Quantum-mechanical operators

I wrote a post on quantum-mechanical operators some while ago but, when re-reading it now, I am not very happy about it, because it tries to cover too much ground in one go. In essence, I regret my attempt to constantly switch between the matrix representation of quantum physics – with the | state 〉 symbols – and the wavefunction approach, so as to show how the operators work for both cases. But then that’s how Feynman approaches this.

However, let’s admit it: while Heisenberg’s matrix approach is equivalent to Schrödinger’s wavefunction approach – and while it’s the only approach that works well for n-state systems – the wavefunction approach is more intuitive, because:

  1. Most practical examples of quantum-mechanical systems (like the description of the electron orbitals of an atomic system) involve continuous coordinate spaces, so we have an infinite number of states and, hence, we need to describe it using the wavefunction approach.
  2. Most of us are much better-versed in using derivatives and integrals, as opposed to matrix operations.
  3. A more intuitive statement of the same argument above is the following: the idea of one state flowing into another, rather than being transformed through some matrix, is much more appealing. 🙂

So let’s stick to the wavefunction approach here. So, while you need to remember that there’s a ‘matrix equivalent’ for each of the equations we’re going to use in this post, we’re not going to talk about it.

The operator idea

In classical physics – high school physics, really – we would describe a pointlike particle traveling in space by a function relating its position (x) to time (t): x = x(t). Its (instantaneous) velocity is, obviously, v(t) = dx/dt. Simple. Obvious. Let’s complicate matters now by saying that the idea of a velocity operator would sort of generalize the v(t) = dx/dt velocity equation by making abstraction of the specifics of the x = x(t) function.

Huh? Yes. We could define a velocity ‘operator’ as:

velocity operator

Now, you may think that’s a rather ridiculous way to describe what an operator does, but – in essence – it’s correct. We have some function – describing an elementary particle, or a system, or an aspect of the system – and then we have some operator, which we apply to our function, to extract the information from it that we want: its velocity, its momentum, its energy. Whatever. Hence, in quantum physics, we have an energy operator, a position operator, a momentum operator, an angular momentum operator and… Well… I guess I listed the most important ones. 🙂

It’s kinda logical. Our velocity operator looks at one particular aspect of whatever it is that’s going on: the time rate of change of position. We do refer to that as the velocity. Our quantum-mechanical operators do the same: they look at one aspect of what’s being described by the wavefunction. [At this point, you may wonder what the other properties of our classical ‘system’ – i.e. other properties than velocity – because we’re just looking at a pointlike particle here, but… Well… Think of electric charge and forces acting on it, so it accelerates and decelerates in all kinds of ways, and we have kinetic and potential energy and all that. Or momentum. So it’s just the same: the x = x(t) function may cover a lot of complexities, just like the wavefunction does!]

The Wikipedia article on the momentum operator is, for a change (I usually find Wikipedia quite abstruse on these matters), quite simple – and, therefore – quite enlightening here. It applies the following simple logic to the elementary wavefunction ψ = ei·(ω·t − k∙x), with the de Broglie relations telling us that ω = E/ħ and k = p/ħ:

mom op 1

Note we forget about the normalization coefficient a here. It doesn’t matter: we can always stuff it in later. The point to note is that we can sort of forget about ψ (or abstract away from it—as mathematicians and physicists would say) by defining the momentum operator, which we’ll write as:

mom op 2

Its three-dimensional equivalent is calculated in very much the same way:

wiki

So this operator, when operating on a particular wavefunction, gives us the (expected) momentum when we would actually catch our particle there, provided the momentum doesn’t vary in time. [Note that it may – and actually is likely to – vary in space!]

So that’s the basic idea of an operator. However, the comparison goes further. Indeed, a superficial reading of what operators are all about gives you the impression we get all these observables (or properties of the system) just by applying the operator to the (wave)function. That’s not the case. There is the randomness. The uncertainty. Actual wavefunctions are superpositions of several elementary waves with various coefficients representing their amplitudes. So we need averages, or expected values: E[X] Even our velocity operator ∂/∂t – in the classical world – gives us an instantaneous velocity only. To get the average velocity (in quantum mechanics, we’ll be interested in the the average momentum, or the average position, or the average energy – rather than the average velocity), we’re going to have the calculate the total distance traveled. Now, that’s going to involve a line integral:

= ∫ds.

The principle is illustrated below.

line integral

You’ll say: this is kids stuff, and it is. Just note how we write the same integral in terms of the x and t coordinate, and using our new velocity operator:

integral

Kids stuff. Yes. But it’s good to think about what it represents really. For example, the simplest quantum-mechanical operator is the position operator. It’s just for the x-coordinate, for the y-coordinate, and z for the z-coordinate. To get the average position of a stationary particle – represented by the wavefunction ψ(r, t) – in three-dimensional space, we need to calculate the following volume integral:

position operator 3D V2

Simple? Yes and no. The r·|ψ(r)|2 integrand is obvious: we multiply each possible position (r) by its probability (or likelihood), which is equal to P(r) = |ψ(r)|2. However, look at the assumptions: we already omitted the time variable. Hence, the particle we’re describing here must be stationary, indeed! So we’ll need to re-visit the whole subject allowing for averages to change with time. We’ll do that later. I just wanted to show you that those integrals – even with very simple operators, like the position operator – can become very complicated. So you just need to make sure you know what you’re looking at.

One wavefunction—or two? Or more?

There is another reason why, with the immeasurable benefit of hindsight, I now feel that my earlier post is confusing: I kept switching between the position and the momentum wavefunction, which gives the impression we have different wavefunctions describing different aspects of the same thing. That’s just not true. The position and momentum wavefunction describe essentially the same thing: we can go from one to the other, and back again, by a simple mathematical manipulation. So I should have stuck to descriptions in terms of ψ(x, t), instead of switching back and forth between the ψ(x, t) and φ(x, t) representations.

In any case, the damage is done, so let’s move forward. The key idea is that, when we know the wavefunction, we know everything. I tried to convey that by noting that the real and imaginary part of the wavefunction must, somehow, represent the total energy of the particle. The structural similarity between the mass-energy equivalence relation (i.e. Einstein’s formula: E = m·c2) and the energy formulas for oscillators and spinning masses is too obvious:

  1. The energy of any oscillator is given by the E = m·ω02/2. We may want to liken the real and imaginary component of our wavefunction to two oscillators and, hence, add them up. The E = m·ω02 formula we get is then identical to the E = m·c2 formula.
  2. The energy of a spinning mass is given by an equivalent formula: E = I·ω2/2 (I is the moment of inertia in this formula). The same 1/2 factor tells us our particle is, somehow, spinning in two dimensions at the same time (i.e. a ‘real’ as well as an ‘imaginary’ space—but both are equally real, because amplitudes interfere), so we get the E = I·ω2 formula. 

Hence, the formulas tell us we should imagine an electron – or an electron orbital – as a very complicated two-dimensional standing wave. Now, when I write two-dimensional, I refer to the real and imaginary component of our wavefunction, as illustrated below. What I am asking you, however, is to not only imagine these two components oscillating up and down, but also spinning about. Hence, if we think about energy as some oscillating mass – which is what the E = m·c2 formula tells us to do, we should remind ourselves we’re talking very complicated motions here: mass oscillates, swirls and spins, and it does so both in real as well as in imaginary space.  rising_circular

What I like about the illustration above is that it shows us – in a very obvious way – why the wavefunction depends on our reference frame. These oscillations do represent something in absolute space, but how we measure it depends on our orientation in that absolute space. But so I am writing this post to talk about operators, not about my grand theory about the essence of mass and energy. So let’s talk about operators now. 🙂

In that post of mine, I showed how the position, momentum and energy operator would give us the average position, momentum and energy of whatever it was that we were looking at, but I didn’t introduce the angular momentum operator. So let me do that now. However, I’ll first recapitulate what we’ve learnt so far in regard to operators.

