Tag Archives: interpretation of quantum mechanics
The metaphysics of physics
I realized that my last posts were just some crude and rude soundbites, so I thought it would be good to briefly summarize them into something more coherent. Please let me know what you think of it.
The Uncertainty Principle: epistemology versus physics
Anyone who has read anything about quantum physics will know that its concepts and principles are very non-intuitive. Several interpretations have therefore emerged. The mainstream interpretation of quantum mechanics is referred to as the Copenhagen interpretation. It mainly distinguishes itself from more frivolous interpretations (such as the many-worlds and the pilot-wave interpretations) because it is… Well… Less frivolous. Unfortunately, the Copenhagen interpretation itself seems to be subject to interpretation.
One such interpretation may be referred to as radical skepticism – or radical empiricism[1]: we can only say something meaningful about Schrödinger’s cat if we open the box and observe its state. According to this rather particular viewpoint, we cannot be sure of its reality if we don’t make the observation. All we can do is describe its reality by a superposition of the two possible states: dead or alive. That’s Hilbert’s logic[2]: the two states (dead or alive) are mutually exclusive but we add them anyway. If a tree falls in the wood and no one hears it, then it is both standing and not standing. Richard Feynman – who may well be the most eminent representative of mainstream physics – thinks this epistemological position is nonsensical, and I fully agree with him:
“A real tree falling in a real forest makes a sound, of course, even if nobody is there. Even if no one is present to hear it, there are other traces left. The sound will shake some leaves, and if we were careful enough we might find somewhere that some thorn had rubbed against a leaf and made a tiny scratch that could not be explained unless we assumed the leaf were vibrating.” (Feynman’s Lectures, III-2-6)
So what is the mainstream physicist’s interpretation of the Copenhagen interpretation of quantum mechanics then? To fully answer that question, I should encourage the reader to read all of Feynman’s Lectures on quantum mechanics. But then you are reading this because you don’t want to do that, so let me quote from his introductory Lecture on the Uncertainty Principle: “Making an observation affects the phenomenon. The point is that the effect cannot be disregarded or minimized or decreased arbitrarily by rearranging the apparatus. When we look for a certain phenomenon we cannot help but disturb it in a certain minimum way.” (ibidem)
It has nothing to do with consciousness. Reality and consciousness are two very different things. After having concluded the tree did make a noise, even if no one was there to hear it, he wraps up the philosophical discussion as follows: “We might ask: was there a sensation of sound? No, sensations have to do, presumably, with consciousness. And whether ants are conscious and whether there were ants in the forest, or whether the tree was conscious, we do not know. Let us leave the problem in that form.” In short, I think we can all agree that the cat is dead or alive, or that the tree is standing or not standing¾regardless of the observer. It’s a binary situation. Not something in-between. The box obscures our view. That’s all. There is nothing more to it.
Of course, in quantum physics, we don’t study cats but look at the behavior of photons and electrons (we limit our analysis to quantum electrodynamics – so we won’t discuss quarks or other sectors of the so-called Standard Model of particle physics). The question then becomes: what can we reasonably say about the electron – or the photon – before we observe it, or before we make any measurement. Think of the Stein-Gerlach experiment, which tells us that we’ll always measure the angular momentum of an electron – along any axis we choose – as either +ħ/2 or, else, as -ħ/2. So what’s its state before it enters the apparatus? Do we have to assume it has some definite angular momentum, and that its value is as binary as the state of our cat (dead or alive, up or down)?
We should probably explain what we mean by a definite angular momentum. It’s a concept from classical physics, and it assumes a precise value (or magnitude) along some precise direction. We may challenge these assumptions. The direction of the angular momentum may be changing all the time, for example. If we think of the electron as a pointlike charge – whizzing around in its own space – then the concept of a precise direction of its angular momentum becomes quite fuzzy, because it changes all the time. And if its direction is fuzzy, then its value will be fuzzy as well. In classical physics, such fuzziness is not allowed, because angular momentum is conserved: it takes an outside force – or torque – to change it. But in quantum physics, we have the Uncertainty Principle: some energy (force over a distance, remember) can be borrowed – so to speak – as long as it’s swiftly being returned – within the quantitative limits set by the Uncertainty Principle: ΔE·Δt = ħ/2.
Mainstream physicists – including Feynman – do not try to think about this. For them, the Stern-Gerlach apparatus is just like Schrödinger’s box: it obscures the view. The cat is dead or alive, and each of the two states has some probability – but they must add up to one – and so they will write the state of the electron before it enters the apparatus as the superposition of the up and down states. I must assume you’ve seen this before:
|ψ〉 = C_{up}|up〉 + C_{down}|down〉
It’s the so-called Dirac or bra-ket notation. C_{up} is the amplitude for the electron spin to be equal to +ħ/2 along the chosen direction – which we refer to as the z-direction because we will choose our reference frame such that the z-axis coincides with this chosen direction – and, likewise, C_{up} is the amplitude for the electron spin to be equal to -ħ/2 (along the same direction, obviously). C_{up} and C_{up} will be functions, and the associated probabilities will vary sinusoidally – with a phase difference so as to make sure both add up to one.
The model is consistent, but it feels like a mathematical trick. This description of reality – if that’s what it is – does not feel like a model of a real electron. It’s like reducing the cat in our box to the mentioned fuzzy state of being alive and dead at the same time. Let’s try to come up with something more exciting. 😊
[1] Academics will immediately note that radical empiricism and radical skepticism are very different epistemological positions but we are discussing some basic principles in physics here rather than epistemological theories.
[2] The reference to Hilbert’s logic refers to Hilbert spaces: a Hilbert space is an abstract vector space. Its properties allow us to work with quantum-mechanical states, which become state vectors. You should not confuse them with the real or complex vectors you’re used to. The only thing state vectors have in common with real or complex vectors is that (1) we also need a base (aka as a representation in quantum mechanics) to define them and (2) that we can make linear combinations.
The ‘flywheel’ electron model
Physicists describe the reality of electrons by a wavefunction. If you are reading this article, you know how a wavefunction looks like: it is a superposition of elementary wavefunctions. These elementary wavefunctions are written as A_{i}·exp(-iθ_{i}), so they have an amplitude A_{i} and an argument θ_{i} = (E_{i}/ħ)·t – (p_{i}/ħ)·x. Let’s forget about uncertainty, so we can drop the index (i) and think of a geometric interpretation of A·exp(-iθ) = A·e^{–i}^{θ}.
Here we have a weird thing: physicists think the minus sign in the exponent (-iθ) should always be there: the convention is that we get the imaginary unit (i) by a 90° rotation of the real unit (1) – but the rotation is counterclockwise rotation. I like to think a rotation in the clockwise direction must also describe something real. Hence, if we are seeking a geometric interpretation, then we should explore the two mathematical possibilities: A·e^{–i}^{θ} and A·e^{+i}^{θ}. I like to think these two wavefunctions describe the same electron but with opposite spin. How should we visualize this? I like to think of A·e^{–i}^{θ} and A·e^{+i}^{θ} as two-dimensional harmonic oscillators:
A·e^{–i}^{θ} = cos(-θ) + i·sin(-θ) = cosθ – i·sinθ
A·e^{+i}^{θ} = cosθ + i·sinθ
So we may want to imagine our electron as a pointlike electric charge (see the green dot in the illustration below) to spin around some center in either of the two possible directions. The cosine keeps track of the oscillation in one dimension, while the sine (plus or minus) keeps track of the oscillation in a direction that is perpendicular to the first one.
Figure 1: A pointlike charge in orbit
So we have a weird oscillator in two dimensions here, and we may calculate the energy in this oscillation. To calculate such energy, we need a mass concept. We only have a charge here, but a (moving) charge has an electromagnetic mass. Now, the electromagnetic mass of the electron’s charge may or may not explain all the mass of the electron (most physicists think it doesn’t) but let’s assume it does for the sake of the model that we’re trying to build up here. The point is: the theory of electromagnetic mass gives us a very simple explanation for the concept of mass here, and so we’ll use it for the time being. So we have some mass oscillating in two directions simultaneously: we basically assume space is, somehow, elastic. We have worked out the V-2 engine metaphor before, so we won’t repeat ourselves here.
Figure 2: A perpetuum mobile?
Previously unrelated but structurally similar formulas may be related here:
- The energy of an oscillator: E = (1/2)·m·a^{2}ω^{2}
- Kinetic energy: E = (1/2)·m·v^{2}
- The rotational (kinetic) energy that’s stored in a flywheel: E = (1/2)·I·ω^{2} = (1/2)·m·r^{2}·ω^{2}
- Einstein’s energy-mass equivalence relation: E = m·c^{2}
Of course, we are mixing relativistic and non-relativistic formulas here, and there’s the 1/2 factor – but these are minor issues. For example, we were talking not one but two oscillators, so we should add their energies: (1/2)·m·a^{2}·ω^{2} + (1/2)·m·a^{2}·ω^{2} = m·a^{2}·ω^{2}. Also, one can show that the classical formula for kinetic energy (i.e. E = (1/2)·m·v^{2}) morphs into E = m·c^{2} when we use the relativistically correct force equation for an oscillator. So, yes, our metaphor – or our suggested physical interpretation of the wavefunction, I should say – makes sense.
If you know something about physics, then you know the concept of the electromagnetic mass – its mathematical derivation, that is – gives us the classical electron radius, aka as the Thomson radius. It’s the smallest of a trio of radii that are relevant when discussing electrons: the other two radii are the Bohr radius and the Compton scattering radius respectively. The Thomson radius is used in the context of elastic scattering: the frequency of the incident particle (usually a photon), and the energy of the electron itself, do not change. In contrast, Compton scattering does change the frequency of the photon that is being scattered, and also impacts the energy of our electron. [As for the Bohr radius, you know that’s the radius of an electron orbital, roughly speaking – or the size of a hydrogen atom, I should say.]
Now, if we combine the E = m·a^{2}·ω^{2} and E = m·c^{2} equations, then a·ω must be equal to c, right? Can we show this? Maybe. It is easy to see that we get the desired equality by substituting the amplitude of the oscillation (a) for the Compton scattering radius r = ħ/(m·c), and ω (the (angular) frequency of the oscillation) by using the Planck relation (ω = E/ħ):
a·ω = [ħ/(m·c)]·[E/ħ] = E/(m·c) = m·c^{2}/(m·c) = c
We get a wonderfully simple geometric model of an electron here: an electric charge that spins around in a plane. Its radius is the Compton electron radius – which makes sense – and the radial velocity of our spinning charge is the speed of light – which may or may not make sense. Of course, we need an explanation of why this spinning charge doesn’t radiate its energy away – but then we don’t have such explanation anyway. All we can say is that the electron charge seems to be spinning in its own space – that it’s racing along a geodesic. It’s just like mass creates its own space here: according to Einstein’s general relativity theory, gravity becomes a pseudo-force—literally: no real force. How? I am not sure: the model here assumes the medium – empty space – is, somehow, perfectly elastic: the electron constantly borrows energy from one direction and then returns it to the other – so to speak. A crazy model, yes – but is there anything better? We only want to present a metaphor here: a possible visualization of quantum-mechanical models.
However, if this model is to represent anything real, then many more questions need to be answered. For starters, let’s think about an interpretation of the results of the Stern-Gerlach experiment.
Precession
A spinning charge is a tiny magnet – and so it’s got a magnetic moment, which we need to explain the Stern-Gerlach experiment. But it doesn’t explain the discrete nature of the electron’s angular momentum: it’s either +ħ/2 or -ħ/2, nothing in-between, and that’s the case along any direction we choose. How can we explain this? Also, space is three-dimensional. Why would electrons spin in a perfect plane? The answer is: they don’t.
Indeed, the corollary of the above-mentioned binary value of the angular momentum is that the angular momentum – or the electron’s spin – is never completely along any direction. This may or may not be explained by the precession of a spinning charge in a field, which is illustrated below (illustration taken from Feynman’s Lectures, II-35-3).
Figure 3: Precession of an electron in a magnetic field
So we do have an oscillation in three dimensions here, really – even if our wavefunction is a two-dimensional mathematical object. Note that the measurement (or the Stein-Gerlach apparatus in this case) establishes a line of sight and, therefore, a reference frame, so ‘up’ and ‘down’, ‘left’ and ‘right’, and ‘in front’ and ‘behind’ get meaning. In other words, we establish a real space. The question then becomes: how and why does an electron sort of snap into place?
The geometry of the situation suggests the logical angle of the angular momentum vector should be 45°. Now, if the value of its z-component (i.e. its projection on the z-axis) is to be equal to ħ/2, then the magnitude of J itself should be larger. To be precise, it should be equal to ħ/√2 ≈ 0.7·ħ (just apply Pythagoras’ Theorem). Is that value compatible with our flywheel model?
Maybe. Let’s see. The classical formula for the magnetic moment is μ = I·A, with I the (effective) current and A the (surface) area. The notation is confusing because I is also used for the moment of inertia, or rotational mass, but… Well… Let’s do the calculation. The effective current is the electron charge (q_{e}) divided by the period (T) of the orbital revolution: : I = q_{e}/T. The period of the orbit is the time that is needed for the electron to complete one loop. That time (T) is equal to the circumference of the loop (2π·a) divided by the tangential velocity (v_{t}). Now, we suggest v_{t} = r·ω = a·ω = c, and the circumference of the loop is 2π·a. For a, we still use the Compton radius a = ħ/(m·c). Now, the formula for the area is A = π·a^{2}, so we get:
μ = I·A = [q_{e}/T]·π·a^{2} = [q_{e}·c/(2π·a)]·[π·a^{2}] = [(q_{e}·c)/2]·a = [(q_{e}·c)/2]·[ħ/(m·c)] = [q_{e}/(2m)]·ħ
In a classical analysis, we have the following relation between angular momentum and magnetic moment:
μ = (q_{e}/2m)·J
Hence, we find that the angular momentum J is equal to ħ, so that’s twice the measured value. We’ve got a problem. We would have hoped to find ħ/2 or ħ/√2. Perhaps it’s because a = ħ/(m·c) is the so-called reduced Compton scattering radius…
Well… No.
Maybe we’ll find the solution one day. I think it’s already quite nice we have a model that’s accurate up to a factor of 1/2 or 1/√2. 😊
Post scriptum: I’ve turned this into a small article which may or may not be more readable. You can link to it here. Comments are more than welcome.
Quantum math: garbage in, garbage out?
This post is basically a continuation of my previous one but – as you can see from its title – it is much more aggressive in its language, as I was inspired by a very thoughtful comment on my previous post. Another advantage is that it avoids all of the math. 🙂 It’s… Well… I admit it: it’s just a rant. 🙂 [Those who wouldn’t appreciate the casual style of what follows, can download my paper on it – but that’s much longer and also has a lot more math in it – so it’s a much harder read than this ‘rant’.]
My previous post was actually triggered by an attempt to re-read Feynman’s Lectures on Quantum Mechanics, but in reverse order this time: from the last chapter to the first. [In case you doubt, I did follow the correct logical order when working my way through them for the first time because… Well… There is no other way to get through them otherwise. 🙂 ] But then I was looking at Chapter 20. It’s a Lecture on quantum-mechanical operators – so that’s a topic which, in other textbooks, is usually tackled earlier on. When re-reading it, I realize why people quickly turn away from the topic of physics: it’s a lot of mathematical formulas which are supposed to reflect reality but, in practice, few – if any – of the mathematical concepts are actually being explained. Not in the first chapters of a textbook, not in its middle ones, and… Well… Nowhere, really. Why? Well… To be blunt: I think most physicists themselves don’t really understand what they’re talking about. In fact, as I have pointed out a couple of times already, Feynman himself admits so much:
“Atomic behavior appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. Even the experts do not understand it the way they would like to.”
So… Well… If you’d be in need of a rather spectacular acknowledgement of the shortcomings of physics as a science, here you have it: if you don’t understand what physicists are trying to tell you, don’t worry about it, because they don’t really understand it themselves. 🙂
Take the example of a physical state, which is represented by a state vector, which we can combine and re-combine using the properties of an abstract Hilbert space. Frankly, I think the word is very misleading, because it actually doesn’t describe an actual physical state. Why? Well… If we look at this so-called physical state from another angle, then we need to transform it using a complicated set of transformation matrices. You’ll say: that’s what we need to do when going from one reference frame to another in classical mechanics as well, isn’t it?
Well… No. In classical mechanics, we’ll describe the physics using geometric vectors in three dimensions and, therefore, the base of our reference frame doesn’t matter: because we’re using real vectors (such as the electric of magnetic field vectors E and B), our orientation vis-á-vis the object – the line of sight, so to speak – doesn’t matter.
In contrast, in quantum mechanics, it does: Schrödinger’s equation – and the wavefunction – has only two degrees of freedom, so to speak: its so-called real and its imaginary dimension. Worse, physicists refuse to give those two dimensions any geometric interpretation. Why? I don’t know. As I show in my previous posts, it would be easy enough, right? We know both dimensions must be perpendicular to each other, so we just need to decide if both of them are going to be perpendicular to our line of sight. That’s it. We’ve only got two possibilities here which – in my humble view – explain why the matter-wave is different from an electromagnetic wave.
I actually can’t quite believe the craziness when it comes to interpreting the wavefunction: we get everything we’d want to know about our particle through these operators (momentum, energy, position, and whatever else you’d need to know), but mainstream physicists still tell us that the wavefunction is, somehow, not representing anything real. It might be because of that weird 720° symmetry – which, as far as I am concerned, confirms that those state vectors are not the right approach: you can’t represent a complex, asymmetrical shape by a ‘flat’ mathematical object!
Huh? Yes. The wavefunction is a ‘flat’ concept: it has two dimensions only, unlike the real vectors physicists use to describe electromagnetic waves (which we may interpret as the wavefunction of the photon). Those have three dimensions, just like the mathematical space we project on events. Because the wavefunction is flat (think of a rotating disk), we have those cumbersome transformation matrices: each time we shift position vis-á-vis the object we’re looking at (das Ding an sich, as Kant would call it), we need to change our description of it. And our description of it – the wavefunction – is all we have, so that’s our reality. However, because that reality changes as per our line of sight, physicists keep saying the wavefunction (or das Ding an sich itself) is, somehow, not real.
Frankly, I do think physicists should take a basic philosophy course: you can’t describe what goes on in three-dimensional space if you’re going to use flat (two-dimensional) concepts, because the objects we’re trying to describe (e.g. non-symmetrical electron orbitals) aren’t flat. Let me quote one of Feynman’s famous lines on philosophers: “These philosophers are always with us, struggling in the periphery to try to tell us something, but they never really understand the subtleties and depth of the problem.” (Feynman’s Lectures, Vol. I, Chapter 16)
Now, I love Feynman’s Lectures but… Well… I’ve gone through them a couple of times now, so I do think I have an appreciation of the subtleties and depth of the problem now. And I tend to agree with some of the smarter philosophers: if you’re going to use ‘flat’ mathematical objects to describe three- or four-dimensional reality, then such approach will only get you where we are right now, and that’s a lot of mathematical mumbo-jumbo for the poor uninitiated. Consistent mumbo-jumbo, for sure, but mumbo-jumbo nevertheless. 🙂 So, yes, I do think we need to re-invent quantum math. 🙂 The description may look more complicated, but it would make more sense.
I mean… If physicists themselves have had continued discussions on the reality of the wavefunction for almost a hundred years now (Schrödinger published his equation in 1926), then… Well… Then the physicists have a problem. Not the philosophers. 🙂 As to how that new description might look like, see my papers on viXra.org. I firmly believe it can be done. This is just a hobby of mine, but… Well… That’s where my attention will go over the coming years. 🙂 Perhaps quaternions are the answer but… Well… I don’t think so either – for reasons I’ll explain later. 🙂
Post scriptum: There are many nice videos on Dirac’s belt trick or, more generally, on 720° symmetries, but this links to one I particularly like. It clearly shows that the 720° symmetry requires, in effect, a special relation between the observer and the object that is being observed. It is, effectively, like there is a leather belt between them or, in this case, we have an arm between the glass and the person who is holding the glass. So it’s not like we are walking around the object (think of the glass of water) and making a full turn around it, so as to get back to where we were. No. We are turning it around by 360°! That’s a very different thing than just looking at it, walking around it, and then looking at it again. That explains the 720° symmetry: we need to turn it around twice to get it back to its original state. So… Well… The description is more about us and what we do with the object than about the object itself. That’s why I think the quantum-mechanical description is defective.
Should we reinvent wavefunction math?
Preliminary note: This post may cause brain damage. 🙂 If you haven’t worked yourself through a good introduction to physics – including the math – you will probably not understand what this is about. So… Well… Sorry. 😦 But if you have… Then this should be very interesting. Let’s go. 🙂
If you know one or two things about quantum math – Schrödinger’s equation and all that – then you’ll agree the math is anything but straightforward. Personally, I find the most annoying thing about wavefunction math are those transformation matrices: every time we look at the same thing from a different direction, we need to transform the wavefunction using one or more rotation matrices – and that gets quite complicated !
Now, if you have read any of my posts on this or my other blog, then you know I firmly believe the wavefunction represents something real or… Well… Perhaps it’s just the next best thing to reality: we cannot know das Ding an sich, but the wavefunction gives us everything we would want to know about it (linear or angular momentum, energy, and whatever else we have an operator for). So what am I thinking of? Let me first quote Feynman’s summary interpretation of Schrödinger’s equation (Lectures, III-16-1):
“We can think of Schrödinger’s equation as describing the diffusion of the probability amplitude from one point to the next. […] But the imaginary coefficient in front of the derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”
Feynman further formalizes this in his Lecture on Superconductivity (Feynman, III-21-2), in which he refers to Schrödinger’s equation as the “equation for continuity of probabilities”. His analysis there is centered on the local conservation of energy, which makes me think Schrödinger’s equation might be an energy diffusion equation. I’ve written about this ad nauseam in the past, and so I’ll just refer you to one of my papers here for the details, and limit this post to the basics, which are as follows.
The wave equation (so that’s Schrödinger’s equation in its non-relativistic form, which is an approximation that is good enough) is written as:The resemblance with the standard diffusion equation (shown below) is, effectively, very obvious:As Feynman notes, it’s just that imaginary coefficient that makes the behavior quite different. How exactly? Well… You know we get all of those complicated electron orbitals (i.e. the various wave functions that satisfy the equation) out of Schrödinger’s differential equation. We can think of these solutions as (complex) standing waves. They basically represent some equilibrium situation, and the main characteristic of each is their energy level. I won’t dwell on this because – as mentioned above – I assume you master the math. Now, you know that – if we would want to interpret these wavefunctions as something real (which is surely what I want to do!) – the real and imaginary component of a wavefunction will be perpendicular to each other. Let me copy the animation for the elementary wavefunction ψ(θ) = a·e^{−i∙θ} = a·e^{−i∙(E/ħ)·t} = a·cos[(E/ħ)∙t] − i·a·sin[(E/ħ)∙t] once more:
So… Well… That 90° angle makes me think of the similarity with the mathematical description of an electromagnetic wave. Let me quickly show you why. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Vψ term – which is just the equivalent of the the sink or source term S in the diffusion equation – disappears. Therefore, Schrödinger’s equation reduces to:
∂ψ(x, t)/∂t = i·(1/2)·(ħ/m_{eff})·∇^{2}ψ(x, t)
Now, the key difference with the diffusion equation – let me write it for you once again: ∂φ(x, t)/∂t = D·∇^{2}φ(x, t) – is that Schrödinger’s equation gives us two equations for the price of one. Indeed, because ψ is a complex-valued function, with a real and an imaginary part, we get the following equations:
- Re(∂ψ/∂t) = −(1/2)·(ħ/m_{eff})·Im(∇^{2}ψ)
- Im(∂ψ/∂t) = (1/2)·(ħ/m_{eff})·Re(∇^{2}ψ)
Huh? Yes. These equations are easily derived from noting that two complex numbers a + i∙b and c + i∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = i∙(ħ/m_{eff})∙∇^{2}ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i^{2} = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i^{2}∙d = − d + i∙c. [Now that we’re getting a bit technical, let me note that the m_{eff} is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write m_{eff} = m.] 🙂 OK. Onwards ! 🙂
The equations above make me think of the equations for an electromagnetic wave in free space (no stationary charges or currents):
- ∂B/∂t = –∇×E
- ∂E/∂t = c^{2}∇×B
Now, these equations – and, I must therefore assume, the other equations above as well – effectively describe a propagation mechanism in spacetime, as illustrated below:
You know how it works for the electromagnetic field: it’s the interplay between circulation and flux. Indeed, circulation around some axis of rotation creates a flux in a direction perpendicular to it, and that flux causes this, and then that, and it all goes round and round and round. 🙂 Something like that. 🙂 I will let you look up how it goes, exactly. The principle is clear enough. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle.
Now, we know the wavefunction consist of a sine and a cosine: the cosine is the real component, and the sine is the imaginary component. Could they be equally real? Could each represent half of the total energy of our particle? I firmly believe they do. The obvious question then is the following: why wouldn’t we represent them as vectors, just like E and B? I mean… Representing them as vectors (I mean real vectors here – something with a magnitude and a direction in a real space – as opposed to state vectors from the Hilbert space) would show they are real, and there would be no need for cumbersome transformations when going from one representational base to another. In fact, that’s why vector notation was invented (sort of): we don’t need to worry about the coordinate frame. It’s much easier to write physical laws in vector notation because… Well… They’re the real thing, aren’t they? 🙂
What about dimensions? Well… I am not sure. However, because we are – arguably – talking about some pointlike charge moving around in those oscillating fields, I would suspect the dimension of the real and imaginary component of the wavefunction will be the same as that of the electric and magnetic field vectors E and B. We may want to recall these:
- E is measured in newton per coulomb (N/C).
- B is measured in newton per coulomb divided by m/s, so that’s (N/C)/(m/s).
The weird dimension of B is because of the weird force law for the magnetic force. It involves a vector cross product, as shown by Lorentz’ formula:
F = qE + q(v×B)
Of course, it is only one force (one and the same physical reality), as evidenced by the fact that we can write B as the following vector cross-product: B = (1/c)∙e_{x}×E, with e_{x} the unit vector pointing in the x-direction (i.e. the direction of propagation of the wave). [Check it, because you may not have seen this expression before. Just take a piece of paper and think about the geometry of the situation.] Hence, we may associate the (1/c)∙e_{x}× operator, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by i also amounts to a rotation by 90° degrees. Hence, if we can agree on a suitable convention for the direction of rotation here, we may boldly write:
B = (1/c)∙e_{x}×E = (1/c)∙i∙E
This is, in fact, what triggered my geometric interpretation of Schrödinger’s equation about a year ago now. I have had little time to work on it, but think I am on the right track. Of course, you should note that, for an electromagnetic wave, the magnitudes of E and B reach their maximum, minimum and zero point simultaneously (as shown below). So their phase is the same.
In contrast, the phase of the real and imaginary component of the wavefunction is not the same, as shown below.
