linear both in space as well in time, so to speak:

As a mental construct, time is always linear – it goes in one direction (and we think of the clock being regular, i.e. not slowing down or speeding up) – and, hence, the mathematical qualities of the time variable in the wavefunction are the same as those of the position variable: it’s a factor in one of its two terms. To be precise, it appears as the t in the E·t term in the argument θ = E·t – p·x. [Note the minus sign appears because we measure angles counter-clockwise when using polar coordinates or complex numbers.]
The trajectory in space is also linear – whether or not space is curved because of the presence of other masses.

OK. I should conclude here, but I want take this conversation one step further. Think of the two graphs below as representing some oscillation in space. Some object that goes back and forth in space: it accelerates and decelerates—and reverses direction. Imagine the g-forces on it as it does so: if you’d be traveling with that object, you would sure feel it’s going back and forth in space! The graph on the left-hand side is our usual perspective on stuff like this: we measure time using some steadily ticking clock, and so the seconds, minutes, hours, days, etcetera just go by. graph 1

The graph on the right-hand side applies our inversion technique. But, frankly, it’s the same thing: it doesn’t give us any new information. It doesn’t look like a well-behaved function but it actually is. It’s just a matter of mathematical convention: if we’d be used to looking at the y-axis as the independent variable (rather than the dependent variable), the function would be acceptable.

This leads me to the idea I started to explore in my previous post, and that’s to try to think of wavefunctions as oscillations of spacetime, rather than oscillations in spacetime. I inserted the following graph in that post—but it doesn’t say all that much, as it suggests we’re doing the same thing here: we’re just swapping axes. The difference is that the θ in the first graph now combines both time and space. We might say it represents spacetime itself. So the wavefunction projects it into some other ‘space’, i.e. the complex space. And then in the second graph, we reflect the whole thing.

dependent independent

So the idea is the following: our functions sort of project one ‘space’ into another ‘space’. In this case: the wavefunction sort of transforms spacetime – i.e. what we like to think of as the ‘physical’ space – into a complex space – which is purely mathematical.

Hmm… This post is becoming way too long, so I need to wrap it up. Look at the graph below, and note the dimension of the axes. We’re looking at an oscillation once more, but an oscillation of time this time around.

Graph 2

Huh? Yes. Imagine that, for some reason, you don’t feel those g-forces while going up and down in space: it’s the rest of the world that’s moving. You think you’re stationary or—what amounts to the same according to the relativity principle—moving in a straight line at constant velocity. The only way how you could explain the rest of the world moving back and forth, accelerating and decelerating, is that time itself is oscillating: objects reverse their direction for no apparent reason—so that’s time reversal—and they do so a varying speeds, so we’ve got a clock going wild!

You’ll nod your head in agreement now and say: that’s Einstein’s intuition in regard to the gravitational force. There’s no force really: mass just bends spacetime in such a way a planet in orbit follows a straight line, in a curved spacetime continuum. What I am saying here is that there must be ways to think of the electromagnetic force in exactly the same way. If the accelerations and decelerations of an electron moving in some electron would really be due to an electromagnetic force in the classical picture of a force (i.e. something pulling or pushing), then it would radiate energy away. We know it doesn’t do that—because otherwise it would spiral down into the nucleus itself. So I’ve been thinking it must be traveling in its own curved spacetime, but then it’s curved because of the electromagnetic force—obviously, as that’s the force that’s much more relevant at this scale.

The underlying thought is simple enough: if gravity curves spacetime, why can’t we look at the other forces as doing the same? Why can’t we think of any force coming ‘with its own space’, so to say? The difference between the various forces is the curvature – which will, obviously, be much more complex (literally) for the electromagnetic force. Just think of the other forces as curving space in more than one dimension. 🙂

I am sure you’ll think I’ve just gone crazy. Perhaps. In any case, I don’t care too much. As mentioned, because the electromagnetic force is different—we don’t have negative masses attracting positive masses when discussing gravity—it’s going to be a much weirder type of curvature, but… Well… That’s probably why we need those ‘two-dimensional’ complex numbers when discussing quantum mechanics! 🙂 So we’ve got some more mathematical dimensions, but the physical principle behind all forces should be the same, no? All forces are measured using Newton’s Law, so we relate them to the motion of some mass. The principle is simple: if force is related to the change in motion of a mass, then the trajectory in the space that’s related to that force will be linear if the force is not acting.

So… Well… Hmm… What?

All of what I write above is a bit of a play with words, isn’t it? An oscillation of spacetime—but then spacetime must oscillate in something else, doesn’it? So in what then is it oscillating?

Great question. You’re right. It must be oscillating in something else or, to be precise, we need some other reference space so as to define what we mean by an oscillation of spacetime. That space is going to be some complex mathematical space—and I use complex both in its mathematical as well as in its everyday meaning here (complicated). Think of, for example, that x-axis representing three-dimensional space. We’d have something similar here: dimensions within dimensions.

There’s some great videos on YouTube that illustrate how one can turn a sphere inside out without punching a hole in it. That’s basically what we’re talking about here: it’s more than just switching the range for the domain of a function, which we can do by that reflection – or mirroring – using the 45º line. Conceptually, it’s really like turning a sphere inside out. Think of the surface of the curve connecting the two spaces.

Huh? Yes. But… Well… You’re right. Stuff like this is for the graduate level, I guess. So I’ll let you think about it—and do watch the videos that follow it. 🙂

In any case, I have to stop my wandering about here. Rather than wrapping up, however, I thought of something else yesterday—and so I’ll quickly jot that down as well, so I can re-visit it some other time. 🙂

Some other thinking on the Uncertainty Principle

I wanted to jot down something else too here. Something about the Uncertainty Principle once more. In my previous post, I noted we should think of Planck’s constant as expressing itself in time or in space, as we have two ways of looking at the dimension of Planck’s constant:

[Planck’s constant] = [ħ] = N∙m∙s = (N∙m)∙s = [energy]∙[time]
[Planck’s constant] = [ħ] = N∙m∙s = (N∙s)∙m = [momentum]∙[distance]

The bracket symbols [ and ] mean: ‘the dimension of what’s between the brackets’. Now, this may look like kids stuff, but the idea is quite fundamental: we’re thinking here of some amount of action (ħ, i.e. the quantum of action) expressing itself in time or, alternatively, expressing itself in space, indeed. In the former case, some amount of energy is expended during some time. In the latter case, some momentum is expended over some distance. We also know ħ can be written in terms of fundamental units, which are referred to as Planck units:

ħ = F_P∙l_P∙t_P = Planck force unit × Planck distance unit × Planck time unit

Finally, we thought of the Planck distance unit and the Planck time unit as the smallest units of time and distance possible. As such, they become countable variables, so we’re talking of a trajectory in terms of discrete steps in space and time here, or discrete states of our particle. As such, the E·t and p·x in the argument (θ) of the wavefunction—remember: θ = (E/ħ)·t − (p/ħ)·x—should be some multiple of ħ as well. We may write:

E·t = m·ħ and p·x = n·ħ, with m and n both positive integers

Of course, there’s uncertainty: Δp·Δx ≥ ħ/2 and ΔE·Δt ≥ ħ/2. Now, if Δx and Δt also become countable variables, so Δx and Δt can only take on values like ±1, ±2, ±3, ±4, etcetera, then we can think of trying to model some kind of random walk through spacetime, combining various values for n and m, as well as various values for Δx and Δt. The relation between E and p, and the related difference between m and n, should determine in what direction our particle should be moving even if it can go along different trajectories. In fact, Feynman’s path integral formulation of quantum mechanics tells us it’s likely to move along different trajectories at the same time, with each trajectory having its own amplitude. Feynman’s formulation uses continuum theory, of course, but a discrete analysis – using a random walk approach – should yield the same result because, when everything is said and done, the fact that physics tells us time and space must become countable at some scale (the Planck scale), suggests that continuum theory may not represent reality, but just be an approximation: a limiting situation, in other words.

Hmmm… Interesting… I’ll need to do something more with this. Unfortunately, I have little time over the coming weeks. Again, I am just writing it down to re-visit it later—probably much later. 😦

The wavefunction as an oscillation of spacetime

Original post:

You probably heard about the experimental confirmation of the existence of gravitational waves by Caltech’s LIGO Lab. Besides further confirming our understanding of the Universe, I also like to think it confirms that the elementary wavefunction represents a propagation mechanism that is common to all forces. However, the fundamental question remains: what is the wavefunction? What are those real and imaginary parts of those complex-valued wavefunctions describing particles and/or photons? [In case you think photons have no wavefunction, see my post on it: it’s fairly straightforward to re-formulate the usual description of an electromagnetic wave (i.e. the description in terms of the electric and magnetic field vectors) in terms of a complex-valued wavefunction. To be precise, in the mentioned post, I showed an electromagnetic wave can be represented as the sum of two wavefunctions whose components reflect each other through a rotation by 90 degrees.]

So what? Well… I’ve started to think that the wavefunction may not only describe some oscillation in spacetime. I’ve started to think the wavefunction—any wavefunction, really (so I am not talking gravitational waves only)—is nothing but an oscillation of spacetime. What makes them different is the geometry of those wavefunctions, and the coefficient(s) representing their amplitude, which must be related to their relative strength—somehow, although I still have to figure out how exactly.

Huh? Yes. Maxwell, after jotting down his equations for the electric and magnetic field vectors, wrote the following back in 1862: “The velocity of transverse undulations in our hypothetical medium, calculated from the electromagnetic experiments of MM. Kohlrausch and Weber, agrees so exactly with the velocity of light calculated from the optical experiments of M. Fizeau, that we can scarcely avoid the conclusion that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena.”

We now know there is no medium – no aether – but physicists still haven’t answered the most fundamental question: what is it that is oscillating? No one has gone beyond the abstract math. I dare to say now that it must be spacetime itself. In order to prove this, I’ll have to study Einstein’s general theory of relativity. But this post will already cover some basics.

The quantum of action and natural units

We can re-write the quantum of action in natural units, which we’ll call Planck units for the time being. They may or may not be the Planck units you’ve heard about, so just think of them as being fundamental, or natural—for the time being, that is. You’ll wonder: what’s natural? What’s fundamental? Well… That’s the question we’re trying to explore in this post, so… Well… Just be patient… 🙂 We’ll denote those natural units as F_P, l_P, and t_P, i.e. the Planck force, Planck distance and Planck time unit respectively. Hence, we write:

ħ = F_P∙l_P∙t_P

Note that F_P, l_P, and t_P are expressed in our old-fashioned SI units here, i.e. in newton (N), meter (m) and seconds (s) respectively. So F_P, l_P, and t_P have a numerical value as well as a dimension, just like ħ. They’re not just numbers. If we’d want to be very explicit, we could write: F_P = F_P [force], or F_P = F_P N, and you could do the same for l_Pand t_P. However, it’s rather tedious to mention those dimensions all the time, so I’ll just assume you understand the symbols we’re using do not represent some dimensionless number. In fact, that’s what distinguishes physical constants from mathematical constants.

Dimensions are also distinguishes physics equations from purely mathematical ones: an equation in physics will always relate some physical quantities and, hence, when you’re dealing with physics equations, you always need to wonder about the dimensions. [Note that the term ‘dimension’ has many definitions… But… Well… I suppose you know what I am talking about here, and I need to move on. So let’s do that.] Let’s re-write that ħ = F_P∙l_P∙t_P formula as follows: ħ/t_P = F_P∙l_P.

F_P∙l_P is, obviously, a force times a distance, so that’s energy. Please do check the dimensions on the left-hand side as well: [ħ/t_P] = [[ħ]/[t_P] = (N·m·s)/s = N·m. In short, we can think of E_P = F_P∙l_P = ħ/t_P as being some ‘natural’ unit as well. But what would it correspond to—physically? What is its meaning? We may be tempted to call it the quantum of energy that’s associated with our quantum of action, but… Well… No. While it’s referred to as the Planck energy, it’s actually a rather large unit, and so… Well… No. We should not think of it as the quantum of energy. We have a quantum of action but no quantum of energy. Sorry. Let’s move on.

In the same vein, we can re-write the ħ = F_P∙l_P∙t_P as ħ/l_P = F_P∙t_P. Same thing with the dimensions—or ‘same-same but different’, as they say in Asia: [ħ/l_P] = [F_P∙t_P] = N·m·s)/m = N·s. Force times time is momentum and, hence, we may now be tempted to think of p_P = F_P∙t_P = ħ/l_P as the quantum of momentum that’s associated with ħ, but… Well… No. There’s no such thing as a quantum of momentum. Not now in any case. Maybe later. 🙂 But, for now, we only have a quantum of action. So we’ll just call ħ/l_P = F_P∙t_P the Planck momentum for the time being.

So now we have two ways of looking at the dimension of Planck’s constant:

[Planck’s constant] = N∙m∙s = (N∙m)∙s = [energy]∙[time]
[Planck’s constant] = N∙m∙s = (N∙s)∙m = [momentum]∙[distance]

In case you didn’t get this from what I wrote above: the brackets here, i.e. the [ and ] symbols, mean: ‘the dimension of what’s between the brackets’. OK. So far so good. It may all look like kids stuff – it actually is kids stuff so far – but the idea is quite fundamental: we’re thinking here of some amount of action (h or ħ, to be precise, i.e. the quantum of action) expressing itself in time or, alternatively, expressing itself in space. In the former case, some amount of energy is expended during some time. In the latter case, some momentum is expended over some distance.

Of course, ideally, we should try to think of action expressing itself in space and time simultaneously, so we should think of it as expressing itself in spacetime. In fact, that’s what the so-called Principle of Least Action in physics is all about—but I won’t dwell on that here, because… Well… It’s not an easy topic, and the diversion would lead us astray. 🙂 What we will do, however, is apply the idea above to the two de Broglie relations: E = ħω and p = ħk. I assume you know these relations by now. If not, just check one of my many posts on them. Let’s see what we can do with them.

**The de Broglie relations**

We can re-write the two de Broglie relations as ħ = E/ω and ħ = p/k. We can immediately derive an interesting property here:

ħ/ħ = 1 = (E/ω)/(p/k) ⇔ E/p = ω/k

So the ratio of the energy and the momentum is equal to the wave velocity. What wave velocity? The group of the phase velocity? We’re talking an elementary wave here, so both are the same: we have only one E and p, and, hence, only one ω and k. The E/p = ω/k identity underscores the following point: the de Broglie equations are a pair of equations here, and one of the key things to learn when trying to understand quantum mechanics is to think of them as an inseparable pair—like an inseparable twin really—as the quantum of action incorporates both a spatial as well as a temporal dimension. Just think of what Minkowski wrote back in 1907, shortly after he had re-formulated Einstein’s special relativity theory in terms of four-dimensional spacetime, and just two years before he died—unexpectely—from an appendicitis: “Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.”

So we should try to think of what that union might represent—and that surely includes looking at the de Broglie equations as a pair of matter-wave equations. Likewise, we should also think of the Uncertainty Principle as a pair of equations: ΔpΔx ≥ ħ/2 and ΔEΔt ≥ ħ/2—but I’ll come back to those later.

The ω in the E = ħω equation and the argument (θ = kx – ωt) of the wavefunction is a frequency in time (or temporal frequency). It’s a frequency expressed in radians per second. You get one radian by dividing one cycle by 2π. In other words, we have 2π radians in one cycle. So ω is related the frequency you’re used to, i.e. f—the frequency expressed in cycles per second (i.e. hertz): we multiply f by 2π to get ω. So we can write: E = ħω = ħ∙2π∙f = h∙f, with h = ħ∙2π (or ħ = h/2π).

Likewise, the k in the p = ħk equation and the argument (θ = kx – ωt) of the wavefunction is a frequency in space (or spatial frequency). Unsurprisingly, it’s expressed in radians per meter.

At this point, it’s good to properly define the radian as a unit in quantum mechanics. We often think of a radian as some distance measured along the circumference, because of the way the unit is constructed (see the illustration below) but that’s right and wrong at the same time. In fact, it’s more wrong than right: the radian is an angle that’s defined using the length of the radius of the unit circle but, when everything is said and done, it’s a unit used to measure some angle—not a distance. That should be obvious from the 2π rad = 360 degrees identity. The angle here is the argument of our wavefunction in quantum mechanics, and so that argument combines both time (t) as well as distance (x): θ = kx – ωt = k(x – c∙t). So our angle (the argument of the wavefunction) integrates both dimensions: space as well as time. If you’re not convinced, just do the dimensional analysis of the kx – ωt expression: both the kx and ωt yield a dimensionless number—or… Well… To be precise, I should say: the kx and ωt products both yield an angle expressed in radians. That angle connects the real and imaginary part of the argument of the wavefunction. Hence, it’s a dimensionless number—but that does not imply it is just some meaningless number. It’s not meaningless at all—obviously!

Let me try to present what I wrote above in yet another way. The θ = kx – ωt = (p/ħ)·x − (E/ħ)·t equation suggests a fungibility: the wavefunction itself also expresses itself in time and/or in space, so to speak—just like the quantum of action. Let me be precise: the p·x factor in the (p/ħ)·x term represents momentum (whose dimension is N·s) being expended over a distance, while the E·t factor in the (E/ħ)·t term represents energy (expressed in N·m) being expended over some time. [As for the minus sign in front of the (E/ħ)·t term, that’s got to do with the fact that the arrow of time points in one direction only while, in space, we can go in either direction: forward or backwards.] Hence, the expression for the argument tells us that both are essentially fungible—which suggests they’re aspects of one and the same thing. So that‘s what Minkowski intuition is all about: spacetime is one, and the wavefunction just connects the physical properties of whatever it is that we are observing – an electron, or a photon, or whatever other elementary particle – to it.

Of course, the corollary to thinking of unified spacetime is thinking of the real and imaginary part of the wavefunction as one—which we’re supposed to do as a complex number is… Well… One complex number. But that’s easier said than actually done, of course. One way of thinking about the connection between the two spacetime ‘dimensions’ – i.e. t and x, with x actually incorporating three spatial dimensions in space in its own right (see how confusing the term ‘dimension’ is?) – and the two ‘dimensions’ of a complex number is going from Cartesian to polar coordinates, and vice versa. You now think of Euler’s formula, of course – if not, you should – but let me insert something more interesting here. 🙂 I took it from Wikipedia. It illustrates how a simple sinusoidal function transforms as we go from Cartesian to polar coordinates.

Cartesian_to_polar

Interesting, isn’t it? Think of the horizontal and vertical axis in the Cartesian space as representing time and… Well… Space indeed. 🙂 The function connects the space and time dimension and might, for example, represent the trajectory of some object in spacetime. Admittedly, it’s a rather weird trajectory, as the object goes back and forth in some box in space, and accelerates and decelerates all of the time, reversing its direction in the process… But… Well… Isn’t that how we think of a an electron moving in some orbital? 🙂 With that in mind, look at how the same movement in spacetime looks like in polar coordinates. It’s also some movement in a box—but both the ‘horizontal’ and ‘vertical’ axis (think of these axes as the real and imaginary part of a complex number) are now delineating our box. So, whereas our box is a one-dimensional box in spacetime only (our object is constrained in space, but time keeps ticking), it’s a two-dimensional box in our ‘complex’ space. Isn’t it just nice to think about stuff this way?

As far as I am concerned, it triggers the same profound thoughts as that E/p = ω/k relation. The left-hand side is a ratio between energy and momentum. Now, one way to look at energy is that it’s action per time unit. Likewise, momentum is action per distance unit. Of course, ω is expressed as some quantity (expressed in radians, to be precise) per time unit, and k is some quantity (again, expressed in radians) per distance unit. Because this is a physical equation, the dimension of both sides of the equation has to be the same—and, of course, it is the same: the action dimension in the numerator and denominator of the ratio on the left-hand side of the E/p = ω/k equation cancel each other. But… What? Well… Wouldn’t it be nice to think of the dimension of the argument of our wavefunction as being the dimension of action, rather than thinking of it as just some mathematical thing, i.e. an angle. I like to think the argument of our wavefunction is more than just an angle. When everything is said and done, it has to be something physical—if onlyh because the wavefunction describes something physical. But… Well… I need to do some more thinking on this, so I’ll just move on here. Otherwise this post risks becoming a book in its own right. 🙂

Let’s get back to the topic we were discussing here. We were talking about natural units. More in particular, we were wondering: what’s natural? What does it mean?

Back to Planck units

Let’s start with time and distance. We may want to think of l_P and t_P as the smallest distance and time units possible—so small, in fact, that both distance and time become countable variables at that scale.

Huh? Yes. I am sure you’ve never thought of time and distance as countable variables but I’ll come back to this rather particular interpretation of the Planck length and time unit later. So don’t worry about it now: just make a mental note of it. The thing is: if t_P and l_P are the smallest time and distance units possible, then the smallest cycle we can possibly imagine will be associated with those two units: we write: ω_P = 1/t_P and k_P = 1/l_P. What’s the ‘smallest possible’ cycle? Well… Not sure. You should not think of some oscillation in spacetime as for now. Just think of a cycle. Whatever cycle. So, as for now, the smallest cycle is just the cycle you’d associate with the smallest time and distance units possible—so we cannot write ω_P = 2/t_P, for example, because that would imply we can imagine a time unit that’s smaller than t_P, as we can associate two cycles with t_P now.

OK. Next step. We can now define the Planck energy and the Planck momentum using the de Broglie relations:

E_P = ħ∙ω_P = ħ/t_P and p_P = ħ∙k_P = ħ/l_P

You’ll say that I am just repeating myself here, as I’ve given you those two equations already. Well… Yes and no. At this point, you should raise the following question: why are we using the angular frequency (ω_P = 2π·f_P) and the reduced Planck constant (ħ = h/2π), rather than f_P or h?

That’s a great question. In fact, it begs the question: what’s the physical constant really? We have two mathematical constants – ħ and h – but they must represent the same physical reality. So is one of the two constants more real than the other? The answer is unambiguously: yes! The Planck energy is defined as E_P = ħ/t_P =(h/2π)/t_P, so we cannot write this as E_P = h/t_P. The difference is that 1/2π factor, and it’s quite fundamental, as it implies we’re actually not associating a full cycle with t_P and l_P but a radian of that cycle only.

Huh? Yes. It’s a rather strange observation, and I must admit I haven’t quite sorted out what this actually means. The fundamental idea remains the same, however: we have a quantum of action, ħ (not h!), that can express itself as energy over the smallest distance unit possible or, alternatively, that expresses itself as momentum over the smallest time unit possible. In the former case, we write it as E_P = F_P∙l_P = ħ/t_P. In the latter, we write it as p_P = F_P∙t_P = ħ/l_P. Both are aspects of the same reality, though, as our particle moves in space as well as in time, i.e. it moves in spacetime. Hence, one step in space, or in time, corresponds to one radian. Well… Sort of… Not sure how to further explain this. I probably shouldn’t try anyway. 🙂

The more fundamental question is: with what speed is is moving? That question brings us to the next point. The objective is to get some specific value for l_P and t_P, so how do we do that? How can we determine these two values? Well… That’s another great question. 🙂

The first step is to relate the natural time and distance units to the wave velocity. Now, we do not want to complicate the analysis and so we’re not going to throw in some rest mass or potential energy here. No. We’ll be talking a theoretical zero-mass particle. So we’re not going to consider some electron moving in spacetime, or some other elementary particle. No. We’re going to think about some zero-mass particle here, or a photon. [Note that a photon is not just a zero-mass particle. It’s similar but different: in one of my previous posts, I showed a photon packs more energy, as you get two wavefunctions for the price of one, so to speak. However, don’t worry about the difference here.]

Now, you know that the wave velocity for a zero-mass particle and/or a photon is equal to the speed of light. To be precise, the wave velocity of a photon is the speed of light and, hence, the speed of any zero-mass particle must be the same—as per the definition of mass in Einstein’s special relativity theory. So we write: l_P/t_P = c ⇔ l_P = c∙t_P and t_P = l_P/c. In fact, we also get this from dividing E_P by p_P, because we know that E/p = c, for any photon (and for any zero-mass particle, really). So we know that E_P/p_P must also equal c. We can easily double-check that by writing: E_P/p_P = (ħ/t_P)/(ħ/l_P) = l_P/t_P = c. Substitution in ħ = F_P∙l_P∙t_P yields ħ = c∙F_P∙t_P² or, alternatively, ħ = F_P∙l_P²/c. So we can now write F_P as a function of l_P and/or t_P:

F_P = ħ∙c/l_P² = ħ/(c∙t_P²)

We can quickly double-check this by dividing F_P = ħ∙c/l_P² by F_P = ħ/(c∙t_P²). We get: 1 = c²∙t_P²/l_P² ⇔ l_P²/t_P² = c² ⇔ l_P/t_P = c.

Nice. However, this does not uniquely define F_P, l_P, and t_P. The problem is that we’ve got only two equations (ħ = F_P∙l_P∙t_P and l_P/t_P = c) for three unknowns (F_P, l_P, and t_P). Can we throw in one or both of the de Broglie equations to get some final answer?

I wish that would help, but it doesn’t—because we get the same ħ = F_P∙l_P∙t_P equation. Indeed, we’re just re-defining the Planck energy (and the Planck momentum) by that E_P = ħ/t_P (and p_P = ħ/l_P) equation here, and so that does not give us a value for E_P (and p_P). So we’re stuck. We need some other formula so we can calculate the third unknown, which is the Planck force unit (F_P). What formula could we possibly choose?

Well… We got a relationship by imposing the condition that l_P/t_P = c, which implies that if we’d measure the velocity of a photon in Planck time and distance units, we’d find that its velocity is one, so c = 1. Can we think of some similar condition involving ħ? The answer is: we can and we can’t. It’s not so simple. Remember we were thinking of the smallest cycle possible? We said it was small because t_P and l_P were the smallest units we could imagine. But how do we define that? The idea is as follows: the smallest possible cycle will pack the smallest amount of action, i.e. h (or, expressed per radian rather than per cycle, ħ).

Now, we usually think of packing energy, or momentum, instead of packing action, but that’s because… Well… Because we’re not good at thinking the way Minkowski wanted us to think: we’re not good at thinking of some kind of union of space and time. We tend to think of something moving in space, or, alternatively, of something moving in time—rather than something moving in spacetime. In short, we tend to separate dimensions. So that’s why we’d say the smallest possible cycle would pack an amount of energy that’s equal to E_P = ħ∙ω_P = ħ/t_P, or an amount of momentum that’s equal to p_P = ħ∙k_P = ħ/l_P. But both are part and parcel of the same reality, as evidenced by the E = m∙c² = m∙c∙c = p∙c equality. [This equation only holds for a zero-mass particle (and a photon), of course. It’s a bit more complicated when we’d throw in some rest mass, but we can do that later. Also note I keep repeating my idea of the smallest cycle, but we’re talking radians of a cycle, really.]

So we have that mass-energy equivalence, which is also a mass-momentum equivalence according to that E = m∙c² = m∙c∙c = p∙c formula. And so now the gravitational force comes into play: there’s a limit to the amount of energy we can pack into a tiny space. Or… Well… Perhaps there’s no limit—but if we pack an awful lot of energy into a really tiny speck of space, then we get a black hole.

However, we’re getting a bit ahead of ourselves here, so let’s first try something else. Let’s throw in the Uncertainty Principle.

The Uncertainty Principle

As mentioned above, we can think of some amount of action expressing itself over some time or, alternatively, over some distance. In the former case, some amount of energy is expended over some time. In the latter case, some momentum is expended over some distance. That’s why the energy and time variables, and the momentum and distance variables, are referred to as complementary. It’s hard to think of both things happening simultaneously (whatever that means in spacetime), but we should try! Let’s now look at the Uncertainty relations once again (I am writing uncertainty with a capital U out of respect—as it’s very fundamental, indeed!):

ΔpΔx ≥ ħ/2 and ΔEΔt ≥ ħ/2.

Note that the ħ/2 factor on the right-hand side quantifies the uncertainty, while the right-hand side of the two equations (ΔpΔx and ΔEΔt) are just an expression of that fundamental uncertainty. In other words, we have two equations (a pair), but there’s only one fundamental uncertainty, and it’s an uncertainty about a movement in spacetime. Hence, that uncertainty expresses itself in both time as well as in space.

Note the use of ħ rather than h, and the fact that the 1/2 factor makes it look like we’re splitting ħ over ΔpΔx and ΔEΔt respectively—which is actually a quite sensible explanation of what this pair of equations actually represent. Indeed, we can add both relations to get the following sum:

ΔpΔx + ΔEΔt ≥ ħ/2 + ħ/2 = ħ

Interesting, isn’t it? It explains that 1/2 factor which troubled us when playing with the de Broglie relations.

Let’s now think about our natural units again—about l_P, and t_P in particular. As mentioned above, we’ll want to think of them as the smallest distance and time units possible: so small, in fact, that both distance and time become countable variables, so we count x and t as 0, 1, 2, 3 etcetera. We may then imagine that the uncertainty in x and t is of the order of one unit only, so we write Δx = l_P and Δt = t_P. So we can now re-write the uncertainty relations as:

Δp·l_P = ħ/2
ΔE·t_P = ħ/2

Hey! Wait a minute! Do we have a solution for the value of l_P and t_P here? What if we equate the natural energy and momentum units to ΔE and Δp here? Well… Let’s try it. First note that we may think of the uncertainty in t, or in x, as being equal to plus or minus one unit, i.e. ±1. So the uncertainty is two units really. [Frankly, I just want to get rid of that 1/2 factor here.] Hence, we can re-write the ΔpΔx = ΔEΔt = ħ/2 equations as:

ΔpΔx = p_P∙l_P = F_P∙t_P∙l_P = ħ
ΔEΔt = E_P∙t_P = F_P∙l_P∙t_P = ħ

Hmm… What can we do with this? Nothing much, unfortunately. We’ve got the same problem: we need a value for F_P (or for p_P, or for E_P) to get some specific value for l_P and t_P, so we’re stuck once again. We have three variables and two equations only, so we have no specific value for either of them. 😦

What to do? Well… I will give you the answer now—the answer you’ve been waiting for, really—but not the technicalities of it. There’s a thing called the Schwarzschild radius, aka as the gravitational radius. Let’s analyze it.

The Schwarzschild radius and the Planck length

The Schwarzschild radius is just the radius of a black hole. Its formal definition is the following: it is the radius of a sphere such that, if all the mass of an object were to be compressed within that sphere, the escape velocity from the surface of the sphere would equal the speed of light (c). The formula for the Schwartzschild radius is the following:

R_S = 2m·G/c²

G is the gravitational constant here: G ≈ 6.674×10⁻¹¹ N⋅m²/kg². [Note that Newton’s F = m·a Law tells us that 1 kg = 1 N·s²/m, as we’ll need to substitute units later.]

But what is the mass (m) in that R_S = 2m·G/c² equation? Using equivalent time and distance units (so c = 1), we wrote the following for a zero-mass particle and for a photon respectively:

E = m = p = ħ/2 (zero-mass particle)
E = m = p = ħ (photon)

How can a zero-mass particle, or a photon, have some mass? Well… Because it moves at the speed of light. I’ve talked about that before, so I’ll just refer you to my post on that. Of course, the dimension of the right-hand side of these equations (i.e. ħ/2 or ħ) symbol has to be the same as the dimension on the left-hand side, so the ‘ħ’ in the E = ħ equation (or E = ħ/2 equation) is a different ‘ħ’ in the p = ħ equation (or p = ħ/2 equation). So we must be careful here. Let’s write it all out, so as to remind ourselves of the dimensions involved:

E [N·m] = ħ [N·m·s/s] = E_P = F_P∙l_P∙t_P/t_P
p [N·s] = ħ [N·m·s/m] = p_P = F_P∙l_P∙t_P/l_P

Now, let’s check this by cheating. I’ll just give you the numerical values—even if we’re not supposed to know them at this point—so you can see I am not writing nonsense here:

E_P= 1.0545718×10⁻³⁴N·m·s/(5.391×10⁻⁴⁴ s) = (1.21×10⁴⁴N)·(1.6162×10⁻³⁵m) = 1.9561×10⁹N·m
p_P=1.0545718×10⁻³⁴N·m·s/(1.6162×10⁻³⁵m) = (1.21×10⁴⁴N)·(5.391×10⁻⁴⁴ s) = 6.52485 N·s

You can google the Planck units, and you’ll see I am not taking you for a ride here. 🙂

The associated Planck mass is m_P = E_P/c² = 1.9561×10⁹N·m/(2.998×10⁸m/s)² = 2.17651×10⁻⁸ N·s²/m = 2.17651×10⁻⁸ kg. So let’s plug that value into R_S = 2m·G/c²equation. We get:

R_S = 2m·G/c²= [(2.17651×10⁻⁸ kg)·(6.674×10⁻¹¹ N⋅m²/kg²)/(8.988×10¹⁶m²·s⁻²)

= 1.6162×10⁻³⁵kg·N⋅m²·kg⁻²·m²·s⁻² = 1.6162×10⁻³⁵kg·N⋅m²·kg⁻²·m²·s⁻²= 1.6162×10⁻³⁵m = l_P

Bingo! You can look it up: 1.6162×10⁻³⁵m is the Planck length indeed, so the Schwarzschild radius is the Planck length. We can now easily calculate the other Planck units:

t_P= l_P/c = 1.6162×10⁻³⁵m/(2.998×10⁸m/s) = 5.391×10⁻⁴⁴s
F_P= ħ/(t_P∙l_P)= (1.0545718×10⁻³⁴N·m·s)/[(1.6162×10⁻³⁵m)·(5.391×10⁻⁴⁴s) = 1.21×10⁻⁴⁴N

Bingo again! 🙂

[…] But… Well… Look at this: we’ve been cheating all the way. First, we just gave you that formula for the Schwarzschild radius. It looks like an easy formula but its derivation involves a profound knowledge of general relativity theory. So we’d need to learn about tensors and what have you. The formula is, in effect, a solution to what is known as Einstein’s field equations, and that’s pretty complicated stuff.

However, my crime is much worse than that: I also gave you those numerical values for the Planck energy and momentum, rather than calculating them. I just couldn’t calculate them with the knowledge we have so far. When everything is said and done, we have more than three unknowns. We’ve got five in total, including the Planck charge (q_P) and, hence, we need five equations. Again, I’ll just copy them from Wikipedia, because… Well… What we’re discussing here is way beyond the undergraduate physics stuff that we’ve been presenting so far. The equations are the following. Just have a look at them and move on. 🙂

Finally, I should note one more thing: I did not use 2m but m in Schwarzschild’s formula. Why? Well… I have no good answer to that. I did it to ensure I got the result we wanted to get. It’s that 1/2 factor again. In fact, the E = m = p = ħ/2 is the correct formula to use, and all would come out alright if we did that and defined the magnitude of the uncertainty as one unit only, but so we used the E = m = p = ħ formula instead, i.e. the equation that’s associated with a photon. You can re-do the calculations as an exercise: you’ll see it comes out alright.

Just to make things somewhat more real, let me note that the Planck energy is very substantial: 1.9561×10⁹N·m ≈ 2×10⁹J is equivalent to the energy that you’d get out of burning 60 liters of gasoline—or the mileage you’d get out of 16 gallons of fuel! In short, it’s huge, and so we’re packing that into a unimaginably small space. To understand how that works, you can think of the E = h∙f ⇔ h = E/f relation once more. The h = E/f ratio implies that energy and frequency are directly proportional to each other, with h the coefficient of proportionality. Shortening the wavelength, amounts to increasing the frequency and, hence, the energy. So, as you think of our cycle becoming smaller and smaller, until it becomes the smallest cycle possible, you should think of the frequency becoming unimaginably large. Indeed, as I explained in one of my other posts on physical constants, we’re talking the the 10⁴³Hz scale here. However, we can always choose our time unit such that we measure the frequency as one cycle per time unit. Because the energy per cycle remains the same, it means the quantum of action (ħ = F_P∙l_P∙t_P) expresses itself over extremely short time spans, which means the E_P = F_P∙l_P product becomes huge, as we’ve shown above. The rest of the story is the same: gravity comes into play, and so our little blob in spacetime becomes a tiny black hole. Again, we should think of both space and time: they are joined in ‘some kind of union’ here, indeed, as they’re intimately connected through the wavefunction, which travels at the speed of light.

The wavefunction as an oscillation in and of spacetime

OK. Now I am going to present the big idea I started with. Let me first ask you a question: when thinking about the Planck-Einstein relation (I am talking about the E = ħ∙ω relation for a photon here, rather than the equivalent de Broglie equation for a matter-particle), aren’t you struck by the fact that the energy of a photon depends on the frequency of the electromagnetic wave only? I mean… It does not depend on its amplitude. The amplitude is mentioned nowhere. The amplitude is fixed, somehow—or considered to be fixed.

Isn’t that strange? I mean… For any other physical wave, the energy would not only depend on the frequency but also on the amplitude of the wave. For a photon, however, it’s just the frequency that counts. Light of the same frequency but higher intensity (read: more energy) is not a photon with higher amplitude, but just more photons. So it’s the photons that add up somehow, and so that explains the amplitude of the electric and magnetic field vectors (i.e. E and B) and, hence, the intensity of the light. However, every photon considered separately has the same amplitude apparently. We can only increase its energy by increasing the frequency. In short, ω is the only variable here.

Let’s look at that angular frequency once more. As you know, it’s expressed in radians per second but, if you multiply ω by 2π, you get the frequency you’re probably more acquainted with: f = 2πω = f cycles per second. The Planck-Einstein relation is then written as E = h∙f. That’s easy enough. But what if we’d change the time unit here? For example, what if our time unit becomes the time that’s needed for a photon to travel one meter? Let’s examine it.

Let’s denote that time unit by t_m, so we write: 1 t_m = 1/c s ⇔ t_m^–¹ = c s^–¹, with c ≈ 3×10⁸. The frequency, as measured using our new time unit, changes, obviously: we have to divide its former value by c now. So, using our little subscript once more, we could write: f_m = f/c. [Why? Just add the dimension to make things more explicit: f s^–¹ = f/c t_m^–¹ = f/c t_m^–¹.] But the energy of the photon should not depend on our time unit, should it?

Don’t worry. It doesn’t: the numerical value of Planck’s constant (h) would also change, as we’d replace the second in its dimension (N∙m∙s) by c times our new time unit t_m. However, Planck’s constant remains what it is: some physical constant. It does not depend on our measurement units: we can use the SI units, or the Planck units (F_P, l_P, and t_P), or whatever unit you can think of. It doesn’t matter: h (or ħ = h/2π) is what is—it’s the quantum of action, and so that’s a physical constant (as opposed to a mathematical constant) that’s associated with one cycle.

Now, I said we do not associate the wavefunction of a photon with an amplitude, but we do associate it with a wavelength. We do so using the standard formula for the velocity of a wave: c = f∙λ ⇔ λ = c/f. We can also write this using the angular frequency and the wavenumber: c = ω/k, with k = 2π/λ. We can double-check this, because we know that, for a photon, the following relation holds: E/p = c. Hence, using the E = ħ∙ω and p = ħ∙k relations, we get: (ħ∙ω)/(ħ∙k) = ω/k = c. So we have options here: h can express itself over a really long wavelength, or it can do so over an extremely short wavelength. We re-write p = ħ∙k as p = E/c = ħ∙2π/λ = h/λ ⇔ E = h∙c/λ ⇔ h∙c = E∙λ. We know this relationship: the energy and the wavelength of a photon (or an electromagnetic wave) are inversely proportional to each other.

Once again, we may want to think of the shortest wavelength possible. As λ gets a zillion times smaller, E gets a zillion times bigger. Is there a limit? There is. As I mentioned above, the gravitational force comes into play here: there’s a limit to the amount of energy we can pack into a tiny space. If we pack an awful lot of energy into a really tiny speck of space, then we get a black hole. In practical terms, that implies our photon can’t travel, as it can’t escape from the black hole it creates. That’s what that calculation of the Schwarzschild radius was all about.

We can—in fact, we should—now apply the same reasoning to the matter-wave. Instead of a photon, we should try to think of a zero-mass matter-particle. You’ll say: that’s a contradiction. Matter-particles – as opposed to force-carrying particles, like photons (or bosons in general) – must have some rest mass, so they can’t be massless. Well… Yes. You’re right. But we can throw the rest mass in later. I first want to focus on the abstract principles, i.e. the propagation mechanism of the matter-wave.