The energy, position and momentum operators

The equation below defines the energy operator, and also shows how we would apply it to the wavefunction:

energy operator

To the purists: sorry for not (always) using the hat symbol. [I explained why in that post of mine: it’s just too cumbersome.] The others 🙂 should note the following:

  • Eaverage is also an expected value: Eav = E[E]
  • The * symbol tells us to take the complex conjugate of the wavefunction.
  • As for the integral, it’s an integral over some volume, so that’s what the d3r shows. Many authors use double or triple integral signs (∫∫ or ∫∫∫) to show it’s a surface or a volume integral, but that makes things look very complicated, and so I don’t that. I could also have written the integral as ∫ψ(r)*·H·ψ(r) dV, but then I’d need to explain that the dV stands for dVolume, not for any (differental) potential energy (V).
  • We must normalize our wavefunction for these formulas to work, so all probabilities over the volume add up to 1.

OK. That’s the energy operator. As you can see, it’s a pretty formidable beast, but then it just reflects Schrödinger’s equation which, as I explained a couple of times already, we can interpret as an energy propagation mechanism, or an energy diffusion equation, so it is actually not that difficult to memorize the formula: if you’re able to remember Schrödinger’s equation, then you’ll also have the operator. If not… Well… Then you won’t pass your undergrad physics exam. 🙂

I already mentioned that the position operator is a much simpler beast. That’s because it’s so intimately related to our interpretation of the wavefunction. It’s the one thing you know about quantum mechanics: the absolute square of the wavefunction gives us the probability density function. So, for one-dimensional space, the position operator is just:

position operator

The equivalent operator for three-dimensional space is equally simple:

position operator 3D V2

Note how the operator, for the one- as well as for the three-dimensional case, gets rid of time as a variable. In fact, the idea itself of an average makes abstraction of the temporal aspect. Well… Here, at least—because we’re looking at some box in space, rather than some box in spacetime. We’ll re-visit that rather particular idea of an average, and allow for averages that change with time, in a short while.

Next, we introduced the momentum operator in that post of mine. For one dimension, Feynman shows this operator is given by the following formula:

momentum operator

Now that does not look very simple. You might think that the ∂/∂x operator reflects our velocity operator, but… Well… No: ∂/∂t gives us a time rate of change, while ∂/∂x gives us the spatial variation. So it’s not the same. Also, that ħ/i factor is quite intriguing, isn’t it? We’ll come back to it in the next section of this post. Let me just give you the three-dimensional equivalent which, remembering that 1/i = −i, you’ll understand to be equal to the following vector operator:

momentum vector operator

Now it’s time to define the operator we wanted to talk about, i.e. the angular momentum operator.

The angular momentum operator

The formula for the angular momentum operator is remarkably simple:

angular momentum operator

Why do I call this a simple formula? Because it looks like the familiar formula of classical mechanics for the z-component of the classical angular momentum L = r × p. I must assume you know how to calculate a vector cross product. If not, check one of my many posts on vector analysis. I must also assume you remember the L = r × p formula. If not, the following animation might bring it all back. If that doesn’t help, check my post on gyroscopes. 🙂

torque_animation-1.gif

Now, spin is a complicated phenomenon, and so, to simplify the analysis, we should think of orbital angular momentum only. This is a simplification, because electron spin is some complicated mix of intrinsic and orbital angular momentum. Hence, the angular momentum operator we’re introducing here is only the orbital angular momentum operator. However, let us not get bogged down in all of the nitty-gritty and, hence, let’s just go along with it for the time being.

I am somewhat hesitant to show you how we get that formula for our operator, but I’ll try to show you using an intuitive approach, which uses only bits and pieces of Feynman’s more detailed derivation. It will, hopefully, give you a bit of an idea of how these differential operators work. Think about a rotation of our reference frame over an infinitesimally small angle – which we’ll denote as ε – as illustrated below.

rotation

Now, the whole idea is that, because of that rotation of our reference frame, our wavefunction will look different. It’s nothing fundamental, but… Well… It’s just because we’re using a different coordinate system. Indeed, that’s where all these complicated transformation rules for amplitudes come in.  I’ve spoken about these at length when we were still discussing n-state systems. In contrast, the transformation rules for the coordinates themselves are very simple:

rotation

Now, because ε is an infinitesimally small angle, we may equate cos(θ) = cos(ε) to 1, and cos(θ) = sin(ε) to ε. Hence, x’ and y’ are then written as x’+ εy and y’− εx, while z‘ remains z. Vice versa, we can also write the old coordinates in terms of the new ones: x = x’ − εy, y = y’ + εx, and zThat’s obvious. Now comes the difficult thing: you need to think about the two-dimensional equivalent of the simple illustration below.

izvod

If we have some function y = f(x), then we know that, for small Δx, we have the following approximation formula for f(x + Δx): f(x + Δx) ≈ f(x) + (dy/dx)·Δx. It’s the formula you saw in high school: you would then take a limit (Δ0), and define dy/dx as the Δy/Δx ratio for Δ0. You would this after re-writing the f(x + Δx) ≈ f(x) + (dy/dx)·Δx formula as:

Δy = Δf = f(x + Δx) − f(x) ≈ (dy/dx)·Δx

Now you need to substitute f for ψ, and Δx for ε. There is only one complication here: ψ is a function of two variables: x and y. In fact, it’s a function of three variables – x, y and z – but we keep constant. So think of moving from and to + εy = + Δand to + Δ− εx. Hence, Δ= εy and Δ= −εx. It then makes sense to write Δψ as:

angular momentum operator v2

If you agree with that, you’ll also agree we can write something like this:

formula 2

Now that implies the following formula for Δψ:

repair

This looks great! You can see we get some sort of differential operator here, which is what we want. So the next step should be simple: we just let ε go to zero and then we’re done, right? Well… No. In quantum mechanics, it’s always a bit more complicated. But it’s logical stuff. Think of the following:

1. We will want to re-write the infinitesimally small ε angle as a fraction of i, i.e. the imaginary unit.

Huh? Yes. This little represents many things. In this particular case, we want to look at it as a right angle. In fact, you know multiplication with i amounts to a rotation by 90 degrees. So we should replace ε by ε·i. It’s like measuring ε in natural units. However, we’re not done.

2. We should also note that Nature measures angles clockwise, rather than counter-clockwise, as evidenced by the fact that the argument of our wavefunction rotates clockwise as time goes by. So our ε is, in fact, a −ε. We will just bring the minus sign inside of the brackets to solve this issue.

Huh? Yes. Sorry. I told you this is a rather intuitive approach to getting what we want to get. 🙂

3. The third modification we’d want to make is to express ε·i as a multiple of Planck’s constant.

Huh? Yes. This is a very weird thing, but it should make sense—intuitively: we’re talking angular momentum here, and its dimension is the same as that of physical action: N·m·s. Therefore, Planck’s quantum of action (ħ = h/2π ≈ 1×10−34 J·s ≈ 6.6×10−16 eV·s) naturally appears as… Well… A natural unit, or a scaling factor, I should say.

To make a long story short, we’ll want to re-write ε as −(i/ħ)·ε. However, there is a thing called mathematical consistency, and so, if we want to do such substitutions and prepare for that limit situation (ε → 0), we should re-write that Δψ equation as follows:

final

So now – finally! – we do have the formula we wanted to find for our angular momentum operator:

final 2

The final substitution, which yields the formula we just gave you when commencing this section, just uses the formula for the linear momentum operator in the x– and y-direction respectively. We’re done! 🙂 Finally! 

Well… No. 🙂 The question, of course, is the same as always: what does it all mean, really? That’s always a great question. 🙂 Unfortunately, the answer is rather boring: we can calculate the average angular momentum in the z-direction, using a similar integral as the one we used to get the average energy, or the average linear momentum in some direction. That’s basically it.

To compensate for that very boring answer, however, I will show you something that is far less boring. 🙂

Quantum-mechanical weirdness

I’ll shameless copy from Feynman here. He notes that many classical equations get carried over into a quantum-mechanical form (I’ll copy some of his illustrations later). But then there are some that don’t. As Feynman puts it—rather humorously: “There had better be some that don’t come out right, because if everything did, then there would be nothing different about quantum mechanics. There would be no new physics.” He then looks at the following super-obvious equation in classical mechanics:

x·p− px·x = 0

In fact, this equation is so super-obvious that it’s almost meaningless. Almost. It’s super-obvious because multiplication is commutative (for real as well for complex numbers). However, when we replace x and pby the position and momentum operator, we get an entirely different result. You can verify the following yourself:

strange

This is plain weird! What does it mean? I am not sure. Feynman’s take on it is nice but leaves us in the dark on it:

Feynman quote 2

He adds: “If Planck’s constant were zero, the classical and quantum results would be the same, and there would be no quantum mechanics to learn!” Hmm… What does it mean, really? Not sure. Let me make two remarks here:

1. We should not put any dot (·) between our operators, because they do not amount to multiplying one with another. We just apply operators successively. Hence, commutativity is not what we should expect.