In fact, because of the Stern-Gerlach experiment, I am actually more thinking of a motion like this:
But that shouldn’t distract you. 🙂 The question here is the following: could we possibly think of a new formulation of Schrödinger’s equation – using vectors (again, real vectors – not these weird state vectors) rather than complex algebra?
I think we can, but then I wonder why the inventors of the wavefunction – Heisenberg, Born, Dirac, and Schrödinger himself, of course – never thought of that. 🙂
Hmm… I need to do some research here. 🙂
Post scriptum: You will, of course, wonder how and why the matter-wave would be different from the electromagnetic wave if my suggestion that the dimension of the wavefunction component is the same is correct. The answer is: the difference lies in the phase difference and then, most probably, the different orientation of the angular momentum. Do we have any other possibilities? 🙂
P.S. 2: I also published this post on my new blog: https://readingeinstein.blog/. However, I thought the followers of this blog should get it first. 🙂
Wavefunctions, perspectives, reference frames, representations and symmetries
Ouff ! This title is quite a mouthful, isn’t it? 🙂 So… What’s the topic of the day? Well… In our previous posts, we developed a few key ideas in regard to a possible physical interpretation of the (elementary) wavefunction. It’s been an interesting excursion, and I summarized it in another pre-publication paper on the open arXiv.org site.
In my humble view, one of the toughest issues to deal with when thinking about geometric (or physical) interpretations of the wavefunction is the fact that a wavefunction does not seem to obey the classical 360° symmetry in space. In this post, I want to muse a bit about this and show that… Well… It does and it doesn’t. It’s got to do with what happens when you change from one representational base (or representation, tout court) to another which is… Well… Like changing the reference frame but, at the same time, it is also more than just a change of the reference frame—and so that explains the weird stuff (like that 720° symmetry of the amplitudes for spin-1/2 particles, for example).
I should warn you before you start reading: I’ll basically just pick up some statements from my paper (and previous posts) and develop some more thoughts on them. As a result, this post may not be very well structured. Hence, you may want to read the mentioned paper first.
The reality of directions
Huh? The reality of directions? Yes. I warned you. This post may cause brain damage. 🙂 The whole argument revolves around a thought experiment—but one whose results have been verified in zillions of experiments in university student labs so… Well… We do not doubt the results and, therefore, we do not doubt the basic mathematical results: we just want to try to understand them better.
So what is the set-up? Well… In the illustration below (Feynman, III, 6-3), Feynman compares the physics of two situations involving rather special beam splitters. Feynman calls them modified or ‘improved’ Stern-Gerlach apparatuses. The apparatus basically splits and then re-combines the two new beams along the z-axis. It is also possible to block one of the beams, so we filter out only particles with their spin up or, alternatively, with their spin down. Spin (or angular momentum or the magnetic moment) as measured along the z-axis, of course—I should immediately add: we’re talking the z-axis of the apparatus here.
The two situations involve a different relative orientation of the apparatuses: in (a), the angle is 0°, while in (b) we have a (right-handed) rotation of 90° about the z-axis. He then proves—using geometry and logic only—that the probabilities and, therefore, the magnitudes of the amplitudes (denoted by C_{+} and C_{−} and C’_{+} and C’_{−} in the S and T representation respectively) must be the same, but the amplitudes must have different phases, noting—in his typical style, mixing academic and colloquial language—that “there must be some way for a particle to tell that it has turned a corner in (b).”
The various interpretations of what actually happens here may shed some light on the heated discussions on the reality of the wavefunction—and of quantum states. In fact, I should note that Feynman’s argument revolves around quantum states. To be precise, the analysis is focused on two-state systems only, and the wavefunction—which captures a continuum of possible states, so to speak—is introduced only later. However, we may look at the amplitude for a particle to be in the up– or down-state as a wavefunction and, therefore (but do note that’s my humble opinion once more), the analysis is actually not all that different.
We know, from theory and experiment, that the amplitudes are different. For example, for the given difference in the relative orientation of the two apparatuses (90°), we know that the amplitudes are given by C’_{+} = e^{i}^{∙φ/2}∙C_{+} = e^{ i}^{∙π/4}∙C_{+} and C’_{−} = e^{−i∙φ/2}∙C_{+} = e^{− i∙π/4}∙C_{−} respectively (the amplitude to go from the down to the up state, or vice versa, is zero). Hence, yes, we—not the particle, Mr. Feynman!—know that, in (b), the electron has, effectively, turned a corner.
The more subtle question here is the following: is the reality of the particle in the two setups the same? Feynman, of course, stays away from such philosophical question. He just notes that, while “(a) and (b) are different”, “the probabilities are the same”. He refrains from making any statement on the particle itself: is or is it not the same? The common sense answer is obvious: of course, it is! The particle is the same, right? In (b), it just took a turn—so it is just going in some other direction. That’s all.
However, common sense is seldom a good guide when thinking about quantum-mechanical realities. Also, from a more philosophical point of view, one may argue that the reality of the particle is not the same: something might—or must—have happened to the electron because, when everything is said and done, the particle did take a turn in (b). It did not in (a). [Note that the difference between ‘might’ and ‘must’ in the previous phrase may well sum up the difference between a deterministic and a non-deterministic world view but… Well… This discussion is going to be way too philosophical already, so let’s refrain from inserting new language here.]
Let us think this through. The (a) and (b) set-up are, obviously, different but… Wait a minute… Nothing is obvious in quantum mechanics, right? How can we experimentally confirm that they are different?
Huh? I must be joking, right? You can see they are different, right? No. I am not joking. In physics, two things are different if we get different measurement results. [That’s a bit of a simplified view of the ontological point of view of mainstream physicists, but you will have to admit I am not far off.] So… Well… We can’t see those amplitudes and so… Well… If we measure the same thing—same probabilities, remember?—why are they different? Think of this: if we look at the two beam splitters as one single tube (an ST tube, we might say), then all we did in (b) was bend the tube. Pursuing the logic that says our particle is still the same even when it takes a turn, we could say the tube is still the same, despite us having wrenched it over a 90° corner.
Now, I am sure you think I’ve just gone nuts, but just try to stick with me a little bit longer. Feynman actually acknowledges the same: we need to experimentally prove (a) and (b) are different. He does so by getting a third apparatus in (U), as shown below, whose relative orientation to T is the same in both (a) and (b), so there is no difference there.
Now, the axis of U is not the z-axis: it is the x-axis in (a), and the y-axis in (b). So what? Well… I will quote Feynman here—not (only) because his words are more important than mine but also because every word matters here:
“The two apparatuses in (a) and (b) are, in fact, different, as we can see in the following way. Suppose that we put an apparatus in front of S which produces a pure +x state. Such particles would be split into +z and −z into beams in S, but the two beams would be recombined to give a +x state again at P_{1}—the exit of S. The same thing happens again in T. If we follow T by a third apparatus U, whose axis is in the +x direction and, as shown in (a), all the particles would go into the + beam of U. Now imagine what happens if T and U are swung around together by 90° to the positions shown in (b). Again, the T apparatus puts out just what it takes in, so the particles that enter U are in a +x state with respect to S, which is different. By symmetry, we would now expect only one-half of the particles to get through.”
I should note that (b) shows the U apparatus wide open so… Well… I must assume that’s a mistake (and should alert the current editors of the Lectures to it): Feynman’s narrative tells us we should also imagine it with the minus channel shut. In that case, it should, effectively, filter approximately half of the particles out, while they all get through in (a). So that’s a measurement result which shows the direction, as we see it, makes a difference.
Now, Feynman would be very angry with me—because, as mentioned, he hates philosophers—but I’d say: this experiment proves that a direction is something real. Of course, the next philosophical question then is: what is a direction? I could answer this by pointing to the experiment above: a direction is something that alters the probabilities between the STU tube as set up in (a) versus the STU tube in (b). In fact—but, I admit, that would be pretty ridiculous—we could use the varying probabilities as we wrench this tube over varying angles to define an angle! But… Well… While that’s a perfectly logical argument, I agree it doesn’t sound very sensical.
OK. Next step. What follows may cause brain damage. 🙂 Please abandon all pre-conceived notions and definitions for a while and think through the following logic.
You know this stuff is about transformations of amplitudes (or wavefunctions), right? [And you also want to hear about those special 720° symmetry, right? No worries. We’ll get there.] So the questions all revolve around this: what happens to amplitudes (or the wavefunction) when we go from one reference frame—or representation, as it’s referred to in quantum mechanics—to another?
Well… I should immediately correct myself here: a reference frame and a representation are two different things. They are related but… Well… Different… Quite different. Not same-same but different. 🙂 I’ll explain why later. Let’s go for it.
Before talking representations, let us first think about what we really mean by changing the reference frame. To change it, we first need to answer the question: what is our reference frame? It is a mathematical notion, of course, but then it is also more than that: it is our reference frame. We use it to make measurements. That’s obvious, you’ll say, but let me make a more formal statement here:
The reference frame is given by (1) the geometry (or the shape, if that sounds easier to you) of the measurement apparatus (so that’s the experimental set-up) here) and (2) our perspective of it.
If we would want to sound academic, we might refer to Kant and other philosophers here, who told us—230 years ago—that the mathematical idea of a three-dimensional reference frame is grounded in our intuitive notions of up and down, and left and right. [If you doubt this, think about the necessity of the various right-hand rules and conventions that we cannot do without in math, and in physics.] But so we do not want to sound academic. Let us be practical. Just think about the following. The apparatus gives us two directions:
(1) The up direction, which we associate with the positive direction of the z-axis, and
(2) the direction of travel of our particle, which we associate with the positive direction of the y-axis.
Now, if we have two axes, then the third axis (the x-axis) will be given by the right-hand rule, right? So we may say the apparatus gives us the reference frame. Full stop. So… Well… Everything is relative? Is this reference frame relative? Are directions relative? That’s what you’ve been told, but think about this: relative to what? Here is where the object meets the subject. What’s relative? What’s absolute? Frankly, I’ve started to think that, in this particular situation, we should, perhaps, not use these two terms. I am not saying that our observation of what physically happens here gives these two directions any absolute character but… Well… You will have to admit they are more than just some mathematical construct: when everything is said and done, we will have to admit that these two directions are real. because… Well… They’re part of the reality that we are observing, right? And the third one… Well… That’s given by our perspective—by our right-hand rule, which is… Well… Our right-hand rule.
Of course, now you’ll say: if you think that ‘relative’ and ‘absolute’ are ambiguous terms and that we, therefore, may want to avoid them a bit more, then ‘real’ and its opposite (unreal?) are ambiguous terms too, right? Well… Maybe. What language would you suggest? 🙂 Just stick to the story for a while. I am not done yet. So… Yes… What is their reality? Let’s think about that in the next section.
Perspectives, reference frames and symmetries
You’ve done some mental exercises already as you’ve been working your way through the previous section, but you’ll need to do plenty more. In fact, they may become physical exercise too: when I first thought about these things (symmetries and, more importantly, asymmetries in space), I found myself walking around the table with some asymmetrical everyday objects and papers with arrows and clocks and other stuff on it—effectively analyzing what right-hand screw, thumb or grip rules actually mean. 🙂
So… Well… I want you to distinguish—just for a while—between the notion of a reference frame (think of the x–y–z reference frame that comes with the apparatus) and your perspective on it. What’s our perspective on it? Well… You may be looking from the top, or from the side and, if from the side, from the left-hand side or the right-hand side—which, if you think about it, you can only define in terms of the various positive and negative directions of the various axes. 🙂 If you think this is getting ridiculous… Well… Don’t. Feynman himself doesn’t think this is ridiculous, because he starts his own “long and abstract side tour” on transformations with a very simple explanation of how the top and side view of the apparatus are related to the axes (i.e. the reference frame) that comes with it. You don’t believe me? This is the very first illustration of his Lecture on this:
He uses it to explain the apparatus (which we don’t do here because you’re supposed to already know how these (modified or improved) Stern-Gerlach apparatuses work). So let’s continue this story. Suppose that we are looking in the positive y-direction—so that’s the direction in which our particle is moving—then we might imagine how it would look like when we would make a 180° turn and look at the situation from the other side, so to speak. We do not change the reference frame (i.e. the orientation) of the apparatus here: we just change our perspective on it. Instead of seeing particles going away from us, into the apparatus, we now see particles coming towards us, out of the apparatus.
What happens—but that’s not scientific language, of course—is that left becomes right, and right becomes left. Top still is top, and bottom is bottom. We are looking now in the negative y-direction, and the positive direction of the x-axis—which pointed right when we were looking in the positive y-direction—now points left. I see you nodding your head now—because you’ve heard about parity inversions, mirror symmetries and what have you—and I hear you say: “That’s the mirror world, right?”
No. It is not. I wrote about this in another post: the world in the mirror is the world in the mirror. We don’t get a mirror image of an object by going around it and looking at its back side. I can’t dwell too much on this (just check that post, and another one who talks about the same), but so don’t try to connect it to the discussions on symmetry-breaking and what have you. Just stick to this story, which is about transformations of amplitudes (or wavefunctions). [If you really want to know—but I know this sounds counterintuitive—the mirror world doesn’t really switch left for right. Your reflection doesn’t do a 180 degree turn: it is just reversed front to back, with no rotation at all. It’s only your brain which mentally adds (or subtracts) the 180 degree turn that you assume must have happened from the observed front to back reversal. So the left to right reversal is only apparent. It’s a common misconception, and… Well… I’ll let you figure this out yourself. I need to move on.] Just note the following:
- The xyz reference frame remains a valid right-handed reference frame. Of course it does: it comes with our beam splitter, and we can’t change its reality, right? We’re just looking at it from another angle. Our perspective on it has changed.
- However, if we think of the real and imaginary part of the wavefunction describing the electrons that are going through our apparatus as perpendicular oscillations (as shown below)—a cosine and sine function respectively—then our change in perspective might, effectively, mess up our convention for measuring angles.
I am not saying it does. Not now, at least. I am just saying it might. It depends on the plane of the oscillation, as I’ll explain in a few moments. Think of this: we measure angles counterclockwise, right? As shown below… But… Well… If the thing below would be some funny clock going backwards—you’ve surely seen them in a bar or so, right?—then… Well… If they’d be transparent, and you’d go around them, you’d see them as going… Yes… Clockwise. 🙂 [This should remind you of a discussion on real versus pseudo-vectors, or polar versus axial vectors, but… Well… We don’t want to complicate the story here.]
Now, if we would assume this clock represents something real—and, of course, I am thinking of the elementary wavefunction e^{i}^{θ} = cosθ + i·sinθ now—then… Well… Then it will look different when we go around it. When going around our backwards clock above and looking at it from… Well… The back, we’d describe it, naively, as… Well… Think! What’s your answer? Give me the formula! 🙂
[…]
We’d see it as e^{−i}^{θ} = cos(−θ) + i·sin(−θ) = cosθ − i·sinθ, right? The hand of our clock now goes clockwise, so that’s the opposite direction of our convention for measuring angles. Hence, instead of e^{i}^{θ}, we write e^{−i}^{θ}, right? So that’s the complex conjugate. So we’ve got a different image of the same thing here. Not good. Not good at all.
You’ll say: so what? We can fix this thing easily, right? You don’t need the convention for measuring angles or for the imaginary unit (i) here. This particle is moving, right? So if you’d want to look at the elementary wavefunction as some sort of circularly polarized beam (which, I admit, is very much what I would like to do, but its polarization is rather particular as I’ll explain in a minute), then you just need to define left- and right-handed angles as per the standard right-hand screw rule (illustrated below). To hell with the counterclockwise convention for measuring angles!
You are right. We could use the right-hand rule more consistently. We could, in fact, use it as an alternative convention for measuring angles: we could, effectively, measure them clockwise or counterclockwise depending on the direction of our particle. But… Well… The fact is: we don’t. We do not use that alternative convention when we talk about the wavefunction. Physicists do use the counterclockwise convention all of the time and just jot down these complex exponential functions and don’t realize that, if they are to represent something real, our perspective on the reference frame matters. To put it differently, the direction in which we are looking at things matters! Hence, the direction is not… Well… I am tempted to say… Not relative at all but then… Well… We wanted to avoid that term, right? 🙂
[…]
I guess that, by now, your brain may suffered from various short-circuits. If not, stick with me a while longer. Let us analyze how our wavefunction model might be impacted by this symmetry—or asymmetry, I should say.
The flywheel model of an electron
In our previous posts, we offered a model that interprets the real and the imaginary part of the wavefunction as oscillations which each carry half of the total energy of the particle. These oscillations are perpendicular to each other, and the interplay between both is how energy propagates through spacetime. Let us recap the fundamental premises:
- The dimension of the matter-wave field vector is force per unit mass (N/kg), as opposed to the force per unit charge (N/C) dimension of the electric field vector. This dimension is an acceleration (m/s^{2}), which is the dimension of the gravitational field.
- We assume this gravitational disturbance causes our electron (or a charged mass in general) to move about some center, combining linear and circular motion. This interpretation reconciles the wave-particle duality: fields interfere but if, at the same time, they do drive a pointlike particle, then we understand why, as Feynman puts it, “when you do find the electron some place, the entire charge is there.” Of course, we cannot prove anything here, but our elegant yet simple derivation of the Compton radius of an electron is… Well… Just nice. 🙂
- Finally, and most importantly in the context of this discussion, we noted that, in light of the direction of the magnetic moment of an electron in an inhomogeneous magnetic field, the plane which circumscribes the circulatory motion of the electron should also comprise the direction of its linear motion. Hence, unlike an electromagnetic wave, the plane of the two-dimensional oscillation (so that’s the polarization plane, really) cannot be perpendicular to the direction of motion of our electron.
Let’s say some more about the latter point here. The illustrations below (one from Feynman, and the other is just open-source) show what we’re thinking of. The direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is traveling—cannot be parallel to the direction of motion. On the contrary, it must be perpendicular to the direction of motion. In other words, if we imagine our electron as spinning around some center (see the illustration on the left-hand side), then the disk it circumscribes (i.e. the plane of the polarization) has to comprise the direction of motion.
Of course, we need to add another detail here. As my readers will know, we do not really have a precise direction of angular momentum in quantum physics. While there is no fully satisfactory explanation of this, the classical explanation—combined with the quantization hypothesis—goes a long way in explaining this: an object with an angular momentum J and a magnetic moment μ that is not exactly parallel to some magnetic field B, will not line up: it will precess—and, as mentioned, the quantization of angular momentum may well explain the rest. [Well… Maybe… We have detailed our attempts in this regard in various posts on this (just search for spin or angular momentum on this blog, and you’ll get a dozen posts or so), but these attempts are, admittedly, not fully satisfactory. Having said that, they do go a long way in relating angles to spin numbers.]
The thing is: we do assume our electron is spinning around. If we look from the up-direction only, then it will be spinning clockwise if its angular momentum is down (so its magnetic moment is up). Conversely, it will be spinning counterclockwise if its angular momentum is up. Let us take the up-state. So we have a top view of the apparatus, and we see something like this:I know you are laughing aloud now but think of your amusement as a nice reward for having stuck to the story so far. Thank you. 🙂 And, yes, do check it yourself by doing some drawings on your table or so, and then look at them from various directions as you walk around the table as—I am not ashamed to admit this—I did when thinking about this. So what do we get when we change the perspective? Let us walk around it, counterclockwise, let’s say, so we’re measuring our angle of rotation as some positive angle. Walking around it—in whatever direction, clockwise or counterclockwise—doesn’t change the counterclockwise direction of our… Well… That weird object that might—just might—represent an electron that has its spin up and that is traveling in the positive y-direction.
When we look in the direction of propagation (so that’s from left to right as you’re looking at this page), and we abstract away from its linear motion, then we could, vaguely, describe this by some wrenched e^{i}^{θ} = cosθ + i·sinθ function, right? The x- and y-axes of the apparatus may be used to measure the cosine and sine components respectively.
Let us keep looking from the top but walk around it, rotating ourselves over a 180° angle so we’re looking in the negative y-direction now. As I explained in one of those posts on symmetries, our mind will want to switch to a new reference frame: we’ll keep the z-axis (up is up, and down is down), but we’ll want the positive direction of the x-axis to… Well… Point right. And we’ll want the y-axis to point away, rather than towards us. In short, we have a transformation of the reference frame here: z’ = z, y’ = − y, and x’ = − x. Mind you, this is still a regular right-handed reference frame. [That’s the difference with a mirror image: a mirrored right-hand reference frame is no longer right-handed.] So, in our new reference frame, that we choose to coincide with our perspective, we will now describe the same thing as some −cosθ − i·sinθ = −e^{i}^{θ} function. Of course, −cosθ = cos(θ + π) and −sinθ = sin(θ + π) so we can write this as:
−cosθ − i·sinθ = cos(θ + π) + i·sinθ = e^{i}^{·(}^{θ+π)} = e^{i}^{π}·e^{i}^{θ} = −e^{i}^{θ}.
Sweet ! But… Well… First note this is not the complex conjugate: e^{−i}^{θ} = cosθ − i·sinθ ≠ −cosθ − i·sinθ = −e^{i}^{θ}. Why is that? Aren’t we looking at the same clock, but from the back? No. The plane of polarization is different. Our clock is more like those in Dali’s painting: it’s flat. 🙂 And, yes, let me lighten up the discussion with that painting here. 🙂 We need to have some fun while torturing our brain, right?
So, because we assume the plane of polarization is different, we get an −e^{i}^{θ} function instead of a e^{−i}^{θ} function.
Let us now think about the e^{i}^{·(}^{θ+π)} function. It’s the same as −e^{i}^{θ} but… Well… We walked around the z-axis taking a full 180° turn, right? So that’s π in radians. So that’s the phase shift here. Hey! Try the following now. Go back and walk around the apparatus once more, but let the reference frame rotate with us, as shown below. So we start left and look in the direction of propagation, and then we start moving about the z-axis (which points out of this page, toward you, as you are looking at this), let’s say by some small angle α. So we rotate the reference frame about the z-axis by α and… Well… Of course, our e^{i}^{·}^{θ} now becomes an our e^{i}^{·(}^{θ+α)} function, right? We’ve just derived the transformation coefficient for a rotation about the z-axis, didn’t we? It’s equal to e^{i}^{·}^{α}, right? We get the transformed wavefunction in the new reference frame by multiplying the old one by e^{i}^{·}^{α}, right? It’s equal to e^{i}^{·}^{α}·e^{i}^{·}^{θ} = e^{i}^{·(}^{θ+α)}, right?
Well…
[…]
No. The answer is: no. The transformation coefficient is not e^{i}^{·}^{α} but e^{i}^{·}^{α/2}. So we get an additional 1/2 factor in the phase shift.
Huh? Yes. That’s what it is: when we change the representation, by rotating our apparatus over some angle α about the z-axis, then we will, effectively, get a new wavefunction, which will differ from the old one by a phase shift that is equal to only half of the rotation angle only.
Huh? Yes. It’s even weirder than that. For a spin down electron, the transformation coefficient is e^{−i·}^{α/2}, so we get an additional minus sign in the argument.
Huh? Yes.
I know you are terribly disappointed, but that’s how it is. That’s what hampers an easy geometric interpretation of the wavefunction. Paraphrasing Feynman, I’d say that, somehow, our electron not only knows whether or not it has taken a turn, but it also knows whether or not it is moving away from us or, conversely, towards us.
[…]
But… Hey! Wait a minute! That’s it, right?
What? Well… That’s it! The electron doesn’t know whether it’s moving away or towards us. That’s nonsense. But… Well… It’s like this:
Our e^{i}^{·}^{α} coefficient describes a rotation of the reference frame. In contrast, the e^{i}^{·}^{α/2} and e^{−i·}^{α/2} coefficients describe what happens when we rotate the T apparatus! Now that is a very different proposition.
Right! You got it! Representations and reference frames are different things. Quite different, I’d say: representations are real, reference frames aren’t—but then you don’t like philosophical language, do you? 🙂 But think of it. When we just go about the z-axis, a full 180°, but we don’t touch that T-apparatus, we don’t change reality. When we were looking at the electron while standing left to the apparatus, we watched the electrons going in and moving away from us, and when we go about the z-axis, a full 180°, looking at it from the right-hand side, we see the electrons coming out, moving towards us. But it’s still the same reality. We simply change the reference frame—from xyz to x’y’z’ to be precise: we do not change the representation.
In contrast, when we rotate the T apparatus over a full 180°, our electron now goes in the opposite direction. And whether that’s away or towards us, that doesn’t matter: it was going in one direction while traveling through S, and now it goes in the opposite direction—relative to the direction it was going in S, that is.
So what happens, really, when we change the representation, rather than the reference frame? Well… Let’s think about that. 🙂
Quantum-mechanical weirdness?
The transformation matrix for the amplitude of a system to be in an up or down state (and, hence, presumably, for a wavefunction) for a rotation about the z-axis is the following one:
Feynman derives this matrix in a rather remarkable intellectual tour de force in the 6th of his Lectures on Quantum Mechanics. So that’s pretty early on. He’s actually worried about that himself, apparently, and warns his students that “This chapter is a rather long and abstract side tour, and it does not introduce any idea which we will not also come to by a different route in later chapters. You can, therefore, skip over it, and come back later if you are interested.”
Well… That’s how I approached it. I skipped it, and didn’t worry about those transformations for quite a while. But… Well… You can’t avoid them. In some weird way, they are at the heart of the weirdness of quantum mechanics itself. Let us re-visit his argument. Feynman immediately gets that the whole transformation issue here is just a matter of finding an easy formula for that phase shift. Why? He doesn’t tell us. Lesser mortals like us must just assume that’s how the instinct of a genius works, right? 🙂 So… Well… Because he knows—from experiment—that the coefficient is e^{i}^{·}^{α/2} instead of e^{i}^{·}^{α}, he just says the phase shift—which he denotes by λ—must be some proportional to the angle of rotation—which he denotes by φ rather than α (so as to avoid confusion with the Euler angle α). So he writes:
λ = m·φ
Initially, he also tries the obvious thing: m should be one, right? So λ = φ, right? Well… No. It can’t be. Feynman shows why that can’t be the case by adding a third apparatus once again, as shown below.
Let me quote him here, as I can’t explain it any better:
“Suppose T is rotated by 360°; then, clearly, it is right back at zero degrees, and we should have C’_{+} = C_{+} and C’_{−} = C_{−} or, what is the same thing, e^{i}^{·m·2π} = 1. We get m = 1. [But no!] This argument is wrong! To see that it is, consider that T is rotated by 180°. If m were equal to 1, we would have C’_{+} = e^{i}^{·π}C_{+} = −C_{+} and C’_{−} = e^{−}^{i}^{·π}C_{−} = −C_{−}. [Feynman works with states here, instead of the wavefunction of the particle as a whole. I’ll come back to this.] However, this is just the original state all over again. Both amplitudes are just multiplied by −1 which gives back the original physical system. (It is again a case of a common phase change.) This means that if the angle between T and S is increased to 180°, the system would be indistinguishable from the zero-degree situation, and the particles would again go through the (+) state of the U apparatus. At 180°, though, the (+) state of the U apparatus is the (−x) state of the original S apparatus. So a (+x) state would become a (−x) state. But we have done nothing to change the original state; the answer is wrong. We cannot have m = 1. We must have the situation that a rotation by 360°, and no smaller angle reproduces the same physical state. This will happen if m = 1/2.”