Using natural units, we know our particle will move in spacetime with velocity Δx/Δt = 1/1 = 1. Of course, it has to have some energy to move, or some momentum. We also showed that, if it’s massless, and the elementary wavefunction is eⁱ^[(p/^ħ)x^{– (E/}^ħ)t), then we know the energy, and the momentum, has to be equal to ħ/2. Where does it get that energy, or momentum? Not sure. I like to think it borrows it from spacetime, as it breaks some potential barrier between those two points, and then it gives it back. Or, if it’s at point x = t = 0, then perhaps it gets it from some other massless particle moving from x = t = −1. In both cases, we’d like to think our particle keeps moving. So if the first description (borrowing) is correct, it needs to keep borrowing and returning energy in some kind of interaction with spacetime itself. If it’s the second description, it’s more like spacetime bumping itself forward.

In both cases, however, we’re actually trying to visualize (or should I say: imagine?) some oscillation of spacetime itself, as opposed to an oscillation in spacetime.

Huh? Yes. The idea is the following here: we like to think of the wavefunction as the dependent variable: both its real as well as its imaginary part are a function of x and t, indeed. But what if we’d think of x and t as dependent variables? In that case, the real and imaginary part of the wavefunction would be the independent variables. It’s just a matter of perspective. We sort of mirror our function: we switch its domain for its range, and its range for its domain, as shown below. It all makes sense, doesn’t it? Space and time appear as separate dimensions to us, but they’re intimately connected through c, ħ and the other fundamental physical constants. Likewise, the real and imaginary part of the wavefunction appear as separate dimensions, but they’re intimately connected through π and Euler’s number, i.e. through mathematical constants. That cannot be a coincidence: the mathematical and physical ‘space’ reflect each other through the wavefunction, just like the domain and range of a function reflect each other through that function. So physics and math must meet in some kind of union—at least in our mind, they do!

So, yes, we can—and probably should—be looking at the wavefunction as an oscillation of spacetime, rather than as an oscillation in spacetime only. As mentioned in my introduction, I’ll need to study general relativity theory—and very much in depth—to convincingly prove that point, but I am sure it can be done.

You’ll probably think I am arrogant when saying that—and I probably am—but then I am very much emboldened by the fact some nuclear scientist told me a photon doesn’t have any wavefunction: it’s just those E and B vectors, he told me—and then I found out he was dead wrong, as I showed in my previous post! So I’d rather think more independently now. I’ll keep you guys posted on progress—but it will probably take a while to figure it all out. In the meanwhile, please do let me know your ideas on this. 🙂

Let me wrap up this little excursion with two small notes:

We have this E/c = p relation. The mass-energy equivalence relation implies momentum must also have an equivalent mass. If E = m∙c², then p = m∙c ⇔ m = p/c. It’s obvious, but I just thought it would be useful to highlight this.
When we studied the ammonia molecule as a two-state system, our space was not a continuum: we allowed just two positions—two points in space, which we defined with respect to the system. So x was a discrete variable. We assumed time to be continuous, however, and so we got those nice sinusoids as a solution to our set of Hamiltonian equations. However, if we look at space as being discrete, or countable, we should probably think of time as being countable as well. So we should, perhaps, think of a particle being at point x = t = 0 first, and, then, being at point x = t = 1. Instead of the nice sinusoids, we get some boxcar function, as illustrated below, but probably varying between 0 and 1—or whatever other normalized values. You get the idea, I hope. 🙂

Post Scriptum on the Principle of Least Action: As noted above, the Principle of Least Action is not very intuitive, even if Feynman’s exposé of it is not as impregnable as it may look at first. To put it simply, the Principle of Least Action says that the average kinetic energy less the average potential energy is as little as possible for the path of an object going from one point to another. So we have a path or line integral here. In a gravitation field, that integral is the following:

Least action

The integral is not all that important. Just note its dimension is the dimension of action indeed, as we multiply energy (the integrand) with time (dt). We can use the Principle of Least Action to re-state Newton’s Law, or whatever other classical law. Among other things, we’ll find that, in the absence of any potential, the trajectory of a particle will just be some straight line.

In quantum mechanics, however, we have uncertainty, as expressed in the ΔpΔx ≥ ħ/2 and ΔEΔt ≥ ħ/2 relations. Now, that uncertainty may express itself in time, or in distance, or in both. That’s where things become tricky. 🙂 I’ve written on this before, but let me copy Feynman himself here, with a more exact explanation of what’s happening (just click on the text to enlarge):

Feynman

The ‘student’ he speaks of above, is himself, of course. 🙂

Too complicated? Well… Never mind. I’ll come back to it later. 🙂

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All what you ever wanted to know about the photon wavefunction…

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately read the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link.

Original post:

This post is, essentially, a continuation of my previous post, in which I juxtaposed the following images:

Both are the same, and then they’re not. The illustration on the right-hand side is a regular quantum-mechanical wavefunction, i.e. an amplitude wavefunction. You’ve seen that one before. In this case, the x-axis represents time, so we’re looking at the wavefunction at some particular point in space. ]You know we can just switch the dimensions and it would all look the same.] The illustration on the left-hand side looks similar, but it’s not an amplitude wavefunction. The animation shows how the electric field vector (E) of an electromagnetic wave travels through space. Its shape is the same. So it’s the same function. Is it also the same reality?

Yes and no. And I would say: more no than yes—in this case, at least. Note that the animation does not show the accompanying magnetic field vector (B). That vector is equally essential in the electromagnetic propagation mechanism according to Maxwell’s equations, which—let me remind you—are equal to:

∂B/∂t = –∇×E
∂E/∂t = ∇×B

In fact, I should write the second equation as ∂E/∂t = c²∇×B, but then I assume we measure time and distance in equivalent units, so c = 1.

You know that E and B are two aspects of one and the same thing: if we have one, then we have the other. To be precise, B is always orthogonal to E in the direction that’s given by the right-hand rule for the following vector cross-product: B = e_x×E, with e_x the unit vector pointing in the x-direction (i.e. the direction of propagation). The reality behind is illustrated below for a linearly polarized electromagnetic wave.

E and b

The B = e_x×E equation is equivalent to writing B= i·E, which is equivalent to:

B = i·E = eⁱ^(π/2)·eⁱ^{(kx − ωt)} = cos(kx − ωt + π/2) + i·sin(kx − ωt + π/2)

= −sin((kx − ωt) + i·cos(kx − ωt)

Now, E and B have only two components: E_yand E_z, and B_yand B_z. That’s only because we’re looking at some ideal or elementary electromagnetic wave here but… Well… Let’s just go along with it. 🙂 It is then easy to prove that the equation above amounts to writing:

B_y= cos(kx − ωt + π/2) = −sin(kx − ωt) = −E_z
B_z= sin(kx − ωt + π/2) = cos(kx − ωt) = E_y

We should now think of E_y and E_zas the real and imaginary part of some wavefunction, which we’ll denote as ψ_E = eⁱ^{(kx − ωt)}. So we write:

E = (E_y, E_z) = E_y + i·E_z= cos(kx − ωt) + i∙sin(kx − ωt) = Re(ψ_E) + i·Im(ψ_E) = ψ_E = eⁱ^{(kx − ωt)}

What about B? We just do the same, so we write:

B = (B_y, B_z) = B_y + i·B_z= ψ_B = i·E = i·ψ_E = −sin(kx − ωt) + i∙sin(kx − ωt) = − Im(ψ_E) + i·Re(ψ_E)

Now we need to prove that ψ_E and ψ_B are regular wavefunctions, which amounts to proving Schrödinger’s equation, i.e. ∂ψ/∂t = i·(ħ/m)·∇²ψ, for both ψ_E and ψ_B. [Note I use the Schrödinger’s equation for a zero-mass spin-zero particle here, which uses the ħ/m factor rather than the ħ/(2m) factor.] To prove that ψ_E and ψ_B are regular wavefunctions, we should prove that:

Re(∂ψ_E/∂t) = −(ħ/m)·Im(∇²ψ_E) and Im(∂ψ_E/∂t) = (ħ/m)·Re(∇²ψ_E), and
Re(∂ψ_B/∂t) = −(ħ/m)·Im(∇²ψ_B) and Im(∂ψ_B/∂t) = (ħ/m)·Re(∇²ψ_B).

Let’s do the calculations for the second pair of equations. The time derivative on the left-hand side is equal to:

∂ψ_B/∂t = −iω·ieⁱ^{(kx − ωt)}= ω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·cos(kx − ωt) + iω·sin(kx − ωt)

The second-order derivative on the right-hand side is equal to:

∇²ψ_B= ∂²ψ_B/∂x²= i·k²·eⁱ^{(kx − ωt)}= k²·cos(kx − ωt) + i·k²·sin(kx − ωt)

So the two equations for ψ_Bare equivalent to writing:

Re(∂ψ_B/∂t) = −(ħ/m)·Im(∇²ψ_B) ⇔ ω·cos(kx − ωt) = k²·(ħ/m)·cos(kx − ωt)
Im(∂ψ_B/∂t) = (ħ/m)·Re(∇²ψ_B) ⇔ ω·sin(kx − ωt) = k²·(ħ/m)·sin(kx − ωt)

So we see that both conditions are fulfilled if, and only if, ω = k²·(ħ/m).

Now, we also demonstrated in that post of mine that Maxwell’s equations imply the following:

∂B_y/∂t = –(∇×E)_y = ∂E_z/∂x = ∂[sin(kx − ωt)]/∂x = k·cos(kx − ωt) = k·E_y
∂B_z/∂t = –(∇×E)_z = – ∂E_y/∂x = – ∂[cos(kx − ωt)]/∂x = k·sin(kx − ωt) = k·E_z

Hence, using those B_y= −E_zand B_z= E_yequations above, we can also calculate these derivatives as:

∂B_y/∂t = −∂E_z/∂t = −∂sin(kx − ωt)/∂t = ω·cos(kx − ωt) = ω·E_y
∂B_z/∂t = ∂E_y/∂t = ∂cos(kx − ωt)/∂t = −ω·[−sin(kx − ωt)] = ω·E_z

In other words, Maxwell’s equations imply that ω = k, which is consistent with us measuring time and distance in equivalent units, so the phase velocity is c = 1 = ω/k.

So far, so good. We basically established that the propagation mechanism for an electromagnetic wave, as described by Maxwell’s equations, is fully coherent with the propagation mechanism—if we can call it like that—as described by Schrödinger’s equation. We also established the following equalities:

ω = k
ω = k²·(ħ/m)

The second of the two de Broglie equations tells us that k = p/ħ, so we can combine these two equations and re-write these two conditions as:

ω/k = 1 = k·(ħ/m) = (p/ħ)·(ħ/m) = p/m ⇔ p = m

What does this imply? The p here is the momentum: p = m·v, so this condition implies v must be equal to 1 too, so the wave velocity is equal to the speed of light. Makes sense, because we actually are talking light here. 🙂 In addition, because it’s light, we also know E/p = c = 1, so we have – once again – the general E = p = m equation, which we’ll need!

OK. Next. Let’s write the Schrödinger wave equation for both wavefunctions:

∂ψ_E/∂t = i·(ħ/m_E)·∇²ψ_E, and
∂ψ_B/∂t = i·(ħ/m_B)·∇²ψ_B.

Huh? What’s m_E and m_E? We should only associate one mass concept with our electromagnetic wave, shouldn’t we? Perhaps. I just want to be on the safe side now. Of course, if we distinguish m_E and m_B, we should probably also distinguish p_E and p_B, and E_E and E_B as well, right? Well… Yes. If we accept this line of reasoning, then the mass factor in Schrödinger’s equations is pretty much like the 1/c² = μ₀ε₀ factor in Maxwell’s (1/c²)·∂E/∂t = ∇×B equation: the mass factor appears as a property of the medium, i.e. the vacuum here! [Just check my post on physical constants in case you wonder what I am trying to say here, in which I explain why and how c defines the (properties of the) vacuum.]

To be consistent, we should also distinguish p_E and p_B, and E_E and E_B, and so we should write ψ_Eand ψ_B as:

ψ_E = eⁱ^{(k_Ex − ω_Et)}, and
ψ_B = eⁱ^{(k_Bx − ω_Bt)}.

Huh? Yes. I know what you think: we’re talking one photon—or one electromagnetic wave—so there can be only one energy, one momentum and, hence, only one k, and one ω. Well… Yes and no. Of course, the following identities should hold: k_E = k_B and, likewise, ω_E= ω_B. So… Yes. They’re the same: one k and one ω. But then… Well… Conceptually, the two k’s and ω’s are different. So we write:

p_E = E_E = m_E, and
p_B = E_B = m_B.

The obvious question is: can we just add them up to find the total energy and momentum of our photon? The answer is obviously positive: E = E_E + E_B, p = p_E + p_B and m = m_E + m_B.

Let’s check a few things now. How does it work for the phase and group velocity of ψ_Eand ψ_B? Simple:

v_g = ∂ω_E/∂k_E = ∂[E_E/ħ]/∂[p_E/ħ] = ∂E_E/∂p_E = ∂p_E/∂p_E = 1
v_p = ω_E/k_E = (E_E/ħ)/(p_E/ħ) = E_E/p_E = p_E/p_E = 1

So we’re fine, and you can check the result for ψ_Bby substituting the subscript E for B. To sum it all up, what we’ve got here is the following:

We can think of a photon having some energy that’s equal to E = p = m (assuming c = 1), but that energy would be split up in an electric and a magnetic wavefunction respectively: ψ_Eand ψ_B.
Schrödinger’s equation applies to both wavefunctions, but the E, p and m in those two wavefunctions are the same and not the same: their numerical value is the same (p_E =E_E = m_E = p_B =E_B = m_B), but they’re conceptually different. They must be: if not, we’d get a phase and group velocity for the wave that doesn’t make sense.

Of course, the phase and group velocity for the sum of the ψ_Eand ψ_Bwaves must also be equal to c. This is obviously the case, because we’re adding waves with the same phase and group velocity c, so there’s no issue with the dispersion relation.

So let’s insert those p_E =E_E = m_E = p_B =E_B = m_B values in the two wavefunctions. For ψ_E, we get:

ψ_E= eⁱ^{[k_Ex − ω_Et)}= eⁱ^{[(p_E/ħ)·x − (E_E/ħ)·t]}

You can do the calculation for ψ_Byourself. Let’s simplify our life a little bit and assume we’re using Planck units, so ħ = 1, and so the wavefunction simplifies to ψ_E= eⁱ^{·(p_E·x − E_E·t)}. We can now add the components of E and B using the summation formulas for sines and cosines:

1. B_y+ E_y = cos(p_B·x − E_B·t + π/2) + cos(p_E·x − E_E·t) = 2·cos[(p·x − E·t + π/2)/2]·cos(π/4) = √2·cos(p·x/2 − E·t/2 + π/4)

2. B_z+ E_z = sin(p_B·x − E_B·t+π/2) + sin(p_E·x − E_E·t) = 2·sin[(p·x − E·t + π/2)/2]·cos(π/4) = √2·sin(p·x/2 − E·t/2 + π/4)

Interesting! We find a composite wavefunction for our photon which we can write as:

E + B = ψ_E+ ψ_B= E + i·E = √2·eⁱ^{(p·x/2 − E·t/2 + π/4)}= √2·eⁱ^(π/4)·eⁱ^{(p·x/2 − E·t/2)}= √2·eⁱ^(π/4)·E

What a great result! It’s easy to double-check, because we can see the E + i·E = √2·eⁱ^(π/4)·E formula implies that 1 + i should equal √2·eⁱ^(π/4). Now that’s easy to prove, both geometrically (just do a drawing) or formally: √2·eⁱ^(π/4) = √2·cos(π/4) + i·sin(π/4eⁱ^(π/4) = (√2/√2) + i·(√2/√2) = 1 + i. We’re bang on! 🙂

We can double-check once more, because we should get the same from adding E and B = i·E, right? Let’s try:

E + B = E + i·E = cos(p_E·x − E_E·t) + i·sin(p_E·x − E_E·t) + i·cos(p_E·x − E_E·t) − sin(p_E·x − E_E·t)

= [cos(p_E·x − E_E·t) – sin(p_E·x − E_E·t)] + i·[sin(p_E·x − E_E·t) – cos(p_E·x − E_E·t)]

Indeed, we can see we’re going to obtain the same result, because the −sinθ in the real part of our composite wavefunction is equal to cos(θ+π/2), and the −cosθ in its imaginary part is equal to sin(θ+π/2). So the sum above is the same sum of cosines and sines that we did already.

So our electromagnetic wavefunction, i.e. the wavefunction for the photon, is equal to:

ψ = ψ_E+ ψ_B= √2·eⁱ^{(p·x/2 − E·t/2 + π/4)} = √2·eⁱ^(π/4)·eⁱ^{(p·x/2 − E·t/2)}

What about the √2 factor in front, and the π/4 term in the argument itself? No sure. It must have something to do with the way the magnetic force works, which is not like the electric force. Indeed, remember the Lorentz formula: the force on some unit charge (q = 1) will be equal to F = E + v×B. So… Well… We’ve got another cross-product here and so the geometry of the situation is quite complicated: it’s not like adding two forces F₁and F₂to get some combined force F = F₁and F₂.

In any case, we need the energy, and we know that its proportional to the square of the amplitude, so… Well… We’re spot on: the square of the √2 factor in the √2·cos product and √2·sin product is 2, so that’s twice… Well… What? Hold on a minute! We’re actually taking the absolute square of the E + B = ψ_E+ ψ_B= E + i·E = √2·eⁱ^{(p·x/2 − E·t/2 + π/4)}wavefunction here. Is that legal? I must assume it is—although… Well… Yes. You’re right. We should do some more explaining here.

We know that we usually measure the energy as some definite integral, from t = 0 to some other point in time, or over the cycle of the oscillation. So what’s the cycle here? Our combined wavefunction can be written as √2·eⁱ^{(p·x/2 − E·t/2 + π/4)}= √2·eⁱ^{(θ/2 + π/4)}, so a full cycle would correspond to θ going from 0 to 4π here, rather than from 0 to 2π. So that explains the √2 factor in front of our wave equation.

Bingo! If you were looking for an interpretation of the Planck energy and momentum, here it is. And, while everything that’s written above is not easy to understand, it’s close to the ‘intuitive’ understanding to quantum mechanics that we were looking for, isn’t it? The quantum-mechanical propagation model explains everything now. 🙂 I only need to show one more thing, and that’s the different behavior of bosons and fermions:

The amplitudes of identitical bosonic particles interfere with a positive sign, so we have Bose-Einstein statistics here. As Feynman writes it: (amplitude direct) + (amplitude exchanged).
The amplitudes of identical fermionic particles interfere with a negative sign, so we have Fermi-Dirac statistics here: (amplitude direct) − (amplitude exchanged).

I’ll think about it. I am sure it’s got something to do with that B= i·E formula or, to put it simply, with the fact that, when bosons are involved, we get two wavefunctions (ψ_Eand ψ_B) for the price of one. The reasoning should be something like this:

I. For a massless particle (i.e. a zero-mass fermion), our wavefunction is just ψ = eⁱ^{(p·x − E·t)}. So we have no √2 or √2·eⁱ^(π/4)factor in front here. So we can just add any number of them – ψ₁+ ψ₂+ ψ₃+ … – and then take the absolute square of the amplitude to find a probability density, and we’re done.

II. For a photon (i.e. a zero-mass boson), our wavefunction is √2·eⁱ^(π/4)·eⁱ^{(p·x − E·t)/2}, which – let’s introduce a new symbol – we’ll denote by φ, so φ = √2·eⁱ^(π/4)·eⁱ^{(p·x − E·t)/2}. Now, if we add any number of these, we get a similar sum but with that √2·eⁱ^(π/4)factor in front, so we write: φ₁+ φ₂+ φ₃+ … = √2·eⁱ^(π/4)·(ψ₁+ ψ₂+ ψ₃+ …). If we take the absolute square now, we’ll see the probability density will be equal to twice the density for the ψ₁+ ψ₂+ ψ₃+ … sum, because

|√2·eⁱ^(π/4)·(ψ₁+ ψ₂+ ψ₃+ …)|² = |√2·eⁱ^(π/4)|²·|ψ₁+ ψ₂+ ψ₃+ …)|² = 2·|ψ₁+ ψ₂+ ψ₃+ …)|²

So… Well… I still need to connect this to Feynman’s (amplitude direct) ± (amplitude exchanged) formula, but I am sure it can be done.

Now, we haven’t tested the complete √2·eⁱ^(π/4)·eⁱ^{(p·x − E·t)/2}wavefunction. Does it respect Schrödinger’s ∂ψ/∂t = i·(1/m)·∇²ψ or, including the 1/2 factor, the ∂ψ/∂t = i·[1/2m)]·∇²ψ equation? [Note we assume, once again, that ħ = 1, so we use Planck units once more.] Let’s see. We can calculate the derivatives as:

∂ψ/∂t = −√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(i·E/2)
∇²ψ = ∂²[√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}]/∂x²= √2·eⁱ^(π/4)·∂[√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(i·p/2)]/∂x = −√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(p²/4)

So Schrödinger’s equation becomes:

−i·√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(i·E/2) = −i·(1/m)·√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(p²/4) ⇔ 1/2 = 1/4!?

That’s funny ! It doesn’t work ! The E and m and p² are OK because we’ve got that E = m = p equation, but we’ve got problems with yet another factor 2. It only works when we use the 2/m coefficient in Schrödinger’s equation.

So… Well… There’s no choice. That’s what we’re going to do. The Schrödinger equation for the photon is ∂ψ/∂t = i·(2/m)·∇²ψ !

It’s a very subtle point. This is all great, and very fundamental stuff! Let’s now move on to Schrödinger’s actual equation, i.e. the ∂ψ/∂t = i·(ħ/2m)·∇²ψ equation.

Post scriptum on the Planck units:

If we measure time and distance in equivalent units, say seconds, we can re-write the quantum of action as:

1.0545718×10⁻³⁴N·m·s = (1.21×10⁴⁴N)·(1.6162×10⁻³⁵m)·(5.391×10⁻⁴⁴ s)

⇔ (1.0545718×10⁻³⁴/2.998×10⁸) N·s² = (1.21×10⁴⁴N)·(1.6162×10⁻³⁵/2.998×10⁸ s)(5.391×10⁻⁴⁴ s)

⇔ (1.21×10⁴⁴N) = [(1.0545718×10⁻³⁴/2.998×10⁸)]/[(1.6162×10⁻³⁵/2.998×10⁸ s)(5.391×10⁻⁴⁴ s)] N·s²/s²

You’ll say: what’s this? Well… Look at it. We’ve got a much easier formula for the Planck force—much easier than the standard formulas you’ll find on Wikipedia, for example. If we re-interpret the symbols ħ and c so they denote the numerical value of the quantum of action and the speed of light in standard SI units (i.e. newton, meter and second)—so ħ and c become dimensionless, or mathematical constants only, rather than physical constants—then the formula above can be written as:

F_P newton = (ħ/c)/[(l_P/c)·t_P] newton ⇔ F_P = ħ/(l_P·t_P)

Just double-check it: 1.0545718×10⁻³⁴/(1.6162×10⁻³⁵·5.391×10⁻⁴⁴) = 1.21×10⁴⁴. Bingo!

You’ll say: what’s the point? The point is: our model is complete. We don’t need the other physical constants – i.e. the Coulomb, Boltzmann and gravitational constant – to calculate the Planck units we need, i.e. the Planck force, distance and time units. It all comes out of our elementary wavefunction! All we need to explain the Universe – or, let’s be more modest, quantum mechanics – is two numerical constants (c and ħ) and Euler’s formula (which uses π and e, of course). That’s it.

If you don’t think that’s a great result, then… Well… Then you’re not reading this. 🙂

The photon wavefunction

Original post:

In my previous posts, I juxtaposed the following images:

Both are the same, and then they’re not. The illustration on the left-hand side shows how the electric field vector (E) of an electromagnetic wave travels through space, but it does not show the accompanying magnetic field vector (B), which is as essential in the electromagnetic propagation mechanism according to Maxwell’s equations:

∂B/∂t = –∇×E
∂E/∂t = c²∇×B = ∇×B for c = 1

The second illustration shows a wavefunction eⁱ^{(kx − ωt)}= cos(kx − ωt) + i∙sin(kx − ωt). Its propagation mechanism—if we can call it like that—is Schrödinger’s equation:

∂ψ/∂t = i·(ħ/2m)·∇²ψ

We already drew attention to the fact that an equation like this models some flow. To be precise, the Laplacian on the right-hand side is the second derivative with respect to x here, and, therefore, expresses a flux density: a flow per unit surface area, i.e. per square meter. To be precise: the Laplacian represents the flux density of the gradient flow of ψ.

On the left-hand side of Schrödinger’s equation, we have a time derivative, so that’s a flow per second. The ħ/2m factor is like a diffusion constant. In fact, strictly speaking, that ħ/2m factor is a diffusion constant, because it does exactly the same thing as the diffusion constant D in the diffusion equation ∂φ/∂t = D·∇²φ, i.e:

As a constant of proportionality, it quantifies the relationship between both derivatives.
As a physical constant, it ensures the dimensions on both sides of the equation are compatible.

So our diffusion constant here is ħ/2m. Because of the Uncertainty Principle, m is always going to be some integer multiple of ħ/2, so ħ/2m = 1, 1/2, 1/3, 1/4 etcetera. In other words, the ħ/2m term is the inverse of the mass measured in units of ħ/2. We get the terms of the harmonic series here. How convenient! 🙂

In our previous posts, we studied the wavefunction for a zero-mass particle. Such particle has zero rest mass but – because of its movement – does have some energy, and, therefore, some mass and momentum. In fact, measuring time and distance in equivalent units (so c = 1), we found that E = m = p = ħ/2 for the zero-mass particle. It had to be. If not, our equations gave us nonsense. So Schrödinger’s equation was reduced to:

∂ψ/∂t = i·∇²ψ

How elegant! We only need to explain that imaginary unit (i) in the equation. It does a lot of things. First, it gives us two equations for the price of one—thereby providing a propagation mechanism indeed. It’s just like the E and B vectors. Indeed, we can write that ∂ψ/∂t = i·∇²ψ equation as:

Re(∂ψ/∂t) = −Im(∇²ψ)
Im(∂ψ/∂t) = Re(∇²ψ)

You should be able to show that the two equations above are effectively equivalent to Schrödinger’s equation. If not… Well… Then you should not be reading this stuff.] The two equations above show that the real part of the wavefunction feeds into its imaginary part, and vice versa. Both are as essential. Let me say this one more time: the so-called real and imaginary part of a wavefunction are equally real—or essential, I should say!

Second, i gives us the circle. Huh? Yes. Writing the wavefunction as ψ = a + i·b is not just like writing a vector in terms of its Cartesian coordinates, even if it looks very much that way. Why not? Well… Never forget: i²= −1, and so—let me use mathematical lingo here—the introduction of i makes our metric space complete. To put it simply: we can now compute everything. In short, the introduction of the imaginary unit gives us that wonderful mathematical construct, eⁱ^{(kx − ωt)}, which allows us to model everything. In case you wonder, I mean: everything! Literally. 🙂

However, we’re not going to impose any pre-conditions here, and so we’re not going to make that E = m = p = ħ/2 assumption now. We’ll just re-write Schrödinger’s equation as we did last time—so we’re going to keep our ‘diffusion constant’ ħ/2m as for now:

Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ)

So we have two pairs of equations now. Can they be related? Well… They look the same, so they had better be related! 🙂 Let’s explore it. First note that, if we’d equate the direction of propagation with the x-axis, we can write the E vector as the sum of two y- and z-components: E = (E_y, E_z). Using complex number notation, we can write E as:

E = (E_y, E_z) = E_y + i·E_z

In case you’d doubt, just think of this simple drawing:

The next step is to imagine—funny word when talking complex numbers—that E_y and E_zare the real and imaginary part of some wavefunction, which we’ll denote as ψ_E = eⁱ^{(kx − ωt)}. So now we can write:

E = (E_y, E_z) = E_y + i·E_z= cos(kx − ωt) + i∙sin(kx − ωt) = Re(ψ_E) + i·Im(ψ_E)

What’s k and ω? Don’t worry about it—for the moment, that is. We’ve done nothing special here. In fact, we’re used to representing waves as some sine or cosine function, so that’s what we are doing here. Nothing more. Nothing less. We just need two sinusoids because of the circular polarization of our electromagnetic wave.

What’s next? Well… If ψ_E is a regular wavefunction, then we should be able to check if it’s a solution to Schrödinger’s equation. So we should be able to write:

Re(∂ψ_E/∂t) = −(ħ/2m)·Im(∇²ψ_E)
Im(∂ψ_E/∂t) = (ħ/2m)·Re(∇²ψ_E)

Are we? How does that work? The time derivative on the left-hand side is equal to:

∂ψ_E/∂t = −iω·eⁱ^{(kx − ωt)}= −iω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·sin(kx − ωt) − iω·cos(kx − ωt)

The second-order derivative on the right-hand side is equal to:

∇²ψ_E= ∂²ψ_E/∂x²= −k²·eⁱ^{(kx − ωt)}= −k²·cos(kx − ωt) − ik²·sin(kx − ωt)

So the two equations above are equivalent to writing:

Re(∂ψ_E/∂t) = −(ħ/2m)·Im(∇²ψ_E) ⇔ ω·sin(kx − ωt) = k²·(ħ/2m)·sin(kx − ωt)
Im(∂ψ_E/∂t) = (ħ/2m)·Re(∇²ψ_E) ⇔ −ω·cos(kx − ωt) = −k²·(ħ/2m)·cos(kx − ωt)

Both conditions are fulfilled if, and only if, ω = k²·(ħ/2m). Now, assuming we measure time and distance in equivalent units (c = 1), we can calculate the phase velocity of the electromagnetic wave as being equal to c = ω/k = 1. We also have the de Broglie equation for the matter-wave, even if we’re not quite sure whether or not we should apply that to an electromagnetic wave. In any case, the de Broglie equation tells us that k = p/ħ. So we can re-write this condition as:

ω/k = 1 = k·(ħ/2m) = (p/ħ)·(ħ/2m) = p/2m ⇔ p = 2m ⇔ m = p/2

So that’s different from the E = m = p equality we imposed when discussing the wavefunction of the zero-mass particle: we’ve got that 1/2 factor which bothered us so much once again! And it’s causing us the same trouble: how do we interpret that m = p/2 equation? It leads to nonsense once more! E = m·c²= m, but E is also supposed to be equal to p·c = p. Here, however, we find that E = p/2! We also get strange results when calculating the group and phase velocity. So… Well… What’s going on here?

I am not quite sure. It’s that damn 1/2 factor. Perhaps it’s got something to do with our definition of mass. The m in the Schrödinger equation was referred to as the effective or reduced mass of the electron wavefunction that it was supposed to model. Now that concept is something funny: it sure allows for some gymnastics, as you’ll see when going through the Wikipedia article on it! I promise I’ll dig into it—but not now and here, as I’ve got no time for that. 😦

However, the good news is that we also get a magnetic field vector with an electromagnetic wave: B. We know B is always orthogonal to E, and in the direction that’s given by the right-hand rule for the vector cross-product. Indeed, we can write B as B = e_x×E/c, with e_x the unit vector pointing in the x-direction (i.e. the direction of propagation), as shown below.

E and b

So we can do the same analysis: we just substitute E for B everywhere, and we’ll find the same condition: m = p/2. To distinguish the two wavefunctions, we used the E and B subscripts for our wavefunctions, so we wrote ψ_Eand ψ_B. We can do the same for that m = p/2 condition:

m_E= p_E/2
m_B= p_B/2

Should we just add m_Eand m_E to get a total momentum and, hence, a total energy, that’s equal to E = m = p for the whole wave? I believe we should, but I haven’t quite figured out how we should interpret that summation!

So… Well… Sorry to disappoint you. I haven’t got the answer here. But I do believe my instinct tells me the truth: the wavefunction for an electromagnetic wave—so that’s the wavefunction for a photon, basically—is essentially the same as our wavefunction for a zero-mass particle. It’s just that we get two wavefunctions for the price of one. That’s what distinguishes bosons from fermions! And so I need to figure out how they differ exactly! And… Well… Yes. That might take me a while!

In the meanwhile, we should play some more with those E and B vectors, as that’s going to help us to solve the riddle—no doubt!

Fiddling with E and B

The B = e_x×E/c equation is equivalent to saying that we’ll get B when rotating E by 90 degrees which, in turn, is equivalent to multiplication by the imaginary unit i. Huh? Yes. Sorry. Just google the meaning of the vector cross product and multiplication by i. So we can write B = i·E, which amounts to writing:

B = i·E = eⁱ^(π/2)·eⁱ^{(kx − ωt)} = eⁱ^{(kx − ωt + π/2)} = cos(kx − ωt + π/2) + i·sin(kx − ωt + π/2)

So we can now associate a wavefunction ψ_B with the field magnetic field vector B, which is the same wavefunction as ψ_E except for a phase shift equal to π/2. You’ll say: so what? Well… Nothing much. I guess this observation just concludes this long digression on the wavefunction of a photon: it’s the same wavefunction as that of a zero-mass particle—except that we get two for the price of one!

It’s an interesting way of looking at things. Let’s look at the equations we started this post with, i.e. Maxwell’s equations in free space—i.e. no stationary charges, and no currents (i.e. moving charges) either! So we’re talking those ∂B/∂t = –∇×E and ∂E/∂t = ∇×B equations now.

Note that they actually give you four equations, because they’re vector equations:

∂B/∂t = –∇×E ⇔ ∂B_y/∂t = –(∇×E)_y and ∂B_z/∂t = –(∇×E)_z
∂E/∂t = ∇×B ⇔ ∂E_y/∂t = (∇×B)_y and ∂E_z/∂t = (∇×B)_z

To figure out what that means, we need to remind ourselves of the definition of the curl operator, i.e. the ∇× operator. For E, the components of ∇×E are the following:

(∇×E)_z = ∇_xE_y– ∇_yE_x= ∂E_y/∂x – ∂E_x/∂y
(∇×E)_x = ∇_yE_z– ∇_zE_y= ∂E_z/∂y – ∂E_y/∂z
(∇×E)_y = ∇_zE_x– ∇_xE_z= ∂E_x/∂z – ∂E_z/∂x

So the four equations above can now be written as:

∂B_y/∂t = –(∇×E)_y = –∂E_x/∂z + ∂E_z/∂x
∂B_z/∂t = –(∇×E)_z = –∂E_y/∂x + ∂E_x/∂y
∂E_y/∂t = (∇×B)_y = ∂B_x/∂z – ∂B_z/∂x
∂E_z/∂t = (∇×B)_z= ∂B_y/∂x – ∂B_x/∂y

What can we do with this? Well… The x-component of E and B is zero, so one of the two terms in the equations simply disappears. We get:

∂B_y/∂t = –(∇×E)_y = ∂E_z/∂x
∂B_z/∂t = –(∇×E)_z = – ∂E_y/∂x
∂E_y/∂t = (∇×B)_y = – ∂B_z/∂x
∂E_z/∂t = (∇×B)_z= ∂B_y/∂x

Interesting: only the derivatives with respect to x remain! Let’s calculate them:

∂B_y/∂t = –(∇×E)_y = ∂E_z/∂x = ∂[sin(kx − ωt)]/∂x = k·cos(kx − ωt) = k·E_y
∂B_z/∂t = –(∇×E)_z = – ∂E_y/∂x = – ∂[cos(kx − ωt)]/∂x = k·sin(kx − ωt) = k·E_z
∂E_y/∂t = (∇×B)_y = – ∂B_z/∂x = – ∂[sin(kx − ωt + π/2)]/∂x = – k·cos(kx − ωt + π/2) = – k·B_y
∂E_z/∂t = (∇×B)_z= ∂B_y/∂x = ∂[cos(kx − ωt + π/2)]/∂x = − k·sin(kx − ωt + π/2) = – k·B_z

What wonderful results! The time derivatives of the components of B and E are equal to ±k times the components of E and B respectively! So everything is related to everything, indeed! 🙂

Let’s play some more. Using the cos(θ + π/2) = −sin(θ) and sin(θ + π/2) = cos(θ) identities, we know that B_y and B_z= sin(kx − ωt + π/2) are equal to:

B_y= cos(kx − ωt + π/2) = −sin(kx − ωt) = −E_z
B_z= sin(kx − ωt + π/2) = cos(kx − ωt) = E_y

Let’s calculate those derivatives once more now:

∂B_y/∂t = −∂E_z/∂t = −∂sin(kx − ωt)/∂t = ω·cos(kx − ωt) = ω·E_y
∂B_z/∂t = ∂E_y/∂t = ∂cos(kx − ωt)/∂t = −ω·sin(kx − ωt) = −ω·E_z

This result can, obviously, be true only if ω = k, which we assume to be the case, as we’re measuring time and distance in equivalent units, so the phase velocity is c = 1 = ω/k.

Hmm… I am sure it won’t be long before I’ll be able to prove what I want to prove. I just need to figure out the math. It’s pretty obvious now that the wavefunction—any wavefunction, really—models the flow of energy. I just need to show how it works for the zero-mass particle—and then I mean: how it works exactly. We must be able to apply the concept of the Poynting vector to wavefunctions. We must be. I’ll find how. One day. 🙂

As for now, however, I feel we’ve played enough with those wavefunctions now. It’s time to do what we promised to do a long time ago, and that is to use Schrödinger’s equation to calculate electron orbitals—and other stuff, of course! Like… Well… We hardly ever talked about spin, did we? That comes with huge complexities. But we’ll get through it. Trust me. 🙂

The quantum of time and distance

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.

Original post:

In my previous post, I introduced the elementary wavefunction of a particle with zero rest mass in free space (i.e. the particle also has zero potential). I wrote that wavefunction as eⁱ^{(kx − ωt)}= eⁱ^{(x/2 − t/2)}= cos[(x−t)/2] + i∙sin[(x−t)/2], and we can represent that function as follows:

5d_euler_f

If the real and imaginary axis in the image above are the y- and z-axis respectively, then the x-axis here is time, so here we’d be looking at the shape of the wavefunction at some fixed point in space.

Now, we could also look at its shape at some fixed in point in time, so the x-axis would then represent the spatial dimension. Better still, we could animate the illustration to incorporate both the temporal as well as the spatial dimension. The following animation does the trick quite well:

Animation

Please do note that space is one-dimensional here: the y- and z-axis represent the real and imaginary part of the wavefunction, not the y- or z-dimension in space.

You’ve seen this animation before, of course: I took it from Wikipedia, and it actually represents the electric field vector (E) for a circularly polarized electromagnetic wave. To get a complete picture of the electromagnetic wave, we should add the magnetic field vector (B), which is not shown here. We’ll come back to that later. Let’s first look at our zero-mass particle denuded of all properties, so that’s not an electromagnetic wave—read: a photon. No. We don’t want to talk charges here.

OK. So far so good. A zero-mass particle in free space. So we got that eⁱ^{(x/2 − t/2)}= cos[(x−t)/2] + i∙sin[(x−t)/2] wavefunction. We got that function assuming the following:

Time and distance are measured in equivalent units, so c = 1. Hence, the classical velocity (v) of our zero-mass particle is equal to 1, and we also find that the energy (E), mass (m) and momentum (p) of our particle are numerically the same. We wrote: E = m = p, using the p = m·v (for v = c) and the E = m∙c² formulas.
We also assumed that the quantum of energy (and, hence, the quantum of mass, and the quantum of momentum) was equal to ħ/2, rather than ħ. The de Broglie relations (k = p/ħ and ω = E/ħ) then gave us the rather particular argument of our wavefunction: kx − ωt = x/2 − t/2.

The latter hypothesis (E = m = p = ħ/2) is somewhat strange at first but, as I showed in that post of mine, it avoids an apparent contradiction: if we’d use ħ, then we would find two different values for the phase and group velocity of our wavefunction. To be precise, we’d find v for the group velocity, but v/2 for the phase velocity. Using ħ/2 solves that problem. In addition, using ħ/2 is consistent with the Uncertainty Principle, which tells us that ΔxΔp = ΔEΔt = ħ/2.

OK. Take a deep breath. Here I need to say something about dimensions. If we’re saying that we’re measuring time and distance in equivalent units – say, in meter, or in seconds – then we are not saying that they’re the same. The dimension of time and space is fundamentally different, as evidenced by the fact that, for example, time flows in one direction only, as opposed to x. To be precise, we assumed that x and t become countable variables themselves at some point in time. However, if we’re at t = 0, then we’d count time as t = 1, 2, etcetera only. In contrast, at the point x = 0, we can go to x = +1, +2, etcetera but we may also go to x = −1, −2, etc.