2. Note that Feynman forgot to put the subscript in that quote. When doing the same calculations for the equivalent of the x·p− py·x expression, we do get zero, as shown below:

not strange

These equations – zero or not – are referred to as ‘commutation rules’. [Again, I should not have used any dot between x and py, because there is no multiplication here. It’s just a separation mark.] Let me quote Feynman on it, so the matter is dealt with:

quote

OK. So what do we conclude? What are we talking about?

Conclusions

Some of the stuff above was really intriguing. For example, we found that the linear and angular momentum operators are differential operators in the true sense of the word. The angular momentum operator shows us what happens to the wavefunction if we rotate our reference frame over an infinitesimally small angle ε. That’s what’s captured by the formulas we’ve developed, as summarized below:

angular momentum

Likewise, the linear momentum operator captures what happens to the wavefunction for an infinitesimally small displacement of the reference frame, as shown by the equivalent formulas below:

linear momentum

What’s the interpretation for the position operator, and the energy operator? Here we are not so sure. The integrals above make sense, but these integrals are used to calculate averages values, as opposed to instantaneous values. So… Well… There is not all that much I can say about the position and energy operator right now, except… Well… We now need to explore the question of how averages could possibly change over time. Let’s do that now.

Averages that change with time

I know: you are totally quantum-mechanicked out by now. So am I. But we’re almost there. In fact, this is Feynman’s last Lecture on quantum mechanics and, hence, I think I should let the Master speak here. So just click on the link and read for yourself. It’s a really interesting chapter, as he shows us the equivalent of Newton’s Law in quantum mechanics, as well as the quantum-mechanical equivalent of other standard equations in classical mechanics. However, I need to warn you: Feynman keeps testing the limits of our intellectual absorption capacity by switching back and forth between matrix and wave mechanics. Interesting, but not easy. For example, you’ll need to remind yourself of the fact that the Hamiltonian matrix is equal to its own complex conjugate (or – because it’s a matrix – its own conjugate transpose.

Having said that, it’s all wonderful. The time rate of change of all those average values is denoted by using the over-dot notation. For example, the time rate of change of the average position is denoted by:

p1

Once you ‘get’ that new notation, you will quickly understand the derivations. They are not easy (what derivations are in quantum mechanics?), but we get very interesting results. Nice things to play with, or think about—like this identity:

formula2

It takes a while, but you suddenly realize this is the equivalent of the classical dx/dtv = p/m formula. 🙂

Another sweet result is the following one:

formula3

This is the quantum-mechanical equivalent of Newton’s force law: F = m·a. Huh? Yes. Think of it: the spatial derivative of the (potential) energy is the force. Now just think of the classical dp/dt = d(m·v) = m·dv/dt = m·a formula. […] Can you see it now? Isn’t this just Great Fun?

Note, however, that these formulas also show the limits of our analysis so far, because they treat m as some constant. Hence, we’ll need to relativistically correct them. But that’s complicated, and so we’ll postpone that to another day.

[…]

Well… That’s it, folks! We’re really through! This was the last of the last of Feynman’s Lectures on Physics. So we’re totally done now. Isn’t this great? What an adventure! I hope that, despite the enormous mental energy that’s required to digest all this stuff, you enjoyed it as much as I did. 🙂

Post scriptum 1: I just love Feynman but, frankly, I think he’s sometimes somewhat sloppy with terminology. In regard to what these operators really mean, we should make use of better terminology: an average is something else than an expected value. Our momentum operator, for example, as such returns an expected value – not an average momentum. We need to deepen the analysis here somewhat, but I’ll also leave that for later.

Post scriptum 2: There is something really interesting about that i·ħ or −(i/ħ) scaling factor – or whatever you want to call it – appearing in our formulas. Remember the Schrödinger equation can also be written as:

i·ħ·∂ψ/∂t = −(1/2)·(ħ2/m)∇2ψ + V·ψ = Hψ

This is interesting in light of our interpretation of the Schrödinger equation as an energy propagation mechanism. If we write Schrödinger’s equation like we write it here, then we have the energy on the right-hand side – which is time-independent. How do we interpret the left-hand side now? Well… It’s kinda simple, but we just have the time rate of change of the real and imaginary part of the wavefunction here, and the i·ħ factor then becomes a sort of unit in which we measure the time rate of change. Alternatively, you may think of ‘splitting’ Planck’s constant in two: Planck’s energy, and Planck’s time unit, and then you bring the Planck energy unit to the other side, so we’d express the energy in natural units. Likewise, the time rate of change of the components of our wavefunction would also be measured in natural time units if we’d do that.

I know this is all very abstract but, frankly, it’s crystal clear to me. This formula tells us that the energy of the particle that’s being described by the wavefunction is being carried by the oscillations of the wavefunction. In fact, the oscillations are the energy. You can play with the mass factor, by moving it to the left-hand side too, or by using Einstein’s mass-energy equivalence relation. The interpretation remains consistent.

In fact, there is something really interesting here. You know that we usually separate out the spatial and temporal part of the wavefunction, so we write: ψ(r, t) = ψ(rei·(E/ħ)·t. In fact, it is quite common to refer to ψ(r) – rather than to ψ(r, t) – as the wavefunction, even if, personally, I find that quite confusing and misleading (see my page onSchrödinger’s equation). Now, we may want to think of what happens when we’d apply the energy operator to ψ(r) rather than to ψ(r, t). We may think that we’d get a time-independent value for the energy at that point in space, so energy is some function of position only, not of time. That’s an interesting thought, and we should explore it. For example, we then may think of energy as an average that changes with position—as opposed to the (average) position and momentum, which we like to think of as averages than change with time, as mentioned above. I will come back to this later – but perhaps in another post or so. Not now. The only point I want to mention here is the following: you cannot use ψ(r) in Schrödinger’s equation. Why? Well… Schrödinger’s equation is no longer valid when substituting ψ for ψ(r), because the left-hand side is always zero, as ∂ψ(r)/∂t is zero – for any r.

There is another, related, point to this observation. If you think that Schrödinger’s equation implies that the operators on both sides of Schrödinger’s equation must be equivalent (i.e. the same), you’re wrong:

i·ħ·∂/∂t ≠ H = −(1/2)·(ħ2/m)∇2 + V

It’s a basic thing, really: Schrödinger’s equation is not valid for just any function. Hence, it does not work for ψ(r). Only ψ(r, t) makes it work, because… Well… Schrödinger’s equation gave us ψ(r, t)!

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The energy and 1/2 factor in Schrödinger’s equation

Schrödinger’s equation, for a particle moving in free space (so we have no external force fields acting on it, so V = 0 and, therefore, the Vψ term disappears) is written as:

∂ψ(x, t)/∂t = i·(1/2)·(ħ/meff)·∇2ψ(x, t)

We already noted and explained the structural similarity with the ubiquitous diffusion equation in physics:

∂φ(x, t)/∂t = D·∇2φ(x, t) with x = (x, y, z)

The big difference between the wave equation and an ordinary diffusion equation is that the wave equation gives us two equations for the price of one: ψ is a complex-valued function, with a real and an imaginary part which, despite their name, are both equally fundamental, or essential. Whatever word you prefer. 🙂 That’s also what the presence of the imaginary unit (i) in the equation tells us. But for the rest it’s the same: the diffusion constant (D) in Schrödinger’s equation is equal to (1/2)·(ħ/meff).

Why the 1/2 factor? It’s ugly. Think of the following: If we bring the (1/2)·(ħ/meff) to the other side, we can write it as meff/(ħ/2). The ħ/2 now appears as a scaling factor in the diffusion constant, just like ħ does in the de Broglie equations: ω = E/ħ and k = p/ħ, or in the argument of the wavefunction: θ = (E·t − p∙x)/ħ. Planck’s constant is, effectively, a physical scaling factor. As a physical scaling constant, it usually does two things:

  1. It fixes the numbers (so that’s its function as a mathematical constant).
  2. As a physical constant, it also fixes the physical dimensions. Note, for example, how the 1/ħ factor in ω = E/ħ and k = p/ħ ensures that the ω·t = (E/ħ)·t and k·x = (p/ħ)·x terms in the argument of the wavefunction are both expressed as some dimensionless number, so they can effectively be added together. Physicists don’t like adding apples and oranges.