The result, of course, is this weird 720° symmetry. While we get the same physics after a 360° rotation of the T apparatus, we do not get the same amplitudes. We get the opposite (complex) number: C’_{+} = e^{i}^{·2π/2}C_{+} = −C_{+} and C’_{−} = e^{−}^{i}^{·2π/2}C_{−} = −C_{−}. That’s OK, because… Well… It’s a common phase shift, so it’s just like changing the origin of time. Nothing more. Nothing less. Same physics. Same reality. But… Well… C’_{+} ≠ −C_{+} and C’_{−} ≠ −C_{−}, right? We only get our original amplitudes back if we rotate the T apparatus two times, so that’s by a full 720 degrees—as opposed to the 360° we’d expect.
Now, space is isotropic, right? So this 720° business doesn’t make sense, right?
Well… It does and it doesn’t. We shouldn’t dramatize the situation. What’s the actual difference between a complex number and its opposite? It’s like x or −x, or t and −t. I’ve said this a couple of times already again, and I’ll keep saying it many times more: Nature surely can’t be bothered by how we measure stuff, right? In the positive or the negative direction—that’s just our choice, right? Our convention. So… Well… It’s just like that −e^{i}^{θ} function we got when looking at the same experimental set-up from the other side: our e^{i}^{θ} and −e^{i}^{θ} functions did not describe a different reality. We just changed our perspective. The reference frame. As such, the reference frame isn’t real. The experimental set-up is. And—I know I will anger mainstream physicists with this—the representation is. Yes. Let me say it loud and clear here:
A different representation describes a different reality.
In contrast, a different perspective—or a different reference frame—does not.
Conventions
While you might have had a lot of trouble going through all of the weird stuff above, the point is: it is not all that weird. We can understand quantum mechanics. And in a fairly intuitive way, really. It’s just that… Well… I think some of the conventions in physics hamper such understanding. Well… Let me be precise: one convention in particular, really. It’s that convention for measuring angles. Indeed, Mr. Leonhard Euler, back in the 18th century, might well be “the master of us all” (as Laplace is supposed to have said) but… Well… He couldn’t foresee how his omnipresent formula—e^{i}^{θ} = cosθ + i·sinθ—would, one day, be used to represent something real: an electron, or any elementary particle, really. If he would have known, I am sure he would have noted what I am noting here: Nature can’t be bothered by our conventions. Hence, if e^{i}^{θ} represents something real, then e^{−i}^{θ} must also represent something real. [Coz I admire this genius so much, I can’t resist the temptation. Here’s his portrait. He looks kinda funny here, doesn’t he? :-)]
Frankly, he would probably have understood quantum-mechanical theory as easily and instinctively as Dirac, I think, and I am pretty sure he would have noted—and, if he would have known about circularly polarized waves, probably agreed to—that alternative convention for measuring angles: we could, effectively, measure angles clockwise or counterclockwise depending on the direction of our particle—as opposed to Euler’s ‘one-size-fits-all’ counterclockwise convention. But so we did not adopt that alternative convention because… Well… We want to keep honoring Euler, I guess. 🙂
So… Well… If we’re going to keep honoring Euler by sticking to that ‘one-size-fits-all’ counterclockwise convention, then I do believe that e^{i}^{θ} and e^{−i}^{θ} represent two different realities: spin up versus spin down.
Yes. In our geometric interpretation of the wavefunction, these are, effectively, two different spin directions. And… Well… These are real directions: we see something different when they go through a Stern-Gerlach apparatus. So it’s not just some convention to count things like 0, 1, 2, etcetera versus 0, −1, −2 etcetera. It’s the same story again: different but related mathematical notions are (often) related to different but related physical possibilities. So… Well… I think that’s what we’ve got here. Think of it. Mainstream quantum math treats all wavefunctions as right-handed but… Well… A particle with up spin is a different particle than one with down spin, right? And, again, Nature surely cannot be bothered about our convention of measuring phase angles clockwise or counterclockwise, right? So… Well… Kinda obvious, right? 🙂
Let me spell out my conclusions here:
1. The angular momentum can be positive or, alternatively, negative: J = +ħ/2 or −ħ/2. [Let me note that this is not obvious. Or less obvious than it seems, at first. In classical theory, you would expect an electron, or an atomic magnet, to line up with the field. Well… The Stern-Gerlach experiment shows they don’t: they keep their original orientation. Well… If the field is weak enough.]
2. Therefore, we would probably like to think that an actual particle—think of an electron, or whatever other particle you’d think of—comes in two variants: right-handed and left-handed. They will, therefore, either consist of (elementary) right-handed waves or, else, (elementary) left-handed waves. An elementary right-handed wave would be written as: ψ(θ_{i}) = e^{i}^{θi} = a_{i}·(cosθ_{i} + i·sinθ_{i}). In contrast, an elementary left-handed wave would be written as: ψ(θ_{i}) = e^{−i}^{θi} = a_{i}·(cosθ_{i} − i·sinθ_{i}). So that’s the complex conjugate.
So… Well… Yes, I think complex conjugates are not just some mathematical notion: I believe they represent something real. It’s the usual thing: Nature has shown us that (most) mathematical possibilities correspond to real physical situations so… Well… Here you go. It is really just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! [As for the differences—different polarization plane and dimensions and what have you—I’ve already summed those up, so I won’t repeat myself here.] The point is: if we have two different physical situations, we’ll want to have two different functions to describe it. Think of it like this: why would we have two—yes, I admit, two related—amplitudes to describe the up or down state of the same system, but only one wavefunction for it? You tell me.
[…]
Authors like me are looked down upon by the so-called professional class of physicists. The few who bothered to react to my attempts to make sense of Einstein’s basic intuition in regard to the nature of the wavefunction all said pretty much the same thing: “Whatever your geometric (or physical) interpretation of the wavefunction might be, it won’t be compatible with the isotropy of space. You cannot imagine an object with a 720° symmetry. That’s geometrically impossible.”
Well… Almost three years ago, I wrote the following on this blog: “As strange as it sounds, a spin-1/2 particle needs two full rotations (2×360°=720°) until it is again in the same state. Now, in regard to that particularity, you’ll often read something like: “There is nothing in our macroscopic world which has a symmetry like that.” Or, worse, “Common sense tells us that something like that cannot exist, that it simply is impossible.” [I won’t quote the site from which I took this quotes, because it is, in fact, the site of a very respectable research center!] Bollocks! The Wikipedia article on spin has this wonderful animation: look at how the spirals flip between clockwise and counterclockwise orientations, and note that it’s only after spinning a full 720 degrees that this ‘point’ returns to its original configuration after spinning a full 720 degrees.
So… Well… I am still pursuing my original dream which is… Well… Let me re-phrase what I wrote back in January 2015:
Yes, we can actually imagine spin-1/2 particles, and we actually do not need all that much imagination!
In fact, I am tempted to think that I’ve found a pretty good representation or… Well… A pretty good image, I should say, because… Well… A representation is something real, remember? 🙂
Post scriptum (10 December 2017): Our flywheel model of an electron makes sense, but also leaves many unanswered questions. The most obvious one question, perhaps, is: why the up and down state only?
I am not so worried about that question, even if I can’t answer it right away because… Well… Our apparatus—the way we measure reality—is set up to measure the angular momentum (or the magnetic moment, to be precise) in one direction only. If our electron is captured by some harmonic (or non-harmonic?) oscillation in multiple dimensions, then it should not be all that difficult to show its magnetic moment is going to align, somehow, in the same or, alternatively, the opposite direction of the magnetic field it is forced to travel through.
Of course, the analysis for the spin up situation (magnetic moment down) is quite peculiar: if our electron is a mini-magnet, why would it not line up with the magnetic field? We understand the precession of a spinning top in a gravitational field, but… Hey… It’s actually not that different. Try to imagine some spinning top on the ceiling. 🙂 I am sure we can work out the math. 🙂 The electron must be some gyroscope, really: it won’t change direction. In other words, its magnetic moment won’t line up. It will precess, and it can do so in two directions, depending on its state. 🙂 […] At least, that’s why my instinct tells me. I admit I need to work out the math to convince you. 🙂
The second question is more important. If we just rotate the reference frame over 360°, we see the same thing: some rotating object which we, vaguely, describe by some e^{+i}^{·θ} function—to be precise, I should say: by some Fourier sum of such functions—or, if the rotation is in the other direction, by some e^{−i}^{·θ} function (again, you should read: a Fourier sum of such functions). Now, the weird thing, as I tried to explain above is the following: if we rotate the object itself, over the same 360°, we get a different object: our e^{i}^{·θ} and e^{−i}^{·θ} function (again: think of a Fourier sum, so that’s a wave packet, really) becomes a −e^{±i}^{·θ} thing. We get a minus sign in front of it. So what happened here? What’s the difference, really?
Well… I don’t know. It’s very deep. If I do nothing, and you keep watching me while turning around me, for a full 360°, then you’ll end up where you were when you started and, importantly, you’ll see the same thing. Exactly the same thing: if I was an e^{+i}^{·θ} wave packet, I am still an an e^{+i}^{·θ} wave packet now. Or if I was an e^{−i}^{·θ} wave packet, then I am still an an e^{−i}^{·θ} wave packet now. Easy. Logical. Obvious, right?
But so now we try something different: I turn around, over a full 360° turn, and you stay where you are. When I am back where I was—looking at you again, so to speak—then… Well… I am not quite the same any more. Or… Well… Perhaps I am but you see me differently. If I was e^{+i}^{·θ} wave packet, then I’ve become a −e^{+i}^{·θ} wave packet now. Not hugely different but… Well… That minus sign matters, right? Or If I was wave packet built up from elementary a·e^{−i}^{·θ} waves, then I’ve become a −e^{−i}^{·θ} wave packet now. What happened?
It makes me think of the twin paradox in special relativity. We know it’s a paradox—so that’s an apparent contradiction only: we know which twin stayed on Earth and which one traveled because of the gravitational forces on the traveling twin. The one who stays on Earth does not experience any acceleration or deceleration. Is it the same here? I mean… The one who’s turning around must experience some force.
Can we relate this to the twin paradox? Maybe. Note that a minus sign in front of the e^{−±i}^{·θ} functions amounts a minus sign in front of both the sine and cosine components. So… Well… The negative of a sine and cosine is the sine and cosine but with a phase shift of 180°: −cosθ = cos(θ ± π) and −sinθ = sin(θ ± π). Now, adding or subtracting a common phase factor to/from the argument of the wavefunction amounts to changing the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360° and 720° symmetries are, effectively, related. 🙂
The reality of the wavefunction
If you haven’t read any of my previous posts on the geometry of the wavefunction (this link goes to the most recent one of them), then don’t attempt to read this one. It brings too much stuff together to be comprehensible. In fact, I am not even sure if I am going to understand what I write myself. 🙂 [OK. Poor joke. Acknowledged.]
Just to recap the essentials, I part ways with mainstream physicists in regard to the interpretation of the wavefunction. For mainstream physicists, the wavefunction is just some mathematical construct. Nothing real. Of course, I acknowledge mainstream physicists have very good reasons for that, but… Well… I believe that, if there is interference, or diffraction, then something must be interfering, or something must be diffracting. I won’t dwell on this because… Well… I have done that too many times already. My hypothesis is that the wavefunction is, in effect, a rotating field vector, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).
Of course, it must be different, and it is. First, the (physical) dimension of the field vector of the matter-wave must be different. So what is it? Well… I am tempted to associate the real and imaginary component of the wavefunction with a force per unit mass (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s^{2}), so that’s the dimension of a gravitational field.
Second, I also am tempted to think that this gravitational disturbance causes an electron (or any matter-particle) to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves do not involve any mass: they’re just an oscillating field. Nothing more. Nothing less. Why would I believe there must still be some pointlike particle involved? Well… As Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s Lectures, III-21-4) So… Well… That’s why.
The third difference is one that I thought of only recently: the plane of the oscillation cannot be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus. I am more explicit on that in the mentioned post, so you may want to check there. 🙂
I wish I mastered the software to make animations such as the one above (for which I have to credit Wikipedia), but so I don’t. You’ll just have to imagine it. That’s great mental exercise, so… Well… Just try it. 🙂
Let’s now think about rotating reference frames and transformations. If the z-direction is the direction along which we measure the angular momentum (or the magnetic moment), then the up-direction will be the positive z-direction. We’ll also assume the y-direction is the direction of travel of our elementary particle—and let’s just consider an electron here so we’re more real. 🙂 So we’re in the reference frame that Feynman used to derive the transformation matrices for spin-1/2 particles (or for two-state systems in general). His ‘improved’ Stern-Gerlach apparatus—which I’ll refer to as a beam splitter—illustrates this geometry.
So I think the magnetic moment—or the angular momentum, really—comes from an oscillatory motion in the x– and y-directions. One is the real component (the cosine function) and the other is the imaginary component (the sine function), as illustrated below.
So the crucial difference with the animations above (which illustrate left- and a right-handed polarization respectively) is that we, somehow, need to imagine the circular motion is not in the xz-plane, but in the yz-plane. Now what happens if we change the reference frame?
Well… That depends on what you mean by changing the reference frame. Suppose we’re looking in the positive y-direction—so that’s the direction in which our particle is moving—, then we might imagine how it would look like when we would make a 180° turn and look at the situation from the other side, so to speak. Now, I did a post on that earlier this year, which you may want to re-read. When we’re looking at the same thing from the other side (from the back side, so to speak), we will want to use our familiar reference frame. So we will want to keep the z-axis as it is (pointing upwards), and we will also want to define the x– and y-axis using the familiar right-hand rule for defining a coordinate frame. So our new x-axis and our new y-axis will the same as the old x- and y-axes but with the sign reversed. In short, we’ll have the following mini-transformation: (1) z‘ = z, (2) x’ = −x, and (3) y’ = −y.
So… Well… If we’re effectively looking at something real that was moving along the y-axis, then it will now still be moving along the y’-axis, but in the negative direction. Hence, our elementary wavefunction e^{i}^{θ} = cosθ + i·sinθ will transform into −cosθ − i·sinθ = −cosθ − i·sinθ = cosθ − i·sinθ. It’s the same wavefunction. We just… Well… We just changed our reference frame. We didn’t change reality.
Now you’ll cry wolf, of course, because we just went through all that transformational stuff in our last post. To be specific, we presented the following transformation matrix for a rotation along the z-axis:
Now, if φ is equal to 180° (so that’s π in radians), then these e^{i}^{φ/2} and e^{−i}^{φ/2}/√2 factors are equal to e^{i}^{π/2} = +i and e^{−i}^{π/2} = −i respectively. Hence, our e^{i}^{θ} = cosθ + i·sinθ becomes…
Hey ! Wait a minute ! We’re talking about two very different things here, right? The e^{i}^{θ} = cosθ + i·sinθ is an elementary wavefunction which, we presume, describes some real-life particle—we talked about an electron with its spin in the up-direction—while these transformation matrices are to be applied to amplitudes describing… Well… Either an up– or a down-state, right?
Right. But… Well… Is it so different, really? Suppose our e^{i}^{θ} = cosθ + i·sinθ wavefunction describes an up-electron, then we still have to apply that e^{i}^{φ/2} = e^{i}^{π/2} = +i factor, right? So we get a new wavefunction that will be equal to e^{i}^{φ/2}·e^{i}^{θ} = e^{i}^{π/2}·e^{i}^{θ} = +i·e^{i}^{θ} = i·cosθ + i^{2}·sinθ = sinθ − i·cosθ, right? So how can we reconcile that with the cosθ − i·sinθ function we thought we’d find?
We can’t. So… Well… Either my theory is wrong or… Well… Feynman can’t be wrong, can he? I mean… It’s not only Feynman here. We’re talking all mainstream physicists here, right?
Right. But think of it. Our electron in that thought experiment does, effectively, make a turn of 180°, so it is going in the other direction now ! That’s more than just… Well… Going around the apparatus and looking at stuff from the other side.
Hmm… Interesting. Let’s think about the difference between the sinθ − i·cosθ and cosθ − i·sinθ functions. First, note that they will give us the same probabilities: the square of the absolute value of both complex numbers is the same. [It’s equal to 1 because we didn’t bother to put a coefficient in front.] Secondly, we should note that the sine and cosine functions are essentially the same. They just differ by a phase factor: cosθ = sin(θ + π/2) and −sinθ = cos(θ + π/2). Let’s see what we can do with that. We can write the following, for example:
sinθ − i·cosθ = −cos(θ + π/2) − i·sin(θ + π/2) = −[cos(θ + π/2) + i·sin(θ + π/2)] = −e^{i}^{·(θ + π/2)}
Well… I guess that’s something at least ! The e^{i}^{·θ} and −e^{i}^{·(θ + π/2)} functions differ by a phase shift and a minus sign so… Well… That’s what it takes to reverse the direction of an electron. 🙂 Let us mull over that in the coming days. As I mentioned, these more philosophical topics are not easily exhausted. 🙂
The geometry of the wavefunction, electron spin and the form factor
Our previous posts showed how a simple geometric interpretation of the elementary wavefunction yielded the (Compton scattering) radius of an elementary particle—for an electron, at least: for the proton, we only got the order of magnitude right—but then a proton is not an elementary particle. We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between the E = m·a^{2}·ω^{2} and E = m·c^{2} relations that we… Well… We need to be more specific about it.
Indeed, I’ve been ambiguous here and there—oscillating between various interpretations, so to speak. 🙂 In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the form factor problem. So… Well… That explains the title of my post. But so… Well… I do want to be somewhat more conclusive in this post. So let’s go and see where we end up. 🙂
To help focus our mind, let us recall the metaphor of the V-2 perpetuum mobile, as illustrated below. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down. It provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring: it is described by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs. Of course, instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft, but then that’s not fancy enough for me. 🙂
At first sight, the analogy between our flywheel model of an electron and the V-twin engine seems to be complete: the 90 degree angle of our V-2 engine makes it possible to perfectly balance the pistons and we may, therefore, think of the flywheel as a (symmetric) rotating mass, whose angular momentum is given by the product of the angular frequency and the moment of inertia: L = ω·I. Of course, the moment of inertia (aka the angular mass) will depend on the form (or shape) of our flywheel:
- I = m·a^{2} for a rotating point mass m or, what amounts to the same, for a circular hoop of mass m and radius r = a.
- For a rotating (uniformly solid) disk, we must add a 1/2 factor: I = m·a^{2}/2.
How can we relate those formulas to the E = m·a^{2}·ω^{2} formula? The kinetic energy that is being stored in a flywheel is equal E_{kinetic} = I·ω^{2}/2, so that is only half of the E = m·a^{2}·ω^{2} product if we substitute I for I = m·a^{2}. [For a disk, we get a factor 1/4, so that’s even worse!] However, our flywheel model of an electron incorporates potential energy too. In fact, the E = m·a^{2}·ω^{2} formula just adds the (kinetic and potential) energy of two oscillators: we do not really consider the energy in the flywheel itself because… Well… The essence of our flywheel model of an electron is not the flywheel: the flywheel just transfers energy from one oscillator to the other, but so… Well… We don’t include it in our energy calculations. The essence of our model is that two-dimensional oscillation which drives the electron, and which is reflected in Einstein’s E = m·c^{2} formula. That two-dimensional oscillation—the a^{2}·ω^{2} = c^{2} equation, really—tells us that the resonant (or natural) frequency of the fabric of spacetime is given by the speed of light—but measured in units of a. [If you don’t quite get this, re-write the a^{2}·ω^{2} = c^{2} equation as ω = c/a: the radius of our electron appears as a natural distance unit here.]
Now, we were extremely happy with this interpretation not only because of the key results mentioned above, but also because it has lots of other nice consequences. Think of our probabilities as being proportional to energy densities, for example—and all of the other stuff I describe in my published paper on this. But there is even more on the horizon: a follower of this blog (a reader with an actual PhD in physics, for a change) sent me an article analyzing elementary particles as tiny black holes because… Well… If our electron is effectively spinning around, then its tangential velocity is equal to v = a·ω = c. Now, recent research suggest black holes are also spinning at (nearly) the speed of light. Interesting, right? However, in order to understand what she’s trying to tell me, I’ll first need to get a better grasp of general relativity, so I can relate what I’ve been writing here and in previous posts to the Schwarzschild radius and other stuff.
Let me get back to the lesson here. In the reference frame of our particle, the wavefunction really looks like the animation below: it has two components, and the amplitude of the two-dimensional oscillation is equal to a, which we calculated as a = ħ·/(m·c) = 3.8616×10^{−13} m, so that’s the (reduced) Compton scattering radius of an electron.
In my original article on this, I used a more complicated argument involving the angular momentum formula, but I now prefer a more straightforward calculation:
c = a·ω = a·E/ħ = a·m·c^{2}/ħ ⇔ a = ħ/(m·c)
The question is: what is that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’t prove anything in this regard. But my hypothesis is that it is, in effect, a rotating field vector, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).
There are a number of crucial differences though:
- The (physical) dimension of the field vector of the matter-wave is different: I associate the real and imaginary component of the wavefunction with a force per unit mass (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s^{2}), so that’s the dimension of a gravitational field.
- I do believe this gravitational disturbance, so to speak, does cause an electron to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves do not involve any mass: they’re just an oscillating field. Nothing more. Nothing less. In contrast, as Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s Lectures, III-21-4)
- The third difference is one that I thought of only recently: the plane of the oscillation cannot be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus.
I mentioned that in my previous post but, for your convenience, I’ll repeat what I wrote there. The basic idea here is illustrated below (credit for this illustration goes to another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to a YouTube video from the Quantum Made Simple site.
The point is: the direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is traveling—cannot be parallel to the direction of motion. On the contrary, it is perpendicular to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will comprise the direction of motion.
However, we need to add an interesting detail here. As you know, we don’t really have a precise direction of angular momentum in quantum physics. [If you don’t know this… Well… Just look at one of my many posts on spin and angular momentum in quantum physics.] Now, we’ve explored a number of hypotheses but, when everything is said and done, a rather classical explanation turns out to be the best: an object with an angular momentum J and a magnetic moment μ (I used bold-face because these are vector quantities) that is parallel to some magnetic field B, will not line up, as you’d expect a tiny magnet to do in a magnetic field—or not completely, at least: it will precess. I explained that in another post on quantum-mechanical spin, which I advise you to re-read if you want to appreciate the point that I am trying to make here. That post integrates some interesting formulas, and so one of the things on my ‘to do’ list is to prove that these formulas are, effectively, compatible with the electron model we’ve presented in this and previous posts.
Indeed, when one advances a hypothesis like this, it’s not enough to just sort of show that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever we think our electron—or its wavefunction—might be, it needs to be compatible with stuff like the observed precession frequency of an electron in a magnetic field.
Our model also needs to be compatible with the transformation formulas for amplitudes. I’ve been talking about this for quite a while now, and so it’s about time I get going on that.
Last but not least, those articles that relate matter-particles to (quantum) gravity—such as the one I mentioned above—are intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? 🙂 Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years ahead—I hope. 🙂
Post scriptum: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But it is that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillators drive the flywheel but, without the flywheel, nothing is happening. It is really the transfer of energy—through the flywheel—which explains why our flywheel goes round and round.”
It may or may not be useful to remind ourselves of the math in this regard. The motion of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) change in kinetic energy at any point in time (as a function of the angle θ) is equal to: d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ. Now, the motion of the second oscillator (just look at that second piston going up and down in the V-2 engine) is given by the sinθ function, which is equal to cos(θ − π /2). Hence, its kinetic energy is equal to sin^{2}(θ − π /2), and how it changes (as a function of θ again) is equal to 2∙sin(θ − π /2)∙cos(θ − π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ. So here we have our energy transfer: the flywheel organizes the borrowing and returning of energy, so to speak. That’s the crux of the matter.
So… Well… What if the relevant energy formula is E = m·a^{2}·ω^{2}/2 instead of E = m·a^{2}·ω^{2}? What are the implications? Well… We get a √2 factor in our formula for the radius a, as shown below.
Now that is not so nice. For the tangential velocity, we get v = a·ω = √2·c. This is also not so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precession—the wobbling of our flywheel in a magnetic field. Remember we may think of J_{z}—the angular momentum or, to be precise, its component in the z-direction (the direction in which we measure it—as the projection of the real angular momentum J. Let me insert Feynman’s illustration here again (Feynman’s Lectures, II-34-3), so you get what I am talking about.
Now, all depends on the angle (θ) between J_{z} and J, of course. We did a rather obscure post on these angles, but the formulas there come in handy now. Just click the link and review it if and when you’d want to understand the following formulas for the magnitude of the presumed actual momentum:In this particular case (spin-1/2 particles), j is equal to 1/2 (in units of ħ, of course). Hence, J is equal to √0.75 ≈ 0.866. Elementary geometry then tells us cos(θ) = (1/2)/√(3/4) = = 1/√3. Hence, θ ≈ 54.73561°. That’s a big angle—larger than the 45° angle we had secretly expected because… Well… The 45° angle has that √2 factor in it: cos(45°) = sin(45°) = 1/√2.
Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? 🙂 We’ll solve the puzzle. One day… But not today, I am afraid. I’ll call it the form factor problem… Because… Well… It sounds better than the 1/2 or √2 problem, right? 🙂
Note: If you’re into quantum math, you’ll note a = ħ/(m·c) is the reduced Compton scattering radius. The standard Compton scattering radius is equal to a·2π = (2π·ħ)/(m·c) = h/(m·c) = h/(m·c). It doesn’t solve the √2 problem. Sorry. The form factor problem. 🙂
To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of two circular oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensional oscillation as an oscillation of a polar and azimuthal angle. It’s just a thought but… Well… I am sure it’s going to keep me busy for a while. 🙂They are oscillations, still, so I am not thinking of two flywheels that keep going around in the same direction. No. More like a wobbling object on a spring. Something like the movement of a bobblehead on a spring perhaps. 🙂
Electron spin and the geometry of the wavefunction
In our previous posts, we interpreted the elementary wavefunction ψ = a·e^{−i∙θ} = a·cosθ − i·a·sinθ as a two-dimensional oscillation in spacetime. In addition to assuming the two directions of the oscillation were perpendicular to each other, we also assumed they were perpendicular to the direction of motion. While the first assumption is essential in our interpretation, the second assumption is solely based on an analogy with a circularly polarized electromagnetic wave. We also assumed the matter wave could be right-handed as well as left-handed (as illustrated below), and that these two physical possibilities corresponded to the angular momentum being equal to plus or minus ħ/2 respectively.