I have to stress this point, because what follows will require some mental flexibility. In fact, we often talk about natural units, such as Planck units, which we get from equating fundamental constants, such as c, or ħ, to 1, but then we often struggle to interpret those units, because we fail to grasp what it means to write c = 1, or ħ = 1. For example, writing c = 1 implies we can measure distance in seconds, or time in meter, but it does not imply that distance becomes time, or vice versa. We still need to keep track of whether or not we’re talking a second in time, or a second in space, i.e. c meter, or, conversely, whether we’re talking a meter in space, or a meter in time, i.e. 1/c seconds. We can make the distinction in various ways. For example, we could mention the dimension of each equation between brackets, so we’d write: t = 1×10⁻¹⁵ s [t] ≈ 299.8×10⁻⁹ m [t]. Alternatively, we could put a little subscript (like _t, or _d), so as to make sure it’s clear our meter is a a ‘light-meter’, so we’d write: t = 1×10⁻¹⁵ s ≈ 299.8×10⁻⁹ m_t. Likewise, we could add a little subscript when measuring distance in light-seconds, so we’d write x = 3×10⁸m ≈ 1 s_d, rather than x = 3×10⁸m [x] ≈ 1 s [x].

If you wish, we could refer to the ‘light-meter’ as a ‘time-meter’ (or a meter of time), and to the light-second as a ‘distance-second’ (or a second of distance). It doesn’t matter what you call it, or how you denote it. In fact, you will never hear of a meter of time, nor will you ever see those subscripts or brackets. But that’s because physicists always keep track of the dimensions of an equation, and so they know. They know, for example, that the dimension of energy combines the dimensions of both force as well as distance, so we write: [energy] = [force]·[distance]. Read: energy amounts to applying a force over a distance. Likewise, momentum amounts to applying some force over some time, so we write: [momentum] = [force]·[time]. Using the usual symbols for energy, momentum, force, distance and time respectively, we can write this as [E] = [F]·[x] and [p] = [F]·[t]. Using the units you know, i.e. joule, newton, meter and seconds, we can also write this as: 1 J = 1 N·m and 1…

Hey! Wait a minute! What’s that N·s unit for momentum? Momentum is mass times velocity, isn’t it? It is. But it amounts to the same. Remember that mass is a measure for the inertia of an object, and so mass is measured with reference to some force (F) and some acceleration (a): F = m·a ⇔ m = F/a. Hence, [m] = kg = [F/a] = N/(m/s²) = N·s²/m. [Note that the m in the brackets is symbol for mass but the other m is a meter!] So the unit of momentum is (N·s²/m)·(m/s) = N·s = newton·second.

Now, the dimension of Planck’s constant is the dimension of action, which combines all dimensions: force, time and distance. We write: ħ ≈ 1.0545718×10⁻³⁴N·m·s (newton·meter·second). That’s great, and I’ll show why in a moment. But, at this point, you should just note that when we write that E = m = p = ħ/2, we’re just saying they are numerically the same. The dimensions of E, m and p are not the same. So what we’re really saying is the following:

The quantum of energy is ħ/2 newton·meter ≈ 0.527286×10⁻³⁴N·m.
The quantum of momentum is ħ/2 newton·second ≈ 0.527286×10⁻³⁴N·s.

What’s the quantum of mass? That’s where the equivalent units come in. We wrote: 1 kg = 1 N·s²/m. So we could substitute the distance unit in this equation (m) by s_d/c = s_d/(3×10⁸). So we get: 1 kg = 3×10⁸ N·s²/s_d. Can we scrap both ‘seconds’ and say that the quantum of mass (ħ/2) is equal to the quantum of momentum? Think about it.

[…]

The answer is… Yes and no—but much more no than yes! The two sides of the equation are only numerically equal, but we’re talking a different dimension here. If we’d write that 1 kg = 0.527286×10⁻³⁴N·s²/s_d = 0.527286×10⁻³⁴N·s, you’d be equating two dimensions that are fundamentally different: space versus time. To reinforce the point, think of it the other way: think of substituting the second (s) for 3×10⁸m. Again, you’d make a mistake. You’d have to write 0.527286×10⁻³⁴N·(m_t)²/m, and you should not assume that a time-meter is equal to a distance-meter. They’re equivalent units, and so you can use them to get some number right, but they’re not equal: what they measure, is fundamentally different. A time-meter measures time, while a distance-meter measure distance. It’s as simple as that. So what is it then? Well… What we can do is remember Einstein’s energy-mass equivalence relation once more: E = m·c² (and m is the mass here). Just check the dimensions once more: [m]·[c²] = (N·s²/m)·(m²/s²) = N·m. So we should think of the quantum of mass as the quantum of energy, as energy and mass are equivalent, really.

Back to the wavefunction

The beauty of the construct of the wavefunction resides in several mathematical properties of this construct. The first is its argument:

θ = kx − ωt, with k = p/ħ and ω = E/ħ

Its dimension is the dimension of an angle: we express in it in radians. What’s a radian? You might think that a radian is a distance unit because… Well… Look at how we measure an angle in radians below:

But you’re wrong. An angle’s measurement in radians is numerically equal to the length of the corresponding arc of the unit circle but… Well… Numerically only. 🙂 Just do a dimensional analysis of θ = kx − ωt = (p/ħ)·x − (E/ħ)·t. The dimension of p/ħ is (N·s)/(N·m·s) = 1/m = m⁻¹, so we get some quantity expressed per meter, which we then multiply by x, so we get a pure number. No dimension whatsoever! Likewise, the dimension of E/ħ is (N·m)/(N·m·s) = 1/s = s⁻¹, which we then multiply by t, so we get another pure number, which we then add to get our argument θ. Hence, Planck’s quantum of action (ħ) does two things for us:

It expresses p and E in units of ħ.
It sorts out the dimensions, ensuring our argument is a dimensionless number indeed.

In fact, I’d say the ħ in the (p/ħ)·x term in the argument is a different ħ than the ħ in the (E/ħ)·t term. Huh? What? Yes. Think of the distinction I made between s and s_d, or between m and m_t. Both were numerically the same: they captured a magnitude, but they measured different things. We’ve got the same thing here:

The meter (m) in ħ ≈ 1.0545718×10⁻³⁴N·m·s in (p/ħ)·x is the dimension of x, and so it gets rid of the distance dimension. So the m in ħ ≈ 1.0545718×10⁻³⁴N·m·s goes, and what’s left measures p in terms of units equal to 1.0545718×10⁻³⁴N·s, so we get a pure number indeed.
Likewise, the second (s) in ħ ≈ 1.0545718×10⁻³⁴N·m·s in (E/ħ)·t is the dimension of t, and so it gets rid of the time dimension. So the s in ħ ≈ 1.0545718×10⁻³⁴N·m·s goes, and what’s left measures E in terms of units equal to 1.0545718×10⁻³⁴N·m, so we get another pure number.
Adding both gives us the argument θ: a pure number that measures some angle.

That’s why you need to watch out when writing θ = (p/ħ)·x − (E/ħ)·t as θ = (p·x − E·t)/ħ or – in the case of our elementary wavefunction for the zero-mass particle – as θ = (x/2 − t/2) = (x − t)/2. You can do it – in fact, you should do when trying to calculate something – but you need to be aware that you’re making abstraction of the dimensions. That’s quite OK, as you’re just calculating something—but don’t forget the physics behind!

You’ll immediately ask: what are the physics behind here? Well… I don’t know. Perhaps nobody knows. As Feynman once famously said: “I think I can safely say that nobody understands quantum mechanics.” But then he never wrote that, and I am sure he didn’t really mean that. And then he said that back in 1964, which is 50 years ago now. 🙂 So let’s try to understand it at least. 🙂

Planck’s quantum of action – 1.0545718×10⁻³⁴N·m·s – comes to us as a mysterious quantity. A quantity is more than a a number. A number is something like π or e, for example. It might be a complex number, like eⁱ^θ, but that’s still a number. In contrast, a quantity has some dimension, or some combination of dimensions. A quantity may be a scalar quantity (like distance), or a vector quantity (like a field vector). In this particular case (Planck’s ħ or h), we’ve got a physical constant combining three dimensions: force, time and distance—or space, if you want. It’s a quantum, so it comes as a blob—or a lump, if you prefer that word. However, as I see it, we can sort of project it in space as well as in time. In fact, if this blob is going to move in spacetime, then it will move in space as well as in time: t will go from 0 to 1, and x goes from 0 to ± 1, depending on what direction we’re going. So when I write that E = p = ħ/2—which, let me remind you, are two numerical equations, really—I sort of split Planck’s quantum over E = m and p respectively.

You’ll say: what kind of projection or split is that? When projecting some vector, we’ll usually have some sine and cosine, or a 1/√2 factor—or whatever, but not a clean 1/2 factor. Well… I have no answer to that, except that this split fits our mathematical construct. Or… Well… I should say: my mathematical construct. Because what I want to find is this clean Schrödinger equation:

∂ψ/∂t = i·(ħ/2m)·∇²ψ = i·∇²ψ for m = ħ/2

Now I can only get this equation if (1) E = m = p and (2) if m = ħ/2 (which amounts to writing that E = p = m = ħ/2). There’s also the Uncertainty Principle. If we are going to consider the quantum vacuum, i.e. if we’re going to look at space (or distance) and time as count variables, then Δx and Δt in the ΔxΔp = ΔEΔt = ħ/2 equations are ± 1 and, therefore, Δp and ΔE must be ± ħ/2. In any case, I am not going to try to justify my particular projection here. Let’s see what comes out of it.

The quantum vacuum

Schrödinger’s equation for my zero-mass particle (with energy E = m = p = ħ/2) amounts to writing:

Re(∂ψ/∂t) = −Im(∇²ψ)
Im(∂ψ/∂t) = Re(∇²ψ)

Now that reminds of the propagation mechanism for the electromagnetic wave, which we wrote as ∂B/∂t = –∇×E and ∂E/∂t = ∇×B, also assuming we measure time and distance in equivalent units. However, we’ll come back to that later. Let’s first study the equation we have, i.e.

eⁱ^{(kx − ωt)} = eⁱ^{(ħ·x/2 − ħ·t/2)/ħ} = eⁱ^{(x/2 − t/2)}= cos[(x−t)/2] + i∙sin[(x−t)/2]

Let’s think some more. What is that eⁱ^{(x/2 − t/2)}function? It’s subject to conceiving time and distance as countable variables, right? I am tempted to say: as discrete variables, but I won’t go that far—not now—because the countability may be related to a particular interpretation of quantum physics. So I need to think about that. In any case… The point is that x can only take on values like 0, 1, 2, etcetera. And the same goes for t. To make things easy, we’ll not consider negative values for x right now (and, obviously, not for t either). But you can easily check it doesn’t make a difference: if you think of the propagation mechanism – which is what we’re trying to model here – then x is always positive, because we’re moving away from some source that caused the wave. In any case, we’ve got a infinite set of points like:

eⁱ^{(0/2 − 0/2)}= eⁱ⁽⁰⁾= cos(0) + i∙sin(0)
eⁱ^{(1/2 − 0/2)}= eⁱ^(1/2)= cos(1/2) + i∙sin(1/2)
eⁱ^{(0/2 − 1/2)}= eⁱ^(−1/2)= cos(−1/2) + i∙sin(−1/2)
eⁱ^{(1/2 − 1/2)}= eⁱ⁽⁰⁾= cos(0) + i∙sin(0)
…

In my previous post, I calculated the real and imaginary part of this wavefunction for x going from 0 to 14 (as mentioned, in steps of 1) and for t doing the same (also in steps of 1), and what we got looked pretty good:

I also said that, if you wonder how the quantum vacuum could possibly look like, you should probably think of these discrete spacetime points, and some complex-valued wave that travels as illustrated above. In case you wonder what’s being illustrated here: the right-hand graph is the cosine value for all possible x = 0, 1, 2,… and t = 0, 1, 2,… combinations, and the left-hand graph depicts the sine values, so that’s the imaginary part of our wavefunction. Taking the absolute square of both gives 1 for all combinations. So it’s obvious we’d need to normalize and, more importantly, we’d have to localize the particle by adding several of these waves with the appropriate contributions. But so that’s not our worry right now. I want to check whether those discrete time and distance units actually make sense. What’s their size? Is it anything like the Planck length (for distance) and/or the Planck time?

Let’s see. What are the implications of our model? The question here is: if ħ/2 is the quantum of energy, and the quantum of momentum, what’s the quantum of force, and the quantum of time and/or distance?

Huh? Yep. We treated distance and time as countable variables above, but now we’d like to express the difference between x = 0 and x = 1 and between t = 0 and t = 1 in the units we know, this is in meter and in seconds. So how do we go about that? Do we have enough equations here? Not sure. Let’s see…

We obviously need to keep track of the various dimensions here, so let’s refer to that discrete distance and time unit as t_Pand l_P respectively. The subscript (P) refers to Planck, and the l refers to a length, but we’re likely to find something else than Planck units. I just need placeholder symbols here. To be clear: t_Pand l_Pare expressed in meter and seconds respectively, just like the actual Planck time and distance, which are equal to 5.391×10⁻⁴⁴ s (more or less) and 1.6162×10⁻³⁵m (more or less) respectively. As I mentioned above, we get these Planck units by equating fundamental physical constants to 1. Just check it: (1.6162×10⁻³⁵m)/(5.391×10⁻⁴⁴ s) = c ≈ 3×10⁸m/s. So the following relation must be true: l_P = c·t_P, or l_P/t_P= c.

Now, as mentioned above, there must be some quantum of force as well, which we’ll write as F_P, and which is – obviously – expressed in newton (N). So we have:

E = ħ/2 ⇒ 0.527286×10⁻³⁴N·m = F_P·l_PN·m
p = ħ/2 ⇒ 0.527286×10⁻³⁴N·s = F_P·t_PN·s

Let’s try to divide both formulas: E/p = (F_P·l_PN·m)/(F_P·t_PN·s) = l_P/t_Pm/s = l_P/t_Pm/s = c m/s. That’s consistent with the E/p = c equation. Hmm… We found what we knew already. My model is not fully determined, it seems. 😦

What about the following simplistic approach? E is numerically equal to 0.527286×10⁻³⁴, and its dimension is [E] = [F]·[x], so we write: E = 0.527286×10⁻³⁴·[E] = 0.527286×10⁻³⁴·[F]·[x]. Hence, [x] = [E]/[F] = (N·m)/N = m. That just confirms what we already know: the quantum of distance (i.e. our fundamental unit of distance) can be expressed in meter. But our model does not give that fundamental unit. It only gives us its dimension (meter), which is stuff we knew from the start. 😦

Let’s try something else. Let’s just accept that Planck length and time, so we write:

l_P = 1.6162×10⁻³⁵m
t_P= 5.391×10⁻⁴⁴ s

Now, if the quantum of action is equal to ħ N·m·s = F_P·l_P·t_P N·m·s = 1.0545718×10⁻³⁴N·m·s, and if the two definitions of l_Pand t_P above hold, then 1.0545718×10⁻³⁴N·m·s = (F_PN)×(1.6162×10⁻³⁵m)×(5.391×10⁻⁴⁴ s) ≈ F_P8.713×10⁻⁷⁹N·m·s ⇔ F_P≈ 1.21×10⁴⁴N.

Does that make sense? It does according to Wikipedia, but how do we relate this to our E = p = m = ħ/2 equations? Let’s try this:

E_P = (1.0545718×10⁻³⁴N·m·s)/(5.391×10⁻⁴⁴ s) = 1.956×10⁹ J. That corresponds to the regular Planck energy.
p_P = (1.0545718×10⁻³⁴N·m·s)/(1.6162×10⁻³⁵m) = 0.6525 N·s. That corresponds to the regular Planck momentum.

Is E_P = p_P? Let’s substitute: 1.956×10⁹ N·m = 1.956×10⁹ N·(s/c) = 1.956×10⁹/2.998×10⁹N·s = 0.6525 N·s. So, yes, it comes out alright. In fact, I omitted the 1/2 factor in the calculations, but it doesn’t matter: it does come out alright. So I did not prove that the difference between my x = 0 and x = 1 points (or my t = 0 and t = 1 points) is equal to the Planck length (or the Planck time unit), but I did show my theory is, at the very least, compatible with those units. That’s more than enough for now. And I’ll come surely come back to it in my next post. 🙂

Post Scriptum: One must solve the following equations to get the fundamental Planck units:

We have five fundamental equations for five fundamental quantities respectively: t_P, l_P, F_P, m_P, and E_P respectively, so that’s OK: it’s a fully determined system alright! But where do the expressions with G, k_B (the Boltzmann constant) and ε₀ come from? What does it mean to equate those constants to 1? Well… I need to think about that, and I’ll get back to you on it. 🙂

The wavefunction of a zero-mass particle

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.

Original post:

I hope you find the title intriguing. A zero-mass particle? So I am talking a photon, right? Well… Yes and no. Just read this post and, more importantly, think about this story for yourself. 🙂

One of my acquaintances is a retired nuclear physicist. We mail every now and then—but he has little or no time for my questions: he usually just tells me to keep studying. I once asked him why there is never any mention of the wavefunction of a photon in physics textbooks. He bluntly told me photons don’t have a wavefunction—not in the sense I was talking at least. Photons are associated with a traveling electric and a magnetic field vector. That’s it. Full stop. Photons do not have a ψ or φ function. [I am using ψ and φ to refer to position or momentum wavefunction. You know both are related: if we have one, we have the other.] But then I never give up, of course. I just can’t let go out of the idea of a photon wavefunction. The structural similarity in the propagation mechanism of the electric and magnetic field vectors E and B just looks too much like the quantum-mechanical wavefunction. So I kept trying and, while I don’t think I fully solved the riddle, I feel I understand it much better now. Let me show you the why and how.

I. An electromagnetic wave in free space is fully described by the following two equations:

∂B/∂t = –∇×E
∂E/∂t = c²∇×B

We’re making abstraction here of stationary charges, and we also do not consider any currents here, so no moving charges either. So I am omitting the ∇·E = ρ/ε₀ equation (i.e. the first of the set of four equations), and I am also omitting the j/ε₀ in the second equation. So, for all practical purposes (i.e. for the purpose of this discussion), you should think of a space with no charges: ρ = 0 and j = 0. It’s just a traveling electromagnetic wave. To make things even simpler, we’ll assume our time and distance units are chosen such that c = 1, so the equations above reduce to:

∂B/∂t = –∇×E
∂E/∂t = ∇×B

Perfectly symmetrical! But note the minus sign in the first equation. As for the interpretation, I should refer you to previous posts but, briefly, the ∇× operator is the curl operator. It’s a vector operator: it describes the (infinitesimal) rotation of a (three-dimensional) vector field. We discussed heat flow a couple of times, or the flow of a moving liquid. So… Well… If the vector field represents the flow velocity of a moving fluid, then the curl is the circulation density of the fluid. The direction of the curl vector is the axis of rotation as determined by the ubiquitous right-hand rule, and its magnitude of the curl is the magnitude of rotation. OK. Next step.

II. For the wavefunction, we have Schrödinger’s equation, ∂ψ/∂t = i·(ħ/2m)·∇²ψ, which relates two complex-valued functions (∂ψ/∂t and ∇²ψ). Complex-valued functions consist of a real and an imaginary part, and you should be able to verify this equation is equivalent to the following set of two equations:

Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ)

[Two complex numbers a + ib and c + id are equal if, and only if, their real and imaginary parts are the same. However, note the −i factor in the right-hand side of the equation, so we get: a + ib = −i·(c + id) = d −ic.] The Schrödinger equation above also assumes free space (i.e. zero potential energy: V = 0) but, in addition – see my previous post – they also assume a zero rest mass of the elementary particle (E₀ = 0). So just assume E₀= V = 0 in de Broglie’s elementary ψ(θ) = ψ(x, t) = e⁻ⁱ^θ = a·e^{−i[(E₀+ p2/(2m) + V)·t − p∙x]/ħ} wavefunction. So, in essence, we’re looking at the wavefunction of a massless particle here. Sounds like nonsense, doesn’t it? But… Well… That should be the wavefunction of a photon in free space then, right? 🙂

Maybe. Maybe not. Let’s go as far as we can.

The energy of a zero-mass particle

What m would we use for a photon? It’s rest mass is zero, but it’s got energy and, hence, an equivalent mass. That mass is given by the m = E/c²mass-energy equivalence. We also know a photon has momentum, and it’s equal to its energy divided by c: p = m·c = E/c. [I know the notation is somewhat confusing: E is, obviously, not the magnitude of E here: it’s energy!] Both yield the same result. We get: m·c = E/c ⇔ m = E/c²⇔ E = m·c².

OK. Next step. Well… I’ve always been intrigued by the fact that the kinetic energy of a photon, using the E = m·v²/2 = E = m·c²/2 formula, is only half of its total energy E = m·c². Half: 1/2. That 1/2 factor is intriguing. Where’s the rest of the energy? It’s really a contradiction: our photon has no rest mass, and there’s no potential here, but its total energy is still twice its kinetic energy. Quid?

There’s only one conclusion: just because of its sheer existence, it must have some hidden energy, and that hidden energy is also equal to E = m·c²/2, and so the kinetic and hidden energy add up to E = m·c².

Huh? Hidden energy? I must be joking, right?

Well… No. No joke. I am tempted to call it the imaginary energy, because it’s linked to the imaginary part of the wavefunction—but then it’s everything but imaginary: it’s as real as the imaginary part of the wavefunction. [I know that sounds a bit nonsensical, but… Well… Think about it: it does make sense.]

Back to that factor 1/2. You may or may not remember it popped up when we were calculating the group and the phase velocity of the wavefunction respectively, again assuming zero rest mass, and zero potential. [Note that the rest mass term is mathematically equivalent to the potential term in both the wavefunction as well as in Schrödinger’s equation: (E₀·t +V·t = (E₀+ V)·t, and V·ψ + E₀·ψ = (V+E₀)·ψ—obviously!]

In fact, let me quickly show you that calculation again: the de Broglie relations tell us that the k and the ω in the eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction (i.e. the spatial and temporal frequency respectively) are equal to k = p/ħ, and ω = E/ħ. If we would now use the kinetic energy formula E = m·v²/2 – which we can also write as E = m·v·v/2 = p·v/2 = p·p/2m = p²/2m, with v = p/m the classical velocity of the elementary particle that Louis de Broglie was thinking of – then we can calculate the group velocity of our eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) as:

v_g = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂[p²/2m]/∂p = 2p/2m = p/m = v

[Don’t tell me I can’t treat m as a constant when calculating ∂ω/∂k: I can. Think about it.] Now the phase velocity. The phase velocity of our eⁱ^{(kx − ωt)} is only half of that. Again, we get that 1/2 factor:

v_p = ω/k = (E/ħ)/(p/ħ) = E/p = (p²/2m)/p = p/2m = v/2

Strange, isn’t it? Why would we get a different value for the phase velocity here? It’s not like we have two different frequencies here, do we? You may also note that the phase velocity turns out to be smaller than the group velocity, which is quite exceptional as well! So what’s the matter?

Well… The answer is: we do seem to have two frequencies here while, at the same time, it’s just one wave. There is only one k and ω here but, as I mentioned a couple of times already, that eⁱ^{(kx − ωt)} wavefunction seems to give you two functions for the price of one—one real and one imaginary: eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt). So are we adding waves, or are we not? It’s a deep question. In my previous post, I said we were adding separate waves, but now I am thinking: no. We’re not. That sine and cosine are part of one and the same whole. Indeed, the apparent contradiction (i.e. the different group and phase velocity) gets solved if we’d use the E = m∙v² formula rather than the kinetic energy E = m∙v²/2. Indeed, assuming that E = m∙v² formula also applies to our zero-mass particle (I mean zero rest mass, of course), and measuring time and distance in natural units (so c = 1), we have:

E = m∙c² = m and p = m∙c²= m, so we get: E = m = p

Waw! What a weird combination, isn’t it? But… Well… It’s OK. [You tell me why it wouldn’t be OK. It’s true we’re glossing over the dimensions here, but natural units are natural units, and so c = c²= 1. So… Well… No worries!] The point is: that E = m = p equality yields extremely simple but also very sensible results. For the group velocity of our eⁱ^{(kx − ωt)} wavefunction, we get:

v_g = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂p/∂p = 1

So that’s the velocity of our zero-mass particle (c, i.e. the speed of light) expressed in natural units once more—just like what we found before. For the phase velocity, we get:

v_p = ω/k = (E/ħ)/(p/ħ) = E/p = p/p = 1

Same result! No factor 1/2 here! Isn’t that great? My ‘hidden energy theory’ makes a lot of sense. 🙂 In fact, I had mentioned a couple of times already that the E = m∙v² relation comes out of the de Broglie relations if we just multiply the two and use the v = f·λ relation:

f·λ = (E/h)·(h/p) = E/p
v = f·λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v) ⇒ E = m·v²

But so I had no good explanation for this. I have one now: the E = m·v²is the correct energy formula for our zero-mass particle. 🙂

The quantization of energy and the zero-mass particle

Let’s now think about the quantization of energy. What’s the smallest value for E that we could possible think of? That’s h, isn’t it? That’s the energy of one cycle of an oscillation according to the Planck-Einstein relation (E = h·f). Well… Perhaps it’s ħ? Because… Well… We saw energy levels were separated by ħ, rather than h, when studying the blackbody radiation problem. So is it ħ = h/2π? Is the natural unit a radian (i.e. a unit distance), rather than a cycle?

Neither is natural, I’d say. We also have the Uncertainty Principle, which suggests the smallest possible energy value is ħ/2, because ΔxΔp = ΔtΔE = ħ/2.

Huh? What’s the logic here?

Well… I am not quite sure but my intuition tells me the quantum of energy must be related to the quantum of time, and the quantum of distance.

Huh? The quantum of time? The quantum of distance? What’s that? The Planck scale?

No. Or… Well… Let me correct that: not necessarily. I am just thinking in terms of logical concepts here. Logically, as we think of the smallest of smallest, then our time and distance variables must become count variables, so they can only take on some integer value n = 0, 1, 2 etcetera. So then we’re literally counting in time and/or distance units. So Δx and Δt are then equal to 1. Hence, Δp and ΔE are then equal to Δp = ΔE = ħ/2. Just think of the radian (i.e. the unit in which we measure θ) as measuring both time as well as distance. Makes sense, no?

No? Well… Sorry. I need to move on. So the smallest possible value for m = E = p would be ħ/2. Let’s substitute that in Schrödinger’s equation, or in that set of equations Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ) and Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ). We get:

Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ) = −(2ħ/2ħ)·Im(∇²ψ) = −Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ) = (2ħ/2ħ)·Re(∇²ψ) = Re(∇²ψ)

Bingo! The Re(∂ψ/∂t) = −Im(∇²ψ) and Im(∂ψ/∂t) = Re(∇²ψ) equations were what I was looking for. Indeed, I wanted to find something that was structurally similar to the ∂B/∂t = –∇×E and ∂E/∂t = ∇×B equations—and something that was exactly similar: no coefficients in front or anything. 🙂

What about our wavefunction? Using the de Broglie relations once more (k = p/ħ, and ω = E/ħ), our eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) now becomes:

eⁱ^{(kx − ωt)} = eⁱ^{(ħ·x/2 − ħ·t/2)/ħ} = eⁱ^{(x/2 − t/2)}= cos[(x−t)/2] + i∙sin[(x−t)/2]

Hmm… Interesting! So we’ve got that 1/2 factor now in the argument of our wavefunction! I really feel I am close to squaring the circle here. 🙂 Indeed, it must be possible to relate the ∂B/∂t = –∇×E and ∂E/∂t = c²∇×B to the Re(∂ψ/∂t) = −Im(∇²ψ) and Im(∂ψ/∂t) = Re(∇²ψ) equations. I am sure it’s a complicated exercise. It’s likely to involve the formula for the Lorentz force, which says that the force on a unit charge is equal to $E + v \times B$ $, with v the velocity of the charge. Why? Note the vector cross-product. Also note that \partial B /\partialt and \partial E /\partialt are vector-valued functions, not scalar-valued functions. Hence, in that sense, \partial B /\partialt and \partial E /\partialt and not like the Re (\partialψ/\partialt) and/or Im (\partialψ/\partialt) function. But\dots Well\dots For the rest, think of it:$ $E and B are orthogonal vectors, and that’s how we usually interpret the real and imaginary part of a complex number as well: the real and imaginary axis are orthogonal too!$

$So I am almost there.$ Who can help me prove what I want to prove here? The two propagation mechanisms are the “same-same but different”, as they say in Asia. The difference between the two propagation mechanisms must also be related to that fundamental dichotomy in Nature: the distinction between bosons and fermions. Indeed, when combining two directional quantities (i.e. two vectors), we like to think there are four different ways of doing that, as shown below. However, when we’re only interested in the magnitude of the result (and not in its direction), then the first and third result below are really the same, as are the second and fourth combination. Now, we’ve got pretty much the same in quantum math: we can, in theory, combine complex-valued amplitudes in four different ways but, in practice, we only have two (rather than four) types of behavior only: photons versus bosons.

Is our zero-mass particle just the electric field vector?

Let’s analyze that eⁱ^{(x/2 − t/2)}= cos[(x−t)/2] + i∙sin[(x−t)/2] wavefunction some more. It’s easy to represent it graphically. The following animation does the trick:

Animation

I am sure you’ve seen this animation before: it represents a circularly polarized electromagnetic wave… Well… Let me be precise: it presents the electric field vector (E) of such wave only. The B vector is not shown here, but you know where and what it is: orthogonal to the E vector, as shown below—for a linearly polarized wave.

eⁱ^{(0/2 − 0/2)}= cos(0) + i∙sin(0)
eⁱ^{(1/2 − 0/2)}= cos(1/2) + i∙sin(1/2)
eⁱ^{(0/2 − 1/2)}= cos(−1/2) + i∙sin(−1/2)
eⁱ^{(1/2 − 1/2)}= cos(0) + i∙sin(0)
…

Now, I quickly opened Excel and calculated those cosine and sine values for x and t going from 0 to 14 below. It’s really easy. Just five minutes of work. You should do yourself as an exercise. The result is shown below. Both graphs connect 14×14 = 196 data points, but you can see what’s going on: this does effectively, represent the elementary wavefunction of a particle traveling in spacetime. In fact, you can see its speed is equal to 1, i.e. it effectively travels at the speed of light, as it should: the wave velocity is v = f·λ = (ω/2π)·(2π/k) = ω/k = (1/2)·(1/2) = 1. The amplitude of our wave doesn’t change along the x = t diagonal. As the Last Samurai puts it, just before he moves to the Other World: “Perfect! They are all perfect!” 🙂

graph imaginary graph real

In fact, in case you wonder how the quantum vacuum could possibly look like, you should probably think of these discrete spacetime points, and some complex-valued wave that travels as it does in the illustration above.

Of course, that elementary wavefunction above does not localize our particle. For that, we’d have to add a potentially infinite number of such elementary wavefunctions, so we’d write the wavefunction as ∑ a_je^−iθ_j functions. [I use the j symbol here for the subscript, rather than the more conventional i symbol for a subscript, so as to avoid confusion with the symbol used for the imaginary unit.] The a_jcoefficients are the contribution that each of these elementary wavefunctions would make to the composite wave. What could they possibly be? Not sure. Let’s first look at the argument of our elementary component wavefunctions. We’d inject uncertainty in it. So we’d say that m = E = p is equal to

m = E = p = ħ/2 + j·ħ with j = 0, 1, 2,…

That amounts to writing: m = E = p = ħ/2, ħ, 3ħ/2, 2ħ, 5/2ħ, etcetera. Waw! That’s nice, isn’t it? My intuition tells me that our a_jcoefficients will be smaller for higher j, so the a_j(j) function would be some decreasing function. What shape? Not sure. Let’s first sum up our thoughts so far:

The elementary wavefunction of a zero-mass particle (again, I mean zero rest mass) in free space is associated with an energy that’s equal to ħ/2.
The zero-mass particle travels at the speed of light, obviously (because it has zero rest mass), and its kinetic energy is equal to E = m·v²/2= m·c²/2.
However, its total energy is equal to E = m·v²= m·c²: it has some hidden energy. Why? Just because it exists.
We may associate its kinetic energy with the real part of its wavefunction, and the hidden energy with its imaginary part. However, you should remember that the imaginary part of the wavefunction is as essential as its real part, so the hidden energy is equally real. 🙂

So… Well… Isn’t this just nice?

I think it is. Another obvious advantage of this way of looking at the elementary wavefunction is that – at first glance at least – it provides an intuitive understanding of why we need to take the (absolute) square of the wavefunction to find the probability of our particle being at some point in space and time. The energy of a wave is proportional to the square of its amplitude. Now, it is reasonable to assume the probability of finding our (point) particle would be proportional to the energy and, hence, to the square of the amplitude of the wavefunction, which is given by those a_j(j) coefficients.

Huh?

OK. You’re right. I am a bit too fast here. It’s a bit more complicated than that, of course. The argument of probability being proportional to energy being proportional to the square of the amplitude of the wavefunction only works for a single wave a·e^−iθ. The argument does not hold water for a sum of functions ∑ a_je^−iθ_j. Let’s write it all out. Taking our m = E = p = ħ/2 + j·ħ = ħ/2, ħ, 3ħ/2, 2ħ, 5/2ħ,… formula into account, this sum would look like:

a₁eⁱ^{(x − t)(1/2)}+ a₂eⁱ^{(x − t)(2/2)}+ a₃eⁱ^{(x − t)(3/2)}+ a₄eⁱ^{(x − t)(4/2)}+ …

But—Hey! We can write this as some power series, can’t we? We just need to add a₀eⁱ^{(x − t)(0/2)}= a₀, and then… Well… It’s not so easy, actually. Who can help me? I am trying to find something like this:

power series

Or… Well… Perhaps something like this:

power series 2

Whatever power series it is, we should be able to relate it to this one—I’d hope:

Hmm… […] It looks like I’ll need to re-visit this, but I am sure it’s going to work out. Unfortunately, I’ve got no more time today, I’ll let you have some fun now with all of this. 🙂 By the way, note that the result of the first power series is only valid for |x| < 1. 🙂

Note 1: What we should also do now is to re-insert mass in the equations. That should not be too difficult. It’s consistent with classical theory: the total energy of some moving mass is E = m·c², out of which m·v²/2 is the classical kinetic energy. All the rest – i.e. m·c² − m·v²/2 – is potential energy, and so that includes the energy that’s ‘hidden’ in the imaginary part of the wavefunction. 🙂

Note 2: I really didn’t pay much attentions to dimensions when doing all of these manipulations above but… Well… I don’t think I did anything wrong. Just to give you some more feel for that wavefunction eⁱ^{(kx − ωt)}, please do a dimensional analysis of its argument. I mean, k = p/ħ, and ω = E/ħ, so check the dimensions:

Momentum is expressed in newton·second, and we divide it by the quantum of action, which is expressed in newton·meter·second. So we get something per meter. But then we multiply it with x, so we get a dimensionless number.
The same is true for the ωt term. Energy is expressed in joule, i.e. newton·meter, and so we divide it by ħ once more, so we get something per second. But then we multiply it with t, so… Well… We do get a dimensionless number: a number that’s expressed in radians, to be precise. And so the radian does, indeed, integrate both the time as well as the distance dimension. 🙂

Schrödinger’s equation and the two de Broglie relations

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.

Original post:

I’ve re-visited the de Broglie equations a couple of times already. In this post, however, I want to relate them to Schrödinger’s equation. Let’s start with the de Broglie equations first. Equations. Plural. Indeed, most popularizing books on quantum physics will give you only one of the two de Broglie equations—the one that associates a wavelength (λ) with the momentum (p) of a matter-particle:

λ = h/p

In fact, even the Wikipedia article on the ‘matter wave’ starts off like that and is, therefore, very confusing, because, for a good understanding of quantum physics, one needs to realize that the λ = h/p equality is just one of a pair of two ‘matter wave’ equations:

λ = h/p
f = E/h

These two equations give you the spatial and temporal frequency of the wavefunction respectively. Now, those two frequencies are related – and I’ll show you how in a minute – but they are not the same. It’s like space and time: they are related, but they are definitely not the same. Now, because any wavefunction is periodic, the argument of the wavefunction – which we’ll introduce shortly – will be some angle and, hence, we’ll want to express it in radians (or – if you’re really old-fashioned – degrees). So we’ll want to express the frequency as an angular frequency (i.e. in radians per second, rather than in cycles per second), and the wavelength as a wave number (i.e. in radians per meter). Hence, you’ll usually see the two de Broglie equations written as:

k = p/ħ
ω = E/ħ

It’s the same: ω = 2π∙f and f = 1/T (T is the period of the oscillation), and k = 2π/λ and then ħ = h/2π, of course! [Just to remove all ambiguities: stop thinking about degrees. They’re a Babylonian legacy, who thought the numbers 6, 12, and 60 had particular religious significance. So that’s why we have twelve-hour nights and twelve-hour days, with each hour divided into sixty minutes and each minute divided into sixty seconds, and – particularly relevant in this context – why ‘once around’ is divided into 6×60 = 360 degrees. Radians are the unit in which we should measure angles because… Well… Google it. They measure an angle in distance units. That makes things easier—a lot easier! Indeed, when studying physics, the last thing you want is artificial units, like degrees.]

So… Where were we? Oh… Yes. The de Broglie relation. Popular textbooks usually commit two sins. One is that they forget to say we have two de Broglie relations, and the other one is that the E = h∙f relationship is presented as the twin of the Planck-Einstein relation for photons, which relates the energy (E) of a photon to its frequency (ν): E = h∙ν = ħ∙ω. The former is criminal neglect, I feel. As for the latter… Well… It’s true and not true: it’s incomplete, I’d say, and, therefore, also very confusing.

Why? Because both things lead one to try to relate the two equations, as momentum and energy are obviously related. In fact, I’ve wasted days, if not weeks, on this. How are they related? What formula should we use? To answer that question, we need to answer another one: what energy concept should we use? Potential energy? Kinetic energy? Should we include the equivalent energy of the rest mass?

One quickly gets into trouble here. For example, one can try the kinetic energy, K.E. = m∙v²/2, and use the definition of momentum (p = m∙v), to write E = p²/(2m), and then we could relate the frequency f to the wavelength λ using the general rule that the traveling speed of a wave is equal to the product of its wavelength and its frequency (v = λ∙f). But if E = p²/(2m) and f = v/λ, we get:

p²/(2m) = h∙v/λ ⇔ λ = 2∙h/p

So that is almost right, but not quite: that factor 2 should not be there. In fact, it’s easy to see that we’d get de Broglie’s λ = h/p equation from his E = h∙f equation if we’d use E = m∙v² rather than E = m∙v²/2. In fact, the E = m∙v² relation comes out of them if we just multiply the two and, yes, use that v = f·λ relation once again:

f·λ = (E/h)·(h/p) = E/p
v = f·λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v) ⇒ E = m·v²

But… Well… E = m∙v²? How could we possibly justify the use of that formula?

The answer is simple: our v = f·λ equation is wrong. It’s just something one shouldn’t apply to the complex-valued wavefunction. The ‘correct’ velocity formula for the complex-valued wavefunction should have that 1/2 factor, so we’d write 2·f·λ = v to make things come out alright. But where would this formula come from?

Well… Now it’s time to introduce the wavefunction.

The wavefunction

You know the elementary wavefunction:

ψ = ψ(x, t) = e^{−i(ωt − kx)} = eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt)

As for terminology, note that the term ‘wavefunction’ refers to what I write above, while the term ‘wave equation’ usually refers to Schrödinger’s equation, which I’ll introduce in a minute. Also note the use of boldface indicates we’re talking vectors, so we’re multiplying the wavenumber vector k with the position vector x = (x, y, z) here, although we’ll often simplify and assume one-dimensional space. In any case…

So the question is: why can’t we use the v = f·λ formula for this wave? The period of cosθ + isinθ is the same as that of the sine and cosine function considered separately: cos(θ+2π) + isin(θ+2π) = cosθ + isinθ, so T = 2π and f = 1/T = 1/2π do not change. So the f, T and λ should be the same, no?

No. We’ve got two oscillations for the price of one here: one ‘real’ and one ‘imaginary’—but both are equally essential and, hence, equally ‘real’. So we’re actually combining two waves. So it’s just like adding other waves: when adding waves, one gets a composite wave that has (a) a phase velocity and (b) a group velocity.