The question is: why did Schrödinger use ħ/2, rather than ħ, as a scaling factor? Let’s explore the question.

The 1/2 factor

We may want to think that 1/2 factor just echoes the 1/2 factor in the Uncertainty Principle, which we should think of as a pair of relations: σx·σp ≥ ħ/2 and σE·σ≥ ħ/2. However, the 1/2 factor in those relations only makes sense because we chose to equate the fundamental uncertainty (Δ) in x, p, E and t with the mathematical concept of the standard deviation (σ), or the half-width, as Feynman calls it in his wonderfully clear exposé on it in one of his Lectures on quantum mechanics (for a summary with some comments, see my blog post on it). We may just as well choose to equate Δ with the full-width of those probability distributions we get for x and p, or for E and t. If we do that, we get σx·σp ≥ ħ and σE·σ≥ ħ.

It’s a bit like measuring the weight of a person on an old-fashioned (non-digital) bathroom scale with 1 kg marks only: do we say this person is x kg ± 1 kg, or x kg ± 500 g? Do we take the half-width or the full-width as the margin of error? In short, it’s a matter of appreciation, and the 1/2 factor in our pair of uncertainty relations is not there because we’ve got two relations. Likewise, it’s not because I mentioned we can think of Schrödinger’s equation as a pair of relations that, taken together, represent an energy propagation mechanism that’s quite similar in its structure to Maxwell’s equations for an electromagnetic wave (as shown below), that we’d insert (or not) that 1/2 factor: either of the two representations below works. It just depends on our definition of the concept of the effective mass.

The 1/2 factor is really a matter of choice, because the rather peculiar – and flexible – concept of the effective mass takes care of it. However, we could define some new effective mass concept, by writing: meffNEW = 2∙meffOLD, and then Schrödinger’s equation would look more elegant:

∂ψ/∂t = i·(ħ/meffNEW)·∇2ψ

Now you’ll want the definition, of course! What is that effective mass concept? Feynman talks at length about it, but his exposé is embedded in a much longer and more general argument on the propagation of electrons in a crystal lattice, which you may not necessarily want to go through right now. So let’s try to answer that question by doing something stupid: let’s substitute ψ in the equation for ψ = a·ei·[E·t − p∙x]/ħ (which is an elementary wavefunction), calculate the time derivative and the Laplacian, and see what we get. If we do that, the ∂ψ/∂t = i·(1/2)·(ħ/meff)·∇2ψ equation becomes:

i·a·(E/ħei∙(E·t − p∙x)/ħ = i·a·(1/2)·(ħ/meff)(p2/ħ2ei∙(E·t − p∙x) 

⇔ E = (1/2)·p2/meff = (1/2)·(m·v)2/meff ⇔ meff = (1/2)·(m/E)·m·v2

⇔ meff = (1/c2)·(m·v2/2) = m·β2/2

Hence, the effective mass appears in this equation as the equivalent mass of the kinetic energy (K.E.) of the elementary particle that’s being represented by the wavefunction. Now, you may think that sounds good – and it does – but you should note the following:

1. The K.E. = m·v2/2 formula is only correct for non-relativistic speeds. In fact, it’s the kinetic energy formula if, and only if, if m ≈ m0. The relativistically correct formula for the kinetic energy calculates it as the difference between (1) the total energy (which is given by the E = m·c2 formula, always) and (2) its rest energy, so we write:

K.E. = E − E0 = mv·c2 − m0·c2 = m0·γ·c2 − m0·c2 = m0·c2·(γ − 1)

2. The energy concept in the wavefunction ψ = a·ei·[E·t − p∙x]/ħ is, obviously, the total energy of the particle. For non-relativistic speeds, the kinetic energy is only a very small fraction of the total energy. In fact, using the formula above, you can calculate the ratio between the kinetic and the total energy: you’ll find it’s equal to 1 − 1/γ = 1 − √(1−v2/c2), and its graph goes from 0 to 1.

graph

Now, if we discard the 1/2 factor, the calculations above yield the following:

i·a·(E/ħ)·ei∙(E·t − p∙x)/ħ = −i·a·(ħ/meff)(p22ei∙(E·t − p∙x)/ħ 

⇔ E = p2/meff = (m·v)2/meff ⇔ meff = (m/E)·m·v2

⇔ meff = m·v2/c= m·β2

In fact, it is fair to say that both definitions are equally weird, even if the dimensions come out alright: the effective mass is measured in old-fashioned mass units, and the βor β2/2 factor appears as a sort of correction factor, varying between 0 and 1 (for β2) or between 0 and 1/2 (for β2/2). I prefer the new definition, as it ensures that meff becomes equal to m in the limit for the velocity going to c. In addition, if we bring the ħ/meff or (1/2)∙ħ/meff factor to the other side of the equation, the choice becomes one between a meffNEW/ħ or a 2∙meffOLD/ħ coefficient.

It’s a choice, really. Personally, I think the equation without the 1/2 factor – and, hence, the use of ħ rather than ħ/2 as the scaling factor – looks better, but then you may argue that – if half of the energy of our particle is in the oscillating real part of the wavefunction, and the other is in the imaginary part – then the 1/2 factor should stay, because it ensures that meff becomes equal to m/2 as v goes to c (or, what amounts to the same, β goes to 1). But then that’s the argument about whether or not we should have a 1/2 factor because we get two equations for the price of one, like we did for the Uncertainty Principle.

So… What to do? Let’s first ask ourselves whether that derivation of the effective mass actually makes sense. Let’s therefore look at both limit situations.

1. For v going to c (or β = v/c going to 1), we do not have much of a problem: meff just becomes the total mass of the particle that we’re looking at, and Schrödinger’s equation can easily be interpreted as an energy propagation mechanism. Our particle has zero rest mass in that case ( we may also say that the concept of a rest mass is meaningless in this situation) and all of the energy – and, therefore, all of the equivalent mass – is kinetic: m = E/cand the effective mass is just the mass: meff = m·c2/c= m. Hence, our particle is everywhere and nowhere. In fact, you should note that the concept of velocity itself doesn’t make sense in this rather particular case. It’s like a photon (but note it’s not a photon: we’re talking some theoretical particle here with zero spin and zero rest mass): it’s a wave in its own frame of reference, but as it zips by at the speed of light, we think of it as a particle.

2. Let’s look at the other limit situation. For v going to 0 (or β = v/c going to 0), Schrödinger’s equation no longer makes sense, because the diffusion constant goes to zero, so we get a nonsensical equation. Huh? What’s wrong with our analysis?

Well… I must be honest. We started off on the wrong foot. You should note that it’s hard – in fact, plain impossible – to reconcile our simple a·ei·[E·t − p∙x]/ħ function with the idea of the classical velocity of our particle. Indeed, the classical velocity corresponds to a group velocity, or the velocity of a wave packet, and so we just have one wave here: no group. So we get nonsense. You can see the same when equating p to zero in the wave equation: we get another nonsensical equation, because the Laplacian is zero! Check it. If our elementary wavefunction is equal to ψ = a·ei·(E/ħ)·t, then that Laplacian is zero.

Hence, our calculation of the effective mass is not very sensical. Why? Because the elementary wavefunction is a theoretical concept only: it may represent some box in space, that is uniformly filled with energy, but it cannot represent any actual particle. Actual particles are always some superposition of two or more elementary waves, so then we’ve got a wave packet (as illustrated below) that we can actually associate with some real-life particle moving in space, like an electron in some orbital indeed. 🙂

wave-packet

I must credit Oregon State University for the animation above. It’s quite nice: a simple particle in a box model without potential. As I showed on my other page (explaining various models), we must add at least two waves – traveling in opposite directions – to model a particle in a box. Why? Because we represent it by a standing wave, and a standing wave is the sum of two waves traveling in opposite directions.