This allowed us to derive the Compton scattering radius of an elementary particle. Indeed, we interpreted the rotating vector as a resultant vector, which we get by adding the sine and cosine motions, which represent the real and imaginary components of our wavefunction. The energy of this two-dimensional oscillation is twice the energy of a one-dimensional oscillator and, therefore, equal to E = m·a^{2}·ω^{2}. Now, the angular frequency is given by ω = E/ħ and E must, obviously, also be equal to E = m·c^{2}. Substitition and re-arranging the terms gives us the Compton scattering radius:
The value given above is the (reduced) Compton scattering radius for an electron. For a proton, we get a value of about 2.1×10^{−16} m, which is about 1/4 of the radius of a proton as measured in scattering experiments. Hence, for a proton, our formula does not give us the exact (i.e. experimentally verified) value but it does give us the correct order of magnitude—which is fine because we know a proton is not an elementary particle and, hence, the motion of its constituent parts (quarks) is… Well… It complicates the picture hugely.
If we’d presume the electron charge would, effectively, be rotating about the center, then its tangential velocity is given by v = a·ω = [ħ·/(m·c)]·(E/ħ) = c, which is yet another wonderful implication of our hypothesis. Finally, the c = a·ω formula allowed us to interpret the speed of light as the resonant frequency of the fabric of space itself, as illustrated when re-writing this equality as follows:
This gave us a natural and forceful interpretation of Einstein’s mass-energy equivalence formula: the energy in the E = m·c^{2}· equation is, effectively, a two-dimensional oscillation of mass.
However, while toying with this and other results (for example, we may derive a Poynting vector and show probabilities are, effectively, proportional to energy densities), I realize the plane of our two-dimensional oscillation cannot be perpendicular to the direction of motion of our particle. In fact, the direction of motion must lie in the same plane. This is a direct consequence of the direction of the angular momentum as measured by, for example, the Stern-Gerlach experiment. The basic idea here is illustrated below (credit for this illustration goes to another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to a YouTube video from the Quantum Made Simple site.
The point is: the direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogenous) magnetic field through which our electron is traveling—cannot be parallel to the direction of motion. On the contrary, it is perpendicular to the direction of motion. In other words, if we imagine our electron as some rotating disk or a flywheel, then it will actually comprise the direction of motion.
What are the implications? I am not sure. I will definitely need to review whatever I wrote about the de Broglie wavelength in previous posts. We will also need to look at those transformations of amplitudes once again. Finally, we will also need to relate this to the quantum-mechanical formulas for the angular momentum and the magnetic moment.
Post scriptum: As in previous posts, I need to mention one particularity of our model. When playing with those formulas, we contemplated two different formulas for the angular mass: one is the formula for a rotating mass (I = m·r^{2}/2), and the other is the one for a rotating mass (I = m·r^{2}). The only difference between the two is a 1/2 factor, but it turns out we need it to get a sensical result. For a rotating mass, the angular momentum is equal to the radius times the momentum, so that’s the radius times the mass times the velocity: L = m·v·r. [See also Feynman, Vol. II-34-2, in this regard)] Hence, for our model, we get L = m·v·r = m·c·a = m·c·ħ/(m·c) = ħ. Now, we know it’s equal to ±ħ/2, so we need that 1/2 factor in the formula.
Can we relate this 1/2 factor to the g-factor for the electron’s magnetic moment, which is (approximately) equal to 2? Maybe. We’d need to look at the formula for a rotating charged disk. That’s for a later post, however. It’s been enough for today, right? 🙂
I would just like to signal another interesting consequence of our model. If we would interpret the radius of our disk (a)—so that’s the Compton radius of our electron, as opposed to the Thomson radius—as the uncertainty in the position of our electron, then our L = m·v·r = m·c·a = p·a = ħ/2 formula as a very particular expression of the Uncertainty Principle: p·Δx= ħ/2. Isn’t that just plain nice? 🙂
Re-visiting the Complementarity Principle: the field versus the flywheel model of the matter-wave
Note: I have published a paper that is very coherent and fully explains what’s going on. There is nothing magical about it these things. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.
Jean Louis Van Belle, 23 December 2018
Original post:
This post is a continuation of the previous one: it is just going to elaborate the questions I raised in the post scriptum of that post. Let’s first review the basics once more.
The geometry of the elementary wavefunction
In the reference frame of the particle itself, the geometry of the wavefunction simplifies to what is illustrated below: an oscillation in two dimensions which, viewed together, form a plane that would be perpendicular to the direction of motion—but then our particle doesn’t move in its own reference frame, obviously. Hence, we could be looking at our particle from any direction and we should, presumably, see a similar two-dimensional oscillation. That is interesting because… Well… If we rotate this circle around its center (in whatever direction we’d choose), we get a sphere, right? It’s only when it starts moving, that it loses its symmetry. Now, that is very intriguing, but let’s think about that later.
Let’s assume we’re looking at it from some specific direction. Then we presumably have some charge (the green dot) moving about some center, and its movement can be analyzed as the sum of two oscillations (the sine and cosine) which represent the real and imaginary component of the wavefunction respectively—as we observe it, so to speak. [Of course, you’ve been told you can’t observe wavefunctions so… Well… You should probably stop reading this. :-)] We write:
ψ = = a·e^{−i∙θ} = a·e^{−i∙E·t/ħ} = a·cos(−E∙t/ħ) + i·a·sin(−E∙t/ħ) = a·cos(E∙t/ħ) − i·a·sin(E∙t/ħ)
So that’s the wavefunction in the reference frame of the particle itself. When we think of it as moving in some direction (so relativity kicks in), we need to add the p·x term to the argument (θ = E·t − p∙x). It is easy to show this term doesn’t change the argument (θ), because we also get a different value for the energy in the new reference frame: E_{v }= γ·E_{0} and so… Well… I’ll refer you to my post on this, in which I show the argument of the wavefunction is invariant under a Lorentz transformation: the way E_{v} and p_{v} and, importantly, the coordinates x and t relativistically transform ensures the invariance.
In fact, I’ve always wanted to read de Broglie‘s original thesis because I strongly suspect he saw that immediately. If you click this link, you’ll find an author who suggests the same. Having said that, I should immediately add this does not imply there is no need for a relativistic wave equation: the wavefunction is a solution for the wave equation and, yes, I am the first to note the Schrödinger equation has some obvious issues, which I briefly touch upon in one of my other posts—and which is why Schrödinger himself and other contemporaries came up with a relativistic wave equation (Oskar Klein and Walter Gordon got the credit but others (including Louis de Broglie) also suggested a relativistic wave equation when Schrödinger published his). In my humble opinion, the key issue is not that Schrödinger’s equation is non-relativistic. It’s that 1/2 factor again but… Well… I won’t dwell on that here. We need to move on. So let’s leave the wave equation for what it is and go back to our wavefunction.
You’ll note the argument (or phase) of our wavefunction moves clockwise—or counterclockwise, depending on whether you’re standing in front of behind the clock. Of course, Nature doesn’t care about where we stand or—to put it differently—whether we measure time clockwise, counterclockwise, in the positive, the negative or whatever direction. Hence, I’ve argued we can have both left- as well as right-handed wavefunctions, as illustrated below (for p ≠ 0). Our hypothesis is that these two physical possibilities correspond to the angular momentum of our electron being either positive or negative: J_{z} = +ħ/2 or, else, J_{z} = −ħ/2. [If you’ve read a thing or two about neutrinos, then… Well… They’re kinda special in this regard: they have no charge and neutrinos and antineutrinos are actually defined by their helicity. But… Well… Let’s stick to trying to describing electrons for a while.]
The line of reasoning that we followed allowed us to calculate the amplitude a. We got a result that tentatively confirms we’re on the right track with our interpretation: we found that a = ħ/m_{e}·c, so that’s the Compton scattering radius of our electron. All good ! But we were still a bit stuck—or ambiguous, I should say—on what the components of our wavefunction actually are. Are we really imagining the tip of that rotating arrow is a pointlike electric charge spinning around the center? [Pointlike or… Well… Perhaps we should think of the Thomson radius of the electron here, i.e. the so-called classical electron radius, which is equal to the Compton radius times the fine-structure constant: r_{Thomson} = α·r_{Compton} ≈ 3.86×10^{−13}/137.]
So that would be the flywheel model.
In contrast, we may also think the whole arrow is some rotating field vector—something like the electric field vector, with the same or some other physical dimension, like newton per charge unit, or newton per mass unit? So that’s the field model. Now, these interpretations may or may not be compatible—or complementary, I should say. I sure hope they are but… Well… What can we reasonably say about it?
Let us first note that the flywheel interpretation has a very obvious advantage, because it allows us to explain the interaction between a photon and an electron, as I demonstrated in my previous post: the electromagnetic energy of the photon will drive the circulatory motion of our electron… So… Well… That’s a nice physical explanation for the transfer of energy. However, when we think about interference or diffraction, we’re stuck: flywheels don’t interfere or diffract. Only waves do. So… Well… What to say?
I am not sure, but here I want to think some more by pushing the flywheel metaphor to its logical limits. Let me remind you of what triggered it all: it was the mathematical equivalence of the energy equation for an oscillator (E = m·a^{2}·ω^{2}) and Einstein’s formula (E = m·c^{2}), which tells us energy and mass are equivalent but… Well… They’re not the same. So what are they then? What is energy, and what is mass—in the context of these matter-waves that we’re looking at. To be precise, the E = m·a^{2}·ω^{2} formula gives us the energy of two oscillators, so we need a two-spring model which—because I love motorbikes—I referred to as my V-twin engine model, but it’s not an engine, really: it’s two frictionless pistons (or springs) whose direction of motion is perpendicular to each other, so they are in a 90° degree angle and, therefore, their motion is, effectively, independent. In other words: they will not interfere with each other. It’s probably worth showing the illustration just one more time. And… Well… Yes. I’ll also briefly review the math one more time.
If the magnitude of the oscillation is equal to a, then the motion of these piston (or the mass on a spring) will be described by x = a·cos(ω·t + Δ). Needless to say, Δ is just a phase factor which defines our t = 0 point, and ω is the natural angular frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by x = a·cos(ω·t), while the motion of the other is given by x = a·cos(ω·t–π/2) = a·sin(ω·t). The kinetic and potential energy of one oscillator – think of one piston or one spring only – can then be calculated as:
- K.E. = T = m·v^{2}/2 = (1/2)·m·ω^{2}·a^{2}·sin^{2}(ω·t + Δ)
- P.E. = U = k·x^{2}/2 = (1/2)·k·a^{2}·cos^{2}(ω·t + Δ)
The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω^{2}. Hence, the total energy—for one piston, or one spring—is equal to:
E = T + U = (1/2)· m·ω^{2}·a^{2}·[sin^{2}(ω·t + Δ) + cos^{2}(ω·t + Δ)] = m·a^{2}·ω^{2}/2
Hence, adding the energy of the two oscillators, we have a perpetuum mobile storing an energy that is equal to twice this amount: E = m·a^{2}·ω^{2}. It is a great metaphor. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. However, we still have to prove this engine is, effectively, a perpetuum mobile: we need to prove the energy that is being borrowed or returned by one piston is the energy that is being returned or borrowed by the other. That is easy to do, but I won’t bother you with that proof here: you can double-check it in the referenced post or – more formally – in an article I posted on viXra.org.
It is all beautiful, and the key question is obvious: if we want to relate the E = m·a^{2}·ω^{2} and E = m·c^{2} formulas, we need to explain why we could, potentially, write c as c = a·ω = a·√(k/m). We’ve done that already—to some extent at least. The tangential velocity of a pointlike particle spinning around some axis is given by v = r·ω. Now, the radius r is given by a = ħ/(m·c), and ω = E/ħ = m·c^{2}/ħ, so v is equal to to v = [ħ/(m·c)]·[m·c^{2}/ħ] = c. Another beautiful result, but what does it mean? We need to think about the meaning of the ω = √(k/m) formula here. In the mentioned article, we boldly wrote that the speed of light is to be interpreted as the resonant frequency of spacetime, but so… Well… What do we really mean by that? Think of the following.
Einstein’s E = mc^{2} equation implies the ratio between the energy and the mass of any particle is always the same:
This effectively reminds us of the ω^{2} = C^{–}^{1}/L or ω^{2} = k/m formula for harmonic oscillators. The key difference is that the ω^{2}= C^{–}^{1}/L and ω^{2} = k/m formulas introduce two (or more) degrees of freedom. In contrast, c^{2}= E/m for any particle, always. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in one physical space only: our spacetime. Hence, the speed of light (c) emerges here as the defining property of spacetime: the resonant frequency, so to speak. We have no further degrees of freedom here.
Let’s think about k. [I am not trying to avoid the ω^{2}= 1/LC formula here. It’s basically the same concept: the ω^{2}= 1/LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor, an inductor, and a capacitor. Writing the formula as ω^{2}= C^{−1}/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring, so… Well… You get it, right? The ω^{2}= C^{–}^{1}/L and ω^{2} = k/m sort of describe the same thing: harmonic oscillation. It’s just… Well… Unlike the ω^{2}= C^{–}^{1}/L, the ω^{2} = k/m is directly compatible with our V-twin engine metaphor, because it also involves physical distances, as I’ll show you here.] The k in the ω^{2} = k/m is, effectively, the stiffness of the spring. It is defined by Hooke’s Law, which states that the force that is needed to extend or compress a spring by some distance x is linearly proportional to that distance, so we write: F = k·x.
Now that is interesting, isn’t it? We’re talking exactly the same thing here: spacetime is, presumably, isotropic, so it should oscillate the same in any direction—I am talking those sine and cosine oscillations now, but in physical space—so there is nothing imaginary here: all is real or… Well… As real as we can imagine it to be. 🙂
We can elaborate the point as follows. The F = k·x equation implies k is a force per unit distance: k = F/x. Hence, its physical dimension is newton per meter (N/m). Now, the x in this equation may be equated to the maximum extension of our spring, or the amplitude of the oscillation, so that’s the radius r = a in the metaphor we’re analyzing here. Now look at how we can re-write the c = a·ω = a·√(k/m) equation:
In case you wonder about the E = F·a substitution: just remember that energy is force times distance. [Just do a dimensional analysis: you’ll see it works out.] So we have a spectacular result here, for several reasons. The first, and perhaps most obvious reason, is that we can actually derive Einstein’s E = m·c^{2} formula from our flywheel model. Now, that is truly glorious, I think. However, even more importantly, this equation suggests we do not necessarily need to think of some actual mass oscillating up and down and sideways at the same time: the energy in the oscillation can be thought of a force acting over some distance, regardless of whether or not it is actually acting on a particle. Now, that energy will have an equivalent mass which is—or should be, I’d say… Well… The mass of our electron or, generalizing, the mass of the particle we’re looking at.
Huh? Yes. In case you wonder what I am trying to get at, I am trying to convey the idea that the two interpretations—the field versus the flywheel model—are actually fully equivalent, or compatible, if you prefer that term. In Asia, they would say: they are the “same-same but different” 🙂 but, using the language that’s used when discussing the Copenhagen interpretation of quantum physics, we should actually say the two models are complementary.
You may shrug your shoulders but… Well… It is a very deep philosophical point, really. 🙂 As far as I am concerned, I’ve never seen a better illustration of the (in)famous Complementarity Principle in quantum physics because… Well… It goes much beyond complementarity. This is about equivalence. 🙂 So it’s just like Einstein’s equation. 🙂
Post scriptum: If you read my posts carefully, you’ll remember I struggle with those 1/2 factors here and there. Textbooks don’t care about them. For example, when deriving the size of an atom, or the Rydberg energy, even Feynman casually writes that “we need not trust our answer [to questions like this] within factors like 2, π, etcetera.” Frankly, that’s disappointing. Factors like 2, 1/2, π or 2π are pretty fundamental numbers, and so they need an explanation. So… Well… I do loose sleep over them. Let me advance some possible explanation here.
As for Feynman’s model, and the derivation of electron orbitals in general, I think it’s got to do with the fact that electrons do want to pair up when thermal motion does not come into play: think of the Cooper pairs we use to explain superconductivity (so that’s the BCS theory). The 1/2 factor in Schrödinger’s equation also has weird consequences (when you plug in the elementary wavefunction and do the derivatives, you get a weird energy concept: E = m·v^{2}, to be precise). This problem may also be solved when assuming we’re actually calculating orbitals for a pair of electrons, rather than orbitals for just one electron only. [We’d get twice the mass (and, presumably, the charge, so… Well… It might work—but I haven’t done it yet. It’s on my agenda—as so many other things, but I’ll get there… One day. :-)]
So… Well… Let’s get back to the lesson here. In this particular context (i.e. in the context of trying to find some reasonable physical interpretation of the wavefunction), you may or may not remember (if not, check my post on it) ‘ll remember I had to use the I = m·r^{2}/2 formula for the angular momentum, as opposed to the I = m·r^{2} formula. I = m·r^{2}/2 (with the 1/2 factor) gives us the angular momentum of a disk with radius r, as opposed to a point mass going around some circle with radius r. I noted that “the addition of this 1/2 factor may seem arbitrary”—and it totally is, of course—but so it gave us the result we wanted: the exact (Compton scattering) radius of our electron.
Now, the arbitrary 1/2 factor may or may be explained as follows. In the field model of our electron, the force is linearly proportional to the extension or compression. Hence, to calculate the energy involved in stretching it from x = 0 to x = a, we need to calculate it as the following integral:
So… Well… That will give you some food for thought, I’d guess. 🙂 If it racks your brain too much—or if you’re too exhausted by this point (which is OK, because it racks my brain too!)—just note we’ve also shown that the energy is proportional to the square of the amplitude here, so that’s a nice result as well… 🙂
Talking food for thought, let me make one final point here. The c^{2} = a^{2}·k/m relation implies a value for k which is equal to k = m·c^{2}/a = E/a. What does this tell us? In one of our previous posts, we wrote that the radius of our electron appeared as a natural distance unit. We wrote that because of another reason: the remark was triggered by the fact that we can write the c/ω ratio as c/ω = a·ω/ω = a. This implies the tangential and angular velocity in our flywheel model of an electron would be the same if we’d measure distance in units of a. Now, the E = a·k = a·F/x (just re-writing…) implies that the force is proportional to the energy— F = (x/a)·E — and the proportionality coefficient is… Well… x/a. So that’s the distance measured in units of a. So… Well… Isn’t that great? The radius of our atom appearing as a natural distance unit does fit in nicely with our geometric interpretation of the wavefunction, doesn’t it? I mean… Do I need to say more?
I hope not because… Well… I can’t explain any better for the time being. I hope I sort of managed to convey the message. Just to make sure, in case you wonder what I was trying to do here, it’s the following: I told you c appears as a resonant frequency of spacetime and, in this post, I tried to explain what that really means. I’d appreciate if you could let me know if you got it. If not, I’ll try again. 🙂 When everything is said and done, one only truly understands stuff when one is able to explain it to someone else, right? 🙂 Please do think of more innovative or creative ways if you can! 🙂
OK. That’s it but… Well… I should, perhaps, talk about one other thing here. It’s what I mentioned in the beginning of this post: this analysis assumes we’re looking at our particle from some specific direction. It could be any direction but… Well… It’s some direction. We have no depth in our line of sight, so to speak. That’s really interesting, and I should do some more thinking about it. Because the direction could be any direction, our analysis is valid for any direction. Hence, if our interpretation would happen to be some true—and that’s a big if, of course—then our particle has to be spherical, right? Why? Well… Because we see this circular thing from any direction, so it has to be a sphere, right?
Well… Yes. But then… Well… While that logic seems to be incontournable, as they say in French, I am somewhat reluctant to accept it at face value. Why? I am not sure. Something inside of me says I should look at the symmetries involved… I mean the transformation formulas for wavefunction when doing rotations and stuff. So… Well… I’ll be busy with that for a while, I guess. 😦
Post scriptum 2: You may wonder whether this line of reasoning would also work for a proton. Well… Let’s try it. Because its mass is so much larger than that of an electron (about 1835 times), the a = ħ/(m·c) formula gives a much smaller radius: 1835 times smaller, to be precise, so that’s around 2.1×10^{−16} m, which is about 1/4 of the so-called charge radius of a proton, as measured by scattering experiments. So… Well… We’re not that far off, but… Well… We clearly need some more theory here. Having said that, a proton is not an elementary particle, so its mass incorporates other factors than what we’re considering here (two-dimensional oscillations).
The flywheel model of an electron
One of my readers sent me the following question on the geometric (or even physical) interpretation of the wavefunction that I’ve been offering in recent posts:
“Does this mean that the wave function is merely describing excitations in a matter field; or is this unsupported?“
My reply was very short: “Yes. In fact, we can think of a matter-particle as a tiny flywheel that stores energy.”
However, I realize this answer answers the question only partially. Moreover, I now feel I’ve been quite ambiguous in my description. When looking at the geometry of the elementary wavefunction (see the animation below, which shows us a left- and right-handed wave respectively), two obvious but somewhat conflicting interpretations readily come to mind:
(1) One is that the components of the elementary wavefunction represent an oscillation (in two dimensions) of a field. We may call it a matter field (yes, think of the scalar Higgs field here), but we could also think of it as an oscillation of the spacetime fabric itself: a tiny gravitational wave, in effect. All we need to do here is to associate the sine and cosine component with a physical dimension. The analogy here is the electromagnetic field vector, whose dimension is force per unit charge (newton/coulomb). So we may associate the sine and cosine components of the wavefunction with, say, the force per unit mass dimension (newton/kg) which, using Newton’s Law (F = m·a) reduces to the dimension of acceleration (m/s^{2}), which is the dimension of gravitational fields. I’ll refer to this interpretation as the field interpretation of the matter wave (or wavefunction).
(2) The other interpretation is what I refer to as the flywheel interpretation of the electron. If you google this, you won’t find anything. However, you will probably stumble upon the so-called Zitterbewegung interpretation of quantum mechanics, which is a more elaborate theory based on the same basic intuition. The Zitterbewegung (a term which was coined by Erwin Schrödinger himself, and which you’ll see abbreviated as zbw) is, effectively, a local circulatory motion of the electron, which is presumed to be the basis of the electron’s spin and magnetic moment. All that I am doing, is… Well… I think I do push the envelope of this interpretation quite a bit. 🙂
The flywheel interpretation
A light beam—an electromagnetic wave—consists of a large number of photons. These photons are thought of as being circularly polarized: look at those animations above again. The Planck-Einstein equation tells us the energy of each photon is equal to E = ħ·ω = h·f. [I should, perhaps, quickly note that the frequency f is, obviously, the frequency of the electromagnetic wave. It, therefore, is not to be associated with a matter wave: the de Broglie wavelength and the wavelength of light are very different concepts, even if the Planck-Einstein equation looks the same for both.]
Now, if our beam consists of N photons, the total energy of our beam will be equal to W = N·E = N·ħ·ω. It is crucially important to note that this energy is to be interpreted as the energy that is carried by the beam in a certain time: we should think of the beam as being finite, somehow, in time and in space. Otherwise, our reasoning doesn’t make sense.
The photons carry angular momentum. Just look at those animations (above) once more. It doesn’t matter much whether or not we think of light as particles or as a wave: you can see there is angular momentum there. Photons are spin-1 particles, so the angular momentum will be equal to ± ħ. Hence, then the total angular momentum J_{z} (the direction of propagation is supposed to be the z-axis here) will be equal to J_{z} = N·ħ. [This, of course, assumes all photons are polarized in the same way, which may or may not be the case. You should just go along with the argument right now.] Combining the W = N·ħ·ω and J_{z} = N·ħ equations, we get:
J_{z} = N·ħ = W/ω
For a photon, we do accept the field interpretation, as illustrated below. As mentioned above, the z-axis here is the direction of propagation (so that’s the line of sight when looking at the diagram). So we have an electric field vector, which we write as ε (epsilon) so as to not cause any confusion with the Ε we used for the energy. [You may wonder if we shouldn’t also consider the magnetic field vector, but then we know the magnetic field vector is, basically, a relativistic effect which vanishes in the reference frame of the charge itself.] The phase of the electric field vector is φ = ω·t.
Now, a charge (so that’s our electron now) will experience a force which is equal to F = q·ε. We use bold letters here because F and ε are vectors. We now need to look at our electron which, in our interpretation of the elementary wavefunction, we think of as rotating about some axis. So that’s what’s represented below. [Both illustrations are Feynman’s, not mine. As for the animations above, I borrowed them from Wikipedia.]
Now, in previous posts, we calculated the radius r based on a similar argument as the one Feynman used to get that J_{z} = N·ħ = W/ω equation. I’ll refer you those posts and just mention the result here: r is the Compton scattering radius for an electron, which is equal to:
An equally spectacular implication of our flywheel model of the electron was the following: we found that the angular velocity v was equal to v = r·ω = [ħ·/(m·c)]·(E/ħ) = c. Hence, in our flywheel model of an electron, it is effectively spinning around at the speed of light. Note that the angular frequency (ω) in the v = r·ω equation is not the angular frequency of our photon: it’s the frequency of our electron. So we use the same Planck-Einstein equation (ω = E/ħ) but the energy E is the (rest) energy of our electron, so that’s about 0.511 MeV (so that’s an order of magnitude which is 100,000 to 300,000 times that of photons in the visible spectrum). Hence, the angular frequencies of our electron and our photon are very different. Feynman casually reflects this difference by noting the phases of our electron and our photon will differ by a phase factor, which he writes as φ_{0}.
Just to be clear here, at this point, our analysis here diverges from Feynman’s. Feynman had no intention whatsoever to talk about Schrödinger’s Zitterbewegung hypothesis when he wrote what he wrote back in the 1960s. In fact, Feynman is very reluctant to venture into physical interpretations of the wavefunction in all his Lectures on quantum mechanics—which is surprising. Because he comes so tantalizing close at many occasions—as he does here: he describes the motion of the electron here as that of a harmonic oscillator which can be driven by an external electric field. Now that is a physical interpretation, and it is totally consistent with the one I’ve advanced in my recent posts. Indeed, Feynman also describes it as an oscillation in two dimensions—perpendicular to each other and to the direction of motion, as we do— in both the flywheel as well as the field interpretation of the wavefunction!
This point is important enough to quote Feynman himself in this regard:
“We have often described the motion of the electron in the atom as a harmonic oscillator which can be driven into oscillation by an external electric field. We’ll suppose that the atom is isotropic, so that it can oscillate equally well in the x– or y- directions. Then in the circularly polarized light, the x displacement and the y displacement are the same, but one is 90° behind the other. The net result is that the electron moves in a circle.”
Right on! But so what happens really? As our light beam—the photons, really—are being absorbed by our electron (or our atom), it absorbs angular momentum. In other words, there is a torque about the central axis. Let me remind you of the formulas for the angular momentum and for torque respectively: L = r×p and τ = r×F. Needless to say, we have two vector cross-products here. Hence, if we use the τ = r×F formula, we need to find the tangential component of the force (F_{t}), whose magnitude will be equal to F_{t} = q·ε_{t}. Now, energy is force over some distance so… Well… You may need to think about it for a while but, if you’ve understood all of the above, you should also be able to understand the following formula:
dW/dt = q·ε_{t}·v
[If you have trouble, remember v is equal to ds/dt = Δs/Δt for Δt → 0, and re-write the equation above as dW = q·ε_{t}·v·dt = q·ε_{t}·ds = F_{t}·ds. Capito?]