Huh? Yes. It’s quite interesting. When adding waves, we usually have a different ω and k for each of the component waves, and the phase and group velocity will depend on the relation between those ω’s and k’s. That relation is referred to as the dispersion relation. To be precise, if you’re adding waves, then the phase velocity of the composite wave will be equal to v_p = ω/k, and its group velocity will be equal to v_g = dω/dk. We’ll usually be interested in the group velocity, and so to calculate that derivative, we need to express ω as a function of k, of course, so we write ω as some function of k, i.e. ω = ω(k). There are number of possibilities then:

ω and k may be directly proportional, so we can write ω as ω = a∙k: in that case, we find that v_p = v_g = a.
ω and k are not directly proportional but have a linear relationship, so we can write write ω as ω = a∙k + b. In that case, we find that v_g = a and… Well… We’ve got a problem calculating v_p, because we don’t know what k to use!
ω and k may be non-linearly related, in which case… Well… One does has to do the calculation and see what comes out. 🙂

Let’s now look back at our eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) function. You’ll say that we’ve got only one ω and one k here, so we’re not adding waves with different ω’s and k’s. So… Well… What?

That’s where the de Broglie equations come in. Look: k = p/ħ, and ω = E/ħ. If we now use the correct energy formula, i.e. the kinetic energy formula E = m·v²/2 (rather than that nonsensical E = m·v² equation) – which we can also write as E = m·v·v/2 = p·v/2 = p·p/2m = p²/2m, with v = p/m the classical velocity of the elementary particle that Louis de Broglie was thinking of – then we can calculate the group velocity of our eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) as:

v_g = dω/dk = d[E/ħ]/d[p/ħ] = dE/dp = d[p²/2m]/dp = 2p/2m = p/m = v

However, the phase velocity of our eⁱ^{(kx − ωt)} is:

v_p = ω/k = (E/ħ)/(p/ħ) = E/p = (p²/2m)/p = p/2m = v/2

So that factor 1/2 only appears for the phase velocity. Weird, isn’t it? We find that the group velocity (v_g) of the eⁱ^{(kx − ωt)} function is equal to the classical velocity of our particle (i.e. v), but that its phase velocity (v_p) is equal to v divided by 2.

Hmm… What to say? Well… Nothing much—except that it makes sense, and very much so, because it’s the group velocity of the wavefunction that’s associated with the classical velocity of a particle, not the phase velocity. In fact, if we include the rest mass in our energy formula, so if we’d use the relativistic E = γm₀c² and p = γm₀v formulas (with γ the Lorentz factor), then we find that v_p = ω/k = E/p = (γm₀c²)/(γm₀v) = c²/v, and so that’s a superluminal velocity, because v is always smaller than c!

What? That’s even weirder! If we take the kinetic energy only, we find a phase velocity equal to v/2, but if we include the rest energy, then we get a superluminal phase velocity. It must be one or the other, no? Yep! You’re right! So that makes us wonder: is E = m·v²/2 really the right energy concept to use? The answer is unambiguous: no! It isn’t! And, just for the record, our young nobleman didn’t use the kinetic energy formula when he postulated his equations in his now famous PhD thesis.

So what did he use then? Where did he get his equations?

I am not sure. 🙂 A stroke of genius, it seems. According to Feynman, that’s how Schrödinger got his equation too: intuition, brilliance. In short, a stroke of genius. 🙂 Let’s relate these these two gems.

**Schrödinger’s equation and the two de Broglie relations**

Erwin Schrödinger and Louis de Broglie published their equations in 1924 and 1926 respectively. Can they be related? The answer is: yes—of course! Let’s first look at de Broglie‘s energy concept, however. Louis de Broglie was very familiar with Einsteins’ work and, hence, he knew that the energy of a particle consisted of three parts:

The particle’s rest energy m₀c², which de Broglie referred to as internal energy (E_int): this ‘internal energy’ includes the rest mass of the ‘internal pieces’, as he put it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’interaction energy);
Any potential energy it may have because of some field (so de Broglie was not assuming the particle was traveling in free space), which we’ll denote by V: the field(s) can be anything—gravitational, electromagnetic—you name it: whatever changes the energy because of the position of the particle;
The particle’s kinetic energy, which we wrote in terms of its momentum p: K.E. = m·v²/2 = m²·v²/(2m) = (m·v)²/(2m) = p²/(2m).

Indeed, in my previous posts, I would write the wavefunction as de Broglie wrote it, which is as follows:

ψ(θ) = ψ(x, t) = a·e⁻ⁱ^θ = a·e^{−i[(E_int+ p²/(2m) + V)·t − p∙x]/ħ}

In those post – such as my post on virtual particles – I’d also note how a change in potential energy plays out: a change in potential energy, when moving from one place to another, would change the wavefunction, but through the momentum only—so it would impact the spatial frequency only. So the change in potential would not change the temporal frequencies ω₁= E_int+ p₁²/(2m) + V₁ and ω₁= E_int+ p₂²/(2m) + V₂. Why? Or why not, I should say? Because of the energy conservation principle—or its equivalent in quantum mechanics. The temporal frequency f or ω, i.e. the time-rate of change of the phase of the wavefunction, does not change: all of the change in potential, and the corresponding change in kinetic energy, goes into changing the spatial frequency, i.e. the wave number k or the wavelength λ, as potential energy becomes kinetic or vice versa.

So is that consistent with what we wrote above, that E = m·v²? Maybe. Let’s think about it. Let’s first look at Schrödinger’s equation in free space (i.e. a space with zero potential) once again:

If we insert our ψ = e^{i(kx − ωt)} formula in Schrödinger’s free-space equation, we get the following nice result. [To keep things simple, we’re just assuming one-dimensional space for the calculations, so ∇²ψ = ∂²ψ/∂x². But the result can easily be generalized.] The time derivative on the left-hand side is ∂ψ/∂t = −iω·e^{i(kx − ωt)}. The second-order derivative on the right-hand side is ∂²ψ/∂x² = (ik)·(ik)·e^{i(kx − ωt)}= −k²·e^{i(kx − ωt)} . The e^{i(kx − ωt)} factor on both sides cancels out and, hence, equating both sides gives us the following condition:

−iω = −(iħ/2m)·k² ⇔ ω = (ħ/2m)·k²

Substituting ω = E/ħ and k = p/ħ yields:

E/ħ = (ħ/2m)·p²/ħ²= m²·v²/(2m·ħ) = m·v²/(2ħ) ⇔ E = m·v²/2

Bingo! We get that kinetic energy formula! But now… What if we’d not be considering free space? In other words: what if there is some potential? Well… We’d use the complete Schrödinger equation, which is:

Huh? Why is there a minus sign now? Look carefully: I moved the iħ factor on the left-hand side to the other when writing the free space version. If we’d do that for the complete equation, we’d get:

I like that representation a lot more—if only because it makes it a lot easier to interpret the equation—but, for some reason I don’t quite understand, you won’t find it like that in textbooks. Now how does it work when using the complete equation, so we add the −(i/ħ)·V·ψ term? It’s simple: the e^{i(kx − ωt)}factor also cancels out, and so we get:

−iω = −(iħ/2m)·k²−(i/ħ)·V ⇔ ω = (ħ/2m)·k²+ V/ħ

Substituting ω = E/ħ and k = p/ħ once more now yields:

E/ħ = (ħ/2m)·p²/ħ²+ V/ħ = m²·v²/(2m·ħ) + V/ħ = m·v²/(2ħ) + V/ħ ⇔ E = m·v²/2 + V

Bingo once more!

The only thing that’s missing now is the particle’s rest energy m₀c², which de Broglie referred to as internal energy (E_int). That includes everything, i.e. not only the rest mass of the ‘internal pieces’ (as said, now we call those ‘internal pieces’ quarks) but also their binding energy (i.e. the quarks’interaction energy). So how do we get that energy concept out of Schrödinger’s equation? There’s only one answer to that: that energy is just like V. We can, quite simply, just add it.

That brings us to the last and final question: what about our v_g = v result if we do not use the kinetic energy concept, but the E = m·v²/2 + V + E_int concept? The answer is simple: nothing. We still get the same, because we’re taking a derivative and the V and E_intjust appear as constants, and so their derivative with respect to p is zero. Check it:

v_g = dω/dk = d[E/ħ]/d[p/ħ] = dE/dp = d[p²/2m + V + E_int ]/dp = 2p/2m = p/m = v

It’s now pretty clear how this thing works. To localize our particle, we just superimpose a zillion of these e^{−i(ωt − kx)}equations. The only condition is that we’ve got that fixed v_g = dω/dk = v relationhip, but so we do have such fixed relationship—as you can see above. In fact, the Wikipedia article on the dispersion relation mentions that the de Broglie equations imply the following relation between ω and k: ω = ħk²/2m. As you can see, that’s not entirely correct: the author conveniently forgets the potential (V) and the rest energy (E_int) in the energy formula here!

What about the phase velocity? That’s a different story altogether. You can think about that for yourself. 🙂

I should make one final point here. As said, in order to localize a particle (or, to be precise, its wavefunction), we’re going to add a zillion elementary wavefunctions, each of which will make its own contribution to the composite wave. That contribution is captured by some coefficient a_i in front of every e^−iθ_i function, so we’ll have a zillion a_ie^−iθ_i functions, really. [Yep. Bit confusing: I use i here as subscript, as well as imaginary unit.] In case you wonder how that works out with Schrödinger’s equation, the answer is – once again – very simple: both the time derivative (which is just a first-order derivative) and the Laplacian are linear operators, so Schrödinger’s equation, for a composite wave, can just be re-written as the sum of a zillion ‘elementary’ wave equations.

So… Well… We’re all set now to effectively use Schrödinger’s equation to calculate the orbitals for a hydrogen atom, which is what we’ll do in our next post.

In the meanwhile, you can amuse yourself with reading a nice Wikibook article on the Laplacian, which gives you a nice feel for what Schrödinger’s equation actually represents—even if I gave you a good feel for that too on my Essentials page. Whatever. You choose. Just let me know what you liked best. 🙂

Oh… One more point: the v_g = dω/dk = d[p²/2m]/dp = p/m = v calculation obviously assumes we can treat m as a constant. In fact, what we’re actually doing is a rather complicated substitution of variables: you should write it all out—but that’s not the point here. The point is that we’re actually doing a non-relativistic calculation. Now, that does not mean that the wavefunction isn’t consistent with special relativity. It is. In fact, in one of my posts, I show how we can explain relativistic length contraction using the wavefunction. But it does mean that our calculation of the group velocity is not relativistically correct. But that’s a minor point: I’ll leave it for you as an exercise to calculate the relativistically correct formula for the group velocity. Have fun with it! 🙂

Note: Notations are often quite confusing. One should, generally speaking, denote a frequency by ν (nu), rather than by f, so as to not cause confusion with any function f, but then… Well… You create a new problem when you do that, because that Greek letter nu (ν) looks damn similar to the v of velocity, so that’s why I’ll often use f when I should be using nu (ν). As for the units, a frequency is expressed in cycles per second, while the angular frequency ω is expressed in radians per second. One cycle covers 2π radians and, therefore, we can write: ν = ω/2π. Hence, h∙ν = h∙ω/2π = ħ∙ω. Both ν as well as ω measure the time-rate of change of the phase of the wave function, as opposed to k, i.e. the spatial frequency of the wave function, which depends on the speed of the wave. Physicists also often use the symbol v for the speed of a wave, which is also hugely confusing, because it’s also used to denote the classical velocity of the particle. And then there’s two wave velocities, of course: the group versus the phase velocity. In any case… I find the use of that other symbol (c) for the wave velocity even more confusing, because this symbol is also used for the speed of light, and the speed of a wave is not necessarily (read: usually not) equal to the speed of light. In fact, both the group as well as the phase velocity of a particle wave are very different from the speed of light. The speed of a wave and the speed of light only coincide for electromagnetic waves and, even then, it should be noted that photons also have amplitudes to travel faster or slower than the speed of light.

Schrödinger’s equation as an energy conservation law

Original post:

In the movie about Stephen Hawking’s life, The Theory of Everything, there is talk about a single unifying equation that would explain everything in the universe. I must assume the real Stephen Hawking is familiar with Feynman’s unworldliness equation: U = 0, which – as Feynman convincingly demonstrates – effectively integrates all known equations in physics. It’s one of Feynman’s many jokes, of course, but an exceptionally clever one, as the argument convincingly shows there’s no such thing as one equation that explains all. Or, to be precise, one can, effectively, ‘hide‘ all the equations you want in a single equation, but it’s just a trick. As Feynman puts it: “When you unwrap the whole thing, you get back where you were before.”

Having said that, some equations in physics are obviously more fundamental than others. You can readily think of obvious candidates: Einstein’s mass-energy equivalence (m = E/c²); the wavefunction (ψ = e^{–i(ω·t − k·x)}) and the two de Broglie relations that come with it (ω = E/ħ and k = p/ħ); and, of course, Schrödinger’s equation, which we wrote as:

In my post on quantum-mechanical operators, I drew your attention to the fact that this equation is structurally similar to the heat diffusion equation. Indeed, assuming the heat per unit volume (q) is proportional to the temperature (T) – which is the case when expressing T in degrees Kelvin (K), so we can write q as q = k·T – we can write the heat diffusion equation as:

Moreover, I noted the similarity is not only structural. There is more to it: both equations model energy flows and/or densities. Look at it: the dimension of the left- and right-hand side of Schrödinger’s equation is the energy dimension: both quantities are expressed in joule. [Remember: a time derivative is a quantity expressed per second, and the dimension of Planck’s constant is the joule·second. You can figure out the dimension of the right-hand side yourself.] Now, the time derivative on the left-hand side is expressed in K/s. The constant in front (k) is just the (volume) heat capacity of the substance, which is expressed in J/(m³·K). So the dimension of k·(∂T/∂t) is J/(m³·s). On the right-hand side we have that Laplacian, whose dimension is K/m², multiplied by the thermal conductivity, whose dimension is W/(m·K) = J/(m·s·K). Hence, the dimension of the product is the same as the left-hand side: J/(m³·s).

We can present the thing in various ways: if we bring k to the other side, then we’ve got something expressed per second on the left-hand side, and something expressed per square meter on the right-hand side—but the k/κ factor makes it alright. The point is: both Schrödinger’s equation as well as the diffusion equation are actually an expression of the energy conservation law. They’re both expressions of Gauss’ flux theorem (but in differential form, rather than in integral form) which, as you know, pops up everywhere when talking energy conservation.

Huh?

Yep. I’ll give another example. Let me first remind you that the k·(∂T/∂t) = ∂q/∂t = κ·∇²T equation can also be written as:

The h in this equation is, obviously, not Planck’s constant, but the heat flow vector, i.e. the heat that flows through a unit area in a unit time, and h is, obviously, equal to h = −κ∇T. And, of course, you should remember your vector calculus here: ∇· is the divergence operator. In fact, we used the equation above, with ∇·h rather than ∇²T to illustrate the energy conservation principle. Now, you may or may not remember that we gave you a similar equation when talking about the energy of fields and the Poynting vector:

This immediately triggers the following reflection: if there’s a ‘Poynting vector’ for heat flow (h), and for the energy of fields (S), then there must be some kind of ‘Poynting vector’ for amplitudes too! I don’t know which one, but it must exist! And it’s going to be some complex vector, no doubt! But it should be out there.

It also makes me think of a point I’ve made a couple of times already—about the similarity between the E and B vectors that characterize the traveling electromagnetic field, and the real and imaginary part of the traveling amplitude. Indeed, the similarity between the two illustrations below cannot be a coincidence. In both cases, we’ve got two oscillating magnitudes that are orthogonal to each other, always. As such, they’re not independent: one follows the other, or vice versa.

The only difference is the phase shift. Euler’s formula incorporates a phase shift—remember: sinθ = cos(θ − π/2)—and so you don’t have that with the E and B vectors. But – Hey! – isn’t that why bosons and fermions are different? 🙂

[…]

This is great fun, and I’ll surely come back to it. As for now, however, I’ll let you ponder the matter for yourself. 🙂

Post scriptum: I am sure that all that the questions I raise here will be answered at the Masters’ level, most probably in some course dealing with quantum field theory, of course. 🙂 In any case, I am happy I can already anticipate such questions. 🙂

Oh – and, as for those two illustrations above, the animation below is one that should help you to think things through. It’s the electric field vector of a traveling circularly polarized electromagnetic wave, as opposed to the linearly polarized light that was illustrated above.

Quantum-mechanical operators

We climbed a mountain—step by step, post by post. 🙂 We have reached the top now, and the view is gorgeous. We understand Schrödinger’s equation, which describes how amplitudes propagate through space-time. It’s the quintessential quantum-mechanical expression. Let’s enjoy now, and deepen our understanding by introducing the concept of (quantum-mechanical) operators.

The operator concept

We’ll introduce the operator concept using Schrödinger’s equation itself and, in the process, deepen our understanding of Schrödinger’s equation a bit. You’ll remember we wrote it as:

However, you’ve probably seen it like it’s written on his bust, or on his grave, or wherever, which is as follows:

It’s the same thing, of course. The ‘over-dot’ is Newton’s notation for the time derivative. In fact, if you click on the picture above (and zoom in a bit), then you’ll see that the craftsman who made the stone grave marker, mistakenly, also carved a dot above the psi (ψ) on the right-hand side of the equation—but then someone pointed out his mistake and so the dot on the right-hand side isn’t painted. 🙂 The thing I want to talk about here, however, is the H in that expression above, which is, obviously, the following operator:

That’s a pretty monstrous operator, isn’t it? It is what it is, however: an algebraic operator (it operates on a number—albeit a complex number—unlike a matrix operator, which operates on a vector or another matrix). As you can see, it actually consists of two other (algebraic) operators:

The ∇²operator, which you know: it’s a differential operator. To be specific, it’s the Laplace operator, which is the divergence (∇·) of the gradient (∇) of a function: ∇²= ∇·∇ = (∂/∂x, ∂/∂y, ∂/∂z)·(∂/∂x, ∂/∂y, ∂/∂z) = ∂²/∂x²+ ∂²/∂y²+ ∂²/∂z². This too operates on our complex-valued function wavefunction ψ, and yields some other complex-valued function, which we then multiply by −ħ²/2m to get the first term.
The V(x, y, z) ‘operator’, which—in this particular context—just means: “multiply with V”. Needless to say, V is the potential here, and so it captures the presence of external force fields. Also note that V is a real number, just like −ħ²/2m.

Let me say something about the dimensions here. On the left-hand side of Schrödinger’s equation, we have the product of ħ and a time derivative (i is just the imaginary unit, so that’s just a (complex) number). Hence, the dimension there is [J·s]/[s] (the dimension of a time derivative is something expressed per second). So the dimension of the left-hand side is joule. On the right-hand side, we’ve got two terms. The dimension of that second-order derivative (∇²ψ) is something expressed per square meter, but then we multiply it with −ħ²/2m, whose dimension is [J²·s²]/[J/(m²/s²)]. [Remember: m = E/c².] So that reduces to [J·m²]. Hence, the dimension of (−ħ²/2m)∇²ψ is joule. And the dimension of V is joule too, of course. So it all works out. In fact, now that we’re here, it may or may not be useful to remind you of that heat diffusion equation we discussed when introducing the basic concepts involved in vector analysis:

That equation illustrated the physical significance of the Laplacian. We were talking about the flow of heat in, say, a block of metal, as illustrated below. The q in the equation above is the heat per unit volume, and the h in the illustration below was the heat flow vector (so it’s got nothing to do with Planck’s constant), which depended on the material, and which we wrote as h = –κ∇T, with T the temperature, and κ (kappa) the thermal conductivity. In any case, the point is the following: the equation below illustrates the physical significance of the Laplacian. We let it operate on the temperature (i.e. a scalar function) and its product with some constant (just think of replacing κ by −ħ²/2m gives us the time derivative of q, i.e. the heat per unit volume.

In fact, we know that q is proportional to T, so if we’d choose an appropriate temperature scale – i.e. choose the zero point such that q = k·T (your physics teacher in high school would refer to k as the (volume) specific heat capacity) – then we could simple write:

∂T/∂t = (κ/k)∇²T

From a mathematical point of view, that equation is just the same as ∂ψ/∂t = –(i·ħ/2m)·∇²ψ, which is Schrödinger’s equation for V = 0. In other words, you can – and actually should – also think of Schrödinger’s equation as describing the flow of… Well… What?

Well… Not sure. I am tempted to think of something like a probability density in space, but ψ represents a (complex-valued) amplitude. Having said that, you get the idea—I hope! 🙂 If not, let me paraphrase Feynman on this:

“We can think of Schrödinger’s equation as describing the diffusion of a probability amplitude from one point to another. In fact, the equation looks something like the diffusion equation we introduced when discussing heat flow, or the spreading of a gas. But there is one main difference: the imaginary coefficient in front of the time derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”

That says it all, right? 🙂 In fact, Schrödinger’s equation – as discussed here – was actually being derived when describing the motion of an electron along a line of atoms, i.e. for motion in one direction only, but you can visualize what it represents in three-dimensional space. The real exponential functions Feynman refer to exponential decay function: as the energy is spread over an ever-increasing volume, the amplitude of the wave becomes smaller and smaller. That may be the case for complex-valued exponentials as well. The key difference between a real- and complex-valued exponential decay function is that a complex exponential is a cyclical function. Now, I quickly googled to see how we could visualize that, and I like the following illustration:

The dimensional analysis of Schrödinger’s equation is also quite interesting because… Well… Think of it: that heat diffusion equation incorporates the same dimensions: temperature is a measure of the average energy of the molecules. That’s really something to think about. These differential equations are not only structurally similar but, in addition, they all seem to describe some flow of energy. That’s pretty deep stuff: it relates amplitudes to energies, so we should think in terms of Poynting vectors and all that. But… Well… I need to move on, and so I will move on—so you can re-visit this later. 🙂

Now that we’ve introduced the concept of an operator, let me say something about notations, because that’s quite confusing.

Some remarks on notation

Because it’s an operator, we should actually use the hat symbol—in line with what we did when we were discussing matrix operators: we’d distinguish the matrix (e.g. A) from its use as an operator (Â). You may or may not remember we do the same in statistics: the hat symbol is supposed to distinguish the estimator (â) – i.e. some function we use to estimate a parameter (which we usually denoted by some Greek symbol, like α) – from a specific estimate of the parameter, i.e. the value (a) we get when applying â to a specific sample or observation. However, if you remember the difference, you’ll also remember that hat symbol was quickly forgotten, because the context made it clear what was what, and so we’d just write a(x) instead of â(x). So… Well… I’ll be sloppy as well here, if only because the WordPress editor only offers very few symbols with a hat! 🙂

In any case, this discussion on the use (or not) of that hat is irrelevant. In contrast, what is relevant is to realize this algebraic operator H here is very different from that other quantum-mechanical Hamiltonian operator we discussed when dealing with a finite set of base states: that H was the Hamiltonian matrix, but used in an ‘operation’ on some state. So we have the matrix operator H, and the algebraic operator H.

Confusing?

Yes and no. First, we’ve got the context again, and so you always know whether you’re looking at continuous or discrete stuff:

If your ‘space’ is continuous (i.e. if states are to defined with reference to an infinite set of base states), then it’s the algebraic operator.
If, on the other hand, your states are defined by some finite set of discrete base states, then it’s the Hamiltonian matrix.

There’s another, more fundamental, reason why there should be no confusion. In fact, it’s the reason why physicists use the same symbol H in the first place: despite the fact that they look so different, these two operators (i.e. H the algebraic operator and H the matrix operator) are actually equivalent. Their interpretation is similar, as evidenced from the fact that both are being referred to as the energy operator in quantum physics. The only difference is that one operates on a (state) vector, while the other operates on a continuous function. It’s just the difference between matrix mechanics as opposed to wave mechanics really.

But… Well… I am sure I’ve confused you by now—and probably very much so—and so let’s start from the start. 🙂

Matrix mechanics

Let’s start with the easy thing indeed: matrix mechanics. The matrix-mechanical approach is summarized in that set of Hamiltonian equations which, by now, you know so well:

new

If we have n base states, then we have n equations like this: one for each i = 1, 2,… n. As for the introduction of the Hamiltonian, and the other subscript (j), just think of the description of a state:

essential

So… Well… Because we had used i already, we had to introduce j. 🙂

Let’s think about |ψ〉. It is the state of a system, like the ground state of a hydrogen atom, or one of its many excited states. But… Well… It’s a bit of a weird term, really. It all depends on what you want to measure: when we’re thinking of the ground state, or an excited state, we’re thinking energy. That’s something else than thinking its position in space, for example. Always remember: a state is defined by a set of base states, and so those base states come with a certain perspective: when talking states, we’re only looking at some aspect of reality, really. Let’s continue with our example of energy states, however.

You know that the lifetime of a system in an excited state is usually short: some spontaneous or induced emission of a quantum of energy (i.e. a photon) will ensure that the system quickly returns to a less excited state, or to the ground state itself. However, you shouldn’t think of that here: we’re looking at stable systems here. To be clear: we’re looking at systems that have some definite energy—or so we think: it’s just because of the quantum-mechanical uncertainty that we’ll always measure some other different value. Does that make sense?

If it doesn’t… Well… Stop reading, because it’s only going to get even more confusing. Not my fault, however!

Psi-chology

The ubiquity of that ψ symbol (i.e. the Greek letter psi) is really something psi-chological 🙂 and, hence, very confusing, really. In matrix mechanics, our ψ would just denote a state of a system, like the energy of an electron (or, when there’s only one electron, our hydrogen atom). If it’s an electron, then we’d describe it by its orbital. In this regard, I found the following illustration from Wikipedia particularly helpful: the green orbitals show excitations of copper (Cu) orbitals on a CuO₂plane. [The two big arrows just illustrate the principle of X-ray spectroscopy, so it’s an X-ray probing the structure of the material.]

800px-CuO2-plane_in_high_Tc_superconductor

So… Well… We’d write ψ as |ψ〉 just to remind ourselves we’re talking of some state of the system indeed. However, quantum physicists always want to confuse you, and so they will also use the psi symbol to denote something else: they’ll use it to denote a very particular C_i amplitude (or coefficient) in that |ψ〉 = ∑|i〉C_i formula above. To be specific, they’d replace the base states |i〉 by the continuous position variable x, and they would write the following:

C_i = ψ(i = x) = ψ(x) = C_ψ(x) = C(x) = 〈x|ψ〉

In fact, that’s just like writing:

φ(p) = 〈 mom p | ψ 〉 = 〈p|ψ〉 = C_φ(p) = C(p)

What they’re doing here, is (1) reduce the ‘system‘ to a ‘particle‘ once more (which is OK, as long as you know what you’re doing) and (2) they basically state the following:

If a particle is in some state |ψ〉, then we can associate some wavefunction ψ(x) or φ(p)—with it, and that wavefunction will represent the amplitude for the system (i.e. our particle) to be at x, or to have a momentum that’s equal to p.

So what’s wrong with that? Well… Nothing. It’s just that… Well… Why don’t they use χ(x) instead of ψ(x)? That would avoid a lot of confusion, I feel: one should not use the same symbol (psi) for the |ψ〉 state and the ψ(x) wavefunction.

Huh? Yes. Think about it. The point is: the position or the momentum, or even the energy, are properties of the system, so to speak and, therefore, it’s really confusing to use the same symbol psi (ψ) to describe (1) the state of the system, in general, versus (2) the position wavefunction, which describes… Well… Some very particular aspect (or ‘state’, if you want) of the same system (in this case: its position). There’s no such problem with φ(p), so… Well… Why don’t they use χ(x) instead of ψ(x) indeed? I have only one answer: psi-chology. 🙂

In any case, there’s nothing we can do about it and… Well… In fact, that’s what this post is about: it’s about how to describe certain properties of the system. Of course, we’re talking quantum mechanics here and, hence, uncertainty, and, therefore, we’re going to talk about the average position, energy, momentum, etcetera that’s associated with a particular state of a system, or—as we’ll keep things very simple—the properties of a ‘particle’, really. Think of an electron in some orbital, indeed! 🙂

So let’s now look at that set of Hamiltonian equations once again:

new

Looking at it carefully – so just look at it once again! 🙂 – and thinking about what we did when going from the discrete to the continuous setting, we can now understand we should write the following for the continuous case:

Of course, combining Schrödinger’s equation with the expression above implies the following:

Now how can we relate that integral to the expression on the right-hand side? I’ll have to disappoint you here, as it requires a lot of math to transform that integral. It requires writing H(x, x’) in terms of rather complicated functions, including – you guessed it, didn’t you? – Dirac’s delta function. Hence, I assume you’ll believe me if I say that the matrix- and wave-mechanical approaches are actually equivalent. In any case, if you’d want to check it, you can always read Feynman yourself. 🙂

Now, I wrote this post to talk about quantum-mechanical operators, so let me do that now.

Quantum-mechanical operators

You know the concept of an operator. As mentioned above, we should put a little hat (^) on top of our Hamiltonian operator, so as to distinguish it from the matrix itself. However, as mentioned above, the difference is usually quite clear from the context. Our operators were all matrices so far, and we’d write the matrix elements of, say, some operator A, as:

A_ij ≡ 〈 i | A | j 〉

The whole matrix itself, however, would usually not act on a base state but… Well… Just on some more general state ψ, to produce some new state φ, and so we’d write:

| φ 〉 = A | ψ 〉

Of course, we’d have to describe | φ 〉 in terms of the (same) set of base states and, therefore, we’d expand this expression into something like this:

You get the idea. I should just add one more thing. You know this important property of amplitudes: the 〈 ψ | φ 〉 amplitude is the complex conjugate of the 〈 φ | ψ 〉 amplitude. It’s got to do with time reversibility, because the complex conjugate of e^−iθ= e^{−i(ω·t−k·x)}is equal to e^iθ= e^{i(ω·t−k·x)}, so we’re just reversing the x- and t–direction. We write:

〈 ψ | φ 〉 = 〈 φ | ψ 〉*

Now what happens if we want to take the complex conjugate when we insert a matrix, so when writing 〈 φ | A | ψ 〉 instead of 〈 φ | ψ 〉, this rules becomes:

〈 φ | A | ψ 〉* = 〈 ψ | A† | φ 〉

The dagger symbol denotes the conjugate transpose, so A† is an operator whose matrix elements are equal to A_ij† = A_ji*. Now, it may or may not happen that the A† matrix is actually equal to the original A matrix. In that case – and only in that case – we can write:

〈 ψ | A | φ 〉 = 〈 φ | A | ψ 〉*

We then say that A is a ‘self-adjoint’ or ‘Hermitian’ operator. That’s just a definition of a property, which the operator may or may not have—but many quantum-mechanical operators are actually Hermitian. In any case, we’re well armed now to discuss some actual operators, and we’ll start with that energy operator.

The energy operator (H)

We know the state of a system is described in terms of a set of base states. Now, our analysis of N-state systems showed we can always describe it in terms of a special set of base states, which are referred to as the states of definite energy because… Well… Because they’re associated with some definite energy. In that post, we referred to these energy levels as E_n(n = I, II,… N). We used boldface for the subscript n (so we wrote n instead of n) because of these Roman numerals. With each energy level, we could associate a base state, of definite energy indeed, that we wrote as |n〉. To make a long story short, we summarized our results as follows:

The energies E_I, E_II,…, E_n,…, E_Nare the eigenvalues of the Hamiltonian matrix H.
The state vectors |n〉 that are associated with each energy E_n, i.e. the set of vectors |n〉, are the corresponding eigenstates.

We’ll be working with some more subscripts in what follows, and these Roman numerals and the boldface notation are somewhat confusing (if only because I don’t want you to think of these subscripts as vectors), we’ll just denote E_I, E_II,…, E_n,…, E_Nas E₁, E₂,…, E_i,…, E_N, and we’ll number the states of definite energy accordingly, also using some Greek letter so as to clearly distinguish them from all our Latin letter symbols: we’ll write these states as: |η₁〉, |η₁〉,… |η_N〉. [If I say, ‘we’, I mean Feynman of course. You may wonder why he doesn’t write |E_i〉, or |ε_i〉. The answer is: writing |E_n〉 would cause confusion, because this state will appear in expressions like: |E_i〉E_i, so that’s the ‘product’ of a state (|E_i〉) and the associated scalar (E_i). Too confusing. As for using η (eta) instead of ε (epsilon) to denote something that’s got to do with energy… Well… I guess he wanted to keep the resemblance with the n, and then the Ancient Greek apparently did use this η letter for a sound like ‘e‘ so… Well… Why not? Let’s get back to the lesson.]

Using these base states of definite energy, we can write the state of the system as:

|ψ〉 = ∑ |η_i〉 C_i = ∑ |η_i〉〈η_i|ψ〉 over all i (i = 1, 2,… , N)

Now, we didn’t talk all that much about what these base states actually mean in terms of measuring something but you’ll believe if I say that, when measuring the energy of the system, we’ll always measure one or the other E₁, E₂,…, E_i,…, E_N value. We’ll never measure something in-between: it’s either–or. Now, as you know, measuring something in quantum physics is supposed to be destructive but… Well… Let us imagine we could make a thousand measurements to try to determine the average energy of the system. We’d do so by counting the number of times we measure E₁ (and of course we’d denote that number as N₁), E₂, E₃, etcetera. You’ll agree that we’d measure the average energy as:

However, measurement is destructive, and we actually know what the expected value of this ‘average’ energy will be, because we know the probabilities of finding the system in a particular base state. That probability is equal to the absolute square of that C_icoefficient above, so we can use the P_i= |C_i|² formula to write:

〈E_av〉 = ∑ P_iE_i over all i (i = 1, 2,… , N)

Note that this is a rather general formula. It’s got nothing to do with quantum mechanics: if A_i represents the possible values of some quantity A, and P_i is the probability of getting that value, then (the expected value of) the average A will also be equal to 〈A_av〉 = ∑ P_iA_i. No rocket science here! 🙂 But let’s now apply our quantum-mechanical formulas to that 〈E_av〉 = ∑ P_iE_i formula. [Oh—and I apologize for using the same angle brackets 〈 and 〉 to denote an expected value here—sorry for that! But it’s what Feynman does—and other physicists! You see: they don’t really want you to understand stuff, and so they often use very confusing symbols.] Remembering that the absolute square of a complex number equals the product of that number and its complex conjugate, we can re-write the 〈E_av〉 = ∑ P_iE_i formula as:

Now, you know that Dirac’s bra-ket notation allows numerous manipulations. For example, what we could do is take out that ‘common factor’ 〈ψ|, and so we may re-write that monster above as:

I know: you’re getting tired and you wonder why we need all this stuff. Just hang in there. We’re almost done. I just need to do a few more unpleasant things, one of which is to remind you that this business of the energy states being eigenstates (and the energy levels being eigenvalues) of our Hamiltonian matrix (see my post on N-state systems) comes with a number of interesting properties, including this one:

H |η_i〉 = E_i|η_i〉 = |η_i〉E_i

Just think about what’s written here: on the left-hand side, we’re multiplying a matrix with a (base) state vector, and on the left-hand side we’re multiplying it with a scalar. So our |φ〉 = ∑ |η_i〉E_i〈η_i|ψ〉 sum now becomes:

|φ〉 = ∑ H |η_i〉〈η_i|ψ〉 over all i (i = 1, 2,… , N)

Now we can manipulate that expression some more so as to get the following:

|φ〉 = H ∑|η_i〉〈η_i|ψ〉 = H|ψ〉

Finally, we can re-combine this now with the 〈E_av〉 = 〈ψ|φ〉 equation above, and so we get the fantastic result we wanted:

〈E_av〉 = 〈 ψ | φ 〉 = 〈 ψ | H | ψ 〉

Huh? Yes! To get the average energy, you operate on |ψ〉 with H, and then you multiply the result with 〈ψ|. It’s a beautiful formula. On top of that, the new formula for the average energy is not only pretty but also useful, because now we don’t need to say anything about any particular set of base states. We don’t even have to know all of the possible energy levels. When we have to calculate the average energy of some system, we only need to be able to describe the state of that system in terms of some set of base states, and we also need to know the Hamiltonian matrix for that set, of course. But if we know that, we can calculate its average energy.

You’ll say that’s not a big deal because… Well… If you know the Hamiltonian, you know everything, so… Well… Yes. You’re right: it’s less of a big deal than it seems. Having said that, the whole development above is very interesting because of something else: we can easily generalize it for other physical measurements. I call it the ‘average value’ operator idea, but you won’t find that term in any textbook. 🙂 Let me explain the idea.

The average value operator (A)

The development above illustrates how we can relate a physical observable, like the (average) energy (E), to a quantum-mechanical operator (H). Now, the development above can easily be generalized to any observable that would be proportional to the energy. It’s perfectly reasonable, for example, to assume the angular momentum – as measured in some direction, of course, which we usually refer to as the z-direction – would be proportional to the energy, and so then it would be easy to define a new operator L_z, which we’d define as the operator of the z-component of the angular momentum L. [I know… That’s a bit of a long name but… Well… You get the idea.] So we can write:

〈L_z〉_av = 〈 ψ | L_z| ψ 〉

In fact, further generalization yields the following grand result:

If a physical observable A is related to a suitable quantum-mechanical operator Â, then the average value of A for the state | ψ 〉 is given by:

〈A〉_av = 〈 ψ | Â| ψ 〉 = 〈 ψ | φ 〉 with | φ 〉 = Â| ψ 〉

At this point, you may have second thoughts, and wonder: what state | ψ 〉? The answer is: it doesn’t matter. It can be any state, as long as we’re able to describe in terms of a chosen set of base states. 🙂

OK. So far, so good. The next step is to look at how this works for the continuity case.

The energy operator for wavefunctions (H)

We can start thinking about the continuous equivalent of the 〈E_av〉 = 〈ψ|H|ψ〉 expression by first expanding it. We write:

You know the continuous equivalent of a sum like this is an integral, i.e. an infinite sum. Now, because we’ve got two subscripts here (i and j), we get the following double integral:

Now, I did take my time to walk you through Feynman’s derivation of the energy operator for the discrete case, i.e. the operator when we’re dealing with matrix mechanics, but I think I can simplify my life here by just copying Feynman’s succinct development:

Done! Given a wavefunction ψ(x), we get the average energy by doing that integral above. Now, the quantity in the braces of that integral can be written as that operator we introduced when we started this post:

So now we can write that integral much more elegantly. It becomes:

〈E〉_av = ∫ ψ*(x) H ψ(x) dx

You’ll say that doesn’t look like 〈E_av〉 = 〈 ψ | H | ψ 〉! It does. Remember that 〈 ψ | = | ψ 〉*. 🙂 Done!

I should add one qualifier though: the formula above assumes our wavefunction has been normalized, so all probabilities add up to one. But that’s a minor thing. The only thing left to do now is to generalize to three dimensions. That’s easy enough. Our expression becomes a volume integral:

〈E〉_av = ∫ ψ*(r) H ψ(r) dV

Of course, dV stands for dVolume here, not for any potential energy, and, of course, once again we assume all probabilities over the volume add up to 1, so all is normalized. Done! 🙂

We’re almost done with this post. What’s left is the position and momentum operator. You may think this is going to another lengthy development but… Well… It turns out the analysis is remarkably simple. Just stay with me a few more minutes and you’ll have earned your degree. 🙂

The position operator (x)

The thing we need to solve here is really easy. Look at the illustration below as representing the probability density of some particle being at x. Think about it: what’s the average position?

Well? What? The (expected value of the) average position is just this simple integral: 〈x〉_av = ∫ x P(x) dx, over all the whole range of possible values for x. 🙂 That’s all. Of course, because P(x) = |ψ(x)|² =ψ*(x)·ψ(x), this integral now becomes:

〈x〉_av = ∫ ψ*(x) x ψ(x) dx

That looks exactly the same as 〈E〉_av = ∫ ψ*(x) H ψ(x) dx, and so we can look at x as an operator too!

Huh? Yes. It’s an extremely simple operator: it just means “multiply by x“. 🙂

I know you’re shaking your head now: is it that easy? It is. Moreover, the ‘matrix-mechanical equivalent’ is equally simple but, as it’s getting late here, I’ll refer you to Feynman for that. 🙂

The momentum operator (p_x)

Now we want to calculate the average momentum of, say, some electron. What integral would you use for that? […] Well… What? […] It’s easy: it’s the same thing as for x. We can just substitute replace x for p in that 〈x〉_av = ∫ x P(x) dx formula, so we get:

〈p〉_av = ∫ p P(p) dp, over all the whole range of possible values for p

Now, you might think the rest is equally simple, and… Well… It actually is simple but there’s one additional thing in regard to the need to normalize stuff here. You’ll remember we defined a momentum wavefunction (see my post on the Uncertainty Principle), which we wrote as:

φ(p) = 〈 mom p | ψ 〉

Now, in the mentioned post, we related this momentum wavefunction to the particle’s ψ(x) = 〈x|ψ〉 wavefunction—which we should actually refer to as the position wavefunction, but everyone just calls it the particle’s wavefunction, which is a bit of a misnomer, as you can see now: a wavefunction describes some property of the system, and so we can associate several wavefunctions with the same system, really! In any case, we noted the following there:

The two probability density functions, φ(p) and ψ(x), look pretty much the same, but the half-width (or standard deviation) of one was inversely proportional to the half-width of the other. To be precise, we found that the constant of proportionality was equal to ħ/2, and wrote that relation as follows: σ_p= (ħ/2)/σ_x.
We also found that, when using a regular normal distribution function for ψ(x), we’d have to normalize the probability density function by inserting a (2πσ_x²)^−1/2in front of the exponential.