So, if our derivation above was not very meaningful, then what is the actual concept of the effective mass?

The concept of the effective mass

I am afraid that, at this point, I do have to direct you back to the Grand Master himself for the detail. Let me just try to sum it up very succinctly. If we have a wave packet, there is – obviously – some energy in it, and it’s energy we may associate with the classical concept of the velocity of our particle – because it’s the group velocity of our wave packet. Hence, we have a new energy concept here – and the equivalent mass, of course. Now, Feynman’s analysis – which is Schrödinger’s analysis, really – shows we can write that energy as:

E = meff·v2/2

So… Well… That’s the classical kinetic energy formula. And it’s the very classical one, because it’s not relativistic. 😦 But that’s OK for relatively small-moving electrons! [Remember the typical (relative) velocity is given by the fine-structure constant: α = β = v/c. So that’s impressive (about 2,188 km per second), but it’s only a tiny fraction of the speed of light, so non-relativistic formulas should work.]

Now, the meff factor in this equation is a function of the various parameters of the model he uses. To be precise, we get the following formula out of his model (which, as mentioned above, is a model of electrons propagating in a crystal lattice):

meff = ħ2/(2·A·b2 )

Now, the b in this formula is the spacing between the atoms in the lattice. The A basically represents an energy barrier: to move from one atom to another, the electron needs to get across it. I talked about this in my post on it, and so I won’t explain the graph below – because I did that in that post. Just note that we don’t need that factor 2: there is no reason whatsoever to write E+ 2·A and E2·A. We could just re-define a new A: (1/2)·ANEW = AOLD. The formula for meff then simplifies to ħ2/(2·AOLD·b2) = ħ2/(ANEW·b2). We then get an Eeff = meff·vformula for the extra energy.

energy

Eeff = meff·v2?!? What energy formula is that? Schrödinger must have thought the same thing, and so that’s why we have that ugly 1/2 factor in his equation. However, think about it. Our analysis shows that it is quite straightforward to model energy as a two-dimensional oscillation of mass. In this analysis, both the real and the imaginary component of the wavefunction each store half of the total energy of the object, which is equal to E = m·c2. Remember, indeed, that we compared it to the energy in an oscillator, which is equal to the sum of kinetic and potential energy, and for which we have the T + U = m·ω02/2 formula. But so we have two oscillators here and, hence, twice the energy. Hence, the E = m·c2 corresponds to m·ω0and, hence, we may think of as the natural frequency of the vacuum.

Therefore, the Eeff = meff·v2 formula makes much more sense. It nicely mirrors Einstein’s E = m·c2 formula and, in fact, naturally merges into E = m·c for v approaching c. But, I admit, it is not so easy to interpret. It’s much easier to just say that the effective mass is the mass of our electron as it appears in the kinetic energy formula, or – alternatively – in the momentum formula. Indeed, Feynman also writes the following formula:

meff·v = p = ħ·k

Now, that is something we easily recognize! 🙂

So… Well… What do we do now? Do we use the 1/2 factor or not?

It would be very convenient, of course, to just stick with tradition and use meff as everyone else uses it: it is just the mass as it appears in whatever medium we happen to look it, which may be a crystal lattice (or a semi-conductor), or just free space. In short, it’s the mass of the electron as it appears to us, i.e. as it appears in the (non-relativistic) kinetic energy formula (K.E. = meff·v2/2), the formula for the momentum of an electron (p = meff·v), or in the wavefunction itself (k = p/ħ = (meff·v)/ħ. In fact, in his analysis of the electron orbitals, Feynman (who just follows Schrödinger here) drops the eff subscript altogether, and so the effective mass is just the mass: meff = m. Hence, the apparent mass of the electron in the hydrogen atom serves as a reference point, and the effective mass in a different medium (such as a crystal lattice, rather than free space or, I should say, a hydrogen atom in free space) will also be different.

The thing is: we get the right results out of Schrödinger’s equation, with the 1/2 factor in it. Hence, Schrödinger’s equation works: we get the actual electron orbitals out of it. Hence, Schrödinger’s equation is true – without any doubt. Hence, if we take that 1/2 factor out, then we do need to use the other effective mass concept. We can do that. Think about the actual relation between the effective mass and the real mass of the electron, about which Feynman writes the following: “The effective mass has nothing to do with the real mass of an electron. It may be quite different—although in commonly used metals and semiconductors it often happens to turn out to be the same general order of magnitude: about 0.1 to 30 times the free-space mass of the electron.” Hence, if we write the relation between meff and m as meff = g(m), then the same relation for our meffNEW = 2∙meffOLD becomes meffNEW = 2·g(m), and the “about 0.1 to 30 times” becomes “about 0.2 to 60 times.”

In fact, in the original 1963 edition, Feynman writes that the effective mass is “about 2 to 20 times” the free-space mass of the electron. Isn’t that interesting? I mean… Note that factor 2! If we’d write meff = 2·m, then we’re fine. We can then write Schrödinger’s equation in the following two equivalent ways:

  1. (meff/ħ)·∂ψ/∂t = i·∇2ψ
  2. (2m/ħ)·∂ψ/∂t = i·∇2ψ

Both would be correct, and it explains why Schrödinger’s equation works. So let’s go for that compromise and write Schrödinger’s equation in either of the two equivalent ways. 🙂 The question then becomes: how to interpret that factor 2? The answer to that question is, effectively, related to the fact that we get two waves for the price of one here. So we have two oscillators, so to speak. Now that‘s quite deep, and I will explore that in one of my next posts.

Let me now address the second weird thing in Schrödinger’s equation: the energy factor. I should be more precise: the weirdness arises when solving Schrödinger’s equation. Indeed, in the texts I’ve read, there is this constant switching back and forth between interpreting E as the energy of the atom, versus the energy of the electron. Now, both concepts are obviously quite different, so which one is it really?

The energy factor E

It’s a confusing point—for me, at least and, hence, I must assume for students as well. Let me indicate, by way of example, how the confusion arises in Feynman’s exposé on the solutions to the Schrödinger equation. Initially, the development is quite straightforward. Replacing V by −e2/r, Schrödinger’s equation becomes:

Eq1

As usual, it is then assumed that a solution of the form ψ (r, t) =  e−(i/ħ)·E·t·ψ(r) will work. Apart from the confusion that arises because we use the same symbol, ψ, for two different functions (you will agree that ψ (r, t), a function in two variables, is obviously not the same as ψ(r), a function in one variable only), this assumption is quite straightforward and allows us to re-write the differential equation above as:

de

To get this, you just need to actually to do that time derivative, noting that the ψ in our equation is now ψ(r), not ψ (r, t). Feynman duly notes this as he writes: “The function ψ(rmust solve this equation, where E is some constant—the energy of the atom.” So far, so good. In one of the (many) next steps, we re-write E as E = ER·ε, with E= m·e4/2ħ2. So we just use the Rydberg energy (E≈ 13.6 eV) here as a ‘natural’ atomic energy unit. That’s all. No harm in that.

Then all kinds of complicated but legitimate mathematical manipulations follow, in an attempt to solve this differential equation—attempt that is successful, of course! However, after all these manipulations, one ends up with the grand simple solution for the s-states of the atom (i.e. the spherically symmetric solutions):

En = −ER/nwith 1/n= 1, 1/4, 1/9, 1/16,…, 1

So we get: En = −13.6 eV, −3.4 eV, −1.5 eV, etcetera. Now how is that possible? How can the energy of the atom suddenly be negative? More importantly, why is so tiny in comparison with the rest energy of the proton (which is about 938 mega-electronvolt), or the electron (0.511 MeV)? The energy levels above are a few eV only, not a few million electronvolt. Feynman answers this question rather vaguely when he states the following:

“There is, incidentally, nothing mysterious about negative numbers for the energy. The energies are negative because when we chose to write V = −e2/r, we picked our zero point as the energy of an electron located far from the proton. When it is close to the proton, its energy is less, so somewhat below zero. The energy is lowest (most negative) for n = 1, and increases toward zero with increasing n.”