Now, you may or may not remember that the time rate of change of angular momentum must be equal to the torque that is being applied. Now, the torque is equal to τ = F_{t}·r = q·ε_{t}·r, so we get:
dJ_{z}/dt = q·ε_{t}·v
The ratio of dW/dt and dJ_{z}/dt gives us the following interesting equation:
Now, Feynman tries to relate this to the J_{z} = N·ħ = W/ω formula but… Well… We should remind ourselves that the angular frequency of these photons is not the angular frequency of our electron. So… Well… What can we say about this equation? Feynman suggests to integrate dJ_{z} and dW over some time interval, which makes sense: as mentioned, we interpreted W as the energy that is carried by the beam in a certain time. So if we integrate dW over this time interval, we get W. Likewise, if we integrate dJ_{z} over the same time interval, we should get the total angular momentum that our electron is absorbing from the light beam. Now, because dJ_{z} = dW/ω, we do concur with Feynman’s conclusion: the total angular momentum which is being absorbed by the electron is proportional to the total energy of the beam, and the constant of proportionality is equal to 1/ω.
It’s just… Well… The ω here is the angular frequency of the electron. It’s not the angular frequency of the beam. Not in our flywheel model of the electron which, admittedly, is not the model which Feynman used in his analysis. Feynman’s analysis is simpler: he assumes an electron at rest, so to speak, and then the beam drives it so it goes around in a circle with a velocity that is, effectively, given by the angular frequency of the beam itself. So… Well… Fine. Makes sense. As said, I just pushed the analysis a bit further along here. Both analyses raise an interesting question: how and where is the absorbed energy being stored? What is the mechanism here?
In Feynman’s analysis, the answer is quite simple: the electron did not have any motion before but does spin around after the beam hit it. So it has more energy now: it wasn’t a tiny flywheel before, but it is now!
In contrast, in my interpretation of the matter wave, the electron was spinning around already, so where does the extra energy go now? As its energy increases, ω = E/ħ must increase, right? Right. At the same time, the velocity v = r·ω must still be equal to v = r·ω = [ħ·/(m·c)]·(E/ħ) = c, right? Right. So… If ω increases, but r·ω must equal the speed of light, then r must actually decrease somewhat, right?
Right. It’s a weird but inevitable conclusion, it seems. I’ll let you think about it. 🙂
To conclude this post—which, I hope, the reader who triggered it will find interesting—I would like to quote Feynman on an issue on which most textbooks remain silent: the two-state nature of photons. I will just quote him without trying to comment or alter what he writes, because what he writes is clear enough, I think:
“Now let’s ask the following question: If light is linearly polarized in the x-direction, what is its angular momentum? Light polarized in the x-direction can be represented as the superposition of RHC and LHC polarized light. […] The interference of these two amplitudes produces the linear polarization, but it has equal probabilities to appear with plus or minus one unit of angular momentum. [Macroscopic measurements made on a beam of linearly polarized light will show that it carries zero angular momentum, because in a large number of photons there are nearly equal numbers of RHC and LHC photons contributing opposite amounts of angular momentum—the average angular momentum is zero.]
Now, we have said that any spin-one particle can have three values of J_{z}, namely +1, 0, −1 (the three states we saw in the Stern-Gerlach experiment). But light is screwy; it has only two states. It does not have the zero case. This strange lack is related to the fact that light cannot stand still. For a particle of spin j which is standing still, there must be the 2j+1 possible states with values of J_{z} going in steps of 1 from −j to +j. But it turns out that for something of spin j with zero mass only the states with the components +j and −j along the direction of motion exist. For example, light does not have three states, but only two—although a photon is still an object of spin one.”
In his typical style and frankness—for which he is revered by some (like me) but disliked by others—he admits this is very puzzling, and not obvious at all! Let me quote him once more:
“How is this consistent with our earlier proofs—based on what happens under rotations in space—that for spin-one particles three states are necessary? For a particle at rest, rotations can be made about any axis without changing the momentum state. Particles with zero rest mass (like photons and neutrinos) cannot be at rest; only rotations about the axis along the direction of motion do not change the momentum state. Arguments about rotations around one axis only are insufficient to prove that three states are required. We have tried to find at least a proof that the component of angular momentum along the direction of motion must for a zero mass particle be an integral multiple of ħ/2—and not something like ħ/3. Even using all sorts of properties of the Lorentz transformation and what not, we failed. Maybe it’s not true. We’ll have to talk about it with Prof. Wigner, who knows all about such things.”
The reference to Eugene Wigner is historically interesting. Feynman probably knew him very well—if only because they had both worked together on the Manhattan Project—and it’s true Wigner was not only a great physicist but a mathematical genius as well. However, Feynman probably quotes him here for the 1963 Nobel Prize he got for… Well… Wigner’s “contributions to the theory of the atomic nucleus and elementary particles, particularly through the discovery and application of fundamental symmetry principles.” 🙂 I’ll let you figure out how what I write about in this post, and symmetry arguments, might be related. 🙂
That’s it for today, folks! I hope you enjoyed this. 🙂
Post scriptum: The main disadvantage of the flywheel interpretation is that it doesn’t explain interference: waves interfere—some rotating mass doesn’t. Ultimately, the wave and flywheel interpretation must, somehow, be compatible. One way to think about it is that the electron can only move as it does—in a “local circulatory motion”—if there is a force on it that makes it move the way it does. That force must be gravitational because… Well… There is no other candidate, is there? [We’re not talking some electron orbital here—some negative charge orbiting around a positive nucleus. We’re just considering the electron itself.] So we just need to prove that our rotating arrow will also represent a force, whose components will make our electron move the way it does. That should not be difficult. The analogy of the V-twin engine should do the trick. I’ll deal with that in my next post. If we’re able to provide such proof (which, as mentioned, should not be difficult), it will be a wonderful illustration of the complementarity principle. 🙂
However, just thinking about it does raise some questions already. Circular motion like this can be explained in two equivalent ways. The most obvious way to think about it is to assume some central field. It’s the planetary model (illustrated below). However, that doesn’t suit our purposes because it’s hard – if possible at all – to relate it to the wavefunction oscillation.
The second model is our two-spring or V-twin engine model (illustrated below), but then what is the mass here? One hypothesis that comes to mind is that we’re constantly accelerating and decelerating an electric charge (the electron charge)—against all other charges in the Universe, so to speak. So that’s a force over a distance—energy. And energy has an equivalent mass.
The question which remains open, then, is the following: what is the nature of this force? In previous posts, I suggested it might be gravitational, but so here we’re back to the drawing board: we’re talking an electrical force, but applied to some mass which acquires mass because of… Well… Because of the force—because of the oscillation (the moving charge) itself. Hmm…. I need to think about this.
The speed of light as an angular velocity (2)
My previous post on the speed of light as an angular velocity was rather cryptic. This post will be a bit more elaborate. Not all that much, however: this stuff is and remains quite dense, unfortunately. 😦 But I’ll do my best to try to explain what I am thinking of. Remember the formula (or definition) of the elementary wavefunction:
ψ = a·e^{−i[E·t − p∙x]/ħ} = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
How should we interpret this? We know an actual particle will be represented by a wave packet: a sum of wavefunctions, each with its own amplitude a_{k} and its own argument θ_{k} = (E_{k}∙t − p_{k}∙x)/ħ. But… Well… Let’s see how far we get when analyzing the elementary wavefunction itself only.
According to mathematical convention, the imaginary unit (i) is a 90° angle in the counterclockwise direction. However, Nature surely cannot be bothered about our convention of measuring phase angles – or time itself – clockwise or counterclockwise. Therefore, both right- as well as left-handed polarization may be possible, as illustrated below.
The left-handed elementary wavefunction would be written as:
ψ = a·e^{i[E·t − p∙x]/ħ} = a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ − E∙t/ħ)
In my previous posts, I hypothesized that the two physical possibilities correspond to the angular momentum of our particle – say, an electron – being either positive or negative: J = +ħ/2 or, else, J = −ħ/2. I will come back to this in a moment. Let us first further examine the functional form of the wavefunction.
We should note that both the direction as well as the magnitude of the (linear) momentum (p) are relative: they depend on the orientation and relative velocity of our reference frame – which are, in effect, relative to the reference frame of our object. As such, the wavefunction itself is relative: another observer will obtain a different value for both the momentum (p) as well as for the energy (E). Of course, this makes us think of the relativity of the electric and magnetic field vectors (E and B) but… Well… It’s not quite the same because – as I will explain in a moment – the argument of the wavefunction, considered as a whole, is actually invariant under a Lorentz transformation.
Let me elaborate this point. If we consider the reference frame of the particle itself, then the idea of direction and momentum sort of vanishes, as the momentum vector shrinks to the origin itself: p = 0. Let us now look at how the argument of the wavefunction transforms. The E and p in the argument of the wavefunction (θ = ω∙t – k∙x = (E/ħ)∙t – (p/ħ)∙x = (E∙t – p∙x)/ħ) are, of course, the energy and momentum as measured in our frame of reference. Hence, we will want to write these quantities as E = E_{v} and p = p_{v }= p_{v}∙v. If we then use natural time and distance units (hence, the numerical value of c is equal to 1 and, hence, the (relative) velocity is then measured as a fraction of c, with a value between 0 and 1), we can relate the energy and momentum of a moving object to its energy and momentum when at rest using the following relativistic formulas:
E_{v }= γ·E_{0} and p_{v }= γ·m_{0}∙v = γ·E_{0}∙v/c^{2}
The argument of the wavefunction can then be re-written as:
θ = [γ·E_{0}/ħ]∙t – [(γ·E_{0}∙v/c^{2})/ħ]∙x = (E_{0}/ħ)·(t − v∙x/c^{2})·γ = (E_{0}/ħ)∙t’
The γ in these formulas is, of course, the Lorentz factor, and t’ is the proper time: t’ = (t − v∙x/c^{2})/√(1−v^{2}/c^{2}). Two essential points should be noted here:
1. The argument of the wavefunction is invariant. There is a primed time (t’) but there is no primed θ (θ’): θ = (E_{v}/ħ)·t – (p_{v}/ħ)·x = (E_{0}/ħ)∙t’.
2. The E_{0}/ħ coefficient pops up as an angular frequency: E_{0}/ħ = ω_{0}. We may refer to it as the frequency of the elementary wavefunction.
Now, if you don’t like the concept of angular frequency, we can also write: f_{0} = ω_{0}/2π = (E_{0}/ħ)/2π = E_{0}/h. Alternatively, and perhaps more elucidating, we get the following formula for the period of the oscillation:
T_{0} = 1/f_{0} = h/E_{0}
This is interesting, because we can look at the period as a natural unit of time for our particle. This period is inversely proportional to the (rest) energy of the particle, and the constant of proportionality is h. Substituting E_{0 }for m_{0}·c^{2}, we may also say it’s inversely proportional to the (rest) mass of the particle, with the constant of proportionality equal to h/c^{2}. The period of an electron, for example, would be equal to about 8×10^{−21} s. That’s very small, and it only gets smaller for larger objects ! But what does all of this really tell us? What does it actually mean?
We can look at the sine and cosine components of the wavefunction as an oscillation in two dimensions, as illustrated below.
Look at the little green dot going around. Imagine it is some mass going around and around. Its circular motion is equivalent to the two-dimensional oscillation. Indeed, instead of saying it moves along a circle, we may also say it moves simultaneously (1) left and right and back again (the cosine) while also moving (2) up and down and back again (the sine).
Now, a mass that rotates about a fixed axis has angular momentum, which we can write as the vector cross-product L = r×p or, alternatively, as the product of an angular velocity (ω) and rotational inertia (I), aka as the moment of inertia or the angular mass: L = I·ω. [Note we write L and ω in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (no boldface).]
We can now do some calculations. We already know the angular velocity (ω) is equal to E_{0}/ħ. Now, the magnitude of r in the L = r×p vector cross-product should equal the magnitude of ψ = a·e^{−i∙E·t/ħ}, so we write: r = a. What’s next? Well… The momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = m·v. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = r·ω.
So now we only need to think about what formula we should use for the angular mass. If we’re thinking, as we are doing here, of some point mass going around some center, then the formula to use is I = m·r^{2}. However, we may also want to think that the two-dimensional oscillation of our point mass actually describes the surface of a disk, in which case the formula for I becomes I = m·r^{2}/2. Of course, the addition of this 1/2 factor may seem arbitrary but, as you will see, it will give us a more intuitive result. This is what we get:
L = I·ω = (m·r^{2}/2)·(E/ħ) = (1/2)·a^{2}·(E/c^{2})·(E/ħ) = a^{2}·E^{2}/(2·ħ·c^{2})
Note that our frame of reference is that of the particle itself, so we should actually write ω_{0}, m_{0} and E_{0} instead of ω, m and E. The value of the rest energy of an electron is about 0.510 MeV, or 8.1871×10^{−14} N∙m. Now, this momentum should equal J = ±ħ/2. We can, therefore, derive the (Compton scattering) radius of an electron:Substituting the various constants with their numerical values, we find that a is equal 3.8616×10^{−13} m, which is the (reduced) Compton scattering radius of an electron. The (tangential) velocity (v) can now be calculated as being equal to v = r·ω = a·ω = [ħ·/(m·c)]·(E/ħ) = c. This is an amazing result. Let us think about it.
In our previous posts, we introduced the metaphor of two springs or oscillators, whose energy was equal to E = m·ω^{2}. Is this compatible with Einstein’s E = m·c^{2} mass-energy equivalence relation? It is. The E = m·c^{2} implies E/m = c^{2}. We, therefore, can write the following:
ω = E/ħ = m·c^{2}/ħ = m·(E/m)·/ħ ⇔ ω = E/ħ
Hence, we should actually have titled this and the previous post somewhat differently: the speed of light appears as a tangential velocity. Think of the following: the ratio of c and ω is equal to c/ω = a·ω/ω = a. Hence, the tangential and angular velocity would be the same if we’d measure distance in units of a. In other words, the radius of an electron appears as a natural distance unit here: if we’d measure ω in units of a per second, rather than in radians (which are expressed in the SI unit of distance, i.e. the meter) per second, the two concepts would coincide.
More fundamentally, we may want to look at the radius of an electron as a natural unit of velocity. Huh? Yes. Just re-write the c/ω = a as ω = c/a. What does it say? Exactly what I said, right? As such, the radius of an electron is not only a norm for measuring distance but also for time. 🙂
If you don’t quite get this, think of the following. For an electron, we get an angular frequency that is equal to ω = E/ħ = (8.19×10^{−14} N·m)/(1.05×10^{−34} N·m·s) ≈ 7.76×10^{20} radians per second. That’s an incredible velocity, because radians are expressed in distance units—so that’s in meter. However, our mass is not moving along the unit circle, but along a much tinier orbit. The ratio of the radius of the unit circle and a is equal to 1/a ≈ (1 m)/(3.86×10^{−13} m) ≈ 2.59×10^{12}. Now, if we divide the above-mentioned velocity of 7.76×10^{20} radians per second by this factor, we get… Right ! The speed of light: 2.998×10^{82} m/s. 🙂
Post scriptum: I have no clear answer to the question as to why we should use the I = m·r^{2}/2 formula, as opposed to the I = m·r^{2} formula. It ensures we get the result we want, but this 1/2 factor is actually rather enigmatic. It makes me think of the 1/2 factor in Schrödinger’s equation, which is also quite enigmatic. In my view, the 1/2 factor should not be there in Schrödinger’s equation. Electron orbitals tend to be occupied by two electrons with opposite spin. That’s why their energy levels should be twice as much. And so I’d get rid of the 1/2 factor, solve for the energy levels, and then divide them by two again. Or something like that. 🙂 But then that’s just my personal opinion or… Well… I’ve always been intrigued by the difference between the original printed edition of the Feynman Lectures and the online version, which has been edited on this point. My printed edition is the third printing, which is dated July 1966, and – on this point – it says the following:
“Don’t forget that m_{eff} 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 2 to 20 times the free-space mass of the electron.”
Two to twenty times. Not 1 or 0.5 to 20 times. No. Two times. As I’ve explained a couple of times, if we’d define a new effective mass which would be twice the old concept – so m_{eff}^{NEW} = 2∙m_{eff}^{OLD} – then such re-definition would not only solve a number of paradoxes and inconsistencies, but it will also justify my interpretation of energy as a two-dimensional oscillation of mass.
However, the online edition has been edited here to reflect the current knowledge about the behavior of an electron in a medium. Hence, if you click on the link above, you will read that the effective mass can be “about 0.1 to 30 times” the free-space mass of the electron. Well… This is another topic altogether, and so I’ll sign off here and let you think about it all. 🙂
The speed of light as an angular velocity
Over the weekend, I worked on a revised version of my paper on a physical interpretation of the wavefunction. However, I forgot to add the final remarks on the speed of light as an angular velocity. I know… This post is for my faithful followers only. It is dense, but let me add the missing bits here:
Post scriptum (29 October): Einstein’s view on aether theories probably still holds true: “We may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an aether. According to the general theory of relativity, space without aether is unthinkable – for in such space there not only would be no propagation of light, but also no possibility of existence for standards of space and time (measuring-rods and clocks), nor therefore any space-time intervals in the physical sense. But this aether may not be thought of as endowed with the quality characteristic of ponderable media, as consisting of parts which may be tracked through time. The idea of motion may not be applied to it.”
The above quote is taken from the Wikipedia article on aether theories. The same article also quotes Robert Laughlin, the 1998 Nobel Laureate in Physics, who said this about aether in 2005: “It is ironic that Einstein’s most creative work, the general theory of relativity, should boil down to conceptualizing space as a medium when his original premise [in special relativity] was that no such medium existed. […] The word ‘aether’ has extremely negative connotations in theoretical physics because of its past association with opposition to relativity. This is unfortunate because, stripped of these connotations, it rather nicely captures the way most physicists actually think about the vacuum. […]The modern concept of the vacuum of space, confirmed every day by experiment, is a relativistic aether. But we do not call it this because it is taboo.”
I really love this: a relativistic aether. My interpretation of the wavefunction is very consistent with that.
Wavefunctions as gravitational waves
This is the paper I always wanted to write. It is there now, and I think it is good – and that‘s an understatement. 🙂 It is probably best to download it as a pdf-file from the viXra.org site because this was a rather fast ‘copy and paste’ job from the Word version of the paper, so there may be issues with boldface notation (vector notation), italics and, most importantly, with formulas – which I, sadly, have to ‘snip’ into this WordPress blog, as they don’t have an easy copy function for mathematical formulas.
It’s great stuff. If you have been following my blog – and many of you have – you will want to digest this. 🙂
Abstract : This paper explores the implications of associating the components of the wavefunction with a physical dimension: force per unit mass – which is, of course, the dimension of acceleration (m/s^{2}) and gravitational fields. The classical electromagnetic field equations for energy densities, the Poynting vector and spin angular momentum are then re-derived by substituting the electromagnetic N/C unit of field strength (mass per unit charge) by the new N/kg = m/s^{2} dimension.
The results are elegant and insightful. For example, the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities, which establishes a physical normalization condition. Also, Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy, and the wavefunction itself can be interpreted as a propagating gravitational wave. Finally, as an added bonus, concepts such as the Compton scattering radius for a particle, spin angular momentum, and the boson-fermion dichotomy, can also be explained more intuitively.
While the approach offers a physical interpretation of the wavefunction, the author argues that the core of the Copenhagen interpretations revolves around the complementarity principle, which remains unchallenged because the interpretation of amplitude waves as traveling fields does not explain the particle nature of matter.
Introduction
This is not another introduction to quantum mechanics. We assume the reader is already familiar with the key principles and, importantly, with the basic math. We offer an interpretation of wave mechanics. As such, we do not challenge the complementarity principle: the physical interpretation of the wavefunction that is offered here explains the wave nature of matter only. It explains diffraction and interference of amplitudes but it does not explain why a particle will hit the detector not as a wave but as a particle. Hence, the Copenhagen interpretation of the wavefunction remains relevant: we just push its boundaries.
The basic ideas in this paper stem from a simple observation: the geometric similarity between the quantum-mechanical wavefunctions and electromagnetic waves is remarkably similar. The components of both waves are orthogonal to the direction of propagation and to each other. Only the relative phase differs : the electric and magnetic field vectors (E and B) have the same phase. In contrast, the phase of the real and imaginary part of the (elementary) wavefunction (ψ = a·e^{−i∙θ} = a∙cosθ – a∙sinθ) differ by 90 degrees (π/2).[1] Pursuing the analogy, we explore the following question: if the oscillating electric and magnetic field vectors of an electromagnetic wave carry the energy that one associates with the wave, can we analyze the real and imaginary part of the wavefunction in a similar way?
We show the answer is positive and remarkably straightforward. If the physical dimension of the electromagnetic field is expressed in newton per coulomb (force per unit charge), then the physical dimension of the components of the wavefunction may be associated with force per unit mass (newton per kg).[2] Of course, force over some distance is energy. The question then becomes: what is the energy concept here? Kinetic? Potential? Both?
The similarity between the energy of a (one-dimensional) linear oscillator (E = m·a^{2}·ω^{2}/2) and Einstein’s relativistic energy equation E = m∙c^{2} inspires us to interpret the energy as a two-dimensional oscillation of mass. To assist the reader, we construct a two-piston engine metaphor.[3] We then adapt the formula for the electromagnetic energy density to calculate the energy densities for the wave function. The results are elegant and intuitive: the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities. Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy itself.
As an added bonus, concepts such as the Compton scattering radius for a particle and spin angular, as well as the boson-fermion dichotomy can be explained in a fully intuitive way.[4]
Of course, such interpretation is also an interpretation of the wavefunction itself, and the immediate reaction of the reader is predictable: the electric and magnetic field vectors are, somehow, to be looked at as real vectors. In contrast, the real and imaginary components of the wavefunction are not. However, this objection needs to be phrased more carefully. First, it may be noted that, in a classical analysis, the magnetic force is a pseudovector itself.[5] Second, a suitable choice of coordinates may make quantum-mechanical rotation matrices irrelevant.[6]
Therefore, the author is of the opinion that this little paper may provide some fresh perspective on the question, thereby further exploring Einstein’s basic sentiment in regard to quantum mechanics, which may be summarized as follows: there must be some physical explanation for the calculated probabilities.[7]
We will, therefore, start with Einstein’s relativistic energy equation (E = mc^{2}) and wonder what it could possibly tell us.
I. Energy as a two-dimensional oscillation of mass
The structural similarity between the relativistic energy formula, the formula for the total energy of an oscillator, and the kinetic energy of a moving body, is striking:
- E = mc^{2}
- E = mω^{2}/2
- E = mv^{2}/2
In these formulas, ω, v and c all describe some velocity.[8] Of course, there is the 1/2 factor in the E = mω^{2}/2 formula[9], but that is exactly the point we are going to explore here: can we think of an oscillation in two dimensions, so it stores an amount of energy that is equal to E = 2·m·ω^{2}/2 = m·ω^{2}?
That is easy enough. Think, for example, of a V-2 engine with the pistons at a 90-degree angle, as illustrated below. The 90° angle makes it possible to perfectly balance the counterweight and the pistons, thereby ensuring smooth travel at all times. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down and provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring. Hence, we can describe it by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs.
Figure 1: Oscillations in two dimensions
If we assume there is no friction, we have a perpetuum mobile here. The compressed air and the rotating counterweight (which, combined with the crankshaft, acts as a flywheel[10]) store the potential energy. The moving masses of the pistons store the kinetic energy of the system.[11]
At this point, it is probably good to quickly review the relevant math. If the magnitude of the oscillation is equal to a, then the motion of the piston (or the mass on a spring) will be described by x = a·cos(ω·t + Δ).[12] Needless to say, Δ is just a phase factor which defines our t = 0 point, and ω is the natural angular frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by x = a·cos(ω·t), while the motion of the other is given by x = a·cos(ω·t–π/2) = a·sin(ω·t).
The kinetic and potential energy of one oscillator (think of one piston or one spring only) can then be calculated as:
- K.E. = T = m·v^{2}/2 = (1/2)·m·ω^{2}·a^{2}·sin^{2}(ω·t + Δ)
- P.E. = U = k·x^{2}/2 = (1/2)·k·a^{2}·cos^{2}(ω·t + Δ)
The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω^{2}. Hence, the total energy is equal to:
E = T + U = (1/2)· m·ω^{2}·a^{2}·[sin^{2}(ω·t + Δ) + cos^{2}(ω·t + Δ)] = m·a^{2}·ω^{2}/2
To facilitate the calculations, we will briefly assume k = m·ω^{2} and a are equal to 1. The motion of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:
d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ
Let us look at the second oscillator now. Just think of the second piston going up and down in the V-2 engine. Its motion is given by the sinθ function, which is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin^{2}(θ−π /2), and how it changes – as a function of θ – will be equal to:
2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ
We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the crankshaft will rotate with a constant angular velocity: linear motion becomes circular motion, and vice versa, and the total energy that is stored in the system is T + U = ma^{2}ω^{2}.
We have a great metaphor here. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. We know the wavefunction consist of a sine and a cosine: the cosine is the real component, and the sine is the imaginary component. Could they be equally real? Could each represent half of the total energy of our particle? Should we think of the c in our E = mc^{2} formula as an angular velocity?
These are sensible questions. Let us explore them.
II. The wavefunction as a two-dimensional oscillation
The elementary wavefunction is written as:
ψ = a·e^{−i[E·t − p∙x]/ħ} = a·e^{−i[E·t − p∙x]/ħ} = a·cos(p∙x/ħ – E∙t/ħ) + i·a·sin(p∙x/ħ – E∙t/ħ)
When considering a particle at rest (p = 0) this reduces to:
ψ = a·e^{−i∙E·t/ħ} = a·cos(–E∙t/ħ) + i·a·sin(–E∙t/ħ) = a·cos(E∙t/ħ) – i·a·sin(E∙t/ħ)
Let us remind ourselves of the geometry involved, which is illustrated below. Note that the argument of the wavefunction rotates clockwise with time, while the mathematical convention for measuring the phase angle (ϕ) is counter-clockwise.
Figure 2: Euler’s formula
If we assume the momentum p is all in the x-direction, then the p and x vectors will have the same direction, and p∙x/ħ reduces to p∙x/ħ. Most illustrations – such as the one below – will either freeze x or, else, t. Alternatively, one can google web animations varying both. The point is: we also have a two-dimensional oscillation here. These two dimensions are perpendicular to the direction of propagation of the wavefunction. For example, if the wavefunction propagates in the x-direction, then the oscillations are along the y– and z-axis, which we may refer to as the real and imaginary axis. Note how the phase difference between the cosine and the sine – the real and imaginary part of our wavefunction – appear to give some spin to the whole. I will come back to this.