Now, it’s a bit of a complicated argument, but the upshot is that we cannot just write what we usually write, i.e. P_i = |C_i|²or P(x) = |ψ(x)|². No. We need to put a normalization factor in front, which combines the two factors I mentioned above. To be precise, we have to write:

P(p) = |〈p|ψ〉|²/(2πħ)

So… Well… Our 〈p〉_av = ∫ p P(p) dp integral can now be written as:

〈p〉_av = ∫ 〈ψ|p〉p〈p|ψ〉 dp/(2πħ)

So that integral is totally like what we found for 〈x〉_av and so… We could just leave it at that, and say we’ve solved the problem. In that sense, it is easy. However, having said that, it’s obvious we’d want some solution that’s written in terms of ψ(x), rather than in terms of φ(p), and that requires some more manipulation. I’ll refer you, once more, to Feynman for that, and I’ll just give you the result:

So… Well… I turns out that the momentum operator – which I tentatively denoted as p_x above – is not so simple as our position operator (x). Still… It’s not hugely complicated either, as we can write it as:

p_x ≡ (ħ/i)·(∂/∂x)

Of course, the purists amongst you will, once again, say that I should be more careful and put a hat wherever I’d need to put one so… Well… You’re right. I’ll wrap this all up by copying Feynman’s overview of the operators we just explained, and so he does use the fancy symbols. 🙂

Well, folks—that’s it! Off we go! You know all about quantum physics now! We just need to work ourselves through the exercises that come with Feynman’s Lectures, and then you’re ready to go and bag a degree in physics somewhere. So… Yes… That’s what I want to do now, so I’ll be silent for quite a while now. Have fun! 🙂

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Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:

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Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:

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Schrödinger’s equation: the original approach

Of course, your first question when seeing the title of this post is: what’s original, really? Well… The answer is simple: it’s the historical approach, and it’s original because it’s actually quite intuitive. Indeed, Lecture no. 16 in Feynman’s third Volume of Lectures on Physics is like a trip down memory lane as Feynman himself acknowledges, after presenting Schrödinger’s equation using that very rudimentary model we developed in our previous post:

So… Well… Let’s have a look at it. 🙂 We were looking at some electron we described in terms of its location at one or the other atom in a linear array (think of it as a line). We did so by defining base states |n〉 = |x_n〉, noting that the state of the electron at any point in time could then be written as:

|φ〉 = ∑ |x_n〉C_n(t) = ∑ |x_n〉〈x_n|φ〉 over all n

The C_n(t) = 〈x_n|φ〉 coefficient is the amplitude for the electron to be at x_nat t. Hence, the C_n(t) amplitudes vary with t as well as with x. We’ll re-write them as C_n(t) = C(x_n, t) = C(x_n). Note that the latter notation does not explicitly show the time dependence. The Hamiltonian equation we derived in our previous post is now written as:

iħ·(∂C(x_n)/∂t) = E₀C(x_n) − AC(x_n+b) − AC(x_n−b)

Note that, as part of our move from the C_n(t) to the C(x_n) notation, we write the time derivative dC_n(t)/dt now as ∂C(x_n)/∂t, so we use the partial derivative symbol now (∂). Of course, the other partial derivative will be ∂C(x)/∂x) as we move from the count variable x_nto the continuous variable x, but let’s not get ahead of ourselves here. The solution we found for our C(x_n) functions was the following wavefunction:

C(x_n) = a·e^{i(k∙x_n−ω·t)}= a·e^−i∙ω·t·eⁱ^∙k∙x_n= a·e^{−i·(E/ħ)·t}·e^i·k∙x_n

We also found the following relationship between E and k:

E = E₀ − 2A·cos(kb)

Now, even Feynman struggles a bit with the definition of E₀ and k here, and their relationship with E, which is graphed below.

Indeed, he first writes, as he starts developing the model, that E₀ is, physically, the energy the electron would have if it couldn’t leak away from one of the atoms, but then he also adds: “It represents really nothing but our choice of the zero of energy.”

This is all quite enigmatic because we cannot just do whatever we want when discussing the energy of a particle. As I pointed out in one of my previous posts, when discussing the energy of a particle in the context of the wavefunction, we generally consider it to be the sum of three different energy concepts:

The particle’s rest energy m₀c², which de Broglie referred to as internal energy (E_int), and which includes the rest mass of the ‘internal pieces’, as Feynman puts it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’ interaction energy).
Any potential energy it may have because of some field (i.e. if it is not traveling in free space), which we usually denote by U. This field can be anything—gravitational, electromagnetic: it’s whatever changes the energy of the particle because of its position in space.
The particle’s kinetic energy, which we write in terms of its momentum p: m·v²/2 = m²·v²/(2m) = (m·v)²/(2m) = p²/(2m).

It’s obvious that we cannot just “choose” the zero point here: the particle’s rest energy is its rest energy, and its velocity is its velocity. So it’s not quite clear what the E₀ in our model really is. As far as I am concerned, it represents the average energy of the system really, so it’s just like the E₀ for our ammonia molecule, or the E₀ for whatever two-state system we’ve seen so far. In fact, when Feynman writes that we can “choose our zero of energy so that E₀ − 2A = 0″ (so the minimum of that curve above is at the zero of energy), he actually makes some assumption in regard to the relative magnitude of the various amplitudes involved.

We should probably think about it in this way: −(i/ħ)·E₀ is the amplitude for the electron to just stay where it is, while i·A/ħ is the amplitude to go somewhere else—and note we’ve got two possibilities here: the electron can go to |x_n+1〉, or, alternatively, it can go to |x_n−1〉. Now, amplitudes can be associated with probabilities by taking the absolute square, so I’d re-write the E₀ − 2A = 0 assumption as:

E₀= 2A ⇔ |−(i/ħ)·E₀|²= |(i/ħ)·2A|²

Hence, in my humble opinion, Feynman’s assumption that E₀ − 2A = 0 has nothing to do with ‘choosing the zero of energy’. It’s more like a symmetry assumption: we’re basically saying it’s as likely for the electron to stay where it is as it is to move to the next position. It’s an idea I need to develop somewhat further, as Feynman seems to just gloss over these little things. For example, I am sure it is not a coincidence that the E_I, E_II, E_III and E_IVenergy levels we found when discussing the hyperfine splitting of the hydrogen ground state also add up to 0. In fact, you’ll remember we could actually measure those energy levels (E_I= E_II= E_III= A ≈ 9.23×10⁻⁶ eV, and E_IV= −3A ≈ −27.7×10⁻⁶ eV), so saying that we can “choose” some zero energy point is plain nonsense. The question just doesn’t arise. In any case, as I have to continue the development here, I’ll leave this point for further analysis in the future. So… Well… Just note this E₀ − 2A = 0 assumption, as we’ll need it in a moment.

The second assumption we’ll need concerns the variation in k. As you know, we can only get a wave packet if we allow for uncertainty in k which, in turn, translates into uncertainty for E. We write:

ΔE = Δ[E₀ − 2A·cos(kb)]

Of course, we’d need to interpret the Δ as a variance (σ²) or a standard deviation (σ) so we can apply the usual rules – i.e. var(a) = 0, var(aX) = a²·var(X), and var(aX ± bY) = a²·var(X) + b²·var(Y) ± 2ab·cov(X, Y) – to be a bit more precise about what we’re writing here, but you get the idea. In fact, let me quickly write it out:

var[E₀ − 2A·cos(kb)] = var(E₀) + 4A²·var[cos(kb)] ⇔ var(E) = 4A²·var[cos(kb)]

Now, you should check my post scriptum to my page on the Essentials, to see how the probability density function of the cosine of a randomly distributed variable looks like, and then you should go online to find a formula for its variance, and then you can work it all out yourself, because… Well… I am not going to do it for you. What I want to do here is just show how Feynman gets Schrödinger’s equation out of all of these simplifications.

So what’s the second assumption? Well… As the graph shows, our k can take any value between −π/b and +π/b, and therefore, the kb argument in our cosine function can take on any value between −π and +π. In other words, kb could be any angle. However, as Feynman puts it—we’ll be assuming that kb is ‘small enough’, so we can use the small-angle approximations whenever we see the cos(kb) and/or sin(kb) functions. So we write: sin(kb) ≈ kb and cos(kb) ≈ 1 − (kb)²/2 = 1 − k²b²/2. Now, that assumption led to another grand result, which we also derived in our previous post. It had to do with the group velocity of our wave packet, which we calculated as:

v = dω/dk = (2Ab²/ħ)·k

Of course, we should interpret our k here as “the typical k“. Huh? Yes… That’s how Feynman refers to it, and I have no better term for it. It’s some kind of ‘average’ of the Δk interval, obviously, but… Well… Feynman does not give us any exact definition here. Of course, if you look at the graph once more, you’ll say that, if the typical kb has to be “small enough”, then its expected value should be zero. Well… Yes and no. If the typical kb is zero, or if k is zero, then v is zero, and then we’ve got a stationary electron, i.e. an electron with zero momentum. However, because we’re doing what we’re doing (that is, we’re studying “stuff that moves”—as I put it unrespectfully in a few of my posts, so as to distinguish from our analyses of “stuff that doesn’t move”, like our two-state systems, for example), our “typical k” should not be zero here. OK… We can now calculate what’s referred to as the effective mass of the electron, i.e. the mass that appears in the classical kinetic energy formula: K.E. = m·v²/2. Now, there are two ways to do that, and both are somewhat tricky in their interpretation:

1. Using both the E₀ − 2A = 0 as well as the “small kb” assumption, we find that E = E₀ − 2A·(1 − k²b²/2) = A·k²b². Using that for the K.E. in our formula yields:

m_eff= 2A·k²b²/v²= 2A·k²b²/[(2Ab²/ħ)·k]²= ħ²/(2Ab²)

2. We can use the classical momentum formula (p = m·v), and then the 2nd de Broglie equation, which tells us that each wavenumber (k) is to be associated with a value for the momentum (p) using the p = ħk (so p is proportional to k, with ħ as the factor of proportionality). So we can now calculate m_eff as m_eff = ħk/v. Substituting v again for what we’ve found above, gives us the same:

m_eff= 2A·k²b²/v= ħ·k/[(2Ab²/ħ)·k]= ħ²/(2Ab²)

Of course, we’re not supposed to know the de Broglie relations at this point in time. 🙂 But, now that you’ve seen them anyway, note how we have two formulas for the momentum:

The classical formula (p = m·v) tells us that the momentum is proportional to the classical velocity of our particle, and m is then the factor of proportionality.
The quantum-mechanical formula (p = ħk) tells us that the (typical) momentum is proportional to the (typical) wavenumber, with Planck’s constant (ħ) as the factor of proportionality. Combining both combines the classical and quantum-mechanical perspective of a moving particle:

m·v = ħk

I know… It’s an obvious equation but… Well… Think of it. It’s time to get back to the main story now. Remember we were trying to find Schrödinger’s equation? So let’s get on with it. 🙂

To do so, we need one more assumption. It’s the third major simplification and, just like the others, the assumption is obvious on first, but not on second thought. 😦 So… What is it? Well… It’s easy to see that, in our m_eff= ħ²/(2Ab²) formula, all depends on the value of 2Ab². So, just like we should wonder what happens with that kb factor in the argument of our sine or cosine function if b goes to zero—i.e. if we’re letting the lattice spacing go to zero, so we’re moving from a discrete to a continuous analysis now—we should also wonder what happens with that 2Ab² factor! Well… Think about it. Wouldn’t it be reasonable to assume that the effective mass of our electron is determined by some property of the material, or the medium (so that’s the silicon in our previous post) and, hence, that it’s constant really. Think of it: we’re not changing the fundamentals really—we just have some electron roaming around in some medium and all that we’re doing now is bringing those x_ncloser together. Much closer. It’s only logical, then, that our amplitude to jump from x_n_±1to x_nwould also increase, no? So what we’re saying is that 2Ab² is some constant which we write as ħ²/m_effor, what amounts to the same, that Ab²= ħ²/2·m_eff.

Of course, you may raise two objections here:

The Ab²= ħ²/2·m_eff assumption establishes a very particular relation between A and b, as we can write A as A = [ħ²/(2m_eff)]·b⁻² now. So we’ve got like an y = 1/x² relation here. Where the hell does that come from?
We were talking some real stuff here: a crystal lattice with atoms that, in reality, do have some spacing, so that corresponds to some real value for b. So that spacing gives some actual physical significance to those x_nvalues.

Well… What can I say? I think you should re-read that quote of Feynman when I started this post. We’re going to get Schrödinger’s equation – i.e. the ultimate prize for all of the hard work that we’ve been doing so far – but… Yes. It’s really very heuristic, indeed! 🙂 But let’s get on with it now! We can re-write our Hamiltonian equation as:

iħ·(∂C(x_n)/∂t) = E₀C(x_n) − AC(x_n+b) − AC(x_n−b)]

= (E₀−2A)C(x_n) + A[2C(x_n) − C(x_n+b) − C(x_n−b) = A[2C(x_n) − C(x_n+b) − C(x_n−b)]

Now, I know your brain is about to melt down but, fiddling with this equation as we’re doing right now, Schrödinger recognized a formula for the second-order derivative of a function. I’ll just jot it down, and you can google it so as to double-check where it comes from:

Just substitute f(x) for C(x_n) in the second part of our equation above, and you’ll see we can effectively write that 2C(x_n) − C(x_n+b) − C(x_n−b) factor as:

We’re done. We just iħ·(∂C(x_n)/∂t) on the left-hand side now and multiply the expression above with A, to get what we wanted to get, and that’s – YES! – Schrödinger’s equation:

Whatever your objections to this ‘derivation’, it is the correct equation. For a particle in free space, we just write m instead of m_eff, but it’s exactly the same. I’ll now give you Feynman’s full quote, which is quite enlightening:

“We do not intend to have you think we have derived the Schrödinger equation but only wish to show you one way of thinking about it. When Schrödinger first wrote it down, he gave a kind of derivation based on some heuristic arguments and some brilliant intuitive guesses. Some of the arguments he used were even false, but that does not matter; the only important thing is that the ultimate equation gives a correct description of nature. The purpose of our discussion is then simply to show you that the correct fundamental quantum mechanical equation [i.e. Schrödinger’s equation] has the same form you get for the limiting case of an electron moving along a line of atoms. We can think of it as describing the diffusion of a probability amplitude from one point to the next along the line. That is, if an electron has a certain amplitude to be at one point, it will, a little time later, have some amplitude to be at neighboring points. In fact, the equation looks something like the diffusion equations which we have used in Volume I. But there is one main difference: the imaginary coefficient in front of the time derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”

So… That says it all, I guess. Isn’t it great to be where we are? We’ve really climbed a mountain here. And I think the view is gorgeous. 🙂

Oh—just in case you’d think I did not give you Schrödinger’s equation, let me write it in the form you’ll usually see it:

schrodinger 3

Done! 🙂

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The de Broglie relations, the wave equation, and relativistic length contraction

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. So no use to read this. Read my recent papers instead. 🙂

Original post:

You know the two de Broglie relations, also known as matter-wave equations:

f = E/h and λ = h/p

You’ll find them in almost any popular account of quantum mechanics, and the writers of those popular books will tell you that f is the frequency of the ‘matter-wave’, and λ is its wavelength. In fact, to add some more weight to their narrative, they’ll usually write them in a somewhat more sophisticated form: they’ll write them using ω and k. The omega symbol (using a Greek letter always makes a big impression, doesn’t it?) denotes the angular frequency, while k is the so-called wavenumber. Now, k = 2π/λ and ω = 2π·f and, therefore, using the definition of the reduced Planck constant, i.e. ħ = h/2π, they’ll write the same relations as:

λ = h/p = 2π/k ⇔ k = 2π·p/h
f = E/h = (ω/2π)

⇒ k = p/ħ and ω = E/ħ

They’re the same thing: it’s just that working with angular frequencies and wavenumbers is more convenient, from a mathematical point of view that is: it’s why we prefer expressing angles in radians rather than in degrees (k is expressed in radians per meter, while ω is expressed in radians per second). In any case, the ‘matter wave’ – even Wikipedia uses that term now – is, of course, the amplitude, i.e. the wave-function ψ(x, t), which has a frequency and a wavelength, indeed. In fact, as I’ll show in a moment, it’s got two frequencies: one temporal, and one spatial. I am modest and, hence, I’ll admit it took me quite a while to fully distinguish the two frequencies, and so that’s why I always had trouble connecting these two ‘matter wave’ equations.

Indeed, if they represent the same thing, they must be related, right? But how exactly? It should be easy enough. The wavelength and the frequency must be related through the wave velocity, so we can write: f·λ = v, with v the velocity of the wave, which must be equal to the classical particle velocity, right? And then momentum and energy are also related. To be precise, we have the relativistic energy-momentum relationship: p·c = m_v·v·c = m_v·c²·v/c = E·v/c. So it’s just a matter of substitution. We should be able to go from one equation to the other, and vice versa. Right?

Well… No. It’s not that simple. We can start with either of the two equations but it doesn’t work. Try it. Whatever substitution you try, there’s no way you can derive one of the two equations above from the other. The fact that it’s impossible is evidenced by what we get when we’d multiply both equations. We get:

f·λ = (E/h)·(h/p) = E/p
v = f·λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v)

⇒ E = m·v²

Huh? What kind of formula is that? E = m·v²? That’s a formula you’ve never ever seen, have you? It reminds you of the kinetic energy formula of course—K.E. = m·v²/2—but… That factor 1/2 should not be there. Let’s think about it for a while. First note that this E = m·v²relation makes perfectly sense if v = c. In that case, we get Einstein’s mass-energy equivalence (E = m·c²), but that’s besides the point here. The point is: if v = c, then our ‘particle’ is a photon, really, and then the E = h·f is referred to as the Planck-Einstein relation. The wave velocity is then equal to c and, therefore, f·λ = c, and so we can effectively substitute to find what we’re looking for:

E/p = (h·f)/(h/λ) = f·λ = c ⇒ E = p·c

So that’s fine: we just showed that the de Broglie relations are correct for photons. [You remember that E = p·c relation, no? If not, check out my post on it.] However, while that’s all nice, it is not what the de Broglie equations are about: we’re talking the matter-wave here, and so we want to do something more than just re-confirm that Planck-Einstein relation, which you can interpret as the limit of the de Broglie relations for v = c. In short, we’re doing something wrong here! Of course, we are. I’ll tell you what exactly in a moment: it’s got to do with the fact we’ve got two frequencies really.

Let’s first try something else. We’ve been using the relativistic E = m_v·c² equation above. Let’s try some other energy concept: let’s substitute the E in the f = E/h by the kinetic energy and then see where we get—if anywhere at all. So we’ll use the E_kinetic = m∙v²/2 equation. We can then use the definition of momentum (p = m∙v) to write E = p²/(2m), and then we can relate the frequency f to the wavelength λ using the v = λ∙f formula once again. That should work, no? Let’s do it. We write:

E = p²/(2m)
E = h∙f = h·v/λ

⇒ λ = h·v/E = h·v/(p²/(2m)) = h·v/[m²·v²/(2m)] = h/[m·v/2] = 2∙h/p

So we find λ = 2∙h/p. That is almost right, but not quite: that factor 2 should not be there. Well… Of course you’re smart enough to see it’s just that factor 1/2 popping up once more—but as a reciprocal, this time around. 🙂 So what’s going on? The honest answer is: you can try anything but it will never work, because the f = E/h and λ = h/p equations cannot be related—or at least not so easily. The substitutions above only work if we use that E = m·v² energy concept which, you’ll agree, doesn’t make much sense—at first, at least. Again: what’s going on? Well… Same honest answer: the f = E/h and λ = h/p equations cannot be related—or at least not so easily—because the wave equation itself is not so easy.

Let’s review the basics once again.

The wavefunction

The amplitude of a particle is represented by a wavefunction. If we have no information whatsoever on its position, then we usually write that wavefunction as the following complex-valued exponential:

ψ(x, t) = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} = a·e^{−i·(ω·t − k∙x)}= a·e^{i(k∙x−ω·t)}= a·e^iθ= a·(cosθ + i·sinθ)

θ is the so-called phase of our wavefunction and, as you can see, it’s the argument of a wavefunction indeed, with temporal frequency ω and spatial frequency k (if we choose our x-axis so its direction is the same as the direction of k, then we can substitute the k and x vectors for the k and x scalars, so that’s what we’re doing here). Now, we know we shouldn’t worry too much about a, because that’s just some normalization constant (remember: all probabilities have to add up to one). However, let’s quickly develop some logic here. Taking the absolute square of this wavefunction gives us the probability of our particle being somewhere in space at some point in time. So we get the probability as a function of x and t. We write:

P(x ,t) = |a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]}|²= a²

As all probabilities have to add up to one, we must assume we’re looking at some box in spacetime here. So, if the length of our box is Δx = x₂ − x₁, then (Δx)·a² = (x₂−x₁)·a²= 1 ⇔ Δx = 1/a². [We obviously simplify the analysis by assuming a one-dimensional space only here, but the gist of the argument is essentially correct.] So, freezing time (i.e. equating t to some point t = t₀), we get the following probability density function:

That’s simple enough. The point is: the two de Broglie equations f = E/h and λ = h/p give us the temporal and spatial frequencies in that ψ(x, t) = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} relation. As you can see, that’s an equation that implies a much more complicated relationship between E/ħ = ω and p/ħ = k. Or… Well… Much more complicated than what one would think of at first.

To appreciate what’s being represented here, it’s good to play a bit. We’ll continue with our simple exponential above, which also illustrates how we usually analyze those wavefunctions: we either assume we’re looking at the wavefunction in space at some fixed point in time (t = t₀) or, else, at how the wavefunction changes in time at some fixed point in space (x = x₀). Of course, we know that Einstein told us we shouldn’t do that: space and time are related and, hence, we should try to think of spacetime, i.e. some ‘kind of union’ of space and time—as Minkowski famously put it. However, when everything is said and done, mere mortals like us are not so good at that, and so we’re sort of condemned to try to imagine things using the classical cut-up of things. 🙂 So we’ll just an online graphing tool to play with that a·e^{i(k∙x−ω·t)}= a·e^iθ= a·(cosθ + i·sinθ) formula.

Compare the following two graps, for example. Just imagine we either look at how the wavefunction behaves at some point in space, with the time fixed at some point t = t₀, or, alternatively, that we look at how the wavefunction behaves in time at some point in space x = x₀. As you can see, increasing k = p/ħ or increasing ω = E/ħ gives the wavefunction a higher ‘density’ in space or, alternatively, in time.

That makes sense, intuitively. In fact, when thinking about how the energy, or the momentum, affects the shape of the wavefunction, I am reminded of an airplane propeller: as it spins, faster and faster, it gives the propeller some ‘density’, in space as well as in time, as its blades cover more space in less time. It’s an interesting analogy: it helps—me, at least—to think through what that wavefunction might actually represent.

So as to stimulate your imagination even more, you should also think of representing the real and complex part of that ψ = a·e^{i(k∙x−ω·t)}= a·e^iθ= a·(cosθ + i·sinθ) formula in a different way. In the graphs above, we just showed the sine and cosine in the same plane but, as you know, the real and the imaginary axis are orthogonal, so Euler’s formula a·e^iθ= a·(cosθ + i·sinθ) = a·cosθ + i·a·sinθ = Re(ψ) + i·Im(ψ) may also be graphed as follows:

The illustration above should make you think of yet another illustration you’ve probably seen like a hundred times before: the electromagnetic wave, propagating through space as the magnetic and electric field induce each other, as illustrated below. However, there’s a big difference: Euler’s formula incorporates a phase shift—remember: sinθ = cos(θ − π/2)—and you don’t have that in the graph below. The difference is much more fundamental, however: it’s really hard to see how one could possibly relate the magnetic and electric field to the real and imaginary part of the wavefunction respectively. Having said that, the mathematical similarity makes one think!

Of course, you should remind yourself of what E and B stand for: they represent the strength of the electric (E) and magnetic (B) field at some point x at some time t. So you shouldn’t think of those wavefunctions above as occupying some three-dimensional space. They don’t. Likewise, our wavefunction ψ(x, t) does not occupy some physical space: it’s some complex number—an amplitude that’s associated with each and every point in spacetime. Nevertheless, as mentioned above, the visuals make one think and, as such, do help us as we try to understand all of this in a more intuitive way.

Let’s now look at that energy-momentum relationship once again, but using the wavefunction, rather than those two de Broglie relations.

Energy and momentum in the wavefunction

I am not going to talk about uncertainty here. You know that Spiel. If there’s uncertainty, it’s in the energy or the momentum, or in both. The uncertainty determines the size of that ‘box’ (in spacetime) in which we hope to find our particle, and it’s modeled by a splitting of the energy levels. We’ll say the energy of the particle may be E₀, but it might also be some other value, which we’ll write as E_n = E₀ ± n·ħ. The thing to note is that energy levels will always be separated by some integer multiple of ħ, so ħ is, effectively , the quantum of energy for all practical—and theoretical—purposes. We then super-impose the various wave equations to get a wave function that might—or might not—resemble something like this:

Who knows? 🙂 In any case, that’s not what I want to talk about here. Let’s repeat the basics once more: if we write our wavefunction a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} as a·e^{−i·[ω·t − k∙x]}, we refer to ω = E/ħ as the temporal frequency, i.e. the frequency of our wavefunction in time (i.e. the frequency it has if we keep the position fixed), and to k = p/ħ as the spatial frequency (i.e. the frequency of our wavefunction in space (so now we stop the clock and just look at the wave in space). Now, let’s think about the energy concept first. The energy of a particle is generally thought of to consist of three parts:

The particle’s rest energy m₀c², which de Broglie referred to as internal energy (E_int): it includes the rest mass of the ‘internal pieces’, as Feynman puts it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’ interaction energy);
Any potential energy it may have because of some field (so de Broglie was not assuming the particle was traveling in free space), which we’ll denote by U, and note that the field can be anything—gravitational, electromagnetic: it’s whatever changes the energy because of the position of the particle;
The particle’s kinetic energy, which we write in terms of its momentum p: m·v²/2 = m²·v²/(2m) = (m·v)²/(2m) = p²/(2m).

So we have one energy concept here (the rest energy) that does not depend on the particle’s position in spacetime, and two energy concepts that do depend on position (potential energy) and/or how that position changes because of its velocity and/or momentum (kinetic energy). The two last bits are related through the energy conservation principle. The total energy is E = m_vc², of course—with the little subscript (v) ensuring the mass incorporates the equivalent mass of the particle’s kinetic energy.

So what? Well… In my post on quantum tunneling, I drew attention to the fact that different potentials , so different potential energies (indeed, as our particle travels one region to another, the field is likely to vary) have no impact on the temporal frequency. Let me re-visit the argument, because it’s an important one. Imagine two different regions in space that differ in potential—because the field has a larger or smaller magnitude there, or points in a different direction, or whatever: just different fields, which corresponds to different values for U₁ and U₂, i.e. the potential in region 1 versus region 2. Now, the different potential will change the momentum: the particle will accelerate or decelerate as it moves from one region to the other, so we also have a different p₁ and p₂. Having said that, the internal energy doesn’t change, so we can write the corresponding amplitudes, or wavefunctions, as:

ψ₁(θ₁) = Ψ₁(x, t) = a·e⁻ⁱ^θ^₁ = a·e^{−i[(E_int+ p₁²/(2m) + U₁)·t − p₁∙x]/ħ}
ψ₂(θ₂) = Ψ₂(x, t) = a·e⁻ⁱ^θ^₂ = a·e^{−i[(E_int+ p₂²/(2m) + U₂)·t − p₂∙x]/ħ}

Now how should we think about these two equations? We are definitely talking different wavefunctions. However, their temporal frequencies ω₁= E_int+ p₁²/(2m) + U₁ and ω₁= E_int+ p₂²/(2m) + U₂must be the same. Why? Because of the energy conservation principle—or its equivalent in quantum mechanics, I should say: the temporal frequency f or ω, i.e. the time-rate of change of the phase of the wavefunction, does not change: all of the change in potential, and the corresponding change in kinetic energy, goes into changing the spatial frequency, i.e. the wave number k or the wavelength λ, as potential energy becomes kinetic or vice versa. The sum of the potential and kinetic energy doesn’t change, indeed. So the energy remains the same and, therefore, the temporal frequency does not change. In fact, we need this quantum-mechanical equivalent of the energy conservation principle to calculate how the momentum and, hence, the spatial frequency of our wavefunction, changes. We do so by boldly equating ω₁= E_int+ p₁²/(2m) + U₁and ω₂ = E_int+ p₂²/(2m) + U₂, and so we write:

ω₁= ω₂ ⇔ E_int+ p₁²/(2m) + U₁= E_int+ p₂²/(2m) + U₂

⇔ p₁²/(2m) − p₂²/(2m) = U₂– U₁⇔ p₂²= (2m)·[p₁²/(2m) – (U₂– U₁)]

⇔ p₂= (p₁² – 2m·ΔU)^1/2

We played with this in a previous post, assuming that p₁² is larger than 2m·ΔU, so as to get a positive number on the right-hand side of the equation for p₂², so then we can confidently take the positive square root of that (p₁² – 2m·ΔU) expression to calculate p₂. For example, when the potential difference ΔU = U₂– U₁ was negative, so ΔU < 0, then we’re safe and sure to get some real positive value for p₂.

Having said that, we also contemplated the possibility that p₂²= p₁² – 2m·ΔU was negative, in which case p₂ has to be some pure imaginary number, which we wrote as p₂= i·p’ (so p’ (read: p prime) is a real positive number here). We could work with that: it resulted in an exponentially decreasing factor e^{−p’·x/ħ} that ended up ‘killing’ the wavefunction in space. However, its limited existence still allowed particles to ‘tunnel’ through potential energy barriers, thereby explaining the quantum-mechanical tunneling phenomenon.

This is rather weird—at first, at least. Indeed, one would think that, because of the E/ħ = ω equation, any change in energy would lead to some change in ω. But no! The total energy doesn’t change, and the potential and kinetic energy are like communicating vessels: any change in potential energy is associated with a change in p, and vice versa. It’s a really funny thing. It helps to think it’s because the potential depends on position only, and so it should not have an impact on the temporal frequency of our wavefunction. Of course, it’s equally obvious that the story would change drastically if the potential would change with time, but… Well… We’re not looking at that right now. In short, we’re assuming energy is being conserved in our quantum-mechanical system too, and so that implies what’s described above: no change in ω, but we obviously do have changes in p whenever our particle goes from one region in space to another, and the potentials differ. So… Well… Just remember: the energy conservation principle implies that the temporal frequency of our wave function doesn’t change. Any change in potential, as our particle travels from one place to another, plays out through the momentum.

Now that we know that, let’s look at those de Broglie relations once again.

**Re-visiting the de Broglie relations**

As mentioned above, we usually think in one dimension only: we either freeze time or, else, we freeze space. If we do that, we can derive some funny new relationships. Let’s first simplify the analysis by re-writing the argument of the wavefunction as:

θ = E·t − p·x

Of course, you’ll say: the argument of the wavefunction is not equal to E·t − p·x: it’s (E/ħ)·t − (p/ħ)∙x. Moreover, θ should have a minus sign in front. Well… Yes, you’re right. We should put that 1/ħ factor in front, but we can change units, and so let’s just measure both E as well as p in units of ħ here. We can do that. No worries. And, yes, the minus sign should be there—Nature choose a clockwise direction for θ—but that doesn’t matter for the analysis hereunder.

The E·t − p·x expression reminds one of those invariant quantities in relativity theory. But let’s be precise here. We’re thinking about those so-called four-vectors here, which we wrote as p_μ = (E, p_x, p_y, p_z) = (E, p) and x_μ = (t, x, y, z) = (t, x) respectively. [Well… OK… You’re right. We wrote those four-vectors as p_μ = (E, p_x·c , p_y·c, p_z·c) = (E, p·c) and x_μ = (c·t, x, y, z) = (t, x). So what we write is true only if we measure time and distance in equivalent units so we have c = 1. So… Well… Let’s do that and move on.] In any case, what was invariant was not E·t − p·x·c or c·t − x (that’s a nonsensical expression anyway: you cannot subtract a vector from a scalar), but p_μ² = p_μp_μ= E² − (p·c)² = E² − p²·c²= E² − (p_x² + p_y² + p_z²)·c² and x_μ² = x_μx_μ= (c·t)² − x² = c²·t² − (x² + y² + z²) respectively. [Remember p_μp_μand x_μx_μare four-vector dot products, so they have that +— signature, unlike the p² and x² or a·b dot products, which are just a simple sum of the squared components.] So… Well… E·t − p·x is not an invariant quantity. Let’s try something else.

Let’s re-simplify by equating ħ as well as c to one again, so we write: ħ = c = 1. [You may wonder if it is possible to ‘normalize’ both physical constants simultaneously, but the answer is yes. The Planck unit system is an example.] then our relativistic energy-momentum relationship can be re-written as E/p = 1/v. [If c would not be one, we’d write: E·β = p·c, with β = v/c. So we got E/p = c/β. We referred to β as the relative velocity of our particle: it was the velocity, but measured as a ratio of the speed of light. So here it’s the same, except that we use the velocity symbol v now for that ratio.]

Now think of a particle moving in free space, i.e. without any fields acting on it, so we don’t have any potential changing the spatial frequency of the wavefunction of our particle, and let’s also assume we choose our x-axis such that it’s the direction of travel, so the position vector (x) can be replaced by a simple scalar (x). Finally, we will also choose the origin of our x-axis such that x = 0 zero when t = 0, so we write: x(t = 0) = 0. It’s obvious then that, if our particle is traveling in spacetime with some velocity v, then the ratio of its position x and the time t that it’s been traveling will always be equal to v = x/t. Hence, for that very special position in spacetime (t, x = v·t) – so we’re talking the actual position of the particle in spacetime here – we get: θ = E·t − p·x = E·t − p·v·t = E·t − m·v·v·t= (E − m∙v²)·t. So… Well… There we have the m∙v² factor.

The question is: what does it mean? How do we interpret this? I am not sure. When I first jotted this thing down, I thought of choosing a different reference potential: some negative value such that it ensures that the sum of kinetic, rest and potential energy is zero, so I could write E = 0 and then the wavefunction would reduce to ψ(t) = e^{−i·m∙v²·t}. Feynman refers to that as ‘choosing the zero of our energy scale such that E = 0’, and you’ll find this in many other works too. However, it’s not that simple. Free space is free space: if there’s no change in potential from one region to another, then the concept of some reference point for the potential becomes meaningless. There is only rest energy and kinetic energy, then. The total energy reduces to E = m (because we chose our units such that c = 1 and, therefore, E = mc² = m·1² = m) and so our wavefunction reduces to:

ψ(t) = a·e^{−i·m·(1 − v²)·t}

We can’t reduce this any further. The mass is the mass: it’s a measure for inertia, as measured in our inertial frame of reference. And the velocity is the velocity, of course—also as measured in our frame of reference. We can re-write it, of course, by substituting t for t = x/v, so we get:

ψ(x) = a·e^{−i·m·(1/v − v)·x}

For both functions, we get constant probabilities, but a wavefunction that’s ‘denser’ for higher values of m. The (1 − v²) and (1/v − v) factors are different, however: these factors becomes smaller for higher v, so our wavefunction becomes less dense for higher v. In fact, for v = 1 (so for travel at the speed of light, i.e. for photons), we get that ψ(t) = ψ(x) = e⁰ = 1. [You should use the graphing tool once more, and you’ll see the imaginary part, i.e. the sine of the a·(cosθ + i·sinθ) expression, just vanishes, as sinθ = 0 for θ = 0.]

The wavefunction and relativistic length contraction

Are exercises like this useful? As mentioned above, these constant probability wavefunctions are a bit nonsensical, so you may wonder why I wrote what I wrote. There may be no real conclusion, indeed: I was just fiddling around a bit, and playing with equations and functions. I feel stuff like this helps me to understand what that wavefunction actually is somewhat better. If anything, it does illustrate that idea of the ‘density’ of a wavefunction, in space or in time. What we’ve been doing by substituting x for x = v·t or t for t = x/v is showing how, when everything is said and done, the mass and the velocity of a particle are the actual variables determining that ‘density’ and, frankly, I really like that ‘airplane propeller’ idea as a pedagogic device. In fact, I feel it may be more than just a pedagogic device, and so I’ll surely re-visit it—once I’ve gone through the rest of Feynman’s Lectures, that is. 🙂

That brings me to what I added in the title of this post: relativistic length contraction. You’ll wonder why I am bringing that into a discussion like this. Well… Just play a bit with those (1 − v²) and (1/v − v) factors. As mentioned above, they decrease the density of the wavefunction. In other words, it’s like space is being ‘stretched out’. Also, it can’t be a coincidence we find the same (1 − v²) factor in the relativistic length contraction formula: L = L₀·√(1 − v²), in which L₀ is the so-called proper length (i.e. the length in the stationary frame of reference) and v is the (relative) velocity of the moving frame of reference. Of course, we also find it in the relativistic mass formula: m = m_v = m₀/√(1−v²). In fact, things become much more obvious when substituting m for m₀/√(1−v²) in that ψ(t) = e^{−i·m·(1 − v²)·t}function. We get:

ψ(t) = a·e^{−i·m·(1 − v²)·t}= a·e^{−i·m₀·√(1−v²)·t}

Well… We’re surely getting somewhere here. What if we go back to our original ψ(x, t) = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} function? Using natural units once again, that’s equivalent to:

ψ(x, t) = a·e^{−i·(m·t − p∙x)}= a·e^{−i·[(m₀/√(1−v²))·t − (m₀·v/√(1−v²)∙x)}

= a·e^{−i·[m₀/√(1−v²)]·(t − v∙x)}

Interesting! We’ve got a wavefunction that’s a function of x and t, but with the rest mass (or rest energy) and velocity as parameters! Now that really starts to make sense. Look at the (blue) graph for that 1/√(1−v²) factor: it goes from one (1) to infinity (∞) as v goes from 0 to 1 (remember we ‘normalized’ v: it’s a ratio between 0 and 1 now). So that’s the factor that comes into play for t. For x, it’s the red graph, which has the same shape but goes from zero (0) to infinity (∞) as v goes from 0 to 1.

Now that makes sense: the ‘density’ of the wavefunction, in time and in space, increases as the velocity v increases. In space, that should correspond to the relativistic length contraction effect: it’s like space is contracting, as the velocity increases and, therefore, the length of the object we’re watching contracts too. For time, the reasoning is a bit more complicated: it’s our time that becomes more dense and, therefore, our clock that seems to tick faster.

[…]

I know I need to explore this further—if only so as to assure you I have not gone crazy. Unfortunately, I have no time to do that right now. Indeed, from time to time, I need to work on other stuff besides this physics ‘hobby’ of mine.

Post scriptum 1: As for the E = m·v²formula, I also have a funny feeling that it might be related to the fact that, in quantum mechanics, both the real and imaginary part of the oscillation actually matter. You’ll remember that we’d represent any oscillator in physics by a complex exponential, because it eased our calculations. So instead of writing A = A₀·cos(ωt + Δ), we’d write: A = A₀·e^{i(ωt + Δ)}= A₀·cos(ωt + Δ) + i·A₀·sin(ωt + Δ). When calculating the energy or intensity of a wave, however, we couldn’t just take the square of the complex amplitude of the wave – remembering that E ∼ A². No! We had to get back to the real part only, i.e. the cosine or the sine only. Now the mean (or average) value of the squared cosine function (or a squared sine function), over one or more cycles, is 1/2, so the mean of A² is equal to 1/2 = A₀². cos(ωt + Δ). I am not sure, and it’s probably a long shot, but one must be able to show that, if the imaginary part of the oscillation would actually matter – which is obviously the case for our matter-wave – then 1/2 + 1/2 is obviously equal to 1. I mean: try to think of an image with a mass attached to two springs, rather than one only. Does that make sense? 🙂 […] I know: I am just freewheeling here. 🙂

Post scriptum 2: The other thing that this E = m·v²equation makes me think of is – curiously enough – an eternally expanding spring. Indeed, the kinetic energy of a mass on a spring and the potential energy that’s stored in the spring always add up to some constant, and the average potential and kinetic energy are equal to each other. To be precise: 〈K.E.〉 + 〈P.E.〉 = (1/4)·k·A² + (1/4)·k·A²= k·A²/2. It means that, on average, the total energy of the system is twice the average kinetic energy (or potential energy). You’ll say: so what? Well… I don’t know. Can we think of a spring that expands eternally, with the mass on its end not gaining or losing any speed? In that case, v is constant, and the total energy of the system would, effectively, be equal to E_total = 2·〈K.E.〉 = (1/2)·m·v²/2 = m·v².