We picked our zero point as the energy of an electron located far away from the proton? But we were talking the energy of the atom all along, right? You’re right. Feynman doesn’t answer the question. The solution is OK – well, sort of, at least – but, in one of those mathematical complications, there is a ‘normalization’ – a choice of some constant that pops up when combining and substituting stuff – that is not so innocent. To be precise, at some point, Feynman substitutes the ε variable for the square of another variable – to be even more precise, he writes: ε = −α2. He then performs some more hat tricks – all legitimate, no doubt – and finds that the only sensible solutions to the differential equation require α to be equal to 1/n, which immediately leads to the above-mentioned solution for our s-states.

The real answer to the question is given somewhere else. In fact, Feynman casually gives us an explanation in one of his very first Lectures on quantum mechanics, where he writes the following:

“If we have a “condition” which is a mixture of two different states with different energies, then the amplitude for each of the two states will vary with time according to an equation like a·eiωt, with ħ·ω = E0 = m·c2. Hence, we can write the amplitude for the two states, for example as:

ei(E1/ħ)·t and ei(E2/ħ)·t

And if we have some combination of the two, we will have an interference. But notice that if we added a constant to both energies, it wouldn’t make any difference. If somebody else were to use a different scale of energy in which all the energies were increased (or decreased) by a constant amount—say, by the amount A—then the amplitudes in the two states would, from his point of view, be

ei(E1+A)·t/ħ and ei(E2+A)·t/ħ

All of his amplitudes would be multiplied by the same factor ei(A/ħ)·t, and all linear combinations, or interferences, would have the same factor. When we take the absolute squares to find the probabilities, all the answers would be the same. The choice of an origin for our energy scale makes no difference; we can measure energy from any zero we want. For relativistic purposes it is nice to measure the energy so that the rest mass is included, but for many purposes that aren’t relativistic it is often nice to subtract some standard amount from all energies that appear. For instance, in the case of an atom, it is usually convenient to subtract the energy Ms·c2, where Ms is the mass of all the separate pieces—the nucleus and the electrons—which is, of course, different from the mass of the atom. For other problems, it may be useful to subtract from all energies the amount Mg·c2, where Mg is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom. So, sometimes we may shift our zero of energy by some very large constant, but it doesn’t make any difference, provided we shift all the energies in a particular calculation by the same constant.”

It’s a rather long quotation, but it’s important. The key phrase here is, obviously, the following: “For other problems, it may be useful to subtract from all energies the amount Mg·c2, where Mg is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom.” So that’s what he’s doing when solving Schrödinger’s equation. However, I should make the following point here: if we shift the origin of our energy scale, it does not make any difference in regard to the probabilities we calculate, but it obviously does make a difference in terms of our wavefunction itself. To be precise, its density in time will be very different. Hence, if we’d want to give the wavefunction some physical meaning – which is what I’ve been trying to do all along – it does make a huge difference. When we leave the rest mass of all of the pieces in our system out, we can no longer pretend we capture their energy.

This is a rather simple observation, but one that has profound implications in terms of our interpretation of the wavefunction. Personally, I admire the Great Teacher’s Lectures, but I am really disappointed that he doesn’t pay more attention to this. 😦

The Essence of Reality

I know it’s a crazy title. It has no place in a physics blog, but then I am sure this article will go elsewhere. […] Well… […] Let me be honest: it’s probably gonna go nowhere. Whatever. I don’t care too much. My life is happier than Wittgenstein’s. 🙂

My original title for this post was: discrete spacetime. That was somewhat less offensive but, while being less offensive, it suffered from the same drawback: the terminology was ambiguous. The commonly accepted term for discrete spacetime is the quantum vacuum. However, because I am just an arrogant bastard trying to establish myself in this field, I am telling you that term is meaningless. Indeed, wouldn’t you agree that, if the quantum vacuum is a vacuum, then it’s empty. So it’s nothing. Hence, it cannot have any properties and, therefore, it cannot be discrete – or continuous, or whatever. We need to put stuff in it to make it real.

Therefore, I’d rather distinguish mathematical versus physical space. Of course, you are smart, and so you now you’ll say that my terminology is as bad as that of the quantum vacuumists. And you are right. However, this is a story that am writing, and so I will write it the way want to write it. 🙂 So where were we? Spacetime! Discrete spacetime.

Yes. Thank you! Because relativity tells us we should think in terms of four-vectors, we should not talk about space but about spacetime. Hence, we should distinguish mathematical spacetime from physical spacetime. So what’s the definitional difference?

Mathematical spacetime is just what it is: a coordinate space – Cartesian, polar, or whatever – which we define by choosing a representation, or a base. And all the other elements of the set are just some algebraic combination of the base set. Mathematical space involves numbers. They don’t – let me emphasize that: they do not!– involve the physical dimensions of the variables. Always remember: math shows us the relations, but it doesn’t show us the stuff itself. Think of it: even if we may refer to the coordinate axes as time, or distance, we do not really think of them as something physical. In math, the physical dimension is just a label. Nothing more. Nothing less.

In contrast, physical spacetime is filled with something – with waves, or with particles – so it’s spacetime filled with energy and/or matter. In fact, we should analyze matter and energy as essentially the same thing, and please do carefully re-read what I wrote: I said they are essentially the same. I did not say they are the same. Energy and mass are equivalent, but not quite the same. I’ll tell you what that means in a moment.

These waves, or particles, come with mass, energy and momentum. There is an equivalence between mass and energy, but they are not the same. There is a twist – literally (only after reading the next paragraphs, you’ll realize how literally): even when choosing our time and distance units such that is numerically equal to 1 – e.g. when measuring distance in light-seconds (or time in light-meters), or when using Planck units – the physical dimension of the cfactor in Einstein’s E = mcequation doesn’t vanish: the physical dimension of energy is kg·m2/s2.

Using Newton’s force law (1 N = 1 kg·m/s2), we can easily see this rather strange unit is effectively equivalent to the energy unit, i.e. the joule (1 J = 1 kg·m2/s2 = 1 (N·s2/m)·m2/s= 1 N·m), but that’s not the point. The (m/s)2 factor – i.e. the square of the velocity dimension – reflects the following:

  1. Energy is nothing but mass in motion. To be precise, it’s oscillating mass. [And, yes, that’s what string theory is all about, but I didn’t want to mention that. It’s just terminology once again: I prefer to say ‘oscillating’ rather than ‘vibrating’. :-)]
  2. The rapidly oscillating real and imaginary component of the matter-wave (or wavefunction, we should say) each capture half of the total energy of the object E = mc2.
  3. The oscillation is an oscillation of the mass of the particle (or wave) that we’re looking at.

In the mentioned publication, I explore the structural similarity between:

  1. The oscillating electric and magnetic field vectors (E and B) that represent the electromagnetic wave, and
  2. The oscillating real and imaginary part of the matter-wave.

The story is simple or complicated, depending on what you know already, but it can be told in an abnoxiously easy way. Note that the associated force laws do not differ in their structure:

Coulomb Law

gravitation law

The only difference is the dimension of m versus q: mass – the measure of inertia -versus charge. 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 guess what comes next, but I won’t talk about that here.:-) Just note the absolute distance between two charges (with the same or the opposite sign) is twice the distance between 0 and 1, which must explains the rather mysterious 2 factor I get for the Schrödinger equation for the electromagnetic wave (but I still need to show how that works out exactly).

The point is: remembering that the physical dimension of the electric field is N/C (newton per coulomb, i.e. force per unit of charge) it should not come as a surprise that we find that the physical dimension of the components of the matter-wave is N/kg: newton per kg, i.e. force per unit of mass. For the detail, I’ll refer you to that article of mine (and, because I know you will not want to work your way through it, let me tell you it’s the last chapter that tells you how to do the trick).

So where were we? Strange. I actually just wanted to talk about discrete spacetime here, but I realize I’ve already dealt with all of the metaphysical questions you could possible have, except the (existential) Who Am I? question, which I cannot answer on your behalf. 🙂

I wanted to talk about physical spacetime, so that’s sanitized mathematical space plus something. A date without logistics. Our mind is a lazy host, indeed.

Reality is the guest that brings all of the wine and the food to the party.

In fact, it’s a guest that brings everything to the party: you – the observer – just need to set the time and the place. In fact, in light of what Kant – and many other eminent philosophers – wrote about space and time being constructs of the mind, that’s another statement which you should interpret literally. So physical spacetime is spacetime filled with something – like a wave, or a field. So how does that look like? Well… Frankly, I don’t know! But let me share my idea of it.