Figure 3: Geometric representation of the wavefunction
Hence, if we would say these oscillations carry half of the total energy of the particle, then we may refer to the real and imaginary energy of the particle respectively, and the interplay between the real and the imaginary part of the wavefunction may then describe how energy propagates through space over time.
Let us consider, once again, a particle at rest. Hence, p = 0 and the (elementary) wavefunction reduces to ψ = a·e^{−i∙E·t/ħ}. Hence, the angular velocity of both oscillations, at some point x, is given by ω = -E/ħ. Now, the energy of our particle includes all of the energy – kinetic, potential and rest energy – and is, therefore, equal to E = mc^{2}.
Can we, somehow, relate this to the m·a^{2}·ω^{2} energy formula for our V-2 perpetuum mobile? Our wavefunction has an amplitude too. Now, if the oscillations of the real and imaginary wavefunction store the energy of our particle, then their amplitude will surely matter. In fact, the energy of an oscillation is, in general, proportional to the square of the amplitude: E µ a^{2}. We may, therefore, think that the a^{2} factor in the E = m·a^{2}·ω^{2} energy will surely be relevant as well.
However, here is a complication: an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude a_{k}, and their own ω_{i} = -E_{i}/ħ. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. To calculate the contribution of each wave to the total, both a_{i} as well as E_{i} will matter.
What is E_{i}? E_{i} varies around some average E, which we can associate with some average mass m: m = E/c^{2}. The Uncertainty Principle kicks in here. The analysis becomes more complicated, but a formula such as the one below might make sense:We can re-write this as:What is the meaning of this equation? We may look at it as some sort of physical normalization condition when building up the Fourier sum. Of course, we should relate this to the mathematical normalization condition for the wavefunction. Our intuition tells us that the probabilities must be related to the energy densities, but how exactly? We will come back to this question in a moment. Let us first think some more about the enigma: what is mass?
Before we do so, let us quickly calculate the value of c^{2}ħ^{2}: it is about 1´10^{–}^{51} N^{2}∙m^{4}. Let us also do a dimensional analysis: the physical dimensions of the E = m·a^{2}·ω^{2} equation make sense if we express m in kg, a in m, and ω in rad/s. We then get: [E] = kg∙m^{2}/s^{2} = (N∙s^{2}/m)∙m^{2}/s^{2} = N∙m = J. The dimensions of the left- and right-hand side of the physical normalization condition is N^{3}∙m^{5}.
III. What is mass?
We came up, playfully, with a meaningful interpretation for energy: it is a two-dimensional oscillation of mass. But what is mass? A new aether theory is, of course, not an option, but then what is it that is oscillating? To understand the physics behind equations, it is always good to do an analysis of the physical dimensions in the equation. Let us start with Einstein’s energy equation once again. If we want to look at mass, we should re-write it as m = E/c^{2}:
[m] = [E/c^{2}] = J/(m/s)^{2} = N·m∙s^{2}/m^{2} = N·s^{2}/m = kg
This is not very helpful. It only reminds us of Newton’s definition of a mass: mass is that what gets accelerated by a force. At this point, we may want to think of the physical significance of the absolute nature of the speed of light. Einstein’s E = mc^{2} equation implies we can write the ratio between the energy and the mass of any particle is always the same, so we can write, for example:This reminds us of the ω^{2}= C^{–}^{1}/L or ω^{2} = k/m of harmonic oscillators once again.[13] The key difference is that the ω^{2}= C^{–}^{1}/L and ω^{2} = k/m formulas introduce two or more degrees of freedom.[14] In contrast, c^{2}= E/m for any particle, always. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in one physical space only: our spacetime. Hence, the speed of light c emerges here as the defining property of spacetime – the resonant frequency, so to speak. We have no further degrees of freedom here.
The Planck-Einstein relation (for photons) and the de Broglie equation (for matter-particles) have an interesting feature: both imply that the energy of the oscillation is proportional to the frequency, with Planck’s constant as the constant of proportionality. Now, for one-dimensional oscillations – think of a guitar string, for example – we know the energy will be proportional to the square of the frequency. It is a remarkable observation: the two-dimensional matter-wave, or the electromagnetic wave, gives us two waves for the price of one, so to speak, each carrying half of the total energy of the oscillation but, as a result, we get a proportionality between E and f instead of between E and f^{2}.
However, such reflections do not answer the fundamental question we started out with: what is mass? At this point, it is hard to go beyond the circular definition that is implied by Einstein’s formula: energy is a two-dimensional oscillation of mass, and mass packs energy, and c emerges us as the property of spacetime that defines how exactly.
When everything is said and done, this does not go beyond stating that mass is some scalar field. Now, a scalar field is, quite simply, some real number that we associate with a position in spacetime. The Higgs field is a scalar field but, of course, the theory behind it goes much beyond stating that we should think of mass as some scalar field. The fundamental question is: why and how does energy, or matter, condense into elementary particles? That is what the Higgs mechanism is about but, as this paper is exploratory only, we cannot even start explaining the basics of it.
What we can do, however, is look at the wave equation again (Schrödinger’s equation), as we can now analyze it as an energy diffusion equation.
IV. Schrödinger’s equation as an energy diffusion equation
The interpretation of Schrödinger’s equation as a diffusion equation is straightforward. Feynman (Lectures, III-16-1) briefly summarizes it as follows:
“We can think of Schrödinger’s equation as describing the diffusion of the probability amplitude from one point to the next. […] But the imaginary coefficient in front of the derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”[17]
Let us review the basic math. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Uψ term disappears. Therefore, Schrödinger’s equation reduces to:
∂ψ(x, t)/∂t = i·(1/2)·(ħ/m_{eff})·∇^{2}ψ(x, t)
The ubiquitous diffusion equation in physics is:
∂φ(x, t)/∂t = D·∇^{2}φ(x, t)
The structural similarity is obvious. The key difference between both equations is that the wave equation gives us two equations for the price of one. Indeed, because ψ is a complex-valued function, with a real and an imaginary part, we get the following equations[18]:
- Re(∂ψ/∂t) = −(1/2)·(ħ/m_{eff})·Im(∇^{2}ψ)
- Im(∂ψ/∂t) = (1/2)·(ħ/m_{eff})·Re(∇^{2}ψ)
These equations make us think of the equations for an electromagnetic wave in free space (no stationary charges or currents):
- ∂B/∂t = –∇×E
- ∂E/∂t = c^{2}∇×B
The above equations effectively describe a propagation mechanism in spacetime, as illustrated below.
Figure 4: Propagation mechanisms
The Laplacian operator (∇^{2}), when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m^{2}). In this case, it is operating on ψ(x, t), so what is the dimension of our wavefunction ψ(x, t)? To answer that question, we should analyze the diffusion constant in Schrödinger’s equation, i.e. the (1/2)·(ħ/m_{eff}) factor:
- As a mathematical constant of proportionality, it will quantify the relationship between both derivatives (i.e. the time derivative and the Laplacian);
- As a physical constant, it will ensure the physical dimensions on both sides of the equation are compatible.
Now, the ħ/m_{eff} factor is expressed in (N·m·s)/(N· s^{2}/m) = m^{2}/s. Hence, it does ensure the dimensions on both sides of the equation are, effectively, the same: ∂ψ/∂t is a time derivative and, therefore, its dimension is s^{–}^{1} while, as mentioned above, the dimension of ∇^{2}ψ is m^{–}^{2}. However, this does not solve our basic question: what is the dimension of the real and imaginary part of our wavefunction?
At this point, mainstream physicists will say: it does not have a physical dimension, and there is no geometric interpretation of Schrödinger’s equation. One may argue, effectively, that its argument, (p∙x – E∙t)/ħ, is just a number and, therefore, that the real and imaginary part of ψ is also just some number.
To this, we may object that ħ may be looked as a mathematical scaling constant only. If we do that, then the argument of ψ will, effectively, be expressed in action units, i.e. in N·m·s. It then does make sense to also associate a physical dimension with the real and imaginary part of ψ. What could it be?
We may have a closer look at Maxwell’s equations for inspiration here. The electric field vector is expressed in newton (the unit of force) per unit of charge (coulomb). Now, there is something interesting here. The physical dimension of the magnetic field is N/C divided by m/s.[19] We may write B as the following vector cross-product: B = (1/c)∙e_{x}×E, with e_{x} the unit vector pointing in the x-direction (i.e. the direction of propagation of the wave). Hence, we may associate the (1/c)∙e_{x}× operator, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by i also amounts to a rotation by 90° degrees. Hence, we may boldly write: B = (1/c)∙e_{x}×E = (1/c)∙i∙E. This allows us to also geometrically interpret Schrödinger’s equation in the way we interpreted it above (see Figure 3).[20]
Still, we have not answered the question as to what the physical dimension of the real and imaginary part of our wavefunction should be. At this point, we may be inspired by the structural similarity between Newton’s and Coulomb’s force laws:Hence, if the electric field vector E is expressed in force per unit charge (N/C), then we may want to think of associating the real part of our wavefunction with a force per unit mass (N/kg). We can, of course, do a substitution here, because the mass unit (1 kg) is equivalent to 1 N·s^{2}/m. Hence, our N/kg dimension becomes:
N/kg = N/(N·s^{2}/m)= m/s^{2}
What is this: m/s^{2}? Is that the dimension of the a·cosθ term in the a·e^{−iθ }= a·cosθ − i·a·sinθ wavefunction?
My answer is: why not? Think of it: m/s^{2} is the physical dimension of acceleration: the increase or decrease in velocity (m/s) per second. It ensures the wavefunction for any particle – matter-particles or particles with zero rest mass (photons) – and the associated wave equation (which has to be the same for all, as the spacetime we live in is one) are mutually consistent.
In this regard, we should think of how we would model a gravitational wave. The physical dimension would surely be the same: force per mass unit. It all makes sense: wavefunctions may, perhaps, be interpreted as traveling distortions of spacetime, i.e. as tiny gravitational waves.
V. Energy densities and flows
Pursuing the geometric equivalence between the equations for an electromagnetic wave and Schrödinger’s equation, we can now, perhaps, see if there is an equivalent for the energy density. For an electromagnetic wave, we know that the energy density is given by the following formula:E and B are the electric and magnetic field vector respectively. The Poynting vector will give us the directional energy flux, i.e. the energy flow per unit area per unit time. We write:Needless to say, the ∇∙ operator is the divergence and, therefore, gives us the magnitude of a (vector) field’s source or sink at a given point. To be precise, the divergence gives us the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. In this case, it gives us the volume density of the flux of S.
We can analyze the dimensions of the equation for the energy density as follows:
- E is measured in newton per coulomb, so [E∙E] = [E^{2}] = N^{2}/C^{2}.
- B is measured in (N/C)/(m/s), so we get [B∙B] = [B^{2}] = (N^{2}/C^{2})·(s^{2}/m^{2}). However, the dimension of our c^{2} factor is (m^{2}/s^{2}) and so we’re also left with N^{2}/C^{2}.
- The ϵ_{0} is the electric constant, aka as the vacuum permittivity. As a physical constant, it should ensure the dimensions on both sides of the equation work out, and they do: [ε_{0}] = C^{2}/(N·m^{2}) and, therefore, if we multiply that with N^{2}/C^{2}, we find that u is expressed in J/m^{3}.[21]
Replacing the newton per coulomb unit (N/C) by the newton per kg unit (N/kg) in the formulas above should give us the equivalent of the energy density for the wavefunction. We just need to substitute ϵ_{0} for an equivalent constant. We may to give it a try. If the energy densities can be calculated – which are also mass densities, obviously – then the probabilities should be proportional to them.
Let us first see what we get for a photon, assuming the electromagnetic wave represents its wavefunction. Substituting B for (1/c)∙i∙E or for −(1/c)∙i∙E gives us the following result:Zero!? An unexpected result! Or not? We have no stationary charges and no currents: only an electromagnetic wave in free space. Hence, the local energy conservation principle needs to be respected at all points in space and in time. The geometry makes sense of the result: for an electromagnetic wave, the magnitudes of E and B reach their maximum, minimum and zero point simultaneously, as shown below.[22] This is because their phase is the same.
Figure 5: Electromagnetic wave: E and B
Should we expect a similar result for the energy densities that we would associate with the real and imaginary part of the matter-wave? For the matter-wave, we have a phase difference between a·cosθ and a·sinθ, which gives a different picture of the propagation of the wave (see Figure 3).[23] In fact, the geometry of the suggestion suggests some inherent spin, which is interesting. I will come back to this. Let us first guess those densities. Making abstraction of any scaling constants, we may write:We get what we hoped to get: the absolute square of our amplitude is, effectively, an energy density !
|ψ|^{2 } = |a·e^{−i∙E·t/ħ}|^{2 }= a^{2 }= u
This is very deep. A photon has no rest mass, so it borrows and returns energy from empty space as it travels through it. In contrast, a matter-wave carries energy and, therefore, has some (rest) mass. It is therefore associated with an energy density, and this energy density gives us the probabilities. Of course, we need to fine-tune the analysis to account for the fact that we have a wave packet rather than a single wave, but that should be feasible.
As mentioned, the phase difference between the real and imaginary part of our wavefunction (a cosine and a sine function) appear to give some spin to our particle. We do not have this particularity for a photon. Of course, photons are bosons, i.e. spin-zero particles, while elementary matter-particles are fermions with spin-1/2. Hence, our geometric interpretation of the wavefunction suggests that, after all, there may be some more intuitive explanation of the fundamental dichotomy between bosons and fermions, which puzzled even Feynman:
“Why is it that particles with half-integral spin are Fermi particles, whereas particles with integral spin are Bose particles? We apologize for the fact that we cannot give you an elementary explanation. An explanation has been worked out by Pauli from complicated arguments of quantum field theory and relativity. He has shown that the two must necessarily go together, but we have not been able to find a way of reproducing his arguments on an elementary level. It appears to be one of the few places in physics where there is a rule which can be stated very simply, but for which no one has found a simple and easy explanation. The explanation is deep down in relativistic quantum mechanics. This probably means that we do not have a complete understanding of the fundamental principle involved.” (Feynman, Lectures, III-4-1)
The physical interpretation of the wavefunction, as presented here, may provide some better understanding of ‘the fundamental principle involved’: the physical dimension of the oscillation is just very different. That is all: it is force per unit charge for photons, and force per unit mass for matter-particles. We will examine the question of spin somewhat more carefully in section VII. Let us first examine the matter-wave some more.
VI. Group and phase velocity of the matter-wave
The geometric representation of the matter-wave (see Figure 3) suggests a traveling wave and, yes, of course: the matter-wave effectively travels through space and time. But what is traveling, exactly? It is the pulse – or the signal – only: the phase velocity of the wave is just a mathematical concept and, even in our physical interpretation of the wavefunction, the same is true for the group velocity of our wave packet. The oscillation is two-dimensional, but perpendicular to the direction of travel of the wave. Hence, nothing actually moves with our particle.
Here, we should also reiterate that we did not answer the question as to what is oscillating up and down and/or sideways: we only associated a physical dimension with the components of the wavefunction – newton per kg (force per unit mass), to be precise. We were inspired to do so because of the physical dimension of the electric and magnetic field vectors (newton per coulomb, i.e. force per unit charge) we associate with electromagnetic waves which, for all practical purposes, we currently treat as the wavefunction for a photon. This made it possible to calculate the associated energy densities and a Poynting vector for energy dissipation. In addition, we showed that Schrödinger’s equation itself then becomes a diffusion equation for energy. However, let us now focus some more on the asymmetry which is introduced by the phase difference between the real and the imaginary part of the wavefunction. Look at the mathematical shape of the elementary wavefunction once again:
ψ = a·e^{−i[E·t − p∙x]/ħ} = a·e^{−i[E·t − p∙x]/ħ} = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
The minus sign in the argument of our sine and cosine function defines the direction of travel: an F(x−v∙t) wavefunction will always describe some wave that is traveling in the positive x-direction (with c the wave velocity), while an F(x+v∙t) wavefunction will travel in the negative x-direction. For a geometric interpretation of the wavefunction in three dimensions, we need to agree on how to define i or, what amounts to the same, a convention on how to define clockwise and counterclockwise directions: if we look at a clock from the back, then its hand will be moving counterclockwise. So we need to establish the equivalent of the right-hand rule. However, let us not worry about that now. Let us focus on the interpretation. To ease the analysis, we’ll assume we’re looking at a particle at rest. Hence, p = 0, and the wavefunction reduces to:
ψ = a·e^{−i∙E·t/ħ} = a·cos(−E∙t/ħ) + i·a·sin(−E_{0}∙t/ħ) = a·cos(E_{0}∙t/ħ) − i·a·sin(E_{0}∙t/ħ)
E_{0} is, of course, the rest mass of our particle and, now that we are here, we should probably wonder whose time t we are talking about: is it our time, or is the proper time of our particle? Well… In this situation, we are both at rest so it does not matter: t is, effectively, the proper time so perhaps we should write it as t_{0}. It does not matter. You can see what we expect to see: E_{0}/ħ pops up as the natural frequency of our matter-particle: (E_{0}/ħ)∙t = ω∙t. Remembering the ω = 2π·f = 2π/T and T = 1/f formulas, we can associate a period and a frequency with this wave, using the ω = 2π·f = 2π/T. Noting that ħ = h/2π, we find the following:
T = 2π·(ħ/E_{0}) = h/E_{0} ⇔ f = E_{0}/h = m_{0}c^{2}/h
This is interesting, because we can look at the period as a natural unit of time for our particle. What about the wavelength? That is tricky because we need to distinguish between group and phase velocity here. The group velocity (v_{g}) should be zero here, because we assume our particle does not move. In contrast, the phase velocity is given by v_{p} = λ·f = (2π/k)·(ω/2π) = ω/k. In fact, we’ve got something funny here: the wavenumber k = p/ħ is zero, because we assume the particle is at rest, so p = 0. So we have a division by zero here, which is rather strange. What do we get assuming the particle is not at rest? We write:
v_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = E/(m·v_{g}) = (m·c^{2})/(m·v_{g}) = c^{2}/v_{g}
This is interesting: it establishes a reciprocal relation between the phase and the group velocity, with c as a simple scaling constant. Indeed, the graph below shows the shape of the function does not change with the value of c, and we may also re-write the relation above as:
v_{p}/c = β_{p} = c/v_{p} = 1/β_{g} = 1/(c/v_{p})
Figure 6: Reciprocal relation between phase and group velocity
We can also write the mentioned relationship as v_{p}·v_{g} = c^{2}, which reminds us of the relationship between the electric and magnetic constant (1/ε_{0})·(1/μ_{0}) = c^{2}. This is interesting in light of the fact we can re-write this as (c·ε_{0})·(c·μ_{0}) = 1, which shows electricity and magnetism are just two sides of the same coin, so to speak.[24]
Interesting, but how do we interpret the math? What about the implications of the zero value for wavenumber k = p/ħ. We would probably like to think it implies the elementary wavefunction should always be associated with some momentum, because the concept of zero momentum clearly leads to weird math: something times zero cannot be equal to c^{2}! Such interpretation is also consistent with the Uncertainty Principle: if Δx·Δp ≥ ħ, then neither Δx nor Δp can be zero. In other words, the Uncertainty Principle tells us that the idea of a pointlike particle actually being at some specific point in time and in space does not make sense: it has to move. It tells us that our concept of dimensionless points in time and space are mathematical notions only. Actual particles – including photons – are always a bit spread out, so to speak, and – importantly – they have to move.
For a photon, this is self-evident. It has no rest mass, no rest energy, and, therefore, it is going to move at the speed of light itself. We write: p = m·c = m·c^{2}/c = E/c. Using the relationship above, we get:
v_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = c ⇒ v_{g} = c^{2}/v_{p} = c^{2}/c = c
This is good: we started out with some reflections on the matter-wave, but here we get an interpretation of the electromagnetic wave as a wavefunction for the photon. But let us get back to our matter-wave. In regard to our interpretation of a particle having to move, we should remind ourselves, once again, of the fact that an actual particle is always localized in space and that it can, therefore, not be represented by the elementary wavefunction ψ = a·e^{−i[E·t − p∙x]/ħ} or, for a particle at rest, the ψ = a·e^{−i∙E·t/ħ} function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude a_{i}, and their own ω_{i} = −E_{i}/ħ. Indeed, in section II, we showed that each of these wavefunctions will contribute some energy to the total energy of the wave packet and that, to calculate the contribution of each wave to the total, both a_{i} as well as E_{i} matter. This may or may not resolve the apparent paradox. Let us look at the group velocity.
To calculate a meaningful group velocity, we must assume the v_{g} = ∂ω_{i}/∂k_{i} = ∂(E_{i}/ħ)/∂(p_{i}/ħ) = ∂(E_{i})/∂(p_{i}) exists. So we must have some dispersion relation. How do we calculate it? We need to calculate ω_{i} as a function of k_{i}_{ }here, or E_{i} as a function of p_{i}. How do we do that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as we did, i.e. by distinguishing the real and imaginary parts of the ∂ψ/∂t =i·[ħ/(2m)]·∇^{2}ψ wave equation and, hence, re-write it as the following pair of two equations:
- Re(∂ψ/∂t) = −[ħ/(2m_{eff})]·Im(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·cos(kx − ωt)
- Im(∂ψ/∂t) = [ħ/(2m_{eff})]·Re(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·sin(kx − ωt)
Both equations imply the following dispersion relation:
ω = ħ·k^{2}/(2m_{eff})
Of course, we need to think about the subscripts now: we have ω_{i}, k_{i}, but… What about m_{eff} or, dropping the subscript, m? Do we write it as m_{i}? If so, what is it? Well… It is the equivalent mass of E_{i} obviously, and so we get it from the mass-energy equivalence relation: m_{i} = E_{i}/c^{2}. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too. Here, I should refer back to Section II: E_{i} varies around some average energy E and, therefore, the Uncertainty Principle kicks in.
VII. Explaining spin
The elementary wavefunction vector – i.e. the vector sum of the real and imaginary component – rotates around the x-axis, which gives us the direction of propagation of the wave (see Figure 3). Its magnitude remains constant. In contrast, the magnitude of the electromagnetic vector – defined as the vector sum of the electric and magnetic field vectors – oscillates between zero and some maximum (see Figure 5).
We already mentioned that the rotation of the wavefunction vector appears to give some spin to the particle. Of course, a circularly polarized wave would also appear to have spin (think of the E and B vectors rotating around the direction of propagation – as opposed to oscillating up and down or sideways only). In fact, a circularly polarized light does carry angular momentum, as the equivalent mass of its energy may be thought of as rotating as well. But so here we are looking at a matter-wave.
The basic idea is the following: if we look at ψ = a·e^{−i∙E·t/ħ} as some real vector – as a two-dimensional oscillation of mass, to be precise – then we may associate its rotation around the direction of propagation with some torque. The illustration below reminds of the math here.
Figure 7: Torque and angular momentum vectors
A torque on some mass about a fixed axis gives it angular momentum, which we can write as the vector cross-product L = r×p or, perhaps easier for our purposes here as the product of an angular velocity (ω) and rotational inertia (I), aka as the moment of inertia or the angular mass. We write:
L = I·ω
Note we can write L and ω in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (no boldface). We can now do some calculations. Let us start with the angular velocity. In our previous posts, we showed that the period of the matter-wave is equal to T = 2π·(ħ/E_{0}). Hence, the angular velocity must be equal to:
ω = 2π/[2π·(ħ/E_{0})] = E_{0}/ħ
We also know the distance r, so that is the magnitude of r in the L = r×p vector cross-product: it is just a, so that is the magnitude of ψ = a·e^{−i∙E·t/ħ}. Now, the momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = m·v. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = r·ω. So now we only need to think about what we should use for m or, if we want to work with the angular velocity (ω), the angular mass (I). Here we need to make some assumption about the mass (or energy) distribution. Now, it may or may not sense to assume the energy in the oscillation – and, therefore, the mass – is distributed uniformly. In that case, we may use the formula for the angular mass of a solid cylinder: I = m·r^{2}/2. If we keep the analysis non-relativistic, then m = m_{0}. Of course, the energy-mass equivalence tells us that m_{0} = E_{0}/c^{2}. Hence, this is what we get:
L = I·ω = (m_{0}·r^{2}/2)·(E_{0}/ħ) = (1/2)·a^{2}·(E_{0}/c^{2})·(E_{0}/ħ) = a^{2}·E_{0}^{2}/(2·ħ·c^{2})
Does it make sense? Maybe. Maybe not. Let us do a dimensional analysis: that won’t check our logic, but it makes sure we made no mistakes when mapping mathematical and physical spaces. We have m^{2}·J^{2} = m^{2}·N^{2}·m^{2} in the numerator and N·m·s·m^{2}/s^{2} in the denominator. Hence, the dimensions work out: we get N·m·s as the dimension for L, which is, effectively, the physical dimension of angular momentum. It is also the action dimension, of course, and that cannot be a coincidence. Also note that the E = mc^{2} equation allows us to re-write it as:
L = a^{2}·E_{0}^{2}/(2·ħ·c^{2})
Of course, in quantum mechanics, we associate spin with the magnetic moment of a charged particle, not with its mass as such. Is there way to link the formula above to the one we have for the quantum-mechanical angular momentum, which is also measured in N·m·s units, and which can only take on one of two possible values: J = +ħ/2 and −ħ/2? It looks like a long shot, right? How do we go from (1/2)·a^{2}·m_{0}^{2}/ħ to ± (1/2)∙ħ? Let us do a numerical example. The energy of an electron is typically 0.510 MeV » 8.1871×10^{−14} N∙m, and a… What value should we take for a?
We have an obvious trio of candidates here: the Bohr radius, the classical electron radius (aka the Thompon scattering length), and the Compton scattering radius.
Let us start with the Bohr radius, so that is about 0.×10^{−10} N∙m. We get L = a^{2}·E_{0}^{2}/(2·ħ·c^{2}) = 9.9×10^{−31} N∙m∙s. Now that is about 1.88×10^{4} times ħ/2. That is a huge factor. The Bohr radius cannot be right: we are not looking at an electron in an orbital here. To show it does not make sense, we may want to double-check the analysis by doing the calculation in another way. We said each oscillation will always pack 6.626070040(81)×10^{−34} joule in energy. So our electron should pack about 1.24×10^{−20} oscillations. The angular momentum (L) we get when using the Bohr radius for a and the value of 6.626×10^{−34} joule for E_{0} and the Bohr radius is equal to 6.49×10^{−59} N∙m∙s. So that is the angular momentum per oscillation. When we multiply this with the number of oscillations (1.24×10^{−20}), we get about 8.01×10^{−51} N∙m∙s, so that is a totally different number.
The classical electron radius is about 2.818×10^{−15} m. We get an L that is equal to about 2.81×10^{−39} N∙m∙s, so now it is a tiny fraction of ħ/2! Hence, this leads us nowhere. Let us go for our last chance to get a meaningful result! Let us use the Compton scattering length, so that is about 2.42631×10^{−12} m.