Post scriptum 3: That substitution I made above – substituting x for x = v·t – is kinda weird. Indeed, if that E = m∙v² equation makes any sense, then E − m∙v² = 0, of course, and, therefore, θ = E·t − p·x = E·t − p·v·t = E·t − m·v·v·t= (E − m∙v²)·t = 0·t = 0. So the argument of our wavefunction is 0 and, therefore, we get a·e⁰= a for our wavefunction. It basically means our particle is where it is. 🙂

Post scriptum 4: This post scriptum – no. 4 – was added later—much later. On 29 February 2016, to be precise. The solution to the ‘riddle’ above is actually quite simple. We just need to make a distinction between the group and the phase velocity of our complex-valued wave. The solution came to me when I was writing a little piece on Schrödinger’s equation. I noticed that we do not find that weird E = m∙v² formula when substituting ψ for ψ = e^{i(kx − ωt)}in Schrödinger’s equation, i.e. in:

Let me quickly go over the logic. To keep things simple, we’ll just assume one-dimensional space, so ∇²ψ = ∂²ψ/∂x². The time derivative on the left-hand side is ∂ψ/∂t = −iω·e^{i(kx − ωt)}. The second-order derivative on the right-hand side is ∂²ψ/∂x² = (ik)·(ik)·e^{i(kx − ωt)}= −k²·e^{i(kx − ωt)} . The e^{i(kx − ωt)} factor on both sides cancels out and, hence, equating both sides gives us the following condition:

−iω = −(iħ/2m)·k² ⇔ ω = (ħ/2m)·k²

Substituting ω = E/ħ and k = p/ħ yields:

E/ħ = (ħ/2m)·p²/ħ²= m²·v²/(2m·ħ) = m·v²/(2ħ) ⇔ E = m·v²/2

In short: the E = m·v²/2 is the correct formula. It must be, because… Well… Because Schrödinger’s equation is a formula we surely shouldn’t doubt, right? So the only logical conclusion is that we must be doing something wrong when multiplying the two de Broglie equations. To be precise: our v = f·λ equation must be wrong. Why? Well… It’s just something one shouldn’t apply to our complex-valued wavefunction. The ‘correct’ velocity formula for the complex-valued wavefunction should have that 1/2 factor, so we’d write 2·f·λ = v to make things come out alright. But where would this formula come from? The period of cosθ + isinθ is the period of the sine and cosine function: cos(θ+2π) + isin(θ+2π) = cosθ + isinθ, so T = 2π and f = 1/T = 1/2π do not change.

But so that’s a mathematical point of view. From a physical point of view, it’s clear we got two oscillations for the price of one: one ‘real’ and one ‘imaginary’—but both are equally essential and, hence, equally ‘real’. So the answer must lie in the distinction between the group and the phase velocity when we’re combining waves. Indeed, the group velocity of a sum of waves is equal to v_g = dω/dk. In this case, we have:

v_g = d[E/ħ]/d[p/ħ] = dE/dp

We can now use the kinetic energy formula to write E as E = m·v²/2 = p·v/2. Now, v and p are related through m (p = m·v, so v = p/m). So we should write this as E = m·v²/2 = p²/(2m). Substituting E and p = m·v in the equation above then gives us the following:

dω/dk = d[p²/(2m)]/dp = 2p/(2m) = v_g= v

However, for the phase velocity, we can just use the v_p= ω/k formula, which gives us that 1/2 factor:

v_p= ω/k = (E/ħ)/(p/ħ) = E/p = (m·v²/2)/(m·v) = v/2

Bingo! Riddle solved! 🙂 Isn’t it nice that our formula for the group velocity also applies to our complex-valued wavefunction? I think that’s amazing, really! But I’ll let you think about it. 🙂

An introduction to virtual particles

In one of my posts on the rules of quantum math, I introduced the propagator function, which gives us the amplitude for a particle to go from one place to another. It looks like this:

The r₁and r₂ vectors are, obviously, position vectors describing (1) where the particle is right now, so the initial state is written as |r₁〉, and (2) where it might go, so the final state is |r₂〉. Now we can combine this with the analysis in my previous post to think about what might happen when an electron sort of ‘jumps’ from one state to another. It’s a rather funny analysis, but it will give you some feel of what these so-called ‘virtual’ particles might represent.

Let’s first look at the shape of that function. The e^{(i/ħ)·(p∙r₁₂)}function in the numerator is now familiar to you. Note the r₁₂in the argument, i.e. the vector pointing from r₁ to r₂. The p∙r₁₂ dot product equals |p|∙|r₁₂|·cosθ = p∙r₁₂·cosθ, with θ the angle between p and r₁₂. If the angle is the same, then cosθ is equal to 1. If the angle is π/2, then it’s 0, and the function reduces to 1/r₁₂. So the angle θ, through the cosθ factor, sort of scales the spatial frequency. Let me try to give you some idea of how this looks like by assuming the angle between p and r₁₂ is the same, so we’re looking at the space in the direction of the momentum only and |p|∙|r₁₂|·cosθ = p∙r₁₂. Now, we can look at the p/ħ factor as a scaling factor, and measure the distance x in units defined by that scale, so we write: x = p∙r₁₂/ħ. The whole function, including the denominator, then reduces to (ħ/p)·e^i∙x/x = (ħ/p)·cos(x)/x + i·(ħ/p)·sin(x)/x, and we just need to square this to get the probability. All of the graphs are drawn hereunder: I’ll let you analyze them. [Note that the graphs do not include the ħ/p factor, which you may look at as yet another scaling factor.] You’ll see – I hope! – that it all makes perfect sense: the probability quickly drops off with distance, both in the positive as well as in the negative x-direction, while going to infinity when very near, i.e. for very small x. [Note that the absolute square, using cos(x)/x and sin(x)/xyields the same graph as squaring 1/x—obviously!]

Now, this propagator function is not dependent on time: it’s only the momentum that enters the argument. Of course, we assume p to be some positive real number. Of course?

This is where Feynman starts an interesting conversation. In the previous post, we studied a model in which we had two protons, and one electron jumping from one to another, as shown below.

This model told us the equilibrium state is a stable ionized hydrogen molecule (so that’s an H₂⁺molecule), with an interproton distance that’s equal to 1 Ångstrom – so that’s like twice the size of a hydrogen atom (which we simply write as H) – and an energy that’s 2.72 eV less than the energy of a hydrogen atom and a proton (so that’s not an H₂⁺molecule but a system consisting of a separate hydrogen atom and a proton). The why and how of that equilibrium state is illustrated below. [For more details, see my previous post.]

Now, the model implies there is a sort of attractive force pulling the two protons together even when the protons are at larger distances than 1 Å. One can see that from the graph indeed. Now, we would not associate any molecular orbital with those distances, as the system is, quite simply, not a molecule but a separate hydrogen atom and a proton. Nevertheless, the amplitude A is non-zero, and so we have an electron jumping back and forth.

We know how that works from our post on tunneling: particles can cross an energy barrier and tunnel through. One of the weird things we had to consider when a particle crosses such potential barrier, is that the momentum factor p in its wavefunction was some pure imaginary number, which we wrote as p = i·p’. We then re-wrote that wavefunction as a·e⁻ⁱ^θ = a·e⁻ⁱ^[^{(E/ħ)∙t − (i·p’/ħ)x]} = a·e⁻ⁱ^(E/ħ)∙t·e^{i²·p’·x/ħ} = a·e⁻ⁱ^(E/ħ)∙t·e^{−p’·x/ħ}. The e^{−p’·x/ħ} factor in this formula is a real-valued exponential function, that sort of ‘kills’ our wavefunction as we move across the potential barrier, which is what is illustrated below: if the distance is too large, then the amplitude for tunneling goes to zero.

From a mathematical point of view, the analysis of our electron jumping back and forth is very similar. However, there are differences too. We can’t really analyze this in terms of a potential barrier in space. The barrier is the potential energy of the electron itself: it’s happy when it’s bound, because its energy then contributes to a reduction of the total energy of the hydrogen atomic system that is equal to the ionization energy, or the Rydberg energy as it’s called, which is equal to not less than 13.6 eV (which, as mentioned, is pretty big at the atomic level). Well… We can take that propagator function (1/r)·e^{(i/ħ)·p∙r} (note the argument has no minus sign: it can be quite tricky!), and just fill in the value for the momentum of the electron.

Huh? What momentum? It’s got no momentum to spare. On the contrary, it wants to stay with the proton, so it has no energy whatsoever to escape. Well… Not in quantum mechanics. In quantum mechanics it can use all its potential energy and convert it into kinetic energy, so it can get away from its proton and convert the energy that’s being released into kinetic energy.

But there is no release of energy! The energy is negative!

Exactly! You’re right. So we boldly write: K.E. = m·v²/2 = p²/(2m) = −13.6 eV, and, because we’re working with complex numbers, we can take a square root of negative number, using the definition of the imaginary unit: i = √(−1), so we get a purely imaginary value for the momentum p, which we write as:

p = ±i·√(2m·E_H)

The sign of p is chosen so it makes sense: our electron should go in one direction only. It’s going to be the plus sign. [If you’d take the negative root, you’d get a nonsensical propagator function.] To make a long story short, our propagator function becomes:

(1/r)·e^{(i/ħ)·i·√(2m·E_H)∙r} = (1/r)·e^{(i/ħ)·i·√(2m·E_H)∙r} = (1/r)·e^{i²/ħ·√(2m·E_H)∙r} = (1/r)·e^{−√(2m·E_H)/ħ∙r}

Of course, from a mathematical point of view, that’s the same function as e^{−p’·x/ħ}: it’s a real-valued exponential function that quickly dies. But it’s an amplitude alright, and it’s just like an amplitude for tunneling indeed: if the distance is too large, then the amplitude goes to zero. The final cherry on the cake, of course, is to write:

A ∼ (1/r)·e^{−√(2m·E_H)/ħ∙r}

Well… No. It gets better. This amplitude is an amplitude for an electron bond between the two protons which, as we know, lowers the energy of the system. By how much? Well… By A itself. Now we know that work or energy is an integral or antiderivative of force over distance, so force is the derivative of energy with respect to the distance. So we can just take the derivative of the expression above to get the force. I’ll leave that you as an exercise: don’t forget to use the product rule! 🙂

So are we done? No. First, we didn’t talk about virtual particles yet! Let me do that now. However, first note that we should add one more effect in our two-proton-one-electron system: the coulomb field (ε) caused by the bare proton will cause the hydrogen molecule to take on an induced electric dipole moment (μ), so we should integrate that in our energy equation. Feynman shows how, but I won’t bother you with that here. Let’s talk about those virtual particles. What are they?

Well… There’s various definitions, but Feynman’s definition is this one:

“There is an exchange of a virtual electron when–as here–the electron has to jump across a space where it would have a negative energy. More specifically, a ‘virtual exchange’ means that the phenomenon involves a quantum-mechanical interference between an exchanged state and a non-exchanged state.”

You’ll say: what’s virtual about it? The electron does go from one place to another, doesn’t it? Well… Yes and no. We can’t observe it while it’s supposed to be doing that. Our analysis just tells us it seems to be useful to distinguish two different states and analyze all in terms of those differential equations. Who knows what’s really going on? What’s actual and what’s virtual? We just have some ‘model’ here: a model for the interaction between a hydrogen atom and a proton. It explains the attraction between them in terms of a sort of continuous exchange of an electron, but is it real?

The point is: in physics, it’s assumed that the coulomb interaction, i.e. all of electrostatics really, comes from the exchange of virtual photons: one electron, or proton, emits a photon, and then another absorbs it in the reverse of the same reaction. Furthermore, it is assumed that the amplitude for doing so is like that formula we found for the amplitude to exchange a virtual electron, except that the rest mass of a photon is zero, and so the formula reduces to 1/r. Such simple relationship makes sense, of course, because that’s how the electrostatic potential varies in space!

That, in essence, is all what there is to the quantum-mechanical theory of electromagnetism, which Feynman refers to as the ‘particle point of view’.

So… Yes. It’s that simple. Yes! For a change! 🙂

Post scriptum: Feynman’s Lecture on virtual particles is actually focused on a model for the nuclear forces. Most of it is devoted to a discussion of the virtual ‘pion’, or π-meson, which was then, when Feynman wrote his Lectures, supposed to mediate the force between two nucleons. However, this theory is clearly outdated: nuclear forces are described by quantum chromodynamics. So I’ll just skip the Yukawa theory here. It’s actually kinda strange his theory, which he proposed in 1935, was the theory for nuclear forces for such a long time. Hence, it’s surely all very interesting from a historical point of view.

The math behind the maser

Pre-script (dated 26 June 2020): I have come to the conclusion one does not need all this hocus-pocus to explain masers or lasers (and two-state systems in general): classical physics will do. So no use to read this. Read my papers instead. 🙂

Original post:

As I skipped the mathematical arguments in my previous post so as to focus on the essential results only, I thought it would be good to complement that post by looking at the math once again, so as to ensure we understand what it is that we’re doing. So let’s do that now. We start with the easy situation: free space.

The two-state system in free space

We started with an ammonia molecule in free space, i.e. we assumed there were no external force fields, like a gravitational or an electromagnetic force field. Hence, the picture was as simple as the one below: the nitrogen atom could be ‘up’ or ‘down’ with regard to its spin around its axis of symmetry.

It’s important to note that this ‘up’ or ‘down’ direction is defined in regard to the molecule itself, i.e. not in regard to some external reference frame. In other words, the reference frame is that of the molecule itself. For example, if I flip the illustration above – like below – then we’re still talking the same states, i.e. the molecule is still in state 1 in the image on the left-hand side and it’s still in state 2 in the image on the right-hand side.

We then modeled the uncertainty about its state by associating two different energy levels with the molecule: E₀+ A and E₀− A. The idea is that the nitrogen atom needs to tunnel through a potential barrier to get to the other side of the plane of the hydrogens, and that requires energy. At the same time, we’ll show the two energy levels are effectively associated with an ‘up’ or ‘down’ direction of the electric dipole moment of the molecule. So that resembles the two spin states of an electron, which we associated with the +ħ/2 and −ħ/2 energies respectively. So if E₀would be zero (we can always take another reference point, remember?), then we’ve got the same thing: two energy levels that are separated by some definite amount: that amount is 2A for the ammonia molecule, and ħ when we’re talking quantum-mechanical spin. I should make a last note here, before I move on: note that these energies only make sense in the presence of some external field, because the + and − signs in the E₀+ A and E₀− A and +ħ/2 and −ħ/2 expressions make sense only with regard to some external direction defining what’s ‘up’ and what’s ‘down’ really. But I am getting ahead of myself here. Let’s go back to free space: no external fields, so what’s ‘up’ or ‘down’ is completely random here. 🙂

Now, we also know an energy level can be associated with a complex-valued wavefunction, or an amplitude as we call it. To be precise, we can associate it with the generic a·e^{−(i/ħ)·(E·t − p∙x)}expression which you know so well by now. Of course, as the reference frame is that of the molecule itself, its momentum is zero, so the p∙x term in the a·e^{−(i/ħ)·(E·t − p∙x)}expression vanishes and the wavefunction reduces to a·e^−i·ω·t= a·e^{−(i/ħ)·E·t}, with ω = E/ħ. In other words, the energy level determines the temporal frequency, or the temporal variation (as opposed to the spatial frequency or variation), of the amplitude.

We then had to find the amplitudes C₁(t) = 〈 1 | ψ 〉 and C₂(t) =〈 2 | ψ 〉, so that’s the amplitude to be in state 1 or state 2 respectively. In my post on the Hamiltonian, I explained why the dynamics of a situation like this can be represented by the following set of differential equations:

As mentioned, the C₁and C₂functions evolve in time, and so we should write them as C₁= C₁(t) and C₂= C₂(t) respectively. In fact, our Hamiltonian coefficients may also evolve in time, which is why it may be very difficult to solve those differential equations! However, as I’ll show below, one usually assumes they are constant, and then one makes informed guesses about them so as to find a solution that makes sense.

Now, I should remind you here of something you surely know: if C₁and C₂are solutions to this set of differential equations, then the superposition principle tells us that any linear combination a·C₁+ b·C₂will also be a solution. So we need one or more extra conditions, usually some starting condition, which we can combine with a normalization condition, so we can get some unique solution that makes sense.

The H_ijcoefficients are referred to as Hamiltonian coefficients and, as shown in the mentioned post, the H₁₁and H₂₂coefficients are related to the amplitude of the molecule staying in state 1 and state 2 respectively, while the H₁₂and H₂₁coefficients are related to the amplitude of the molecule going from state 1 to state 2 and vice versa. Because of the perfect symmetry of the situation here, it’s easy to see that H₁₁should equal H₂₂, and that H₁₂and H₂₁should also be equal to each other. Indeed, Nature doesn’t care what we call state 1 or 2 here: as mentioned above, we did not define the ‘up’ and ‘down’ direction with respect to some external direction in space, so the molecule can have any orientation and, hence, switching the i an j indices should not make any difference. So that’s one clue, at least, that we can use to solve those equations: the perfect symmetry of the situation and, hence, the perfect symmetry of the Hamiltonian coefficients—in this case, at least!

The other clue is to think about the solution if we’d not have two states but one state only. In that case, we’d need to solve iħ·[dC₁(t)/dt] = H₁₁·C₁(t). That’s simple enough, because you’ll remember that the exponential function is its own derivative. To be precise, we write: d(a·e^iωt)/dt = a·d(e^iωt)/dt = a·iω·e^iωt, and please note that a can be any complex number: we’re not necessarily talking a real number here! In fact, we’re likely to talk complex coefficients, and we multiply with some other complex number (iω) anyway here! So if we write iħ·[dC₁/dt] = H₁₁·C₁ as dC₁/dt = −(i/ħ)·H₁₁·C₁ (remember: i⁻¹ = 1/i = −i), then it’s easy to see that the C₁= a·e^{–(i/ħ)·H₁₁·t}function is the general solution for this differential equation. Let me write it out for you, just to make sure:

dC₁/dt = d[a·e^{–(i/ħ)H₁₁t}]/dt = a·d[e^{–(i/ħ)H₁₁t}]/dt = –a·(i/ħ)·H₁₁·e^{–(i/ħ)H₁₁t}

= –(i/ħ)·H₁₁·a·e^{–(i/ħ)H₁₁t}= −(i/ħ)·H₁₁·C₁

Of course, that reminds us of our generic wavefunction a·e^{−(i/ħ)·E₀·t} wavefunction: we only need to equate H₁₁ with E₀ and we’re done! Hence, in a one-state system, the Hamiltonian coefficient is, quite simply, equal to the energy of the system. In fact, that’s a result can be generalized, as we’ll see below, and so that’s why Feynman says the Hamiltonian ought to be called the energy matrix.

In fact, we actually may have two states that are entirely uncoupled, i.e. a system in which there is no dependence of C₁ on C₂and vice versa. In that case, the two equations reduce to:

iħ·[dC₁/dt] = H₁₁·C₁ and iħ·[dC₂/dt] = H₂₂·C₂

These do not form a coupled system and, hence, their solutions are independent:

C₁(t) = a·e^{–(i/ħ)·H₁₁·t}and C₂(t)= b·e^{–(i/ħ)·H₂₂·t}

The symmetry of the situation suggests we should equate a and b, and then the normalization condition says that the probabilities have to add up to one, so |C₁(t)|²+ |C₂(t)|²= 1, so we’ll find that a = b = 1/√2.

OK. That’s simple enough, and this story has become quite long, so we should wrap it up. The two ‘clues’ – about symmetry and about the Hamiltonian coefficients being energy levels – lead Feynman to suggest that the Hamiltonian matrix for this particular case should be equal to:

Why? Well… It’s just one of Feynman’s clever guesses, and it yields probability functions that makes sense, i.e. they actually describe something real. That’s all. 🙂 I am only half-joking, because it’s a trial-and-error process indeed and, as I’ll explain in a separate section in this post, one needs to be aware of the various approximations involved when doing this stuff. So let’s be explicit about the reasoning here:

We know that H₁₁= H₂₂= E₀if the two states would be identical. In other words, if we’d have only one state, rather than two – i.e. if H₁₂and H₂₁would be zero – then we’d just plug that in. So that’s what Feynman does. So that’s what we do here too! 🙂
However, H₁₂and H₂₁are not zero, of course, and so assume there’s some amplitude to go from one position to the other by tunneling through the energy barrier and flipping to the other side. Now, we need to assign some value to that amplitude and so we’ll just assume that the energy that’s needed for the nitrogen atom to tunnel through the energy barrier and flip to the other side is equal to A. So we equate H₁₂and H₂₁ with −A.

Of course, you’ll wonder: why minus A? Why wouldn’t we try H₁₂= H₂₁ = A? Well… I could say that a particle usually loses potential energy as it moves from one place to another, but… Well… Think about it. Once it’s through, it’s through, isn’t it? And so then the energy is just E₀again. Indeed, if there’s no external field, the + or − sign is quite arbitrary. So what do we choose? The answer is: when considering our molecule in free space, it doesn’t matter. Using +A or −A yields the same probabilities. Indeed, let me give you the amplitudes we get for H₁₁= H₂₂= E₀and H₁₂and H₂₁ = −A:

C₁(t) = 〈 1 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}+ (1/2)·e^{−(i/ħ)·(E₀+ A)·t}= e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t]
C₂(t) = 〈 2 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}– (1/2)·e^{−(i/ħ)·(E₀+ A)·t}= i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t]

[In case you wonder how we go from those exponentials to a simple sine and cosine factor, remember that the sum of complex conjugates, i.e e^iθ+ e^−iθreduces to 2·cosθ, while e^iθ− e^−iθreduces to 2·i·sinθ.]

Now, it’s easy to see that, if we’d have used +A rather than −A, we would have gotten something very similar:

C₁(t) = 〈 1 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀+ A)·t}+ (1/2)·e^{−(i/ħ)·(E₀− A)·t}= e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t]
C₂(t) = 〈 2 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀+ A)·t}– (1/2)·e^{−(i/ħ)·(E₀− A)·t}= −i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t]

So we get a minus sign in front of our C₂(t) function, because cos(α) = cos(–α) but sin(α) = −sin(α). However, the associated probabilities are exactly the same. For both, we get the same P₁(t) and P₂(t) functions:

P₁(t) = |C₁(t)|² = cos²[(A/ħ)·t]
P₂(t) = |C₂(t)|²= sin²[(A/ħ)·t]

[Remember: the absolute square of i and −i is |i|²= +√1²= +1 and |−i|²= (−1)²|i|²= +1 respectively, so the i and −i in the two C₂(t) formulas disappear.]

You’ll remember the graph:

Of course, you’ll say: that plus or minus sign in front of C₂(t) should matter somehow, doesn’t it? Well… Think about it. Taking the absolute square of some complex number – or some complex function , in this case! – amounts to multiplying it with its complex conjugate. Because the complex conjugate of a product is the product of the complex conjugates, it’s easy to see what happens: the e^{−(i/ħ)·E₀·t} factor in C₁(t) = e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t] and C₂(t) = ±i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t] gets multiplied by e^{+(i/ħ)·E₀·t} and, hence, doesn’t matter: e^{−(i/ħ)·E₀·t}·e^{+(i/ħ)·E₀·t} = e⁰= 1. The cosine factor in C₁(t) = e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t] is real, and so its complex conjugate is the same. Now, the ±i·sin[(A/ħ)·t] factor in C₂(t) = ±i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t] is a pure imaginary number, and so its complex conjugate is its opposite. For some reason, we’ll find similar solutions for all of the situations we’ll describe below: the factor determining the probability will either be real or, else, a pure imaginary number. Hence, from a math point of view, it really doesn’t matter if we take +A or −A for or real factor for those H₁₂and H₂₁ coefficients. We just need to be consistent in our choice, and I must assume that, in order to be consistent, Feynman likes to think of our nitrogen atom borrowing some energy from the system and, hence, temporarily reducing its energy by an amount that’s equal to −A. If you have a better interpretation, please do let me know! 🙂

OK. We’re done with this section… Except… Well… I have to show you how we got those C₁(t) and C₁(t) functions, no? Let me copy Feynman here:

Note that the ‘trick’ involving the addition and subtraction of the differential equations is a trick we’ll use quite often, so please do have a look at it. As for the value of the a and b coefficients – which, as you can see, we’ve equated to 1 in our solutions for C₁(t) and C₁(t) – we get those because of the following starting condition: we assume that at t = 0, the molecule will be in state 1. Hence, we assume C₁(0) = 1 and C₂(0) = 0. In other words: we assume that we start out on that P₁(t) curve in that graph with the probability functions above, so the C₁(0) = 1 and C₂(0) = 0 starting condition is equivalent to P₁(0) = 1 and P₁(0) = 0. Plugging that in gives us a/2 + b/2 = 1 and a/2 − b/2 = 0, which is possible only if a = b = 1.

Of course, you’ll say: what if we’d choose to start out with state 2, so our starting condition is P₁(0) = 0 and P₁(0) = 1? Then a = 1 and b = −1, and we get the solution we got when equating H₁₂and H₂₁ with +A, rather than with −A. So you can think about that symmetry once again: when we’re in free space, then it’s quite arbitrary what we call ‘up’ or ‘down’.

So… Well… That’s all great. I should, perhaps, just add one more note, and that’s on that A/ħ value. We calculated it in the previous post, because we wanted to actually calculate the period of those P₁(t) and P₂(t) functions. Because we’re talking the square of a cosine and a sine respectively, the period is equal to π, rather than 2π, so we wrote: (A/ħ)·T = π ⇔ T = π·ħ/A. Now, the separation between the two energy levels E₀+ A and E₀− A, so that’s 2A, has been measured as being equal, more or less, to 2A ≈ 10⁻⁴eV.

How does one measure that? As mentioned above, I’ll show you, in a moment, that, when applying some external field, the plus and minus sign do matter, and the separation between those two energy levels E₀+ A and E₀− A will effectively represent something physical. More in particular, we’ll have transitions from one energy level to another and that corresponds to electromagnetic radiation being emitted or absorbed, and so there’s a relation between the energy and the frequency of that radiation. To be precise, we can write 2A = h·f₀. The frequency of the radiation that’s being absorbed or emitted is 23.79 GHz, which corresponds to microwave radiation with a wavelength of λ = c/f₀ = 1.26 cm. Hence, 2·A ≈ 25×10⁹ Hz times 4×10⁻¹⁵ eV·s = 10⁻⁴eV, indeed, and, therefore, we can write: T = π·ħ/A ≈ 3.14 × 6.6×10⁻¹⁶eV·s divided by 0.5×10⁻⁴eV, so that’s 40×10⁻¹²seconds = 40 picoseconds. That’s 40 trillionths of a seconds. So that’s very short, and surely much shorter than the time that’s associated with, say, a freely emitting sodium atom, which is of the order of 3.2×10⁻⁸seconds. You may think that makes sense, because the photon energy is so much lower: a sodium light photon is associated with an energy equal to E = h·f = 500×10¹² Hz times 4×10⁻¹⁵ eV·s = 2 eV, so that’s 20,000 times 10⁻⁴eV.

There’s a funny thing, however. An oscillation of a frequency of 500 tera-hertz that lasts 3.2×10⁻⁸seconds is equivalent to 500×10¹² Hz times 3.2×10⁻⁸s ≈ 16 million cycles. However, an oscillation of a frequency of 23.97 giga-hertz that only lasts 40×10⁻¹²seconds is equivalent to 23.97×10⁹ Hz times 40×10⁻¹²s ≈ 1000×10⁻³= 1 ! One cycle only? We’re surely not talking resonance here!

So… Well… I am just flagging it here. We’ll have to do some more thinking about that later. [I’ve added an addendum that may or may not help us in this regard. :-)]

The two-state system in a field

As mentioned above, when there is no external force field, we define the ‘up’ or ‘down’ direction of the nitrogen atom was defined with regard to its its spin around its axis of symmetry, so with regard to the molecule itself. However, when we apply an external electromagnetic field, as shown below, we do have some external reference frame.

Now, the external reference frame – i.e. the physics of the situation, really – may make it more convenient to define the whole system using another set of base states, which we’ll refer to as I and II, rather than 1 and 2. Indeed, you’ve seen the picture below: it shows a state selector, or a filter as we called it. In this case, there’s a filtering according to whether our ammonia molecule is in state I or, alternatively, state II. It’s like a Stern-Gerlach apparatus splitting an electron beam according to the spin state of the electrons, which is ‘up’ or ‘down’ too, but in a totally different way than our ammonia molecule. Indeed, the ‘up’ and ‘down’ spin of an electron has to do with its magnetic moment and its angular momentum. However, there are a lot of similarities here, and so you may want to compare the two situations indeed, i.e. the electron beam in an inhomogeneous magnetic field versus the ammonia beam in an inhomogeneous electric field.

Now, when reading Feynman, as he walks us through the relevant Lecture on all of this, you get the impression that it’s the I and II states only that have some kind of physical or geometric interpretation. That’s not the case. Of course, the diagram of the state selector above makes it very obvious that these new I and II base states make very much sense in regard to the orientation of the field, i.e. with regard to external space, rather than with respect to the position of our nitrogen atom vis-á-vis the hydrogens. But… Well… Look at the image below: the direction of the field (which we denote by ε because we’ve been using the E for energy) obviously matters when defining the old ‘up’ and ‘down’ states of our nitrogen atom too!

In other words, our previous | 1 〉 and | 2 〉 base states acquire a new meaning too: it obviously matters whether or not the electric dipole moment of the molecule is in the same or, conversely, in the opposite direction of the field. To be precise, the presence of the electromagnetic field suddenly gives the energy levels that we’d associate with these two states a very different physical interpretation.

Indeed, from the illustration above, it’s easy to see that the electric dipole moment of this particular molecule in state 1 is in the opposite direction and, therefore, temporarily ignoring the amplitude to flip over (so we do not think of A for just a brief little moment), the energy that we’d associate with state 1 would be equal to E₀+ με. Likewise, the energy we’d associate with state 2 is equal to E₀− με. Indeed, you’ll remember that the (potential) energy of an electric dipole is equal to the vector dot product of the electric dipole moment μ and the field vector ε, but with a minus sign in front so as to get the sign for the energy righ. So the energy is equal to −μ·ε = −|μ|·|ε|·cosθ, with θ the angle between both vectors. Now, the illustration above makes it clear that state 1 and 2 are defined for θ = π and θ = 0 respectively. [And, yes! Please do note that state 1 is the highest energy level, because it’s associated with the highest potential energy: the electric dipole moment μ of our ammonia molecule will – obviously! – want to align itself with the electric field ε ! Just think of what it would imply to turn the molecule in the field!]

Therefore, using the same hunches as the ones we used in the free space example, Feynman suggests that, when some external electric field is involved, we should use the following Hamiltonian matrix:

So we’ll need to solve a similar set of differential equations with this Hamiltonian now. We’ll do that later and, as mentioned above, it will be more convenient to switch to another set of base states, or another ‘representation’ as it’s referred to. But… Well… Let’s not get too much ahead of ourselves: I’ll say something about that before we’ll start solving the thing, but let’s first look at that Hamiltonian once more.

When I say that Feynman uses the same clues here, then… Well.. That’s true and not true. You should note that the diagonal elements in the Hamiltonian above are not the same: E₀+ με ≠ E₀+ με. So we’ve lost that symmetry of free space which, from a math point of view, was reflected in those identical H₁₁= H₂₂= E₀coefficients.

That should be obvious from what I write above: state 1 and state 2 are no longer those 1 and 2 states we described when looking at the molecule in free space. Indeed, the | 1 〉 and | 2 〉 states are still ‘up’ or ‘down’, but the illustration above also makes it clear we’re defining state 1 and state 2 not only with respect to the molecule’s spin around its own axis of symmetry but also vis-á-vis some direction in space. To be precise, we’re defining state 1 and state 2 here with respect to the direction of the electric field ε. Now that makes a really big difference in terms of interpreting what’s going on.

In fact, the ‘splitting’ of the energy levels because of that amplitude A is now something physical too, i.e. something that goes beyond just modeling the uncertainty involved. In fact, we’ll find it convenient to distinguish two new energy levels, which we’ll write as E_I= E₀+ A and E_II= E₀− A respectively. They are, of course, related to those new base states | I 〉 and | II 〉 that we’ll want to use. So the E₀+ A and E₀− A energy levels themselves will acquire some physical meaning, and especially the separation between them, i.e. the value of 2A. Indeed, E_I= E₀+ A and E_II= E₀− A will effectively represent an ‘upper’ and a ‘lower’ energy level respectively.

But, again, I am getting ahead of myself. Let’s first, as part of working towards a solution for our equations, look at what happens if and when we’d switch to another representation indeed.

Switching to another representation

Let me remind you of what I wrote in my post on quantum math in this regard. The actual state of our ammonia molecule – or any quantum-mechanical system really – is always to be described in terms of a set of base states. For example, if we have two possible base states only, we’ll write:

| φ 〉 = | 1 〉 C₁ + | 2 〉 C₂

You’ll say: why? Our molecule is obviously always in either state 1 or state 2, isn’t it? Well… Yes and no. That’s the mystery of quantum mechanics: it is and it isn’t. As long as we don’t measure it, there is an amplitude for it to be in state 1 and an amplitude for it to be in state 2. So we can only make sense of its state by actually calculating 〈 1 | φ 〉 and 〈 2 | φ 〉 which, unsurprisingly are equal to 〈 1 | φ 〉 = 〈 1 | 1 〉 C₁ + 〈 1 | 2 〉 C₂ = C₁(t) and 〈 2 | φ 〉 = 〈 2 | 1 〉 C₁ + 〈 2 | 2 〉 C₂ = C₂(t) respectively, and so these two functions give us the probabilities P₁(t) and P₂(t) respectively. So that’s Schrödinger’s cat really: the cat is dead or alive, but we don’t know until we open the box, and we only have a probability function – so we can say that it’s probably dead or probably alive, depending on the odds – as long as we do not open the box. It’s as simple as that.

Now, the ‘dead’ and ‘alive’ condition are, obviously, the ‘base states’ in Schrödinger’s rather famous example, and we can write them as | DEAD 〉 and | ALIVE 〉 you’d agree it would be difficult to find another representation. For example, it doesn’t make much sense to say that we’ve rotated the two base states over 90 degrees and we now have two new states equal to (1/√2)·| DEAD 〉 – (1/√2)·| ALIVE 〉 and (1/√2)·| DEAD 〉 + (1/√2)·| ALIVE 〉 respectively. There’s no direction in space in regard to which we’re defining those two base states: dead is dead, and alive is alive.

The situation really resembles our ammonia molecule in free space: there’s no external reference against which to define the base states. However, as soon as some external field is involved, we do have a direction in space and, as mentioned above, our base states are now defined with respect to a particular orientation in space. That implies two things. The first is that we should no longer say that our molecule will always be in either state 1 or state 2. There’s no reason for it to be perfectly aligned with or against the field. Its orientation can be anything really, and so its state is likely to be some combination of those two pure base states | 1 〉 and | 2 〉.

The second thing is that we may choose another set of base states, and specify the very same state in terms of the new base states. So, assuming we choose some other set of base states | I 〉 and | II 〉, we can write the very same state | φ 〉 = | 1 〉 C₁ + | 2 〉 C₂as:

| φ 〉 = | I 〉 C_I + | II 〉 C_II

It’s really like what you learned about vectors in high school: one can go from one set of base vectors to another by a transformation, such as, for example, a rotation, or a translation. It’s just that, just like in high school, we need some direction in regard to which we define our rotation or our translation.

For state vectors, I showed how a rotation of base states worked in one of my posts on two-state systems. To be specific, we had the following relation between the two representations:

The (1/√2) factor is there because of the normalization condition, and the two-by-two matrix equals the transformation matrix for a rotation of a state filtering apparatus about the y-axis, over an angle equal to (minus) 90 degrees, which we wrote as:

The y-axis? What y-axis? What state filtering apparatus? Just relax. Think about what you’ve learned already. The orientations are shown below: the S apparatus separates ‘up’ and ‘down’ states along the z-axis, while the T-apparatus does so along an axis that is tilted, about the y-axis, over an angle equal to α, or φ, as it’s written in the table above.

Of course, we don’t really introduce an apparatus at this or that angle. We just introduced an electromagnetic field, which re-defined our | 1 〉 and | 2 〉 base states and, therefore, through the rotational transformation matrix, also defines our | I 〉 and | II 〉 base states.

[…] You may have lost me by now, and so then you’ll want to skip to the next section. That’s fine. Just remember that the representations in terms of | I 〉 and | II 〉 base states or in terms of | 1 〉 and | 2 〉 base states are mathematically equivalent. Having said that, if you’re reading this post, and you want to understand it, truly (because you want to truly understand quantum mechanics), then you should try to stick with me here. 🙂 Indeed, there’s a zillion things you could think about right now, but you should stick to the math now. Using that transformation matrix, we can relate the C_Iand C_IIcoefficients in the | φ 〉 = | I 〉 C_I + | II 〉 C_II expression to the C_Iand C_IIcoefficients in the | φ 〉 = | 1 〉 C₁ + | 2 〉 C₂ expression. Indeed, we wrote:

C_I= 〈 I | ψ 〉 = (1/√2)·(C₁− C₂)
C_II= 〈 II | ψ 〉 = (1/√2)·(C₁+ C₂)

That’s exactly the same as writing:

OK. […] Waw! You just took a huge leap, because we can now compare the two sets of differential equations:

They’re mathematically equivalent, but the mathematical behavior of the functions involved is very different. Indeed, unlike the C₁(t) and C₂(t) amplitudes, we find that the C_I(t) and C_II(t) amplitudes are stationary, i.e. the associated probabilities – which we find by taking the absolute square of the amplitudes, as usual – do not vary in time. To be precise, if you write it all out and simplify, you’ll find that the C_I(t) and C_II(t) amplitudes are equal to:

C_I(t) = 〈 I | ψ 〉 = (1/√2)·(C₁− C₂) = (1/√2)·e^{−(i/ħ)·(E₀+ A)·t} = (1/√2)·e^{−(i/ħ)·E_I·t}
C_II(t) = 〈 II | ψ 〉 = (1/√2)·(C₁+ C₂) = (1/√2)·e^{−(i/ħ)·(E₀− A)·t}= (1/√2)·e^{−(i/ħ)·E_II·t}

As the absolute square of the exponential is equal to one, the associated probabilities, i.e. |C_I(t)|² and |C_II(t)|², are, quite simply, equal to |1/√2|² = 1/2. Now, it is very tempting to say that this means that our ammonia molecule has an equal chance to be in state I or state II. In fact, while I may have said something like that in my previous posts, that’s not how one should interpret this. The chance of our molecule being exactly in state I or state II, or in state 1 or state 2 is varying with time, with the probability being ‘dumped’ from one state to the other all of the time.

I mean… The electric dipole moment can point in any direction, really. So saying that our molecule has a 50/50 chance of being in state 1 or state 2 makes no sense. Likewise, saying that our molecule has a 50/50 chance of being in state I or state II makes no sense either. Indeed, the state of our molecule is specified by the | φ 〉 = | I 〉 C_I + | II 〉 C_II= | 1 〉 C₁ + | 2 〉 C₂equations, and neither of these two expressions is a stationary state. They mix two frequencies, because they mix two energy levels.

Having said that, we’re talking quantum mechanics here and, therefore, an external inhomogeneous electric field will effectively split the ammonia molecules according to their state. The situation is really like what a Stern-Gerlach apparatus does to a beam of electrons: it will split the beam according to the electron’s spin, which is either ‘up’ or, else, ‘down’, as shown in the graph below:

The graph for our ammonia molecule, shown below, is very similar. The vertical axis measures the same: energy. And the horizontal axis measures με, which increases with the strength of the electric field ε. So we see a similar ‘splitting’ of the energy of the molecule in an external electric field.