Because of the unity of Planck’s quantum of action (ħ ≈ 1.0545718×10−34 N·m·s), a wave traveling in spacetime might be represented as a set of discrete spacetime points and the associated amplitudes, as illustrated below. [I just made an easy Excel graph. Nothing fancy.]

spacetime

The space in-between the discrete spacetime points, which are separated by the Planck time and distance units, is not real. It is plain nothingness, or – if you prefer that term – the space in-between in is mathematical space only: a figment of the mind – nothing real, because quantum theory tells us that the real, physical, space is discontinuous.

Why is that so? Well… Smaller time and distance units cannot exist, because we would not be able to pack Planck’s quantum of action in them: a box of the Planck scale, with ħ in it, is just a black hole and, hence, nothing could go from here to there, because all would be trapped. Of course, now you’ll wonder what it means to ‘pack‘ Planck’s quantum of action in a Planck-scale spacetime box. Let me try  to explain this. It’s going to be a rather rudimentary explanation and, hence, it may not satisfy you. But then the alternative is to learn more about black holes and the Schwarzschild radius, which I warmly recommend for two equivalent reasons:

  1. The matter is actually quite deep, and I’d recommend you try to fully understand it by reading some decent physics course.
  2. You’d stop reading this nonsense.

If, despite my warning, you would continue to read what I write, you may want to note that we could also use the logic below to define Planck’s quantum of action, rather than using it to define the Planck time and distance unit. Everything is related to everything in physics. But let me now give the rather naive explanation itself:

  • Planck’s quantum of action (ħ ≈ 1.0545718×10−34 N·m·s) is the smallest thing possible. It may express itself as some momentum (whose physical dimension is N·s) over some distance (Δs), or as some amount of energy (whose dimension is N·m) over some time (Δt).
  • Now, energy is an oscillation of mass (I will repeat that a couple of times, and show you the detail of what that means in the last chapter) and, hence, ħ must necessarily express itself both as momentum as well as energy over some time and some distance. Hence, it is what it is: some force over some distance over some time. This reflects the physical dimension of ħ, which is the product of force, distance and time. So let’s assume some force ΔF, some distance Δs, and some time Δt, so we can write ħ as ħ = ΔF·Δs·Δt.
  • Now let’s pack that into a traveling particle – like a photon, for example – which, as you know (and as I will show in this publication) is, effectively, just some oscillation of mass, or an energy flow. Now let’s think about one cycle of that oscillation. How small can we make it? In spacetime, I mean.
  • If we decrease Δs and/or Δt, then ΔF must increase, so as to ensure the integrity (or unity) of ħ as the fundamental quantum of action. Note that the increase in the momentum (ΔF·Δt) and the energy (ΔF·Δs) is proportional to the decrease in Δt and Δs. Now, in our search for the Planck-size spacetime box, we will obviously want to decrease Δs and Δt simultaneously.
  • Because nothing can exceed the speed of light, we may want to use equivalent time and distance units, so the numerical value of the speed of light is equal to 1 and all velocities become relative velocities. If we now assume our particle is traveling at the speed of light – so it must be a photon, or a (theoretical) matter-particle with zero rest mass (which is something different than a photon) – then our Δs and Δt should respect the following condition: Δs/Δt = c = 1.
  • Now, when Δs = 1.6162×10−35 m and Δt = 5.391×10−44 s, we find that Δs/Δt = c, but ΔF = ħ/(Δs·Δt) = (1.0545718×10−34 N·m·s)/[(1.6162×10−35 m)·(5.391×10−44 s)] ≈ 1.21×1044 N. That force is monstrously huge. Think of it: because of gravitation, a mass of 1 kg in our hand, here on Earth, will exert a force of 9.8 N. Now note the exponent in that 1.21×1044 number.
  • If we multiply that monstrous force with Δs – which is extremely tiny – we get the Planck energy: (1.6162×10−35 m)·(1.21×1044 N) ≈ 1.956×109 joule. Despite the tininess of Δs, we still get a fairly big value for the Planck energy. Just to give you an idea, it’s 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 fact, the equivalent mass of that energy, packed in such tiny space, makes it a black hole.
  • In short, the conclusion is that our particle can’t move (or, thinking of it as a wave, that our wave can’t wave) because it’s caught in the black hole it creates by its own energy: so the energy can’t escape and, hence, it can’t flow. 🙂

Of course, you will now say that we could imagine half a cycle, or a quarter of that cycle. And you are right: we can surely imagine that, but we get the same thing: to respect the unity of ħ, we’ll then have to pack it into half a cycle, or a quarter of a cycle, which just means the energy of the whole cycle is 2·ħ, or 4·ħ. However, our conclusion still stands: we won’t be able to pack that half-cycle, or that quarter-cycle, into something smaller than the Planck-size spacetime box, because it would make it a black hole, and so our wave wouldn’t go anywhere, and the idea of our wave itself – or the particle – just doesn’t make sense anymore.

This brings me to the final point I’d like to make here. When Maxwell or Einstein, or the quantum vacuumists – or I 🙂 – say that the speed of light is just a property of the vacuum, then that’s correct and not correct at the same time. First, we should note that, if we say that, we might also say that ħ is a property of the vacuum. All physical constants are. Hence, it’s a pretty meaningless statement. Still, it’s a statement that helps us to understand the essence of reality. Second, and more importantly, we should dissect that statement. The speed of light combines two very different aspects:

  1. It’s a physical constant, i.e. some fixed number that we will find to be the same regardless of our reference frame. As such, it’s as essential as those immovable physical laws that we find to be the same in each and every reference frame.
  2. However, its physical dimension is the ratio of the distance and the time unit: m/s. We may choose other time and distance units, but we will still combine them in that ratio. These two units represent the two dimensions in our mind that – as Kant noted – structure our perception of reality: the temporal and spatial dimension.

Hence, we cannot just say that is ‘just a property of the vacuum’. In our definition of as a velocity, we mix reality – the ‘outside world’ – with our perception of it. It’s unavoidable. Frankly, while we should obviously try – and we should try very hard! – to separate what’s ‘out there’ versus ‘how we make sense of it’, it is and remains an impossible job because… Well… When everything is said and done, what we observe ‘out there’ is just that: it’s just what we – humans – observe. 🙂

So, when everything is said and done, the essence of reality consists of four things:

  1. Nothing
  2. Mass, i.e. something, or not nothing
  3. Movement (of something), from nowhere to somewhere.
  4. Us: our mind. Or God’s Mind. Whatever. Mind.

The first is like yin and yang, or manicheism, or whatever dualistic religious system. As for Movement and Mind… Hmm… In some very weird way, I feel they must be part of one and the same thing as well. 🙂 In fact, we may also think of those four things as:

  1. 0 (zero)
  2. 1 (one), or as some sine or a cosine, which is anything in-between 0 and 1.
  3. Well… I am not sure! I can’t really separate point 3 and point 4, because they combine point 1 and point 2.

So we’ve don’t have a quadrupality, right? We do have Trinity here, don’t we? […] Maybe. I won’t comment, because I think I just found Unity here. 🙂

The wavefunction and relativity

When reading about quantum theory, and wave mechanics, you will often encounter the rather enigmatic statement that the Schrödinger equation is not relativistically correct. What does that mean?

In my previous post on the wavefunction and relativity, I boldly claimed that relativity theory had been around for quite a while when the young Comte Louis de Broglie wrote his short groundbreaking PhD thesis, back in 1924. Moreover, it is more than likely that he suggested the θ = ω∙t – kx = (E∙t – px)/ħ formula for the argument of the wavefunction exactly because relativity theory had already established the invariance of the four-vector product pμxμ = E∙t – px = pμ‘xμ‘ = E’∙t’ – p’x’. [Note that Planck’s constant, as a physical constant, should obviously not depend on the reference frame either. Hence, if the E∙t – px product is invariant, so is (E∙t – px)/ħ.] However, I didn’t prove that, and I didn’t relate it to Schrödinger’s equation. Hence, let’s explore the matter somewhat further here.