This gives us an L of 2.08×10^{−33} N∙m∙s, which is only 20 times ħ. This is not so bad, but it is good enough? Let us calculate it the other way around: what value should we take for a so as to ensure L = a^{2}·E_{0}^{2}/(2·ħ·c^{2}) = ħ/2? Let us write it out:
In fact, this is the formula for the so-called reduced Compton wavelength. This is perfect. We found what we wanted to find. Substituting this value for a (you can calculate it: it is about 3.8616×10^{−33} m), we get what we should find:
This is a rather spectacular result, and one that would – a priori – support the interpretation of the wavefunction that is being suggested in this paper.^{ }
VIII. The boson-fermion dichotomy
Let us do some more thinking on the boson-fermion dichotomy. Again, we should remind ourselves that an actual particle is localized in space and that it can, therefore, not be represented by the elementary wavefunction ψ = a·e^{−i[E·t − p∙x]/ħ} or, for a particle at rest, the ψ = a·e^{−i∙E·t/ħ} function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude a_{i}, and their own ω_{i} = −E_{i}/ħ. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. Now, we can have another wild but logical theory about this.
Think of the apparent right-handedness of the elementary wavefunction: surely, Nature can’t be bothered about our convention of measuring phase angles clockwise or counterclockwise. Also, the angular momentum can be positive or negative: J = +ħ/2 or −ħ/2. Hence, we would probably like to think that an actual particle – think of an electron, or whatever other particle you’d think of – may consist of right-handed as well as left-handed elementary waves. To be precise, we may think they either consist of (elementary) right-handed waves or, else, of (elementary) left-handed waves. An elementary right-handed wave would be written as:
ψ(θ_{i}) = a_{i}·(cosθ_{i} + i·sinθ_{i})
In contrast, an elementary left-handed wave would be written as:
ψ(θ_{i}) = a_{i}·(cosθ_{i} − i·sinθ_{i})
How does that work out with the E_{0}·t argument of our wavefunction? Position is position, and direction is direction, but time? Time has only one direction, but Nature surely does not care how we count time: counting like 1, 2, 3, etcetera or like −1, −2, −3, etcetera is just the same. If we count like 1, 2, 3, etcetera, then we write our wavefunction like:
ψ = a·cos(E_{0}∙t/ħ) − i·a·sin(E_{0}∙t/ħ)
If we count time like −1, −2, −3, etcetera then we write it as:
ψ = a·cos(−E_{0}∙t/ħ) − i·a·sin(−E_{0}∙t/ħ)= a·cos(E_{0}∙t/ħ) + i·a·sin(E_{0}∙t/ħ)
Hence, it is just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! This, then, should explain why we can have either positive or negative quantum-mechanical spin (+ħ/2 or −ħ/2). It is the usual thing: we have two mathematical possibilities here, and so we must have two physical situations that correspond to it.
It is only natural. If we have left- and right-handed photons – or, generalizing, left- and right-handed bosons – then we should also have left- and right-handed fermions (electrons, protons, etcetera). Back to the dichotomy. The textbook analysis of the dichotomy between bosons and fermions may be epitomized by Richard Feynman’s Lecture on it (Feynman, III-4), which is confusing and – I would dare to say – even inconsistent: how are photons or electrons supposed to know that they need to interfere with a positive or a negative sign? They are not supposed to know anything: knowledge is part of our interpretation of whatever it is that is going on there.
Hence, it is probably best to keep it simple, and think of the dichotomy in terms of the different physical dimensions of the oscillation: newton per kg versus newton per coulomb. And then, of course, we should also note that matter-particles have a rest mass and, therefore, actually carry charge. Photons do not. But both are two-dimensional oscillations, and the point is: the so-called vacuum – and the rest mass of our particle (which is zero for the photon and non-zero for everything else) – give us the natural frequency for both oscillations, which is beautifully summed up in that remarkable equation for the group and phase velocity of the wavefunction, which applies to photons as well as matter-particles:
(v_{phase}·c)·(v_{group}·c) = 1 ⇔ v_{p}·v_{g} = c^{2}
The final question then is: why are photons spin-zero particles? Well… We should first remind ourselves of the fact that they do have spin when circularly polarized.[25] Here we may think of the rotation of the equivalent mass of their energy. However, if they are linearly polarized, then there is no spin. Even for circularly polarized waves, the spin angular momentum of photons is a weird concept. If photons have no (rest) mass, then they cannot carry any charge. They should, therefore, not have any magnetic moment. Indeed, what I wrote above shows an explanation of quantum-mechanical spin requires both mass as well as charge.[26]
IX. Concluding remarks
There are, of course, other ways to look at the matter – literally. For example, we can imagine two-dimensional oscillations as circular rather than linear oscillations. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of a two-dimensional oscillation as an oscillation of a polar and azimuthal angle.
Figure 8: Two-dimensional circular movement
The point of this paper is not to make any definite statements. That would be foolish. Its objective is just to challenge the simplistic mainstream viewpoint on the reality of the wavefunction. Stating that it is a mathematical construct only without physical significance amounts to saying it has no meaning at all. That is, clearly, a non-sustainable proposition.
The interpretation that is offered here looks at amplitude waves as traveling fields. Their physical dimension may be expressed in force per mass unit, as opposed to electromagnetic waves, whose amplitudes are expressed in force per (electric) charge unit. Also, the amplitudes of matter-waves incorporate a phase factor, but this may actually explain the rather enigmatic dichotomy between fermions and bosons and is, therefore, an added bonus.
The interpretation that is offered here has some advantages over other explanations, as it explains the how of diffraction and interference. However, while it offers a great explanation of the wave nature of matter, it does not explain its particle nature: while we think of the energy as being spread out, we will still observe electrons and photons as pointlike particles once they hit the detector. Why is it that a detector can sort of ‘hook’ the whole blob of energy, so to speak?
The interpretation of the wavefunction that is offered here does not explain this. Hence, the complementarity principle of the Copenhagen interpretation of the wavefunction surely remains relevant.
Appendix 1: The de Broglie relations and energy
The 1/2 factor in Schrödinger’s equation is related to the concept of the effective mass (m_{eff}). It is easy to make the wrong calculations. For example, when playing with the famous de Broglie relations – aka as the matter-wave equations – one may be tempted to derive the following energy concept:
- E = h·f and p = h/λ. Therefore, f = E/h and λ = p/h.
- v = f·λ = (E/h)∙(p/h) = E/p
- p = m·v. Therefore, E = v·p = m·v^{2}
E = m·v^{2}? This resembles the E = mc^{2} equation and, therefore, one may be enthused by the discovery, especially because the m·v^{2} also pops up when working with the Least Action Principle in classical mechanics, which states that the path that is followed by a particle will minimize the following integral:Now, we can choose any reference point for the potential energy but, to reflect the energy conservation law, we can select a reference point that ensures the sum of the kinetic and the potential energy is zero throughout the time interval. If the force field is uniform, then the integrand will, effectively, be equal to KE − PE = m·v^{2}.[27]
However, that is classical mechanics and, therefore, not so relevant in the context of the de Broglie equations, and the apparent paradox should be solved by distinguishing between the group and the phase velocity of the matter wave.
Appendix 2: The concept of the effective mass
The effective mass – as used in Schrödinger’s equation – is a rather enigmatic concept. To make sure we are making the right analysis here, I should start by noting you will usually see Schrödinger’s equation written as:This formulation includes a term with the potential energy (U). In free space (no potential), this term disappears, and the equation can be re-written as:
∂ψ(x, t)/∂t = i·(1/2)·(ħ/m_{eff})·∇^{2}ψ(x, t)
We just moved the i·ħ coefficient to the other side, noting that 1/i = –i. Now, in one-dimensional space, and assuming ψ is just the elementary wavefunction (so we substitute a·e^{−i∙[E·t − p∙x]/ħ} for ψ), this implies the following:
−a·i·(E/ħ)·e^{−}i∙^{[E·t − p∙x]/ħ} = −i·(ħ/2m_{eff})·a·(p^{2}/ħ^{2})· e^{−i∙[E·t − p∙x]/ħ }
⇔ E = p^{2}/(2m_{eff}) ⇔ m_{eff} = m∙(v/c)^{2}/2 = m∙β^{2}/2
It is an ugly formula: it resembles the kinetic energy formula (K.E. = m∙v^{2}/2) but it is, in fact, something completely different. The β^{2}/2 factor ensures the effective mass is always a fraction of the mass itself. To get rid of the ugly 1/2 factor, we may re-define m_{eff} as two times the old m_{eff} (hence, m_{eff}^{NEW} = 2∙m_{eff}^{OLD}), as a result of which the formula will look somewhat better:
m_{eff} = m∙(v/c)^{2} = m∙β^{2}
We know β varies between 0 and 1 and, therefore, m_{eff} will vary between 0 and m. Feynman drops the subscript, and just writes m_{eff} as m in his textbook (see Feynman, III-19). On the other hand, the electron mass as used is also the electron mass that is used to calculate the size of an atom (see Feynman, III-2-4). As such, the two mass concepts are, effectively, mutually compatible. It is confusing because the same mass is often defined as the mass of a stationary electron (see, for example, the article on it in the online Wikipedia encyclopedia[28]).
In the context of the derivation of the electron orbitals, we do have the potential energy term – which is the equivalent of a source term in a diffusion equation – and that may explain why the above-mentioned m_{eff} = m∙(v/c)^{2} = m∙β^{2} formula does not apply.
References
This paper discusses general principles in physics only. Hence, references can be limited to references to physics textbooks only. For ease of reading, any reference to additional material has been limited to a more popular undergrad textbook that can be consulted online: Feynman’s Lectures on Physics (http://www.feynmanlectures.caltech.edu). References are per volume, per chapter and per section. For example, Feynman III-19-3 refers to Volume III, Chapter 19, Section 3.
Notes
[1] Of course, an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction ψ = a·e^{−i∙θ} = a·e^{−i[E·t − p∙x]/ħ} = a·(cosθ – i·a·sinθ). We must build a wave packet for that: a sum of wavefunctions, each with its own amplitude a_{k} and its own argument θ_{k} = (E_{k}∙t – p_{k}∙x)/ħ. This is dealt with in this paper as part of the discussion on the mathematical and physical interpretation of the normalization condition.
[2] The N/kg dimension immediately, and naturally, reduces to the dimension of acceleration (m/s^{2}), thereby facilitating a direct interpretation in terms of Newton’s force law.
[3] In physics, a two-spring metaphor is more common. Hence, the pistons in the author’s perpetuum mobile may be replaced by springs.
[4] The author re-derives the equation for the Compton scattering radius in section VII of the paper.
[5] The magnetic force can be analyzed as a relativistic effect (see Feynman II-13-6). The dichotomy between the electric force as a polar vector and the magnetic force as an axial vector disappears in the relativistic four-vector representation of electromagnetism.
[6] For example, when using Schrödinger’s equation in a central field (think of the electron around a proton), the use of polar coordinates is recommended, as it ensures the symmetry of the Hamiltonian under all rotations (see Feynman III-19-3)
[7] This sentiment is usually summed up in the apocryphal quote: “God does not play dice.”The actual quote comes out of one of Einstein’s private letters to Cornelius Lanczos, another scientist who had also emigrated to the US. The full quote is as follows: “You are the only person I know who has the same attitude towards physics as I have: belief in the comprehension of reality through something basically simple and unified… It seems hard to sneak a look at God’s cards. But that He plays dice and uses ‘telepathic’ methods… is something that I cannot believe for a single moment.” (Helen Dukas and Banesh Hoffman, Albert Einstein, the Human Side: New Glimpses from His Archives, 1979)
[8] Of course, both are different velocities: ω is an angular velocity, while v is a linear velocity: ω is measured in radians per second, while v is measured in meter per second. However, the definition of a radian implies radians are measured in distance units. Hence, the physical dimensions are, effectively, the same. As for the formula for the total energy of an oscillator, we should actually write: E = m·a^{2}∙ω^{2}/2. The additional factor (a) is the (maximum) amplitude of the oscillator.
[9] We also have a 1/2 factor in the E = mv^{2}/2 formula. Two remarks may be made here. First, it may be noted this is a non-relativistic formula and, more importantly, incorporates kinetic energy only. Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as K.E. = E − E_{0} = m_{v}c^{2} − m_{0}c^{2} = m_{0}γc^{2} − m_{0}c^{2} = m_{0}c^{2}(γ − 1). As for the exclusion of the potential energy, we may note that we may choose our reference point for the potential energy such that the kinetic and potential energy mirror each other. The energy concept that then emerges is the one that is used in the context of the Principle of Least Action: it equals E = mv^{2}. Appendix 1 provides some notes on that.
[10] Instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft.
[11] It is interesting to note that we may look at the energy in the rotating flywheel as potential energy because it is energy that is associated with motion, albeit circular motion. In physics, one may associate a rotating object with kinetic energy using the rotational equivalent of mass and linear velocity, i.e. rotational inertia (I) and angular velocity ω. The kinetic energy of a rotating object is then given by K.E. = (1/2)·I·ω^{2}.
[12] Because of the sideways motion of the connecting rods, the sinusoidal function will describe the linear motion only approximately, but you can easily imagine the idealized limit situation.
[13] The ω^{2}= 1/LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor (R), an inductor (L), and a capacitor (C). Writing the formula as ω^{2}= C^{–}^{1}/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring.
[14] The resistance in an electric circuit introduces a damping factor. When analyzing a mechanical spring, one may also want to introduce a drag coefficient. Both are usually defined as a fraction of the inertia, which is the mass for a spring and the inductance for an electric circuit. Hence, we would write the resistance for a spring as γm and as R = γL respectively.
[15] Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. Feynman (Lectures, I-33-3) shows us how to calculate the Q of these atomic oscillators: it is of the order of 10^{8}, which means the wave train will last about 10^{–8 }seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/e). For example, for sodium light, the radiation will last about 3.2×10^{–8 }seconds (this is the so-called decay time τ). Now, because the frequency of sodium light is some 500 THz (500×10^{12 }oscillations per second), this makes for some 16 million oscillations. There is an interesting paradox here: the speed of light tells us that such wave train will have a length of about 9.6 m! How is that to be reconciled with the pointlike nature of a photon? The paradox can only be explained by relativistic length contraction: in an analysis like this, one need to distinguish the reference frame of the photon – riding along the wave as it is being emitted, so to speak – and our stationary reference frame, which is that of the emitting atom.
[16] This is a general result and is reflected in the K.E. = T = (1/2)·m·ω^{2}·a^{2}·sin^{2}(ω·t + Δ) and the P.E. = U = k·x^{2}/2 = (1/2)· m·ω^{2}·a^{2}·cos^{2}(ω·t + Δ) formulas for the linear oscillator.
[17] Feynman further formalizes this in his Lecture on Superconductivity (Feynman, III-21-2), in which he refers to Schrödinger’s equation as the “equation for continuity of probabilities”. The analysis is centered on the local conservation of energy, which confirms the interpretation of Schrödinger’s equation as an energy diffusion equation.
[18] The m_{eff} is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write m_{eff} = m. Appendix 2 provides some additional notes on the concept. As for the equations, they are easily derived from noting that two complex numbers a + i∙b and c + i∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = i∙(ħ/m_{eff})∙∇^{2}ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i^{2} = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i^{2}∙d = − d + i∙c.
[19] The dimension of B is usually written as N/(m∙A), using the SI unit for current, i.e. the ampere (A). However, 1 C = 1 A∙s and, hence, 1 N/(m∙A) = 1 (N/C)/(m/s).
[20] Of course, multiplication with i amounts to a counterclockwise rotation. Hence, multiplication by –i also amounts to a rotation by 90 degrees, but clockwise. Now, to uniquely identify the clockwise and counterclockwise directions, we need to establish the equivalent of the right-hand rule for a proper geometric interpretation of Schrödinger’s equation in three-dimensional space: if we look at a clock from the back, then its hand will be moving counterclockwise. When writing B = (1/c)∙i∙E, we assume we are looking in the negative x-direction. If we are looking in the positive x-direction, we should write: B = -(1/c)∙i∙E. Of course, Nature does not care about our conventions. Hence, both should give the same results in calculations. We will show in a moment they do.
[21] In fact, when multiplying C^{2}/(N·m^{2}) with N^{2}/C^{2}, we get N/m^{2}, but we can multiply this with 1 = m/m to get the desired result. It is significant that an energy density (joule per unit volume) can also be measured in newton (force per unit area.
[22] The illustration shows a linearly polarized wave, but the obtained result is general.
[23] The sine and cosine are essentially the same functions, except for the difference in the phase: sinθ = cos(θ−π /2).
[24] I must thank a physics blogger for re-writing the 1/(ε_{0}·μ_{0}) = c^{2} equation like this. See: http://reciprocal.systems/phpBB3/viewtopic.php?t=236 (retrieved on 29 September 2017).
[25] A circularly polarized electromagnetic wave may be analyzed as consisting of two perpendicular electromagnetic plane waves of equal amplitude and 90° difference in phase.
[26] Of course, the reader will now wonder: what about neutrons? How to explain neutron spin? Neutrons are neutral. That is correct, but neutrons are not elementary: they consist of (charged) quarks. Hence, neutron spin can (or should) be explained by the spin of the underlying quarks.
[27] We detailed the mathematical framework and detailed calculations in the following online article: https://readingfeynman.org/2017/09/15/the-principle-of-least-action-re-visited.
[28] https://en.wikipedia.org/wiki/Electron_rest_mass (retrieved on 29 September 2017).
Math, physics and reality
This blog has been nice. It doesn’t get an awful lot of traffic (about a thousand visitors a week) but, from time to time, I do get a response or a question that fires me up, if only because it tells me someone is actually reading what I write.
Looking at the site now, I feel like I need to reorganize it completely. It’s just chaos, right? But then that’s what gets me the positive feedback: my readers are in the same boat. We’re trying to make sense of what physicists tell us is reality. The interference model I presented in my previous post is really nice. It has all the ingredients of quantum mechanics, which I would group under two broad categories: uncertainty and duality. Both are related, obviously. I will not talk about the reality of the wavefunction here, because I am biased: I firmly believe the wavefunction represents something real. Why? Because Einstein’s E = m·c^{2} formula tells us so: energy is a two-dimensional oscillation of mass. Two-dimensional, because it’s got twice the energy of the classroom oscillator (think of a mass on a spring). More importantly, the real and imaginary dimension of the oscillation are both real: they’re perpendicular to the direction of motion of the wave-particle. Photon or electron. It doesn’t matter. Of course, we have all of the transformation formulas, but… Well… These are not real: they are only there to accommodate our perspective: the state of the observer.
The distinction between the group and phase velocity of a wave packet is probably the best example of the failure of ordinary words to describe reality: particles are not waves, and waves are not particles. They are both… Well… Both at the same time. To calculate the action along some path, we assume there is some path, and we assume there is some particle following some path. The path and the particle are just figments of our mind. Useful figments of the mind, but… Well… There is no such thing as an infinitesimally small particle, and the concept of some one-dimensional line in spacetime does not make sense either. Or… Well… They do. Because they help us to make sense of the world. Of what is, whatever it is. 🙂
The mainstream views on the physical significance of the wavefunction are probably best summed up in the Encyclopædia Britannica, which says the wavefunction has no physical significance. Let me quote the relevant extract here:
“The wave function, in quantum mechanics, is a variable quantity that mathematically describes the wave characteristics of a particle. The value of the wave function of a particle at a given point of space and time is related to the likelihood of the particle’s being there at the time. By analogy with waves such as those of sound, a wave function, designated by the Greek letter psi, Ψ, may be thought of as an expression for the amplitude of the particle wave (or de Broglie wave), although for such waves amplitude has no physical significance. The square of the wave function, Ψ^{2}, however, does have physical significance: the probability of finding the particle described by a specific wave function Ψ at a given point and time is proportional to the value of Ψ^{2}.”
Really? First, this is factually wrong: the probability is given by the square of the absolute value of the wave function. These are two very different things:
- The square of a complex number is just another complex number: (a + ib)^{2 }= a^{2 }+ (ib)^{2 }+ 2iab = a^{2 }+ i^{2}b^{2 }+ 2iab = a^{2 }– b^{2 }+ 2iab.
- In contrast, the square of the absolute value always gives us a real number, to which we assign the mentioned physical interpretation:|a + ib|^{2 }= [√(a^{2 }+ b^{2})]^{2} = a^{2 }+ b^{2}.
But it’s not only position: using the right operators, we can also get probabilities on momentum, energy and other physical variables. Hence, the wavefunction is so much more than what the Encyclopædia Britannica suggests.
More fundamentally, what is written there is philosophically inconsistent. Squaring something – the number itself or its norm – is a mathematical operation. How can a mathematical operation suddenly yield something that has physical significance, if none of the elements it operates on, has any. One cannot just go from the mathematical to the physical space. The mathematical space describes the physical space. Always. In physics, at least. 🙂
So… Well… There is too much nonsense around. Disgusting. And the Encyclopædia Britannica should not just present the mainstream view. The truth is: the jury is still out, and there are many guys like me. We think the majority view is plain wrong. In this case, at least. 🙂
Thinking again…
One of the comments on my other blog made me think I should, perhaps, write something on waves again. The animation below shows the elementary wavefunction ψ = a·e^{−iθ }= ψ = a·e^{−i·θ } = a·e^{−i}^{(}^{ω·t−k·x}^{)} = a·e^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)} .We know this elementary wavefunction cannot represent a real-life particle. Indeed, the a·e^{−i·θ }function implies the probability of finding the particle – an electron, a photon, or whatever – would be equal to P(x, t) = |ψ(x, t)|^{2} = |a·e^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)}|^{2} = |a|^{2}·|e^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)}|^{2} = |a|^{2}·1^{2}= a^{2} everywhere. Hence, the particle would be everywhere – and, therefore, nowhere really. We need to localize the wave – or build a wave packet. We can do so by introducing uncertainty: we then add a potentially infinite number of these elementary wavefunctions with slightly different values for E and p, and various amplitudes a. Each of these amplitudes will then reflect the contribution to the composite wave, which – in three-dimensional space – we can write as:
ψ(r, t) = e^{−i·(E/ħ)·t}·f(r)
As I explained in previous posts (see, for example, my recent post on reality and perception), the f(r) function basically provides some envelope for the two-dimensional e^{−i·θ} = e^{−i·(E/ħ)·t} = cosθ + i·sinθ oscillation, with r = (x, y, z), θ = (E/ħ)·t = ω·t and ω = E/ħ.
Note that it looks like the wave propagates from left to right – in the positive direction of an axis which we may refer to as the x-axis. Also note this perception results from the fact that, naturally, we’d associate time with the rotation of that arrow at the center – i.e. with the motion in the illustration, while the spatial dimensions are just what they are: linear spatial dimensions. [This point is, perhaps, somewhat less self-evident than you may think at first.]
Now, the axis which points upwards is usually referred to as the z-axis, and the third and final axis – which points towards us – would then be the y-axis, obviously. Unfortunately, this definition would violate the so-called right-hand rule for defining a proper reference frame: the figures below shows the two possibilities – a left-handed and a right-handed reference frame – and it’s the right-handed reference (i.e. the illustration on the right) which we have to use in order to correctly define all directions, including the direction of rotation of the argument of the wavefunction.Hence, if we don’t change the direction of the y– and z-axes – so we keep defining the z-axis as the axis pointing upwards, and the y-axis as the axis pointing towards us – then the positive direction of the x-axis would actually be the direction from right to left, and we should say that the elementary wavefunction in the animation above seems to propagate in the negative x-direction. [Note that this left- or right-hand rule is quite astonishing: simply swapping the direction of one axis of a left-handed frame makes it right-handed, and vice versa.]
Note my language when I talk about the direction of propagation of our wave. I wrote: it looks like, or it seems to go in this or that direction. And I mean that: there is no real traveling here. At this point, you may want to review a post I wrote for my son, which explains the basic math behind waves, and in which I also explained the animation below.
Note how the peaks and troughs of this pulse seem to move leftwards, but the wave packet (or the group or the envelope of the wave—whatever you want to call it) moves to the right. The point is: the pulse itself doesn’t travel left or right. Think of the horizontal axis in the illustration above as an oscillating guitar string: each point on the string just moves up and down. Likewise, if our repeated pulse would represent a physical wave in water, for example, then the water just stays where it is: it just moves up and down. Likewise, if we shake up some rope, the rope is not going anywhere: we just started some motion that is traveling down the rope. In other words, the phase velocity is just a mathematical concept. The peaks and troughs that seem to be traveling are just mathematical points that are ‘traveling’ left or right. That’s why there’s no limit on the phase velocity: it can – and, according to quantum mechanics, actually will – exceed the speed of light. In contrast, the group velocity – which is the actual speed of the particle that is being represented by the wavefunction – may approach – or, in the case of a massless photon, will actually equal – the speed of light, but will never exceed it, and its direction will, obviously, have a physical significance as it is, effectively, the direction of travel of our particle – be it an electron, a photon (electromagnetic radiation), or whatever.
Hence, you should not think the spin of a particle – integer or half-integer – is somehow related to the direction of rotation of the argument of the elementary wavefunction. It isn’t: Nature doesn’t give a damn about our mathematical conventions, and that’s what the direction of rotation of the argument of that wavefunction is: just some mathematical convention. That’s why we write a·e^{−i}^{(}^{ω·t−k·x}^{)} rather than a·e^{i}^{(}^{ω·t+k·x}^{)} or a·e^{i}^{(}^{ω·t−k·x}^{)}: it’s just because of the right-hand rule for coordinate frames, and also because Euler defined the counter-clockwise direction as the positive direction of an angle. There’s nothing more to it.
OK. That’s obvious. Let me now return to my interpretation of Einstein’s E = m·c^{2} formula (see my previous posts on this). I noted that, in the reference frame of the particle itself (see my basics page), the elementary wavefunction a·e^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)} reduces to a·e^{−}^{(}^{i}^{/ħ)·}^{(E’·}^{t’}^{)}: the origin of the reference frame then coincides with (the center of) our particle itself, and the wavefunction only varies with the time in the inertial reference frame (i.e. the proper time t’), with the rest energy of the object (E’) as the time scale factor. How should we interpret this?
Well… Energy is force times distance, and force is defined as that what causes some mass to accelerate. To be precise, the newton – as the unit of force – is defined as the magnitude of a force which would cause a mass of one kg to accelerate with one meter per second per second. Per second per second. This is not a typo: 1 N corresponds to 1 kg times 1 m/s per second, i.e. 1 kg·m/s^{2}. So… Because energy is force times distance, the unit of energy may be expressed in units of kg·m/s^{2}·m, or kg·m^{2}/s^{2}, i.e. the unit of mass times the unit of velocity squared. To sum it all up:
1 J = 1 N·m = 1 kg·(m/s)^{2}
This reflects the physical dimensions on both sides of the E = m·c^{2} formula again but… Well… How should we interpret this? Look at the animation below once more, and imagine the green dot is some tiny mass moving around the origin, in an equally tiny circle. We’ve got two oscillations here: each packing half of the total energy of… Well… Whatever it is that our elementary wavefunction might represent in reality – which we don’t know, of course.