How should we explain this? It is very tempting to think that the presence of an external force field causes the electrons, or the ammonia molecule, to ‘snap into’ one of the two possible states, which are referred to as state I and state II respectively in the illustration of the ammonia state selector below. But… Well… Here we’re entering the murky waters of actually interpreting quantum mechanics, for which (a) we have no time, and (b) we are not qualified. So you should just believe, or take for granted, what’s being shown here: an inhomogeneous electric field will split our ammonia beam according to their state, which we define as I and II respectively, and which are associated with the energy E₀+ A and E₀− A respectively.

electric field

As mentioned above, you should note that these two states are stationary. The Hamiltonian equations which, as they always do, describe the dynamics of this system, imply that the amplitude to go from state I to state II, or vice versa, is zero. To make sure you ‘get’ that, I reproduce the associated Hamiltonian matrix once again:

Of course, that will change when we start our analysis of what’s happening in the maser. Indeed, we will have some non-zero H_I,II and H_II,Iamplitudes in the resonant cavity of our ammonia maser, in which we’ll have an oscillating electric field and, as a result, induced transitions from state I to II and vice versa. However, that’s for later. While I’ll quickly insert the full picture diagram below, you should, for the moment, just think about those two stationary states and those two zeroes. 🙂

Capito? If not… Well… Start reading this post again, I’d say. 🙂

Intermezzo: on approximations

At this point, I need to say a few things about all of the approximations involved, because it can be quite confusing indeed. So let’s take a closer look at those energy levels and the related Hamiltonian coefficients. In fact, in his Lectures, Feynman shows us that we can always have a general solution for the Hamiltonian equations describing a two-state system whenever we have constant Hamiltonian coefficients. That general solution – which, mind you, is derived assuming Hamiltonian coefficients that do not depend on time – can always be written in terms of two stationary base states, i.e. states with a definite energy and, hence, a constant probability. The equations, and the two definite energy levels are:

That yields the following values for the energy levels for the stationary states:

Now, that’s very different from the E_I= E₀+ A and E_II= E₀− A energy levels for those stationary states we had defined in the previous section: those stationary states had no square root, and no μ²ε², in their energy. In fact, that sort of answers the question: if there’s no external field, then that μ²ε² factor is zero, and the square root in the expression becomes ±√A²= ±A. So then we’re back to our E_I= E₀+ A and E_II= E₀− A formulas. The whole point, however, is that we will actually have an electric field in that cavity. Moreover, it’s going to be a field that varies in time, which we’ll write:

Now, part of the confusion in Feynman’s approach is that he constantly switches between representing the system in terms of the I and II base states and the 1 and 2 base states respectively. For a good understanding, we should compare with our original representation of the dynamics in free space, for which the Hamiltonian was the following one:

That matrix can easily be related to the new one we’re going to have to solve, which is equal to:

The interpretation is easy if we look at that illustration again:

If the direction of the electric dipole moment is opposite to the direction ε, then the associated energy is equal to −μ·ε = −μ·ε = −|μ|·|ε|·cosθ = −μ·ε·cos(π) = +με. Conversely, for state 2, we find −μ·ε·cos(0) = −με for the energy that’s associated with the dipole moment. You can and should think about the physics involved here, because they make sense! Thinking of amplitudes, you should note that the +με and −με terms effectively change the H₁₁ and H₂₂ coefficients, so they change the amplitude to stay in state 1 or state 2 respectively. That, of course, will have an impact on the associated probabilities, and so that’s why we’re talking of induced transitions now.

Having said that, the Hamiltonian matrix above keeps the −A for H₁₂ and H₂₁, so the matrix captures spontaneous transitions too!

Still… You may wonder why Feynman doesn’t use those E_Iand E_IIformulas with the square root because that would give us some exact solution, wouldn’t it? The answer to that question is: maybe it would, but would you know how to solve those equations? We’ll have a varying field, remember? So our Hamiltonian H₁₁ and H₂₂ coefficients will no longer be constant, but time-dependent. As you’re going to see, it takes Feynman three pages to solve the whole thing using the +με and −με approximation. So just imagine how complicated it would be using that square root expression! [By the way, do have a look at those asymptotic curves in that illustration showing the splitting of energy levels above, so you see how that approximation looks like.]

So that’s the real answer: we need to simplify somehow, so as to get any solutions at all!

Of course, it’s all quite confusing because, after Feynman first notes that, for strong fields, the A² in that square root is small as compared to μ²ε², thereby justifying the use of the simplified E_I= E₀+ με = H₁₁ and E_II= E₀− με = H₂₂ coefficients, he continues and bluntly uses the very same square root expression to explain how that state selector works, saying that the electric field in the state selector will be rather weak and, hence, that με will be much smaller than A, so one can use the following approximation for the square root in the expressions above:

The energy expressions then reduce to:

And then we can calculate the force on the molecules as:

So the electric field in the state selector is weak, but the electric field in the cavity is supposed to be strong, and so… Well… That’s it, really. The bottom line is that we’ve a beam of ammonia molecules that are all in state I, and it’s what happens with that beam then, that is being described by our new set of differential equations:

Solving the equations

As all molecules in our ammonia beam are described in terms of the | I 〉 and | II 〉 base states – as evidenced by the fact that we say all molecules that enter the cavity are state I – we need to switch to that representation. We do that by using that transformation above, so we write:

C_I= 〈 I | ψ 〉 = (1/√2)·(C₁− C₂)
C_II= 〈 II | ψ 〉 = (1/√2)·(C₁+ C₂)

Keeping these ‘definitions’ of C_Iand C_IIin mind, you should then add the two differential equations, divide the result by the square root of 2, and you should get the following new equation:

Please! Do it and verify the result! You want to learn something here, no? 🙂

Likewise, subtracting the two differential equations, we get:

We can re-write this as:

Now, the problem is that the Hamiltonian constants here are not constant. To be precise, the electric field ε varies in time. We wrote:

So H_I,IIand H_II,I, which are equal to με, are not constant: we’ve got Hamiltonian coefficients that are a function of time themselves. […] So… Well… We just need to get on with it and try to finally solve this thing. Let me just copy Feynman as he grinds through this:

This is only the first step in the process. Feynman just takes two trial functions, which are really similar to the very general C₁= a·e^{–(i/ħ)·H₁₁·t}function we presented when only one equation was involved, or – if you prefer a set of two equations – those C_I(t) = a·e^{−(i/ħ)·E_I·t}and C_I(t) = b·e^{−(i/ħ)·E_II·}^tequations above. The difference is that the coefficients in front, i.e. γ_I and γ_II are not some (complex) constant, but functions of time themselves. The next step in the derivation is as follows:

One needs to do a bit of gymnastics here as well to follow what’s going on, but please do check and you’ll see it works. Feynman derives another set of differential equations here, and they specify these γ_I = γ_I(t) and γ_II = γ_II(t) functions. These equations are written in terms of the frequency of the field, i.e. ω, and the resonant frequency ω₀, which we mentioned above when calculating that 23.79 GHz frequency from the 2A = h·f₀ equation. So ω₀ is the same molecular resonance frequency but expressed as an angular frequency, so ω₀ = f₀/2π = ħ/2A. He then proceeds to simplify, using assumptions one should check. He then continues:

That gives us what we presented in the previous post:

So… Well… What to say? I explained those probability functions in my previous post, indeed. We’ve got two probabilities here:

P_I= cos²[(με₀/ħ)·t]
P_II= sin²[(με₀/ħ)·t]

So that’s just like the P₁= cos²[(A/ħ)·t] and P₂= sin²[(A/ħ)·t] probabilities we found for spontaneous transitions. But so here we are talking induced transitions.

As you can see, the frequency and, hence, the period, depend on the strength, or magnitude, of the electric field, i.e. the ε₀constant in the ε = 2ε₀cos(ω·t) expression. The natural unit for measuring time would be the period once again, which we can easily calculate as (με₀/ħ)·T = π ⇔ T = π·ħ/με₀.

Now, we had that T = (π·ħ)/(2A) expression above, which allowed us to calculate the period of the spontaneous transition frequency, which we found was like 40 picoseconds, i.e. 40×10⁻¹²seconds. Now, the T = (π·ħ)/(2με₀) is very similar, it allows us to calculate the expected, average, or mean time for an induced transition. In fact, if we write T_induced = (π·ħ)/(2με₀) and T_spontaneous = (π·ħ)/(2A), then we can take ratio to find:

T_induced/T_spontaneous = [(π·ħ)/(2με₀)]/[(π·ħ)/(2A)] = A/με₀

This A/με₀ratio is greater than one, so T_induced/T_spontaneous is greater than one, which, in turn, means that the presence of our electric field – which, let me remind you, dances to the beat of the resonant frequency – causes a slower transition than we would have had if the oscillating electric field were not present.

But – Hey! – that’s the wrong comparison! Remember all molecules enter in a stationary state, as they’ve been selected so as to ensure they’re in state I. So there is no such thing as a spontaneous transition frequency here! They’re all polarized, so to speak, and they would remain that way if there was no field in the cavity. So if there was no oscillating electric field, they would never transition. Nothing would happen! Well… In terms of our particular set of base states, of course! Why? Well… Look at the Hamiltonian coefficients H_I,II= H_II,I= με: these coefficients are zero if ε is zero. So… Well… That says it all.

So that‘s what it’s all about: induced emission and, as I explained in my previous post, because all molecules enter in state I, i.e. the upper energy state, literally, they all ‘dump’ a net amount of energy equal to 2A into the cavity at the occasion of their first transition. The molecules then keep dancing, of course, and so they absorb and emit the same amount as they go through the cavity, but… Well… We’ve got a net contribution here, which is not only enough to maintain the cavity oscillations, but actually also provides a small excess of power that can be drawn from the cavity as microwave radiation of the same frequency.

As Feynman notes, an exact description of what actually happens requires an understanding of the quantum mechanics of the field in the cavity, i.e. quantum field theory, which I haven’t studied yet. But… Well… That’s for later, I guess. 🙂

Post scriptum: The sheer length of this post shows we’re not doing something that’s easy here. Frankly, I feel the whole analysis is still quite obscure, in the sense that – despite looking at this thing again and again – it’s hard to sort of interpret what’s going on, in a physical sense that is. But perhaps one shouldn’t try that. I’ve quoted Feynman’s view on how easy or how difficult it is to ‘understand’ quantum mechanics a couple of times already, so let me do it once more:

“Because atomic behavior is so unlike ordinary experience, it is very difficult to get used to, and it appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. Even the experts do not understand it the way they would like to, and it is perfectly reasonable that they should not, because all of direct, human experience and human intuition applies to large objects.”

So… Well… I’ll grind through the remaining Lectures now – I am halfway through Volume III now – and then re-visit all of this. Despite Feynman’s warning, I want to understand it the way I like to, even if I don’t quite know what way that is right now. 🙂

Addendum: As for those cycles and periods, I noted a couple of times already that the Planck-Einstein equation E = h·f can usefully be re-written as E/f = h, as it gives a physical interpretation to the value of the Planck constant. In fact, I said h is the energy that’s associated with one cycle, regardless of the frequency of the radiation involved. Indeed, the energy of a photon divided by the number of cycles per second, should give us the energy per cycle, no?

Well… Yes and no. Planck’s constant h and the frequency f are both expressed referencing the time unit. However, if we say that a sodium atom emits one photon only as its electron transitions from a higher energy level to a lower one, and if we say that involves a decay time of the order of 3.2×10⁻⁸seconds, then what we’re saying really is that a sodium light photon will ‘pack’ like 16 million cycles, which is what we get when we multiply the number of cycles per second (i.e. the mentioned frequency of 500×10¹²Hz) by the decay time (i.e. 3.2×10⁻⁸seconds): (500×10¹²Hz)·(3.2×10⁻⁸s) = 16 ×10⁶cycles, indeed. So the energy per cycle is 2.068 eV (i.e. the photon energy) divided by 16×10⁶, so that’s 0.129×10⁻⁶eV. Unsurprisingly, that’s what we get when we we divide h by 3.2×10⁻⁸s: (4.13567×10⁻¹⁵)/(3.2×10⁻⁸s) = 1.29×10⁻⁷eV. We’re just putting some values in to the E/(f·T) = h/T equation here.

The logic for that 2A = h·f₀ is the same. The frequency of the radiation that’s being absorbed or emitted is 23.79 GHz, so the photon energy is (23.97×10⁹ Hz)·(4.13567×10⁻¹⁵ eV·s) ≈ 1×10⁻⁴eV. Now, we calculated the transition period T as T = π·ħ/A ≈ (π·6.626×10⁻¹⁶eV·s)/(0.5×10⁻⁴eV) ≈ 41.6×10⁻¹²seconds. Now, an oscillation of a frequency of 23.97 giga-hertz that only lasts 41.6×10⁻¹²seconds is an oscillation of one cycle only. The consequence is that, when we continue this style of reasoning, we’d have a photon that packs all of its energy into one cycle!

Let’s think about what this implies in terms of the density in space. The wavelength of our microwave radiation is 1.25×10⁻²m, so we’ve got a ‘density’ of 1×10⁻⁴eV/1.25×10⁻²m = 0.8×10⁻²eV/m = 0.008 eV/m. The wavelength of our sodium light is 0.6×10⁻⁶m, so we get a ‘density’ of 1.29×10⁻⁷eV/0.6×10⁻⁶m = 2.15×10⁻¹eV/m = 0.215 eV/m. So the energy ‘density’ of our sodium light is 26.875 times that of our microwave radiation. 🙂

Frankly, I am not quite sure if calculations like this make much sense. In fact, when talking about energy densities, I should review my posts on the Poynting vector. However, they may help you think things through. 🙂

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Re-visiting uncertainty…

I re-visited the Uncertainty Principle a couple of times already, but here I really want to get at the bottom of the thing? What’s uncertain? The energy? The time? The wavefunction itself? These questions are not easily answered, and I need to warn you: you won’t get too much wiser when you’re finished reading this. I just felt like freewheeling a bit. [Note that the first part of this post repeats what you’ll find on the Occam page, or my post on Occam’s Razor. But these post do not analyze uncertainty, which is what I will be trying to do here.]

Let’s first think about the wavefunction itself. It’s tempting to think it actually is the particle, somehow. But it isn’t. So what is it then? Well… Nobody knows. In my previous post, I said I like to think it travels with the particle, but then doesn’t make much sense either. It’s like a fundamental property of the particle. Like the color of an apple. But where is that color? In the apple, in the light it reflects, in the retina of our eye, or is it in our brain? If you know a thing or two about how perception actually works, you’ll tend to agree the quality of color is not in the apple. When everything is said and done, the wavefunction is a mental construct: when learning physics, we start to think of a particle as a wavefunction, but they are two separate things: the particle is reality, the wavefunction is imaginary.

But that’s not what I want to talk about here. It’s about that uncertainty. Where is the uncertainty? You’ll say: you just said it was in our brain. No. I didn’t say that. It’s not that simple. Let’s look at the basic assumptions of quantum physics:

Quantum physics assumes there’s always some randomness in Nature and, hence, we can measure probabilities only. We’ve got randomness in classical mechanics too, but this is different. This is an assumption about how Nature works: we don’t really know what’s happening. We don’t know the internal wheels and gears, so to speak, or the ‘hidden variables’, as one interpretation of quantum mechanics would say. In fact, the most commonly accepted interpretation of quantum mechanics says there are no ‘hidden variables’.
However, as Shakespeare has one of his characters say: there is a method in the madness, and the pioneers– I mean Werner Heisenberg, Louis de Broglie, Niels Bohr, Paul Dirac, etcetera – discovered that method: all probabilities can be found by taking the square of the absolute value of a complex-valued wavefunction (often denoted by Ψ), whose argument, or phase (θ), is given by the de Broglie relations ω = E/ħ and k = p/ħ. The generic functional form of that wavefunction is:

Ψ = Ψ(x, t) = a·e^−iθ= a·e^{−i(ω·t − k ∙x)} = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]}

That should be obvious by now, as I’ve written more than a dozens of posts on this. 🙂 I still have trouble interpreting this, however—and I am not ashamed, because the Great Ones I just mentioned have trouble with that too. It’s not that complex exponential. That e^−iφ is a very simple periodic function, consisting of two sine waves rather than just one, as illustrated below. [It’s a sine and a cosine, but they’re the same function: there’s just a phase difference of 90 degrees.]

No. To understand the wavefunction, we need to understand those de Broglie relations, ω = E/ħ and k = p/ħ, and then, as mentioned, we need to understand the Uncertainty Principle. We need to understand where it comes from. Let’s try to go as far as we can by making a few remarks:

Adding or subtracting two terms in math, (E/ħ)·t − (p/ħ)∙x, implies the two terms should have the same dimension: we can only add apples to apples, and oranges to oranges. We shouldn’t mix them. Now, the (E/ħ)·t and (p/ħ)·x terms are actually dimensionless: they are pure numbers. So that’s even better. Just check it: energy is expressed in newton·meter (energy, or work, is force over distance, remember?) or electronvolts (1 eV = 1.6×10⁻¹⁹J = 1.6×10⁻¹⁹N·m); Planck’s constant, as the quantum of action, is expressed in J·s or eV·s; and the unit of (linear) momentum is 1 N·s = 1 kg·m/s = 1 N·s. E/ħ gives a number expressed per second, and p/ħ a number expressed per meter. Therefore, multiplying E/ħ and p/ħ by t and x respectively gives us a dimensionless number indeed.
It’s also an invariant number, which means we’ll always get the same value for it, regardless of our frame of reference. As mentioned above, that’s because the four-vector product p_μx_μ= E·t − p∙x is invariant: it doesn’t change when analyzing a phenomenon in one reference frame (e.g. our inertial reference frame) or another (i.e. in a moving frame).
Now, Planck’s quantum of action h, or ħ – h and ħ only differ in their dimension: h is measured in cycles per second, while ħ is measured in radians per second: both assume we can at least measure one cycle – is the quantum of energy really. Indeed, if “energy is the currency of the Universe”, and it’s real and/or virtual photons who are exchanging it, then it’s good to know the currency unit is h, i.e. the energy that’s associated with one cycle of a photon. [In case you want to see the logic of this, see my post on the physical constants c, h and α.]
It’s not only time and space that are related, as evidenced by the fact that t − x itself is an invariant four-vector, E and p are related too, of course! They are related through the classical velocity of the particle that we’re looking at: E/p = c²/v and, therefore, we can write: E·β = p·c, with β = v/c, i.e. the relative velocity of our particle, as measured as a ratio of the speed of light. Now, I should add that the t − x four-vector is invariant only if we measure time and space in equivalent units. Otherwise, we have to write c·t − x. If we do that, so our unit of distance becomes c meter, rather than one meter, or our unit of time becomes the time that is needed for light to travel one meter, then c = 1, and the E·β = p·c becomes E·β = p, which we also write as β = p/E: the ratio of the energy and the momentum of our particle is its (relative) velocity.

Combining all of the above, we may want to assume that we are measuring energy and momentum in terms of the Planck constant, i.e. the ‘natural’ unit for both. In addition, we may also want to assume that we’re measuring time and distance in equivalent units. Then the equation for the phase of our wavefunctions reduces to:

θ = (ω·t − k ∙x) = E·t − p·x

Now, θ is the argument of a wavefunction, and we can always re-scale such argument by multiplying or dividing it by some constant. It’s just like writing the argument of a wavefunction as v·t–x or (v·t–x)/v = t –x/v with v the velocity of the waveform that we happen to be looking at. [In case you have trouble following this argument, please check the post I did for my kids on waves and wavefunctions.] Now, the energy conservation principle tells us the energy of a free particle won’t change. [Just to remind you, a ‘free particle’ means it’s in a ‘field-free’ space, so our particle is in a region of uniform potential.] So we can, in this case, treat E as a constant, and divide E·t − p·x by E, so we get a re-scaled phase for our wavefunction, which I’ll write as:

φ = (E·t − p·x)/E = t − (p/E)·x = t − β·x

Alternatively, we could also look at p as some constant, as there is no variation in potential energy that will cause a change in momentum, and the related kinetic energy. We’d then divide by p and we’d get (E·t − p·x)/p = (E/p)·t − x) = t/β − x, which amounts to the same, as we can always re-scale by multiplying it with β, which would again yield the same t − β·x argument.

The point is, if we measure energy and momentum in terms of the Planck unit (I mean: in terms of the Planck constant, i.e. the quantum of energy), and if we measure time and distance in ‘natural’ units too, i.e. we take the speed of light to be unity, then our Platonic wavefunction becomes as simple as:

Φ(φ) = a·e^−iφ= a·e^{−i(t − β·x)}

This is a wonderful formula, but let me first answer your most likely question: why would we use a relative velocity?Well… Just think of it: when everything is said and done, the whole theory of relativity and, hence, the whole of physics, is based on one fundamental and experimentally verified fact: the speed of light is absolute. In whatever reference frame, we will always measure it as 299,792,458 m/s. That’s obvious, you’ll say, but it’s actually the weirdest thing ever if you start thinking about it, and it explains why those Lorentz transformations look so damn complicated. In any case, this fact legitimately establishes c as some kind of absolute measure against which all speeds can be measured. Therefore, it is only natural indeed to express a velocity as some number between 0 and 1. Now that amounts to expressing it as the β = v/c ratio.

Let’s now go back to that Φ(φ) = a·e^−iφ= a·e^{−i(t − β·x)}wavefunction. Its temporal frequency ω is equal to one, and its spatial frequency k is equal to β = v/c. It couldn’t be simpler but, of course, we’ve got this remarkably simple result because we re-scaled the argument of our wavefunction using the energy and momentum itself as the scale factor. So, yes, we can re-write the wavefunction of our particle in a particular elegant and simple form using the only information that we have when looking at quantum-mechanical stuff: energy and momentum, because that’s what everything reduces to at that level.

So… Well… We’ve pretty much explained what quantum physics is all about here. You just need to get used to that complex exponential: e^−iφ = cos(−φ) + i·sin(−φ) = cos(φ) −i·sin(φ). It would have been nice if Nature would have given us a simple sine or cosine function. [Remember the sine and cosine function are actually the same, except for a phase difference of 90 degrees: sin(φ) = cos(π/2−φ) = cos(φ+π/2). So we can go always from one to the other by shifting the origin of our axis.] But… Well… As we’ve shown so many times already, a real-valued wavefunction doesn’t explain the interference we observe, be it interference of electrons or whatever other particles or, for that matter, the interference of electromagnetic waves itself, which, as you know, we also need to look at as a stream of photons , i.e. light quanta, rather than as some kind of infinitely flexible aether that’s undulating, like water or air.

However, the analysis above does not include uncertainty. That’s as fundamental to quantum physics as de Broglie‘s equations, so let’s think about that now.

Introducing uncertainty

Our information on the energy and the momentum of our particle will be incomplete: we’ll write E = E₀± σ_E, and p = p₀± σ_p. Huh? No ΔE or ΔE? Well… It’s the same, really, but I am a bit tired of using the Δ symbol, so I am using the σ symbol here, which denotes a standard deviation of some density function. It underlines the probabilistic, or statistical, nature of our approach.

The simplest model is that of a two-state system, because it involves two energy levels only: E = E₀± A, with A some constant. Large or small, it doesn’t matter. All is relative anyway. 🙂 We explained the basics of the two-state system using the example of an ammonia molecule, i.e. an NH₃molecule, so it consists on one nitrogen and three hydrogen atoms. We had two base states in this system: ‘up’ or ‘down’, which we denoted as base state | 1 〉 and base state | 2 〉 respectively. This ‘up’ and ‘down’ had nothing to do with the classical or quantum-mechanical notion of spin, which is related to the magnetic moment. No. It’s much simpler than that: the nitrogen atom could be either beneath or, else, above the plane of the hydrogens, as shown below, with ‘beneath’ and ‘above’ being defined in regard to the molecule’s direction of rotation around its axis of symmetry.

In any case, for the details, I’ll refer you to the post(s) on it. Here I just want to mention the result. We wrote the amplitude to find the molecule in either one of these two states as:

C₁= 〈 1 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}+ (1/2)·e^{−(i/ħ)·(E₀+ A)·t}
C₂= 〈 2 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}– (1/2)·e^{−(i/ħ)·(E₀+ A)·t}

That gave us the following probabilities:

If our molecule can be in two states only, and it starts off in one, then the probability that it will remain in that state will gradually decline, while the probability that it flips into the other state will gradually increase.

Now, the point you should note is that we get these time-dependent probabilities only because we’re introducing two different energy levels: E₀+ A and E₀− A. [Note they separated by an amount equal to 2·A, as I’ll use that information later.] If we’d have one energy level only – which amounts to saying that we know it, and that it’s something definite – then we’d just have one wavefunction, which we’d write as:

a·e^−iθ= a·e⁻⁽^{i/ħ)·(E₀·t − p·x)}= a·e⁻⁽ⁱ^/ħ)·(E^₀·t)·e⁽^{i/ħ)·(p·x)}

Note that we can always split our wavefunction in a ‘time’ and a ‘space’ part, which is quite convenient. In fact, because our ammonia molecule stays where it is, it has no momentum: p = 0. Therefore, its wavefunction reduces to:

a·e^−iθ= a·e⁻⁽^{i/ħ)·(E₀·t)}

As simple as it can be. 🙂 The point is that a wavefunction like this, i.e. a wavefunction that’s defined by a definite energy, will always yield a constant and equal probability, both in time as well as in space. That’s just the math of it: |a·e^−iθ|²= a². Always! If you want to know why, you should think of Euler’s formula and Pythagoras’ Theorem: cos²θ +sin²θ = 1. Always! 🙂

That constant probability is annoying, because our nitrogen atom never ‘flips’, and we know it actually does, thereby overcoming a energy barrier: it’s a phenomenon that’s referred to as ‘tunneling’, and it’s real! The probabilities in that graph above are real! Also, if our wavefunction would represent some moving particle, it would imply that the probability to find it somewhere in space is the same all over space, which implies our particle is everywhere and nowhere at the same time, really.

So, in quantum physics, this problem is solved by introducing uncertainty. Introducing some uncertainty about the energy, or about the momentum, is mathematically equivalent to saying that we’re actually looking at a composite wave, i.e. the sum of a finite or potentially infinite set of component waves. So we have the same ω = E/ħ and k = p/ħ relations, but we apply them to n energy levels, or to some continuous range of energy levels ΔE. It amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ. In our two-state system, n = 2, obviously! So we’ve two energy levels only and so our composite wave consists of two component waves only.

We know what that does: it ensures our wavefunction is being ‘contained’ in some ‘envelope’. It becomes a wavetrain, or a kind of beat note, as illustrated below:

[The animation comes from Wikipedia, and shows the difference between the group and phase velocity: the green dot shows the group velocity, while the red dot travels at the phase velocity.]

So… OK. That should be clear enough. Let’s now apply these thoughts to our ‘reduced’ wavefunction

Φ(φ) = a·e^−iφ= a·e^{−i(t − β·x)}

Thinking about uncertainty

Frankly, I tried to fool you above. If the functional form of the wavefunction is a·e⁻⁽^{i/ħ)·(E·t − p·x)}, then we can measure E and p in whatever unit we want, including h or ħ, but we cannot re-scale the argument of the function, i.e. the phase θ, without changing the functional form itself. I explained that in that post for my kids on wavefunctions:, in which I explained we may represent the same electromagnetic wave by two different functional forms:

F(ct−x) = G(t−x/c)

So F and G represent the same wave, but they are different wavefunctions. In this regard, you should note that the argument of F is expressed in distance units, as we multiply t with the speed of light (so it’s like our time unit is 299,792,458 m now), while the argument of G is expressed in time units, as we divide x by the distance traveled in one second). But F and G are different functional forms. Just do an example and take a simple sine function: you’ll agree that sin(θ) ≠ sin(θ/c) for all values of θ, except 0. Re-scaling changes the frequency, or the wavelength, and it does so quite drastically in this case. 🙂 Likewise, you can see that a·e^−i(φ/E)= [a·e^−iφ]^1/E, so that’s a very different function. In short, we were a bit too adventurous above. Now, while we can drop the 1/ħ in the a·e⁻⁽^{i/ħ)·(E·t − p·x)}function when measuring energy and momentum in units that are numerically equal to ħ, we’ll just revert to our original wavefunction for the time being, which equals

Ψ(θ) = a·e^−iθ= a·e⁻^{i·[(E/ħ)·t − (p/ħ)·x]}

Let’s now introduce uncertainty once again. The simplest situation is that we have two closely spaced energy levels. In theory, the difference between the two can be as small as ħ, so we’d write: E = E₀± ħ/2. [Remember what I said about the ± A: it means the difference is 2A.] However, we can generalize this and write: E = E₀± n·ħ/2, with n = 1, 2, 3,… This does not imply any greater uncertainty – we still have two states only – but just a larger difference between the two energy levels.

Let’s also simplify by looking at the ‘time part’ of our equation only, i.e. a·e⁻^i·(E/ħ)·t. It doesn’t mean we don’t care about the ‘space part’: it just means that we’re only looking at how our function varies in time and so we just ‘fix’ or ‘freeze’ x. Now, the uncertainty is in the energy really but, from a mathematical point of view, we’ve got an uncertainty in the argument of our wavefunction, really. This uncertainty in the argument is, obviously, equal to:

(E/ħ)·t = [(E₀± n·ħ/2)/ħ]·t = (E₀/ħ ± n/2)·t = (E₀/ħ)·t ± (n/2)·t

So we can write:

a·e⁻^i·(E/ħ)·t = a·e⁻^{i·[(E₀/ħ)·t ± (1/2)·t]} = a·e⁻^{i·[(E₀/ħ)·t]}·e^{i·[±(n/2)·t]}

This is valid for any value of t. What the expression says is that, from a mathematical point of view, introducing uncertainty about the energy is equivalent to introducing uncertainty about the wavefunction itself. It may be equal to a·e⁻^{i·[(E₀/ħ)·t]}·e^i·(n/2)·t, but it may also be equal to a·e⁻^{i·[(E₀/ħ)·t]}·e^{−i·(n/2)·t}. The phases of the e^−i·t/2 and e^i·t/2factors are separated by a distance equal to t.

So… Well…

[…]

Hmm… I am stuck. How is this going to lead me to the ΔE·Δt = ħ/2 principle? To anyone out there: can you help? 🙂

[…]

The thing is: you won’t get the Uncertainty Principle by staring at that formula above. It’s a bit more complicated. The idea is that we have some distribution of the observables, like energy and momentum, and that implies some distribution of the associated frequencies, i.e. ω for E, and k for p. The Wikipedia article on the Uncertainty Principle gives you a formal derivation of the Uncertainty Principle, using the so-called Kennard formulation of it. You can have a look, but it involves a lot of formalism—which is what I wanted to avoid here!

I hope you get the idea though. It’s like statistics. First, we assume we know the population, and then we describe that population using all kinds of summary statistics. But then we reverse the situation: we don’t know the population but we do have sample information, which we also describe using all kinds of summary statistics. Then, based on what we find for the sample, we calculate the estimated statistics for the population itself, like the mean value and the standard deviation, to name the most important ones. So it’s a bit the same here, except that, in quantum mechanics, there may not be any real value underneath: the mean and the standard deviation represent something fuzzy, rather than something precise.

Hmm… I’ll leave you with these thoughts. We’ll develop them further as we will be digging into all much deeper over the coming weeks. 🙂

Post scriptum: I know you expect something more from me, so… Well… Think about the following. If we have some uncertainty about the energy E, we’ll have some uncertainty about the momentum p according to that β = p/E. [By the way, please think about this relationship: it says, all other things being equal (such as the inertia, i.e. the mass, of our particle), that more energy will all go into more momentum. More specifically, note that ∂p/∂p = β according to this equation. In fact, if we include the mass of our particle, i.e. its inertia, as potential energy, then we might say that (1−β)·E is the potential energy of our particle, as opposed to its kinetic energy.] So let’s try to think about that.

Let’s denote the uncertainty about the energy as ΔE. As should be obvious from the discussion above, it can be anything: it can mean two separate energy levels E = E₀± A, or a potentially infinite set of values. However, even if the set is infinite, we know the various energy levels need to be separated by ħ, at least. So if the set is infinite, it’s going to be a countable infinite set, like the set of natural numbers, or the set of integers. But let’s stick to our example of two values E = E₀± A only, with A = ħ so E + ΔE = E₀± ħ and, therefore, ΔE = ± ħ. That implies Δp = Δ(β·E) = β·ΔE = ± β·ħ.

Hmm… This is a bit fishy, isn’t it? We said we’d measure the momentum in units of ħ, but so here we say the uncertainty in the momentum can actually be a fraction of ħ. […] Well… Yes. Now, the momentum is the product of the mass, as measured by the inertia of our particle to accelerations or decelerations, and its velocity. If we assume the inertia of our particle, or its mass, to be constant – so we say it’s a property of the object that is not subject to uncertainty, which, I admit, is a rather dicey assumption (if all other measurable properties of the particle are subject to uncertainty, then why not its mass?) – then we can also write: Δp = Δ(m·v) = Δ(m·β) = m·Δβ. [Note that we’re not only assuming that the mass is not subject to uncertainty, but also that the velocity is non-relativistic. If not, we couldn’t treat the particle’s mass as a constant.] But let’s be specific here: what we’re saying is that, if ΔE = ± ħ, then Δv = Δβ will be equal to Δβ = Δp/m = ± (β/m)·ħ. The point to note is that we’re no longer sure about the velocity of our particle. Its (relative) velocity is now:

β ± Δβ = β ± (β/m)·ħ

But, because velocity is the ratio of distance over time, this introduces an uncertainty about time and distance. Indeed, if its velocity is β ± (β/m)·ħ, then, over some time T, it will travel some distance X = [β ± (β/m)·ħ]·T. Likewise, it we have some distance X, then our particle will need a time equal to T = X/[β ± (β/m)·ħ].

You’ll wonder what I am trying to say because… Well… If we’d just measure X and T precisely, then all the uncertainty is gone and we know if the energy is E₀+ ħ or E₀− ħ. Well… Yes and no. The uncertainty is fundamental – at least that’s what’s quantum physicists believe – so our uncertainty about the time and the distance we’re measuring is equally fundamental: we can have either of the two values X = [β ± (β/m)·ħ] T = X/[β ± (β/m)·ħ], whenever or wherever we measure. So we have a ΔX and ΔT that are equal to ± [(β/m)·ħ]·T and X/[± (β/m)·ħ] respectively. We can relate this to ΔE and Δp:

ΔX = (1/m)·T·Δp
ΔT = X/[(β/m)·ΔE]

You’ll grumble: this still doesn’t give us the Uncertainty Principle in its canonical form. Not at all, really. I know… I need to do some more thinking here. But I feel I am getting somewhere. 🙂 Let me know if you see where, and if you think you can get any further. 🙂

The thing is: you’ll have to read a bit more about Fourier transforms and why and how variables like time and energy, or position and momentum, are so-called conjugate variables. As you can see, energy and time, and position and momentum, are obviously linked through the E·t and p·x products in the E₀·t − p·x sum. That says a lot, and it helps us to understand, in a more intuitive way, why the ΔE·Δt and Δp·Δx products should obey the relation they are obeying, i.e. the Uncertainty Principle, which we write as ΔE·Δt ≥ ħ/2 and Δp·Δx ≥ ħ/2. But so proving involves more than just staring at that Ψ(θ) = a·e^−iθ= a·e⁻^{i·[(E/ħ)·t − (p/ħ)·x]}relation.

Having said, it helps to think about how that E·t − p·x sum works. For example, think about two particles, a and b, with different velocity and mass, but with the same momentum, so p_a= p_b ⇔ m_a·v_a= m_a·v_a⇔ m_a/v_b= m_b/v_a. The spatial frequency of the wavefunction would be the same for both but the temporal frequency would be different, because their energy incorporates the rest mass and, hence, because m_a≠ m_b, we also know that E_a≠ E_b. So… It all works out but, yes, I admit it’s all very strange, and it takes a long time and a lot of reflection to advance our understanding.

Occam’s Razor

The analysis of a two-state system (i.e. the rather famous example of an ammonia molecule ‘flipping’ its spin direction from ‘up’ to ‘down’, or vice versa) in my previous post is a good opportunity to think about Occam’s Razor once more. What are we doing? What does the math tell us?

In the example we chose, we didn’t need to worry about space. It was all about time: an evolving state over time. We also knew the answers we wanted to get: if there is some probability for the system to ‘flip’ from one state to another, we know it will, at some point in time. We also want probabilities to add up to one, so we knew the graph below had to be the result we would find: if our molecule can be in two states only, and it starts of in one, then the probability that it will remain in that state will gradually decline, while the probability that it flips into the other state will gradually increase, which is what is depicted below.

However, the graph above is only a Platonic idea: we don’t bother to actually verify what state the molecule is in. If we did, we’d have to ‘re-set’ our t = 0 point, and start all over again. The wavefunction would collapse, as they say, because we’ve made a measurement. However, having said that, yes, in the physicist’s Platonic world of ideas, the probability functions above make perfect sense. They are beautiful. You should note, for example, that P₁ (i.e. the probability to be in state 1) and P₂ (i.e. the probability to be in state 2) add up to 1 all of the time, so we don’t need to integrate over a cycle or something: so it’s all perfect!

These probability functions are based on ideas that are even more Platonic: interfering amplitudes. Let me explain.

Quantum physics is based on the idea that these probabilities are determined by some wavefunction, a complex-valued amplitude that varies in time and space. It’s a two-dimensional thing, and then it’s not. It’s two-dimensional because it combines a sine and cosine, i.e. a real and an imaginary part, but the argument of the sine and the cosine is the same, and the sine and cosine are the same function, except for a phase shift equal to π. We write:

a·e^−iθ= a·cos(θ) – a·sin(−θ) = a·cosθ – a·sinθ

The minus sign is there because it turns out that Nature measures angles, i.e. our phase, clockwise, rather than counterclockwise, so that’s not as per our mathematical convention. But that’s a minor detail, really. [It should give you some food for thought, though.] For the rest, the related graph is as simple as the formula:

Now, the phase of this wavefunction is written as θ = (ω·t − k ∙x). Hence, ω determines how this wavefunction varies in time, and the wavevector k tells us how this wave varies in space. The young Frenchman Comte Louis de Broglie noted the mathematical similarity between the ω·t − k ∙x expression and Einstein’s four-vector product p_μx_μ= E·t − p∙x, which remains invariant under a Lorentz transformation. He also understood that the Planck-Einstein relation E = ħ·ω actually defines the energy unit and, therefore, that any frequency, any oscillation really, in space or in time, is to be expressed in terms of ħ.

[To be precise, the fundamental quantum of energy is h = ħ·2π, because that’s the energy of one cycle. To illustrate the point, think of the Planck-Einstein relation. It gives us the energy of a photon with frequency f: E_γ = h·f. If we re-write this equation as E_γ/f = h, and we do a dimensional analysis, we get: h = E_γ/f ⇔ 6.626×10⁻³⁴ joule·second = [x joule]/[f cycles per second] ⇔ h = 6.626×10⁻³⁴ joule per cycle. It’s only because we are expressing ω and k as angular frequencies (i.e. in radians per second or per meter, rather than in cycles per second or per meter) that we have to think of ħ = h/2π rather than h.]

Louis de Broglie connected the dots between some other equations too. He was fully familiar with the equations determining the phase and group velocity of composite waves, or a wavetrain that actually might represent a wavicle traveling through spacetime. In short, he boldly equated ω with ω = E/ħ and k with k = p/ħ, and all came out alright. It made perfect sense!

I’ve written enough about this. What I want to write about here is how this also makes for the situation on hand: a simple two-state system that depends on time only. So its phase is θ = ω·t = E₀/ħ. What’s E₀? It is the total energy of the system, including the equivalent energy of the particle’s rest mass and any potential energy that may be there because of the presence of one or the other force field. What about kinetic energy? Well… We said it: in this case, there is no translational or linear momentum, so p = 0. So our Platonic wavefunction reduces to:

a·e^−iθ= ae⁻⁽^{i/ħ)·(E₀·t)}

Great! […] But… Well… No! The problem with this wavefunction is that it yields a constant probability. To be precise, when we take the absolute square of this wavefunction – which is what we do when calculating a probability from a wavefunction − we get P = a², always. The ‘normalization’ condition (so that’s the condition that probabilities have to add up to one) implies that P₁ = P₂ = a² = 1/2. Makes sense, you’ll say, but the problem is that this doesn’t reflect reality: these probabilities do not evolve over time and, hence, our ammonia molecule never ‘flips’ its spin direction from ‘up’ to ‘down’, or vice versa. In short, our wavefunction does not explain reality.

The problem is not unlike the problem we’d had with a similar function relating the momentum and the position of a particle. You’ll remember it: we wrote it as a·e^−iθ= ae⁽^{i/ħ)·(p·x)}. [Note that we can write a·e^−iθ= a·e⁻⁽^{i/ħ)·(E₀·t − p·x)}= a·e⁻⁽ⁱ^/ħ)·(E^₀·t)·e⁽^{i/ħ)·(p·x)}, so we can always split our wavefunction in a ‘time’ and a ‘space’ part.] But then we found that this wavefunction also yielded a constant and equal probability all over space, which implies our particle is everywhere (and, therefore, nowhere, really).