I don’t want to do the academic thing, of course – and that is to prove the invariance of the four-vector dot product. If you want such proof, let me just give you a link to some course material that does just that. Here, I will just summarize the conclusions of such course material:

  1. Four-vector dot products – like xμxμ = xμ2, pμpμ = pμ2, the spacetime interval s= (Δr)– Δt2, or our pμxμ product here – are invariant under a Lorentz transformation (aka as a Lorentz boost). To be formally correct, I should write xμxμ, pμpμ, and pμxμ, because the product multiplies a row vector with a column vector, which is what the sub- and superscript indicate.
  2. Four-vector dot products are referred to as Lorentz scalars.
  3. When derivatives are involved, we must use the so-called four-gradient, which is denoted by  or μ and defined as:

 = μ = (∂/∂t, –) = (∂/∂t, –∂/∂x, –∂/∂y, –∂/∂z)

Applying the four-gradient vector operator to the wavefunction, we get:

μψ= (∂ψ/∂t, –ψ) = (∂ψ/∂t, –∂ψ/∂x, –∂ψ/∂y, –∂ψ/∂z)

We wrote about that in the context of electromagnetic theory (see, for instance, my post on the relativistic transformation of fields), so I won’t dwell on it here. Note, however, that that’s the weak spot in Schrödinger’s equation: it’s good, but not good enough. However, in the context in which it’s being used – i.e. to calculate electron orbitals – the approximation works just fine, so you shouldn’t worry about it. The point to remember is that the wavefunction itself is relativistically correct. 🙂

Of course, it is always good to work through a simple example, so let’s do that here. Let me first remind you of that transformation we presented a couple of times already, and that’s how to calculate the argument of the wavefunction in the reference frame of the particle itself, i.e. the inertial frame. It goes like this: when measuring all variables in Planck units, the physical constants ħ and c are numerically equal to one, then we can then re-write the argument of the wavefunction as follows:

  1. ħ = 1 ⇒ θ = (E∙t – p∙x)/ħ = E∙t – p∙x = Ev∙t − (mvv)∙x
  2. E= E0/√(1−v2) and m= m0/√(1−v2)  ⇒ θ = [E0/√(1−v2)]∙t – [m0v/√(1−v2)]∙x
  3. c = 1 ⇒ m0 = E⇒ θ = [E0/√(1−v2)]∙t – [E0v/√(1−v2)]∙x = E0∙(t − v∙x)/√(1−v2)

⇔ θ = E0∙t’ = E’·t’ with t’ = (t − v∙x)/√(1−v2)

The t’ in the θ = E0∙t’ expression is, obviously, the proper time as measured in the inertial reference frame. Needless to say, is the relative velocity, which is usually denoted by β. Note that this derivation uses the numerical m0 = E0 identity, which emerges when using natural time and distance units (c = 1). However, while mass and energy are equivalent, they are different physical concepts and, hence, they still have different physical dimensions. It is interesting to spell out what happens with the dimensions here:

  • The dimension of Evt and/or E0∙t’ is (N∙m)∙s, i.e. the dimension of (physical) action.
  • The dimension of the (mvv)∙x term must be the same, but how is that possible? Despite us using natural units – so the value of is now some number between 0 and 1 – velocity is what it is: velocity. Hence, its dimension is m/s. Hence, the dimension of the mvv∙x term is kg∙m = (N∙s2/m)∙(m/s)∙m = N∙m∙s.
  • Hence, the dimension of the [E0v/√(1−v2)]∙x term only makes sense if we remember the m2/s2 dimension of the c2 factor in the E = m∙c2 equivalence relation. We write: [E0v∙x] = [E0]∙[v]∙[x] = [(N∙m)∙(s2/m2)]∙(m/s)∙m = N∙m∙s. In short, when doing the mv = Ev and/or m0 = E0 substitution, we should not get rid of the physical 1/c2 dimension.

That should be clear enough. Let’s now do the example. The rest energy of an electron, expressed in Planck units, EeP = Ee/EP = (0.511×10eV)/(1.22×1028 eV) = 4.181×10−23. That is a very tiny fraction. However, the numerical value of the Planck time unit is even smaller: about 5.4×10−44 seconds. Hence, as a frequency is expressed as the number of cycles (or, as an angular frequency, as the number of radians) per time unit, the natural frequency of the wavefunction of the electron is 4.181×10−23 rad per Planck time unit, so that’s a frequency in the order of [4.181×10−23/(2π)]/(5.4×10−44 s) ≈ 1×1020 cycles per second (or hertz). The relevant calculations are given hereunder.

Electron
Rest energy (in joule) 8.1871E-14
Planck energy (in joule) 1.9562E+09
Rest energy in Planck units 4.1853E-23
Frequency in cycles per second 1.2356E+20

Because of these rather incredible numbers (like 10–31 or 1020), the calculations are not always very obvious, but the logic is clear enough: a higher rest mass increases the (angular) frequency of the real and imaginary part of the wavefunction, and gives them a much higher density in spacetime. How does a frequency like 1.235×1020 Hz compare to, say, the frequency of gamma rays. The answer may surprise you: they are of the same order, as is their energy! 🙂 However, their nature, as a wave ,is obviously very different: gamma rays are an electromagnetic wave, so they involve an E and B vector, rather than the two components of the matter-wave. As an energy propagation mechanism, they are structurally similar, though, as I showed in my previous post.

Now, the typical speed of an electron is given by of the fine-structure constant (α), which is (also) equal to the  is the (relative) speed of an electron (for the many interpretations of the fine-structure constant, see my post on it). So we write:

α = β = v/c

More importantly, we can use this formula to calculate it, which is done hereunder. As you can see, while the typical electron speed is quite impressive (about 2,188 km per second), it is only a fraction of the speed of light and, therefore, the Lorentz factor is still equal to one for all practical purposes. Therefore, its speed adds hardly anything to its energy.

 

Fine-structure constant 0.007297353
Typical speed of the electron (m/s) 2.1877E+06
Typical speed of the electron (km/s) 2,188 km/s
Lorentz factor (γ) 1.0000266267

But I admit it does have momentum now and, hence, the p∙x term in the θ = E∙t – p∙x comes into play. What is its momentum? That’s calculated below. Remember we calculate all in Planck units here!

Electron energy moving at alpha (in Planck units) 4.1854E-23
Electron mass moving at alpha (in Planck units) 4.1854E-23
Planck momentum (p = m·v = m·α ) 3.0542E-25

The momentum is tiny, but it’s real. Also note the increase in its energy. Now, when substituting x for x = v·t, we get the following formula for the argument of our wavefunction:

θ = E·t – p·x = E·t − p·v·t = mv·t − mv·v·v·t = mv·(1 − v2)·t

Now, how does that compare to our θ = θ = E0∙t’ = E’·t’ expression? Well… The value of the two coefficients is calculated below. You can, effectively, see it hardly matters.

mv·(1 − v2) 4.1852E-23
Rest energy in Planck units 4.1853E-23

With that, we are finally ready to use the non-relativistic Schrödinger equation in a non-relativistic way, i.e. we can start calculating electron orbitals with it now, which is what we did in one of my previous posts, but I will re-visit that post soon – and provide some extra commentary! 🙂

The Poynting vector for the matter-wave

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.

amw propagation

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 afactor is the maximum amplitude – which depends on the initial conditions, i.e. the initial pull or push. The ωin 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, = √(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×bn = |a|·|bsinφ·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:

Cross_product

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:

ab = |a|·|bcosφ

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:

  1. [∂E/∂t] = (N/C)/s = N/(C·s)
  2. [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 = ex×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 = ex×E = i·E formulas, jointly, only make sense if we’d assign the s/m dimension to ex and/or to i? I know you’ll think that’s nonsense because you’ve learned to think of the ex× and/or 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.

circle_cos_sin

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:

energy density

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 cfactor that comes with the B•B product. It does two things here:

  1. As a physical constant, with some dimension of its own, it ensures that the dimensions on both sides of the equation come out alright.
  2. 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:

  1. E is measured in newton per coulomb, so [E•E] = [E2] = N2/C2.
  2. 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.
  3. 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/mwith m/m, and then its dimension becomes N·m/m= 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:

energy flux

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·ei·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:

0] = [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/s= 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 ex 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/s= N/m2

As weird as it is, it all works out. We can calculate 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:

  1. 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.
  2. 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

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. 🙂

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/sidentity, 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:

  1. 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.
  2. 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. 🙂