In quantum physics, this problem is solved by introducing uncertainty. Introducing some uncertainty about the energy, or about the momentum, is mathematically equivalent to saying that we’re actually looking at a composite wave, i.e. the sum of a finite or infinite set of component waves. So we have the same ω = E/ħ and k = p/ħ relations, but we apply them to n energy levels, or to some continuous range of energy levels ΔE. It amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ.

We know what that does: it ensures our wavefunction is being ‘contained’ in some ‘envelope’. It becomes a wavetrain, or a kind of beat note, as illustrated below:

[The animation also shows the difference between the group and phase velocity: the green dot shows the group velocity, while the red dot travels at the phase velocity.]

This begs the following question: what’s the uncertainty really? Is it an uncertainty in the energy, or is it an uncertainty in the wavefunction? I mean: we have a function relating the energy to a frequency. Introducing some uncertainty about the energy is mathematically equivalent to introducing uncertainty about the frequency. Of course, the answer is: the uncertainty is in both, so it’s in the frequency and in the energy and both are related through the wavefunction. So… Well… Yes. In some way, we’re chasing our own tail. 🙂

However, the trick does the job, and perfectly so. Let me summarize what we did in the previous post: we had the ammonia molecule, i.e. an NH₃ molecule, with the nitrogen ‘flipping’ across the hydrogens from time to time, as illustrated below:

This ‘flip’ requires energy, which is why we associate two energy levels with the molecule, rather than just one. We wrote these two energy levels as E₀+ A and E₀− A. That assumption solved all of our problems. [Note that we don’t specify what the energy barrier really consists of: moving the center of mass obviously requires some energy, but it is likely that a ‘flip’ also involves overcoming some electrostatic forces, as shown by the reversal of the electric dipole moment in the illustration above.] To be specific, it gave us the following wavefunctions for the amplitude to be in the ‘up’ or ‘1’ state versus the ‘down’ or ‘2’ state respectivelly:

C₁= (1/2)·e^{−(i/ħ)·(E₀− A)·t}+ (1/2)·e^{−(i/ħ)·(E₀+ A)·t}
C₂= (1/2)·e^{−(i/ħ)·(E₀− A)·t}– (1/2)·e^{−(i/ħ)·(E₀+ A)·t}

Both are composite waves. To be precise, they are the sum of two component waves with a temporal frequency equal to ω₁= (E₀− A)/ħ and ω₁= (E₀+ A)/ħ respectively. [As for the minus sign in front of the second term in the wave equation for C₂, −1 = e^±iπ, so + (1/2)·e^{−(i/ħ)·(E₀+ A)·t}and – (1/2)·e^{−(i/ħ)·(E₀+ A)·t} are the same wavefunction: they only differ because their relative phase is shifted by ±π.] So the so-called base states of the molecule themselves are associated with two different energy levels: it’s not like one state has more energy than the other.

You’ll say: so what?

Well… Nothing. That’s it really. That’s all I wanted to say here. The absolute square of those two wavefunctions gives us those time-dependent probabilities above, i.e. the graph we started this post with. So… Well… Done!

You’ll say: where’s the ‘envelope’? Oh! Yes! Let me tell you. The C₁(t) and C₂(t) equations can be re-written as:

Now, remembering our rules for adding and subtracting complex conjugates (e^iθ + e^–iθ = 2cosθ and e^iθ − e^–iθ = 2sinθ), we can re-write this as:

So there we are! We’ve got wave equations whose temporal variation is basically defined by E₀but, on top of that, we have an envelope here: the cos(A·t/ħ) and sin(A·t/ħ) factor respectively. So their magnitude is no longer time-independent: both the phase as well as the amplitude now vary with time. The associated probabilities are the ones we plotted:

|C₁(t)|²= cos²[(A/ħ)·t], and
|C₂(t)|²= sin²[(A/ħ)·t].

So, to summarize it all once more, allowing the nitrogen atom to push its way through the three hydrogens, so as to flip to the other side, thereby breaking the energy barrier, is equivalent to associating two energy levels to the ammonia molecule as a whole, thereby introducing some uncertainty, or indefiniteness as to its energy, and that, in turn, gives us the amplitudes and probabilities that we’ve just calculated. [And you may want to note here that the probabilities “sloshing back and forth”, or “dumping into each other” – as Feynman puts it – is the result of the varying magnitudes of our amplitudes, so that’s the ‘envelope’ effect. It’s only because the magnitudes vary in time that their absolute square, i.e. the associated probability, varies too.

So… Well… That’s it. I think this and all of the previous posts served as a nice introduction to quantum physics. More in particular, I hope this post made you appreciate the mathematical framework is not as horrendous as it often seems to be.

When thinking about it, it’s actually all quite straightforward, and it surely respects Occam’s principle of parsimony in philosophical and scientific thought, also know as Occam’s Razor: “When trying to explain something, it is vain to do with more what can be done with less.” So the math we need is the math we need, really: nothing more, nothing less. As I’ve said a couple of times already, Occam would have loved the math behind QM: the physics call for the math, and the math becomes the physics.

That’s what makes it beautiful. 🙂

Post scriptum:

One might think that the addition of a term in the argument in itself would lead to a beat note and, hence, a varying probability but, no! We may look at e^{−(i/ħ)·(E₀+ A)·t}as a product of two amplitudes:

e^{−(i/ħ)·(E₀+ A)·t}= e^{−(i/ħ)·E₀·t}·e^{−(i/ħ)·A·t}

But, when writing this all out, one just gets a cos(α·t+β·t)–sin(α·t+β·t), whose absolute square |cos(α·t+β·t)–sin(α·t+β·t)|²= 1. However, writing e^{−(i/ħ)·(E₀+ A)·t}as a product of two amplitudes in itself is interesting. We multiply amplitudes when an event consists of two sub-events. For example, the amplitude for some particle to go from s to x via some point a is written as:

〈 x | s 〉_{via a} = 〈 x | a 〉〈 a | s 〉

Having said that, the graph of the product is uninteresting: the real and imaginary part of the wavefunction are a simple sine and cosine function, and their absolute square is constant, as shown below.

Adding two waves with very different frequencies – A is a fraction of E₀– gives a much more interesting pattern, like the one below, which shows an e^−iαt+e^−iβt= cos(αt)−i·sin(αt)+cos(βt)−i·sin(βt) = cos(αt)+cos(βt)−i·[sin(αt)+sin(βt)] pattern for α = 1 and β = 0.1.

That doesn’t look a beat note, does it? The graphs below, which use 0.5 and 0.01 for β respectively, are not typical beat notes either.

We get our typical ‘beat note’ only when we’re looking at a wave traveling in space, so then we involve the space variable x again, and the relations that come with in, i.e. a phase velocity v_p= ω/k = (E/ħ)/(p/ħ) = E/p = c²/v (read: all component waves travel at the same speed), and a group velocity v_g= dω/dk = v (read: the composite wave or wavetrain travels at the classical speed of our particle, so it travels with the particle, so to speak). That’s what’s I’ve shown numerous times already, but I’ll insert one more animation here, just to make sure you see what we’re talking about. [Credit for the animation goes to another site, one on acoustics, actually!]

So what’s left? Nothing much. The only thing you may want to do is to continue thinking about that wavefunction. It’s tempting to think it actually is the particle, somehow. But it isn’t. So what is it then? Well… Nobody knows, really, but I like to think it does travel with the particle. So it’s like a fundamental property of the particle. We need it every time when we try to measure something: its position, its momentum, its spin (i.e. angular momentum) or, in the example of our ammonia molecule, its orientation in space. So the funny thing is that, in quantum mechanics,

We can measure probabilities only, so there’s always some randomness. That’s how Nature works: we don’t really know what’s happening. We don’t know the internal wheels and gears, so to speak, or the ‘hidden variables’, as one interpretation of quantum mechanics would say. In fact, the most commonly accepted interpretation of quantum mechanics says there are no ‘hidden variables’.
But then, as Polonius famously put, there is a method in this madness, and the pioneers – I mean Werner Heisenberg, Louis de Broglie, Niels Bohr, Paul Dirac, etcetera – discovered. All probabilities can be found by taking the square of the absolute value of a complex-valued wavefunction (often denoted by Ψ), whose argument, or phase (θ), is given by the de Broglie relations ω = E/ħ and k = p/ħ:

θ = (ω·t − k ∙x) = (E/ħ)·t − (p/ħ)·x

That should be obvious by now, as I’ve written dozens of posts on this by now. 🙂 I still have trouble interpreting this, however—and I am not ashamed, because the Great Ones I just mentioned have trouble with that too. But let’s try to go as far as we can by making a few remarks:

Adding two terms in math implies the two terms should have the same dimension: we can only add apples to apples, and oranges to oranges. We shouldn’t mix them. Now, the (E/ħ)·t and (p/ħ)·x terms are actually dimensionless: they are pure numbers. So that’s even better. Just check it: energy is expressed in newton·meter (force over distance, remember?) or electronvolts (1 eV = 1.6×10⁻¹⁹J = 1.6×10⁻¹⁹N·m); Planck’s constant, as the quantum of action, is expressed in J·s or eV·s; and the unit of (linear) momentum is 1 N·s = 1 kg·m/s = 1 N·s. E/ħ gives a number expressed per second, and p/ħ a number expressed per meter. Therefore, multiplying it by t and x respectively gives us a dimensionless number indeed.
It’s also an invariant number, which means we’ll always get the same value for it. As mentioned above, that’s because the four-vector product p_μx_μ= E·t − p∙x is invariant: it doesn’t change when analyzing a phenomenon in one reference frame (e.g. our inertial reference frame) or another (i.e. in a moving frame).
Now, Planck’s quantum of action h or ħ (they only differ in their dimension: h is measured in cycles per second and ħ is measured in radians per second) is the quantum of energy really. Indeed, if “energy is the currency of the Universe”, and it’s real and/or virtual photons who are exchanging it, then it’s good to know the currency unit is h, i.e. the energy that’s associated with one cycle of a photon.
It’s not only time and space that are related, as evidenced by the fact that t − x itself is an invariant four-vector, E and p are related too, of course! They are related through the classical velocity of the particle that we’re looking at: E/p = c²/v and, therefore, we can write: E·β = p·c, with β = v/c, i.e. the relative velocity of our particle, as measured as a ratio of the speed of light. Now, I should add that the t − x four-vector is invariant only if we measure time and space in equivalent units. Otherwise, we have to write c·t − x. If we do that, so our unit of distance becomes c meter, rather than one meter, or our unit of time becomes the time that is needed for light to travel one meter, then c = 1, and the E·β = p·c becomes E·β = p, which we also write as β = p/E: the ratio of the energy and the momentum of our particle is its (relative) velocity.

θ = (ω·t − k ∙x) = E·t − p·x

Now, θ is the argument of a wavefunction, and we can always re-scale such argument by multiplying or dividing it by some constant. It’s just like writing the argument of a wavefunction as v·t–x or (v·t–x)/v = t –x/v with v the velocity of the waveform that we happen to be looking at. [In case you have trouble following this argument, please check the post I did for my kids on waves and wavefunctions.] Now, the energy conservation principle tells us the energy of a free particle won’t change. [Just to remind you, a ‘free particle’ means it is present in a ‘field-free’ space, so our particle is in a region of uniform potential.] You see what I am going to do now: we can, in this case, treat E as a constant, and divide E·t − p·x by E, so we get a re-scaled phase for our wavefunction, which I’ll write as:

φ = (E·t − p·x)/E = t − (p/E)·x = t − β·x

Now that’s the argument of a wavefunction with the argument expressed in distance units. Alternatively, we could also look at p as some constant, as there is no variation in potential energy that will cause a change in momentum, i.e. in kinetic energy. We’d then divide by p and we’d get (E·t − p·x)/p = (E/p)·t − x) = t/β − x, which amounts to the same, as we can always re-scale by multiplying it with β, which would then yield the same t − β·x argument.

Φ(φ) = a·e^−iφ= a·e^{−i(t − β·x)}

Of course, the analysis above does not include uncertainty. Our information on the energy and the momentum of our particle will be incomplete: we’ll write E = E₀± σ_E, and p = p₀± σ_p. [I am a bit tired of using the Δ symbol, so I am using the σ symbol here, which denotes a standard deviation of some density function. It underlines the probabilistic, or statistical, nature of our approach.] But, including that, we’ve pretty much explained what quantum physics is about here.

You just need to get used to that complex exponential: e^−iφ = cos(−φ) + i·sin(−φ) = cos(φ) − i·sin(φ). Of course, it would have been nice if Nature would have given us a simple sine or cosine function. [Remember the sine and cosine function are actually the same, except for a phase difference of 90 degrees: sin(φ) = cos(π/2−φ) = cos(φ+π/2). So we can go always from one to the other by shifting the origin of our axis.] But… Well… As we’ve shown so many times already, a real-valued wavefunction doesn’t explain the interference we observe, be it interference of electrons or whatever other particles or, for that matter, the interference of electromagnetic waves itself, which, as you know, we also need to look at as a stream of photons , i.e. light quanta, rather than as some kind of infinitely flexible aether that’s undulating, like water or air.

So… Well… Just accept that e^−iφ is a very simple periodic function, consisting of two sine waves rather than just one, as illustrated below.

And then you need to think of stuff like this (the animation is taken from Wikipedia), but then with a projection of the sine of those phasors too. It’s all great fun, so I’ll let you play with it now. 🙂

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Quantum math: states as vectors, and apparatuses as operators

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. 🙂

Original post:

I actually wanted to write about the Hamiltonian matrix. However, I realize that, before I can serve the plat de résistance, we need to review or introduce some more concepts and ideas. It all revolves around the same theme: working with states is like working with vectors, but so you need to know how exactly. Let’s go for it. 🙂

In my previous posts, I repeatedly said that a set of base states is like a coordinate system. A coordinate system allows us to describe (i.e. uniquely identify) vectors in an n-dimensional space: we associate a vector with a set of real numbers, like x, y and z, for example. Likewise, we can describe any state in terms of a set of complex numbers – amplitudes, really – once we’ve chosen a set of base states. We referred to this set of base states as a ‘representation’. For example, if our set of base states is +S, 0S and −S, then any state φ can be defined by the amplitudes C₊ = 〈 +S | φ 〉, C₀ = 〈 0S | φ 〉, and C₋ = 〈 −S | φ 〉.

We have to choose some representation (but we are free to choose which one) because, as I demonstrated when doing a practical example (see my description of muon decay in my post on how to work with amplitudes), we’ll usually want to calculate something like the amplitude to go from one state to another – which we denoted as 〈 χ | φ 〉 – and we’ll do that by breaking it up. To be precise, we’ll write that amplitude 〈 χ | φ 〉 – i.e. the amplitude to go from state φ to state χ (you have to read this thing from right to left, like Hebrew or Arab) – as the following sum:

So that’s a sum over a complete set of base states (that’s why I write all i under the summation symbol ∑). We discussed this rule in our presentation of the ‘Laws’ of quantum math.

Now we can play with this. As χ can be defined in terms of the chosen set of base states too, it’s handy to know that 〈 χ | i 〉 and 〈 i | χ 〉 are each other’s complex conjugates – we write this as: 〈 χ | i 〉 = 〈 i | χ 〉* – so if we have one, we have the other (we can also write: 〈 i | χ 〉* = 〈 χ | i 〉). In other words, if we have all C_i = 〈 i | φ 〉 and all D_i = 〈 i | χ 〉, i.e. the ‘components’ of both states in terms of our base states, then we can calculate 〈 χ | φ 〉 as:

〈 χ | φ 〉 = ∑ D_i*C_i = ∑〈 χ | i 〉〈 i | φ 〉,

provided we make sure we do the summation over a complete set of base states. For example, if we’re looking at the angular momentum of a spin-1/2 particle, like an electron or a proton, then we’ll have two base states, +ħ/2 and +ħ/2, so then we’ll have only two terms in our sum, but the spin number (j) of a cobalt nucleus is 7/2, so if we’d be looking at the angular momentum of a cobalt nucleus, we’ll have eight (2·j + 1) base states and, hence, eight terms when doing the sum. So it’s very much like working with vectors, indeed, and that’s why states are often referred to as state vectors. So now you know that term too. 🙂

However, the similarities run even deeper, and we’ll explore all of them in this post. You may or may not remember that your math teacher actually also defined ordinary vectors in three-dimensional space in terms of base vectors e_i, defined as: e₁= [1, 0, 0], e₂= [0, 1, 0] and e₂= [0, 0, 1]. You may also remember that the units along the x, y and z-axis didn’t have to be the same – we could, for example, measure in cm along the x-axis, but in inches along the z-axis, even if that’s not very convenient to calculate stuff – but that it was very important to ensure that the base vectors were a set of orthogonal vectors. In any case, we’d chose our set of orthogonal base vectors and write all of our vectors as:

A = A_x·e₁ + A_y·e₂+ A_z·e₃

That’s simple enough. In fact, one might say that the equation above actually defines coordinates. However, there’s another way of defining them. We can write A_x, A_y, and A_z as vector dot products, aka scalar vector products (as opposed to cross products, or vector products tout court). Check it:

A_x= A·e₁, A_y= A·e₂, and A_z= A·e₃.

This actually allows us to re-write the vector dot product A·B in a way you’ve probably haven’t seen before. Indeed, you’d usually calculate A·B as |A|∙|B|·cosθ = A∙B·cosθ (A and B is the magnitude of the vectors A and B respectively) or, quite simply, as A_xB_x+ A_yB_y+ A_zB_z. However, using the dot products above, we can now also write it as:

We deliberately wrote B·A instead of A∙B because, while the mathematical similarity with the

〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | φ 〉

equation is obvious, B·A = A·B but 〈 χ | φ 〉 ≠ 〈 φ | χ 〉. Indeed, 〈 χ | φ 〉 and 〈 φ | χ 〉 are complex conjugates – so 〈 χ | φ 〉 = 〈 φ | χ 〉* – but they’re not equal. So we’ll have to watch the order when working with those amplitudes. That’s because we’re working with complex numbers instead of real numbers. Indeed, it’s only because the A·B dot product involves real numbers, whose complex conjugate is the same, that we have that commutativity in the real vector space. Apart from that – so apart from having to carefully check the order of our products – the correspondence is complete.

Let me mention another similarity here. As mentioned above, our base vectors e_i had to be orthogonal. We can write this condition as:

e_i·e_j = δ_ij, with δ_ij= 0 if i ≠ j, and 1 if i = j.

Now, our first quantum-mechanical rule says the same:

〈 i | j 〉 = δ_ij, with δ_ij= 0 if i ≠ j, and 1 if i = j.

So our set of base states also has to be ‘orthogonal’, which is the term you’ll find in physics textbooks, although – as evidenced from our discussion on the base states for measuring angular momentum – one should not try to give any geometrical interpretation here: +ħ/2 and +ħ/2 (so that’s spin ‘up’ and ‘down’ respectively) are not ‘orthogonal’ in any geometric sense, indeed. It’s just that pure states, i.e. base states, are separate, which we write as: 〈 ‘up’ | ‘down’ 〉 = 〈 ‘down’ | ‘up’ 〉 = 0 and 〈 ‘up’ | ‘up’ 〉 = 〈 ‘down’ | ‘down’ 〉 = 1. It just means they are just different base states, and so it’s one or the other. For our +S, 0S and −S example, we’d have nine such amplitudes, and we can organize them in a little matrix:

In fact, just like we defined the base vectors e_i as e₁= [1, 0, 0], e₂= [0, 1, 0] and e₂= [0, 0, 1] respectively, we may say that the matrix above, which states exactly the same as the 〈 i | j 〉 = δ_ij rule, can serve as a definition of what base states actually are. [Having said that, it’s obvious we like to believe that base states are more than just mathematical constructs: we’re talking reality here. The angular momentum as measured in the x-, y- or z-direction, or in whatever direction, is more than just a number.]

OK. You get this. In fact, you’re probably getting impatient because this is too simple for you. So let’s take another step. We showed that the 〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | χ 〉 and B·A = ∑(B·e_i)(e_i·A) are structurally equivalent – from a mathematical point of view, that is – but B and A are separate vectors, while 〈 χ | φ 〉 is just a complex number. Right?

Well… No. We can actually analyze the bra and the ket in the 〈 χ | φ 〉 bra-ket as separate pieces too. Moreover, we’ll show they are actually state vectors too, even if the bra, i.e. 〈 χ |, and the ket, i.e. | φ 〉, are ‘unfinished pieces’, so to speak. Let’s be bold. Let’s just cut the 〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | χ 〉 by writing:

Huh?

Yes. That’s the power of Dirac’s bra-ket notation: we can just drop symbols left or right. It’s quite incredible. But, of course, the question is: so what does this actually mean? Well… Don’t rack your brain. I’ll tell you. We define | φ 〉 as a state vector because we define | i 〉 as a (base) state vector. Look at it this way: we wrote the 〈 +S | φ 〉, 〈 0S | φ 〉 and 〈 −S | φ 〉 amplitudes as C₊, C₀, C₋, respectively, so we can write the equation above as:

So we’ve got a sum of products here, and it’s just like A = A_x·e₁+ A_y·e₂ + A_z·e₃. Just substitute the A_icoefficients for C_i and the e_ibase vectors for the | i 〉 base states. We get:

| φ 〉 = |+S〉 C₊ + |0S〉 C₀+ |+S〉 C₋

Of course, you’ll wonder what those terms mean: what does it mean to ‘multiply’ C₊ (remember: C₊ is some complex number) by |+S〉? Be patient. Just wait. You’ll understand when we do some examples, so when you start working with this stuff. You’ll see it all makes sense—later. 🙂

Of course, we’ll have a similar equation for | χ 〉, and so if we write 〈 χ | i 〉 as D_i, then we can write | χ 〉 = ∑ | i 〉〈 χ | i 〉 as | χ 〉 = ∑ | i 〉 D_i.

So what? Again: be patient. We know that 〈 χ | i 〉 = 〈 i | χ 〉*, so our second equation above becomes:

You’ll have two questions now. The first is the same as the one above: what does it mean to ‘multiply’, let’s say, D₀* (i.e. the complex conjugate of D₀, so if D₀= a + ib, then D₀* = a − ib) with 〈0S|? The answer is the same: be patient. 🙂 Your second question is: why do I use another symbol for the index here? Why j instead of i? Well… We’ll have to re-combine stuff, so it’s better to keep things separate by using another symbol for the same index. 🙂

In fact, let’s re-combine stuff right now, in exactly the same way as we took it apart: we just write the two things right next to each other. We get the following:

What? Is that it? So we went through all of this hocus-pocus just to find the same equation as we started out with?

Yes. I had to take you through this so you get used to juggling all those symbols, because that’s what we’ll do in the next post. Just think about it and give yourself some time. I know you’ve probably never ever handled such exercise in symbols before – I haven’t, for sure! – but it all makes sense: we cut and paste. It’s all great! 🙂 [Oh… In case you wonder about the transition from the sum involving i and j to the sum involving i only, think about the Kronecker expression: 〈 j | i 〉 = δ_ij, with δ_ij= 0 if i ≠ j, and 1 if i = j, so most of the terms are zero.]

To summarize the whole discussion, note that the expression above is completely analogous with the B·A = B_xA_x+ B_yA_y+ B_zA_zformula. The only difference is that we’re talking complex numbers here, so we need to watch out. We have to watch the order of stuff, and we can’t use the D_inumbers themselves: we have to use their complex conjugates D_i*. But, for the rest, we’re all set! 🙂 If we’ve got a set of base states, then we can define any state in terms of a set of ‘coordinates’ or ‘coefficients’ – i.e. the C_i or D_i numbers for the φ or χ example above – and we can then calculate the amplitude to go from one state to another as:

In case you’d get confused, just take the original equation:

The two equations are fully equivalent.

[…]

So we just went through all of the shit above so as to show that structural similarity with vector spaces?

Yes. It’s important. You just need to remember that we may have two, three, four, five,… or even an infinite number of base states depending on the situation we’re looking at, and what we’re trying to measure. I am sorry I had to take you through all of this. However, there’s more to come, and so you need this baggage. We’ll take the next step now, and that is to introduce the concept of an operator.

Look at the middle term in that expression above—let me copy it:

We’ve got three terms in that double sum (a double sum is a sum involving two indices, which is what we have here: i and j). When we have two indices like that, one thinks of matrices. That’s easy to do here, because we represented that 〈 i | j 〉 = δ_ij equation as a matrix too! To be precise, we presented it as the identity matrix, and a simple substitution allows us to re-write our equation above as:

I must assume you’re shaking your head in disbelief now: we’ve expanded a simple amplitude into a product of three matrices now. Couldn’t we just stick to that sum, i.e that vector dot product ∑ D_i*C_i? What’s next? Well… I am afraid there’s a lot more to come. For starters, we’ll take that idea of ‘putting something in the middle’ to the next level by going back to our Stern-Gerlach filters and whatever other apparatus we can think of. Let’s assume that, instead of some filter S or T, we’ve got something more complex now, which we’ll denote by A. [Don’t confuse it with our vectors: we’re talking an apparatus now, so you should imagine some beam of particles, polarized or not, entering it, going through, and coming out.]

We’ll stick to the symbols we used already, and so we’ll just assume a particle enters into the apparatus in some state φ, and that it comes out in some state χ. Continuing the example of spin-one particles, and assuming our beam has not been filtered – so, using lingo, we’d say it’s unpolarized – we’d say there’s a probability of 1/3 for being either in the ‘plus’, ‘zero’, or ‘minus’ state with respect to whatever representation we’d happen to be working with, and the related amplitudes would be 1/√3. In other words, we’d say that φ is defined by C₊ = 〈 +S | φ 〉, C₀ = 〈 0S | φ 〉, and C₋ = 〈 −S | φ 〉, with C₊ = C₀ = C₋= 1/√3. In fact, using that | φ 〉 = |+S〉 C₊ + |0S〉 C₀+ |+S〉 C₋expression we invented above, we’d write: | φ 〉 = (1/√3)|+S〉 + (1/√3)|0S〉 C₀+ (1/√3)|+S〉 C₋ or, using ‘matrices’—just a row and a column, really:

However, you don’t need to worry about that now. The new big thing is the following expression:

〈 χ | A | φ〉

It looks simple enough: φ to A to χ. Right? Well… Yes and no. The question is: what do you do with this? How would we take its complex conjugate, for example? And if we know how to do that, would it be equal to 〈 φ | A | χ〉?

You guessed it: we’ll have to take it apart, but how? We’ll do this using another fantastic abstraction. Remember how we took Dirac’s 〈 χ | φ 〉 bra-ket apart by writing | φ 〉 = ∑ | i 〉〈 i | φ 〉? We just dropped the 〈 χ left and right in our 〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | φ 〉 expression. We can go one step further now, and drop the φ 〉 left and right in our | φ 〉 = ∑ | i 〉〈 i | φ 〉 expression. We get the following wonderful thing:

| = ∑ | i 〉〈 i | over all base states i

With characteristic humor, Feynman calls this ‘The Great Law of Quantum Mechanics’ and, frankly, there’s actually more than one grain of truth in this. 🙂

Now, if we apply this ‘Great Law’ to our 〈 χ | A | φ〉 expression – we should apply it twice, actually – we get:

As Feynman points out, it’s easy to add another apparatus in series. We just write:

Just put a | bar between B and A and apply the same trick. The | bar is really like a factor 1 in multiplication. However, that’s all great fun but it doesn’t solve our problem. Our ‘Great Law’ allows us to sort of ‘resolve’ our apparatus A in terms of base states, as we now have 〈 i | A | j 〉 in the middle, rather than 〈 χ | A | φ〉 but, again, how do we work with that?

Well… The answer will surprise you. Rather than trying to break this thing up, we’ll say that the apparatus A is actually being described, or defined, by the nine 〈 i | A | j 〉 amplitudes. [There are nine for this example, but four only for the example involving spin-1/2 particles, of course.] We’ll call those amplitudes, quite simply, the matrix of amplitudes, and we’ll often denote it by A_ij.

Now, I wanted to talk about operators here. The idea of an operator comes up when we’re creative again, and when we drop the 〈 χ | state from the 〈 χ | A | φ〉 expression. We write:

So now we think of the particle entering the ‘apparatus’ A in the state ϕ and coming out of A in some state ψ (‘psi’). We can generalize this and think of it as an ‘operator’, which Feynman intuitively defines as follows:

The symbol A is neither an amplitude, nor a vector; it is a new kind of thing called an operator. It is something which “operates on” a state to produce a new state.”

But… Wait a minute! | ψ 〉 is not the same as 〈 χ |. Why can we do that substitution? We can only do it because any state ψ and χ are related through that other ‘Law’ of quantum math:

Combining the two shows our ‘definition’ of an operator is OK. We should just note that it’s an ‘open’ equation until it is completed with a ‘bra’, i.e. a state like 〈 χ |, so as to give the 〈 χ | ψ〉 = 〈 χ | A | φ〉 type of amplitude that actually means something. In practical terms, that means our operator or our apparatus doesn’t mean much as long as we don’t measure what comes out, so then we choose some set of base states, i.e. a representation, which allows us to describe the final state, i.e. 〈 χ |.

[…]

Well… Folks, that’s it. I know this was mighty abstract, but the next posts should bring things back to earth again. I realize it’s only by working examples and doing exercises that one can get some kind of ‘feel’ for this kind of stuff, so that’s what we’ll have to go through now. 🙂

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Taking the magic out of God’s number: some additional reflections

Note: I have published a paper that is very coherent and fully explains this so-called God-given number. There is nothing magical about it. It is just a scaling constant. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

In my previous post, I explained why the fine-structure constant α is not a ‘magical’ number, even if it relates all fundamental properties of the electron: its mass, its energy, its charge, its radius, its photon scattering cross-section (i.e. the Bohr radius, or the size of the atom really) and, finally, the coupling constant for photon-electron interactions. The key to such understanding of α was the model of an electron as a tiny ball of charge. As such, we have two energy formulas for it. One is the energy that’s needed to assemble the charge from infinitely dispersed infinitesimal charges, which we denoted as U_elec. The other formula is the energy of the field of the tiny ball of charge, which we denoted as E_elec.

The formula for E_elec is calculated using the formula for the field momentum of a moving charge and, using the m = E/c²mas-energy equivalence relationship, is equivalent to the electromagnetic mass. We went through the derivation in our previous post, so let me just jot down the result:

The second formula depends on what ball of charge we’re thinking of, because the formulas for a charged sphere and a spherical shell of charge are different: both have the same structure as the relationship above (so the energy is also proportional to the square of the electron charge and inversely proportional to the radius a), but the constant of proportionality is different. For a sphere of charge, we write:

For a spherical shell of charge we write:

shell

To compare the formulas, you need to note that the square of the electron charge e in the formula for the field energy is equal to e²= q_e²/4πε₀= k_e·q_e². So we multiply the square of the actual electron charge by the Coulomb constant k_e= 1/4πε₀. As you can see, the three formulas have exactly the same form then. It’s just the proportionality constant that’s different: it’s 2/3, 3/5 and 1/2 respectively. It’s interesting to quickly reflect on the dimensions here: [k_e] ≈ 9×10⁹N·m²/C², so e² is expressed in N·m². That makes the units come out alright, as we divide by a (so that’s in meter) and so we get the energy in joule (which is newton·meter). In fact, now that we’re here, let’s quickly calculate the value of e²: it’s that k_e·q_e² product, so it’s equal to 2.3×10⁻²⁸N·m². We can quickly check this value because we know that the classical electron radius is equal to:

So we divide 2.3×10⁻²⁸N·m²by m_ec_²≈ 8.2×10⁻¹⁴J, so we get r₀≈ 2.82×10⁻¹⁵m. So we’re spot on! Why did I do this check? Not really to check what I wrote. It’s more to show what’s going on. We’ve got yet another formula relating the energy and the radius of an electron here, so now we have three. In fact we have more because the formula for U_elec depends on the finer details of our model for the electron (sphere versus shell, uniform versus non-uniform distribution):

E_elec= (2/3)·(e²/a): This is the formula for the energy of the field, so we may all it is external energy.
U_elec= (3/5)·(e²/a), or U_elec= (1/2)·(e²/a): This is the energy needed to assemble our electron, so we might, perhaps, call it its internal energy. The first formula assumes our electron is a uniformly charged sphere. The second assumes all charges sit on the surface of the sphere. If we drop the assumption of the charge having to be uniformly distributed, we’ll find yet another formula.
m_ec²= e²/r₀: This is the energy associated with the so-called classical electron radius (r₀) and the electron’s rest mass (m_e).

In our previous posts, we assumed the last equation was the right one. Why? Because it’s the one that’s been verified experimentally. The discrepancies between the various proportionality coefficients – i.e. the difference between 2/3 and 1, basically – are to be explained because of the binding forces within the electron, without which the electron would just ‘explode’, as the French physicist and polymath Henri Poincaré famously put it. Indeed, if the electron is a little ball of negative charge, the repulsive forces between its parts should rip it apart. So we will not say anything more about this. You can have fun yourself by googling all the various theories that try to model these binding forces. [I may do the same some day, but now I’ve got other priorities: I want to move to Feynman’s third volume of Lectures, which is devoted to quantum physics only, so I look very much forward to that.]

In this post, I just wanted to reflect once more on what constants are really fundamental and what constants are somewhat less fundamental. From all what I wrote in my previous post, I said there were three:

The fine-structure constant α, which is a dimensionless number.
Planck’s constant h, whose dimension is joule·second, so that’s the dimension of action.
The speed of light c, whose dimension is that of a velocity.

The three are related through the following expression:

This is an interesting expression. Let’s first check its dimension. We already explained that e_² is expressed in N·m². That’s rather strange, because it means the dimension of e itself is N^1/2·m: what’s the square root of a force of one newton? In fact, to interpret the formula above, it’s probably better to re-write e²as e²= q_e²/4πε₀= k_e·q_e². That shows you how the electron charge and Coulomb’s constant are related. Of course, they are part and parcel of one and the same force law: Coulomb’s law. We don’t need anything else, except for relativity theory, because we need to explain the magnetic force as well—and that we can do because magnetism is just a relativistic effect. Think of the field momentum indeed: the magnetic field comes into play only when we start to move our electron. The relativity effect is captured by c in that formula for α above. As for ħ, ħ = h/2π comes with the E = h·f equation, which links us to the electron’s Compton wavelength λ through the de Broglie relation λ = h/p.

The point is: we should probably not look at α as a ‘fundamental physical constant’. It’s e² that’s the third fundamental constant, besides h and c. Indeed, it’s from e² that all the rest follows: the electron’s internal energy, its external energy, and its radius, and then all the rest by combining stuff with other stuff.

Now, we took the magic out of α by doing what we did in the previous posts, and that’s to combine stuff with other stuff, and so now you may think I am putting the magic back in with that formula for α, which seems to define α in terms of the three mentioned ‘fundamental’ constants. That’s not the case: this relation comes out of all of the other relationships we found, and so it’s nothing new really. It’s actually not a definition of α: it just does what it does, and that’s to relate α to the ‘fundamental’ physical constants behind.

So… No new magic. In fact, I want to close this post by taking away even more of the magic. If you read my previous post, I said that α was ‘God’s cut-off factor’ 🙂 ensuring our energy functions do not blow up, but I also said it was impossible to say why he chose 0.00729735256 as the cut-off factor. The question is actually easily answered by thinking about those two formulas we had for the internal and external energy respectively. Let’s re-write them in natural units and, temporarily, two different subscripts for α, so we write:

E_elec= α_e/r₀: This is the formula for the energy of the field.
U_elec= α_u/r₀: This is the energy needed to assemble our electron.

Both energies are determined by the above-mentioned laws, i.e. Coulomb’s Law and the theory of relativity, so α has got nothing to do what that. However, both energies have to be the same, and so α_ehas to be equal to α_u. In that sense, α is, quite simply, a proportionality constant that achieves that equality. Now that explains why we can derive α from the three other constants which, as mentioned above, are probably more fundamental. In fact, we’ve got only three degrees of freedom here, so if we chose c, h and e as ‘fundamental’, then α isn’t any more.

The underlying deep question behind it all is why those two energies should be equal. Why would our electron have some internal energy if it’s elementary? The answer to that question is: because it has some non-zero radius, and it has some non-zero radius because we don’t want our formula for the field energy (or the field momentum) to blow up. Now, if it has some radius, then it has to have some internal energy.

You’ll say: that makes sense, but it doesn’t answer the question. Why would it have internal energy, with or without a zero radius? If an electron is an elementary particle, then it’s really elementary, isn’t? And so then we shouldn’t try to ‘assemble’ it from an infinite number of infinitesimally small charges. You’re right, and here we can also note that the fact that the electron doesn’t blow up is firm evidence it’s very elementary indeed.

I should also note that Feynman actually doesn’t talk about the energy that’s needed to assemble a charge: he gets his U_elec= (1/2)·(e²/a) by calculating the external field energy for a spherical shell of charge, and he sticks to it—presumably because it’s the same field for a uniform or non-uniform sphere of charge. He only notes there has to be some radius because, if not, the formula he uses blows up, indeed. So – who knows? – perhaps he doesn’t quite believe that formula for the internal energy is relevant either.

So perhaps there is no internal energy indeed. Perhaps there’s just the energy of the field. So… Well… I can’t say much about this… Except… Well… Perhaps just one more thing. Let me note something that, I hope, you noticed as well: the k_e·q_e²is the numerator in Coulomb’s Law itself. You also know that energy equals force times distance. So if we divide both sides by r₀, we get Coulomb’s Law itself F_elec= k_e·q_e²/r₀². The only thing is: what’s the distance? It’s one charge only, and there is no distance between one charge, is there? Well… Yes and no. I have been thinking that the requirement of the internal and external energies being equal resembles the statement that the forces between two charges are equal and opposite. That ties in with the idea of the internal energy itself: remember we were basically talking forces between infinitesimally small elements of charge within the electron itself? So r₀ is, perhaps, some average distance or so. There must be some way of thinking of it like that. But… Well… Which one exactly?

This kind of reflection may not make sense. Who knows? I obviously need to think all of this through and so this post is, indeed, just a bunch of reflections for which I will have more time later—hopefully. 🙂 Perhaps we’re all just pushing the matter too far. Perhaps we should just accept that the external energy has that 2/3 factor but that the actual energy of the electron should also include the equivalent energy of some binding force that holds the electron together. Well… In any case. That’s all I am going to do on this extremely complicated matter. It’s time to move indeed! So the point to take home here is probably just this:

When calculating the radius of an electron using classical theory, we get in trouble: not only do we find different radii, but the radii that we find do not respect the E = m_ec²law. It’s only the m_ec²= e²/r₀ that’s relativistically correct.
That suggests the electron also has some non-electric mass, which are referred to as ‘binding forces’ or ‘Poincaré stresses’, but which remain to be explained convincingly.
All of this shouldn’t surprise us: for all we know, the electron is something fuzzy. 🙂

So my next posts will focus on the ‘essentials’ preparing for Feynman’s Volume on quantum mechanics. Those ‘essentials’ will still involve some classical stuff but, as you will see, even more contradictions, that – hopefully! – will then be solved in the quantum-mechanical picture of it all. 🙂

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Basics

The Poynting vector for the matter-wave

Foreword

The argument of the wavefunction as the proper time

What’s wrong with Schrödinger’s equation?

The missing energy

The hidden energy as imaginary energy

The imaginary energy space

So what about the 1/2 factor?

The quantum of action and natural units

The de Broglie relations

Back to Planck units

The Uncertainty Principle

The Schwarzschild radius and the Planck length

The wavefunction as an oscillation in and of spacetime

Fiddling with E and B

Back to the wavefunction

The quantum vacuum

The energy of a zero-mass particle

The quantization of energy and the zero-mass particle

Is our zero-mass particle just the electric field vector?

The wavefunction

Schrödinger’s equation and the two de Broglie relations

The operator concept

Some remarks on notation

Matrix mechanics

Psi-chology

Quantum-mechanical operators

The energy operator (H)

The average value operator (A)

The energy operator for wavefunctions (H)

The position operator (x)

The momentum operator (px)

The wavefunction

Energy and momentum in the wavefunction

Re-visiting the de Broglie relations

The wavefunction and relativistic length contraction

The two-state system in free space

The two-state system in a field

Switching to another representation

Intermezzo: on approximations

Solving the equations

Introducing uncertainty

Thinking about uncertainty

**The de Broglie relations**

**Schrödinger’s equation and the two de Broglie relations**

The momentum operator (p_x)

**Re-visiting the de Broglie relations**