Now, the blue and the red dot – i.e. the horizontal and vertical projection of the green dot – accelerate up and down. If we look carefully, we see these dots accelerate towards the zero point and, once they’ve crossed it, they decelerate, so as to allow for a reversal of direction: the blue dot goes up, and then down. Likewise, the red dot does the same. The interplay between the two oscillations, because of the 90° phase difference, is interesting: if the blue dot is at maximum speed (near or at the origin), the red dot reverses speed (its speed is, therefore, (almost) nil), and vice versa. The metaphor of our frictionless V-2 engine, our perpetuum mobile, comes to mind once more.
The question is: what’s going on, really?
My answer is: I don’t know. I do think that, somehow, energy should be thought of as some two-dimensional oscillation of something – something which we refer to as mass, but we didn’t define mass very clearly either. It also, somehow, combines linear and rotational motion. Each of the two dimensions packs half of the energy of the particle that is being represented by our wavefunction. It is, therefore, only logical that the physical unit of both is to be expressed as a force over some distance – which is, effectively, the physical dimension of energy – or the rotational equivalent of them: torque over some angle. Indeed, the analogy between linear and angular movement is obvious: the kinetic energy of a rotating object is equal to K.E. = (1/2)·I·ω^{2}. In this formula, I is the rotational inertia – i.e. the rotational equivalent of mass – and ω is the angular velocity – i.e. the rotational equivalent of linear velocity. Noting that the (average) kinetic energy in any system must be equal to the (average) potential energy in the system, we can add both, so we get a formula which is structurally similar to the E = m·c^{2} formula. But is it the same? Is the effective mass of some object the sum of an almost infinite number of quanta that incorporate some kind of rotational motion? And – if we use the right units – is the angular velocity of these infinitesimally small rotations effectively equal to the speed of light?
I am not sure. Not at all, really. But, so far, I can’t think of any explanation of the wavefunction that would make more sense than this one. I just need to keep trying to find better ways to articulate or imagine what might be going on. 🙂 In this regard, I’d like to add a point – which may or may not be relevant. When I talked about that guitar string, or the water wave, and wrote that each point on the string – or each water drop – just moves up and down, we should think of the physicality of the situation: when the string oscillates, its length increases. So it’s only because our string is flexible that it can vibrate between the fixed points at its ends. For a rope that’s not flexible, the end points would need to move in and out with the oscillation. Look at the illustration below, for example: the two kids who are holding rope must come closer to each other, so as to provide the necessary space inside of the oscillation for the other kid. 🙂The next illustration – of how water waves actually propagate – is, perhaps, more relevant. Just think of a two-dimensional equivalent – and of the two oscillations as being transverse waves, as opposed to longitudinal. See how string theory starts making sense? 🙂
The most fundamental question remains the same: what is it, exactly, that is oscillating here? What is the field? It’s always some force on some charge – but what charge, exactly? Mass? What is it? Well… I don’t have the answer to that. It’s the same as asking: what is electric charge, really? So the question is: what’s the reality of mass, of electric charge, or whatever other charge that causes a force to act on it?
If you know, please let me know. 🙂
Post scriptum: The fact that we’re talking some two-dimensional oscillation here – think of a surface now – explains the probability formula: we need to square the absolute value of the amplitude to get it. And normalize, of course. Also note that, when normalizing, we’d expect to get some factor involving π somewhere, because we’re talking some circular surface – as opposed to a rectangular one. But I’ll let you figure that out. 🙂
Reality and perception
It’s quite easy to get lost in all of the math when talking quantum mechanics. In this post, I’d like to freewheel a bit. I’ll basically try to relate the wavefunction we’ve derived for the electron orbitals to the more speculative posts I wrote on how to interpret the wavefunction. So… Well… Let’s go. 🙂
If there is one thing you should remember from all of the stuff I wrote in my previous posts, then it’s that the wavefunction for an electron orbital – ψ(x, t), so that’s a complex-valued function in two variables (position and time) – can be written as the product of two functions in one variable:
ψ(x, t) = e^{−i·(E/ħ)·t}·f(x)
In fact, we wrote f(x) as ψ(x), but I told you how confusing that is: the ψ(x) and ψ(x, t) functions are, obviously, very different. To be precise, the f(x) = ψ(x) function basically provides some envelope for the two-dimensional e^{iθ} = e^{−i·(E/ħ)·t} = cosθ + i·sinθ oscillation – as depicted below (θ = −(E/ħ)·t = ω·t with ω = −E/ħ).When analyzing this animation – look at the movement of the green, red and blue dots respectively – one cannot miss the equivalence between this oscillation and the movement of a mass on a spring – as depicted below.The e^{−i·(E/ħ)·t} function just gives us two springs for the price of one. 🙂 Now, you may want to imagine some kind of elastic medium – Feynman’s famous drum-head, perhaps 🙂 – and you may also want to think of all of this in terms of superimposed waves but… Well… I’d need to review if that’s really relevant to what we’re discussing here, so I’d rather not make things too complicated and stick to basics.
First note that the amplitude of the two linear oscillations above is normalized: the maximum displacement of the object from equilibrium, in the positive or negative direction, which we may denote by x = ±A, is equal to one. Hence, the energy formula is just the sum of the potential and kinetic energy: T + U = (1/2)·A^{2}·m·ω^{2} = (1/2)·m·ω^{2}. But so we have two springs and, therefore, the energy in this two-dimensional oscillation is equal to E = 2·(1/2)·m·ω^{2} = m·ω^{2}.
This formula is structurally similar to Einstein’s E = m·c^{2} formula. Hence, one may want to assume that the energy of some particle (an electron, in our case, because we’re discussing electron orbitals here) is just the two-dimensional motion of its mass. To put it differently, we might also want to think that the oscillating real and imaginary component of our wavefunction each store one half of the total energy of our particle.
However, the interpretation of this rather bold statement is not so straightforward. First, you should note that the ω in the E = m·ω^{2} formula is an angular velocity, as opposed to the c in the E = m·c^{2} formula, which is a linear velocity. Angular velocities are expressed in radians per second, while linear velocities are expressed in meter per second. However, while the radian measures an angle, we know it does so by measuring a length. Hence, if our distance unit is 1 m, an angle of 2π rad will correspond to a length of 2π meter, i.e. the circumference of the unit circle. So… Well… The two velocities may not be so different after all.
There are other questions here. In fact, the other questions are probably more relevant. First, we should note that the ω in the E = m·ω^{2} can take on any value. For a mechanical spring, ω will be a function of (1) the stiffness of the spring (which we usually denote by k, and which is typically measured in newton (N) per meter) and (2) the mass (m) on the spring. To be precise, we write: ω^{2} = k/m – or, what amounts to the same, ω = √(k/m). Both k and m are variables and, therefore, ω can really be anything. In contrast, we know that c is a constant: c equals 299,792,458 meter per second, to be precise. So we have this rather remarkable expression: c = √(E/m), and it is valid for any particle – our electron, or the proton at the center, or our hydrogen atom as a whole. It is also valid for more complicated atoms, of course. In fact, it is valid for any system.
Hence, we need to take another look at the energy concept that is used in our ψ(x, t) = e^{−i·(E/ħ)·t}·f(x) wavefunction. You’ll remember (if not, you should) that the E here is equal to E_{n }= −13.6 eV, −3.4 eV, −1.5 eV and so on, for n = 1, 2, 3, etc. Hence, this energy concept is rather particular. As Feynman puts it: “The energies are negative because 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.”
Now, this is the one and only issue I have with the standard physics story. I mentioned it in one of my previous posts and, just for clarity, let me copy what I wrote at the time:
Feynman gives us a rather casual explanation [on choosing a zero point for measuring energy] 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·e^{−iωt}, with ħ·ω = E = m·c^{2}. Hence, we can write the amplitude for the two states, for example as:
e^{−i(E1/ħ)·t} and e^{−i(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:
e^{−i(E1+A)·t/ħ} and e^{−i(E2+A)·t/ħ}
All of his amplitudes would be multiplied by the same factor e^{−i(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 M_{s}·c^{2}, where M_{s} 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 M_{g}·c^{2}, where M_{g} 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 M_{g}·c^{2}, where M_{g} 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.
So… Well… There you go. If we’d want to try to interpret our ψ(x, t) = e^{−i·(En/ħ)·t}·f(x) function as a two-dimensional oscillation of the mass of our electron, the energy concept in it – so that’s the E_{n }in it – should include all pieces. Most notably, it should also include the electron’s rest energy, i.e. its energy when it is not in a bound state. This rest energy is equal to 0.511 MeV. […] Read this again: 0.511 mega-electronvolt (10^{6} eV), so that’s huge as compared to the tiny energy values we mentioned so far (−13.6 eV, −3.4 eV, −1.5 eV,…).
Of course, this gives us a rather phenomenal order of magnitude for the oscillation that we’re looking at. Let’s quickly calculate it. We need to convert to SI units, of course: 0.511 MeV is about 8.2×10^{−14} joule (J), and so the associated frequency is equal to ν = E/h = (8.2×10^{−14} J)/(6.626×10^{−34} J·s) ≈ 1.23559×10^{20} cycles per second. Now, I know such number doesn’t say all that much: just note it’s the same order of magnitude as the frequency of gamma rays and… Well… No. I won’t say more. You should try to think about this for yourself. [If you do, think – for starters – about the difference between bosons and fermions: matter-particles are fermions, and photons are bosons. Their nature is very different.]
The corresponding angular frequency is just the same number but multiplied by 2π (one cycle corresponds to 2π radians and, hence, ω = 2π·ν = 7.76344×10^{20} rad per second. Now, if our green dot would be moving around the origin, along the circumference of our unit circle, then its horizontal and/or vertical velocity would approach the same value. Think of it. We have this e^{iθ} = e^{−i·(E/ħ)·t} = e^{i·ω·t} = cos(ω·t) + i·sin(ω·t) function, with ω = E/ħ. So the cos(ω·t) captures the motion along the horizontal axis, while the sin(ω·t) function captures the motion along the vertical axis. Now, the velocity along the horizontal axis as a function of time is given by the following formula:
v(t) = d[x(t)]/dt = d[cos(ω·t)]/dt = −ω·sin(ω·t)
Likewise, the velocity along the vertical axis is given by v(t) = d[sin(ω·t)]/dt = ω·cos(ω·t). These are interesting formulas: they show the velocity (v) along one of the two axes is always less than the angular velocity (ω). To be precise, the velocity v approaches – or, in the limit, is equal to – the angular velocity ω when ω·t is equal to ω·t = 0, π/2, π or 3π/2. So… Well… 7.76344×10^{20} meter per second!? That’s like 2.6 trillion times the speed of light. So that’s not possible, of course!
That’s where the amplitude of our wavefunction comes in – our envelope function f(x): the green dot does not move along the unit circle. The circle is much tinier and, hence, the oscillation should not exceed the speed of light. In fact, I should probably try to prove it oscillates at the speed of light, thereby respecting Einstein’s universal formula:
c = √(E/m)
Written like this – rather than as you know it: E = m·c^{2} – this formula shows the speed of light is just a property of spacetime, just like the ω = √(k/m) formula (or the ω = √(1/LC) formula for a resonant AC circuit) shows that ω, the natural frequency of our oscillator, is a characteristic of the system.
Am I absolutely certain of what I am writing here? No. My level of understanding of physics is still that of an undergrad. But… Well… It all makes a lot of sense, doesn’t it? 🙂
Now, I said there were a few obvious questions, and so far I answered only one. The other obvious question is why energy would appear to us as mass in motion in two dimensions only. Why is it an oscillation in a plane? We might imagine a third spring, so to speak, moving in and out from us, right? Also, energy densities are measured per unit volume, right?
Now that‘s a clever question, and I must admit I can’t answer it right now. However, I do suspect it’s got to do with the fact that the wavefunction depends on the orientation of our reference frame. If we rotate it, it changes. So it’s like we’ve lost one degree of freedom already, so only two are left. Or think of the third direction as the direction of propagation of the wave. 🙂 Also, we should re-read what we wrote about the Poynting vector for the matter wave, or what Feynman wrote about probability currents. Let me give you some appetite for that by noting that we can re-write joule per cubic meter (J/m^{3}) as newton per square meter: J/m^{3} = N·m/m^{3} = N/m^{2}. [Remember: the unit of energy is force times distance. In fact, looking at Einstein’s formula, I’d say it’s kg·m^{2}/s^{2} (mass times a squared velocity), but that simplifies to the same: kg·m^{2}/s^{2} = [N/(m/s^{2})]·m^{2}/s^{2}.]
I should probably also remind you that there is no three-dimensional equivalent of Euler’s formula, and the way the kinetic and potential energy of those two oscillations works together is rather unique. Remember I illustrated it with the image of a V-2 engine in previous posts. There is no such thing as a V-3 engine. [Well… There actually is – but not with the third cylinder being positioned sideways.]
But… Then… Well… Perhaps we should think of some weird combination of two V-2 engines. The illustration below shows the superposition of two one-dimensional waves – I think – one traveling east-west and back, and the other one traveling north-south and back. So, yes, we may to think of Feynman’s drum-head again – but combining two-dimensional waves – two waves that both have an imaginary as well as a real dimension
Hmm… Not sure. If we go down this path, we’d need to add a third dimension – so w’d have a super-weird V-6 engine! As mentioned above, the wavefunction does depend on our reference frame: we’re looking at stuff from a certain direction and, therefore, we can only see what goes up and down, and what goes left or right. We can’t see what comes near and what goes away from us. Also think of the particularities involved in measuring angular momentum – or the magnetic moment of some particle. We’re measuring that along one direction only! Hence, it’s probably no use to imagine we’re looking at three waves simultaneously!
In any case… I’ll let you think about all of this. I do feel I am on to something. I am convinced that my interpretation of the wavefunction as an energy propagation mechanism, or as energy itself – as a two-dimensional oscillation of mass – makes sense. 🙂
Of course, I haven’t answered one key question here: what is mass? What is that green dot – in reality, that is? At this point, we can only waffle – probably best to just give its standard definition: mass is a measure of inertia. A resistance to acceleration or deceleration, or to changing direction. But that doesn’t say much. I hate to say that – in many ways – all that I’ve learned so far has deepened the mystery, rather than solve it. The more we understand, the less we understand? But… Well… That’s all for today, folks ! Have fun working through it for yourself. 🙂
Post scriptum: I’ve simplified the wavefunction a bit. As I noted in my post on it, the complex exponential is actually equal to e^{−i·[(E/ħ)·t − }^{m·φ]}, so we’ve got a phase shift because of m, the quantum number which denotes the z-component of the angular momentum. But that’s a minor detail that shouldn’t trouble or worry you here.
Re-visiting electron orbitals (III)
In my previous post, I mentioned that it was not so obvious (both from a physical as well as from a mathematical point of view) to write the wavefunction for electron orbitals – which we denoted as ψ(x, t), i.e. a function of two variables (or four: one time coordinate and three space coordinates) – as the product of two other functions in one variable only.
[…] OK. The above sentence is difficult to read. Let me write in math. 🙂 It is not so obvious to write ψ(x, t) as:
ψ(x, t) = e^{−i·(E/ħ)·t}·ψ(x)
As I mentioned before, the physicists’ use of the same symbol (ψ, psi) for both the ψ(x, t) and ψ(x) function is quite confusing – because the two functions are very different:
- ψ(x, t) is a complex-valued function of two (real) variables: x and t. Or four, I should say, because x = (x, y, z) – but it’s probably easier to think of x as one vector variable – a vector-valued argument, so to speak. And then t is, of course, just a scalar variable. So… Well… A function of two variables: the position in space (x), and time (t).
- In contrast, ψ(x) is a real-valued function of one (vector) variable only: x, so that’s the position in space only.
Now you should cry foul, of course: ψ(x) is not necessarily real-valued. It may be complex-valued. You’re right. You know the formula:Note the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from x = (x, y, z) to r = (r, θ, φ), and that the function is also a function of the two quantum numbers l and m now, i.e. the orbital angular momentum (l) and its z-component (m) respectively. In my previous post(s), I gave you the formulas for Y_{l,m}(θ, φ) and F_{l,m}(r) respectively. F_{l,m}(r) was a real-valued function alright, but the Y_{l,m}(θ, φ) had that e^{i·m·φ} factor in it. So… Yes. You’re right: the Y_{l,m}(θ, φ) function is real-valued if – and only if – m = 0, in which case e^{i·m·φ} = 1. Let me copy the table from Feynman’s treatment of the topic once again:The P_{l}^{m}(cosθ) functions are the so-called (associated) Legendre polynomials, and the formula for these functions is rather horrible:Don’t worry about it too much: just note the P_{l}^{m}(cosθ) is a real-valued function. The point is the following:the ψ(x, t) is a complex-valued function because – and only because – we multiply a real-valued envelope function – which depends on position only – with e^{−i·(E/ħ)·t}·e^{i·m·φ} = e^{−i·[(E/ħ)·t − }^{m·φ]}.
[…]
Please read the above once again and – more importantly – think about it for a while. 🙂 You’ll have to agree with the following:
- As mentioned in my previous post, the e^{i·m·φ} factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak, and the whole e^{−i·[(E/ħ)·t − }^{m·φ]} factor just disappears when we’re calculating probabilities.
- The envelope function gives us the basic amplitude – in the classical sense of the word: the maximum displacement from the zero value. And so it’s that e^{−i·[(E/ħ)·t − }^{m·φ]} that ensures the whole expression somehow captures the energy of the oscillation.
Let’s first look at the envelope function again. Let me copy the illustration for n = 5 and l = 2 from a Wikimedia Commons article. Note the symmetry planes:
- Any plane containing the z-axis is a symmetry plane – like a mirror in which we can reflect one half of the shape to get the other half. [Note that I am talking the shape only here. Forget about the colors for a while – as these reflect the complex phase of the wavefunction.]
- Likewise, the plane containing both the x– and the y-axis is a symmetry plane as well.
The first symmetry plane – or symmetry line, really (i.e. the z-axis) – should not surprise us, because the azimuthal angle φ is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge the e^{i·m·φ} factor with the e^{−i·(E/ħ)·t}, so it’s just part and parcel of what the author of the illustrations above refers to as the ‘complex phase’ of our wavefunction. OK. Clear enough – I hope. 🙂 But why is the the xy-plane a symmetry plane too? We need to look at that monstrous formula for the P_{l}^{m}(cosθ) function here: just note the cosθ argument in it is being squared before it’s used in all of the other manipulation. Now, we know that cosθ = sin(π/2 − θ). So we can define some new angle – let’s just call it α – which is measured in the way we’re used to measuring angle, which is not from the z-axis but from the xy-plane. So we write: cosθ = sin(π/2 − θ) = sinα. The illustration below may or may not help you to see what we’re doing here.So… To make a long story short, we can substitute the cosθ argument in the P_{l}^{m}(cosθ) function for sinα = sin(π/2 − θ). Now, if the xy-plane is a symmetry plane, then we must find the same value for P_{l}^{m}(sinα) and P_{l}^{m}[sin(−α)]. Now, that’s not obvious, because sin(−α) = −sinα ≠ sinα. However, because the argument in that P_{l}^{m}(x) function is being squared before any other operation (like subtracting 1 and exponentiating the result), it is OK: [−sinα]^{2} = [sinα]^{2 }= sin^{2}α. […] OK, I am sure the geeks amongst my readers will be able to explain this more rigorously. In fact, I hope they’ll have a look at it, because there’s also that d^{l+m}/dx^{l+m} operator, and so you should check what happens with the minus sign there. 🙂
[…] Well… By now, you’re probably totally lost, but the fact of the matter is that we’ve got a beautiful result here. Let me highlight the most significant results:
- A definite energy state of a hydrogen atom (or of an electron orbiting around some nucleus, I should say) appears to us as some beautifully shaped orbital – an envelope function in three dimensions, really – which has the z-axis – i.e. the vertical axis – as a symmetry line and the xy-plane as a symmetry plane.
- The e^{−i·[(E/ħ)·t − }^{m·φ]} factor gives us the oscillation within the envelope function. As such, it’s this factor that, somehow, captures the energy of the oscillation.
It’s worth thinking about this. Look at the geometry of the situation again – as depicted below. We’re looking at the situation along the x-axis, in the direction of the origin, which is the nucleus of our atom.
The e^{i·m·φ} factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the e^{−i·[(E/ħ)·t} factor, which gives us the two-dimensional oscillation that captures the energy of the state.
Now, the obvious question is: the oscillation of what, exactly? I am not quite sure but – as I explained in my Deep Blue page – the real and imaginary part of our wavefunction are really like the electric and magnetic field vector of an oscillating electromagnetic field (think of electromagnetic radiation – if that makes it easier). Hence, just like the electric and magnetic field vector represent some rapidly changing force on a unit charge, the real and imaginary part of our wavefunction must also represent some rapidly changing force on… Well… I am not quite sure on what though. The unit charge is usually defined as the charge of a proton – rather than an electron – but then forces act on some mass, right? And the mass of a proton is hugely different from the mass of an electron. The same electric (or magnetic) force will, therefore, give a hugely different acceleration to both.
So… Well… My guts instinct tells me the real and imaginary part of our wavefunction just represent, somehow, a rapidly changing force on some unit of mass, but then I am not sure how to define that unit right now (it’s probably not the kilogram!).
Now, there is another thing we should note here: we’re actually sort of de-constructing a rotation (look at the illustration above once again) in two linearly oscillating vectors – one along the z-axis and the other along the y-axis. Hence, in essence, we’re actually talking about something that’s spinning. In other words, we’re actually talking some torque around the x-axis. In what direction? I think that shouldn’t matter – that we can write E or −E, in other words, but… Well… I need to explore this further – as should you! 🙂
Let me just add one more note on the e^{i·m·φ} factor. It sort of defines the geometry of the complex phase itself. Look at the illustration below. Click on it to enlarge it if necessary – or, better still, visit the magnificent Wikimedia Commons article from which I get these illustrations. These are the orbitals n = 4 and l = 3. Look at the red hues in particular – or the blue – whatever: focus on one color only, and see how how – for m = ±1, we’ve got one appearance of that color only. For m = ±1, the same color appears at two ends of the ‘tubes’ – or tori (plural of torus), I should say – just to sound more professional. 🙂 For m = ±2, the torus consists of three parts – or, in mathematical terms, we’d say the order of its rotational symmetry is equal to 3. Check that Wikimedia Commons article for higher values of n and l: the shapes become very convoluted, but the observation holds. 🙂
Have fun thinking all of this through for yourself – and please do look at those symmetries in particular. 🙂
Post scriptum: You should do some thinking on whether or not these m = ±1, ±2,…, ±l orbitals are really different. As I mentioned above, a phase difference is just what it is: a re-set of the t = 0 point. Nothing more, nothing less. So… Well… As far as I am concerned, that’s not a real difference, is it? 🙂 As with other stuff, I’ll let you think about this for yourself.
An interpretation of the wavefunction
This is my umpteenth post on the same topic. 😦 It is obvious that this search for a sensible interpretation is consuming me. Why? I am not sure. Studying physics is frustrating. As a leading physicist puts it:
“The teaching of quantum mechanics these days usually follows the same dogma: firstly, the student is told about the failure of classical physics at the beginning of the last century; secondly, the heroic confusions of the founding fathers are described and the student is given to understand that no humble undergraduate student could hope to actually understand quantum mechanics for himself; thirdly, a deus ex machina arrives in the form of a set of postulates (the Schrödinger equation, the collapse of the wavefunction, etc); fourthly, a bombardment of experimental verifications is given, so that the student cannot doubt that QM is correct; fifthly, the student learns how to solve the problems that will appear on the exam paper, hopefully with as little thought as possible.”
That’s obviously not the way we want to understand quantum mechanics. [With we, I mean, me, of course, and you, if you’re reading this blog.] Of course, that doesn’t mean I don’t believe Richard Feynman, one of the greatest physicists ever, when he tells us no one, including himself, understands physics quite the way we’d like to understand it. Such statements should not prevent us from trying harder. So let’s look for better metaphors. The animation below shows the two components of the archetypal wavefunction – a simple sine and cosine. They’re the same function actually, but their phases differ by 90 degrees (π/2).
It makes me think of a V-2 engine with the pistons at a 90-degree angle. Look at the illustration below, which I took from a rather simple article on cars and engines that has nothing to do with quantum mechanics. Think of the moving pistons as harmonic oscillators, like springs.
We will also think of the center of each cylinder as the zero point: think of that point as a point where – if we’re looking at one cylinder alone – the internal and external pressure balance each other, so the piston would not move… Well… If it weren’t for the other piston, because the second piston is not at the center when the first is. In fact, it is easy to verify and compare the following positions of both pistons, as well as the associated dynamics of the situation:
Piston 1 |
Piston 2 |
Motion of Piston 1 |
Motion Piston 2 |
Top |
Center |
Compressed air will push piston down |
Piston moves down against external pressure |
Center |
Bottom |
Piston moves down against external pressure |
External air pressure will push piston up |
Bottom |
Center |
External air pressure will push piston up |
Piston moves further up and compresses the air |
Center |
Top |
Piston moves further up and compresses the air |
Compressed air will push piston down |
When the pistons move, their linear motion will be described by a sinusoidal function: a sine or a cosine. In fact, the 90-degree V-2 configuration ensures that the linear motion of the two pistons will be exactly the same, except for a phase difference of 90 degrees. [Of course, because of the sideways motion of the connecting rods, our sine and cosine function describes the linear motion only approximately, but you can easily imagine the idealized limit situation. If not, check Feynman’s description of the harmonic oscillator.]
The question is: if we’d have a set-up like this, two springs – or two harmonic oscillators – attached to a shaft through a crank, would this really work as a perpetuum mobile? We obviously talk energy being transferred back and forth between the rotating shaft and the moving pistons… So… Well… Let’s model this: the total energy, potential and kinetic, in each harmonic oscillator is constant. Hence, the piston only delivers or receives kinetic energy from the rotating mass of the shaft.
Now, in physics, that’s a bit of an oxymoron: we don’t think of negative or positive kinetic (or potential) energy in the context of oscillators. We don’t think of the direction of energy. But… Well… If we’ve got two oscillators, our picture changes, and so we may have to adjust our thinking here.
Let me start by giving you an authoritative derivation of the various formulas involved here, taking the example of the physical spring as an oscillator—but the formulas are basically the same for any harmonic oscillator.
The first formula is a general description of the motion of our oscillator. The coefficient in front of the cosine function (a) is the maximum amplitude. Of course, you will also recognize ω_{0} as the natural frequency of the oscillator, and Δ as the phase factor, which takes into account our t = 0 point. In our case, for example, we have two oscillators with a phase difference equal to π/2 and, hence, Δ would be 0 for one oscillator, and –π/2 for the other. [The formula to apply here is sinθ = cos(θ – π/2).] Also note that we can equate our θ argument to ω_{0}·t. Now, if a = 1 (which is the case here), then these formulas simplify to:
- K.E. = T = m·v^{2}/2 = m·ω_{0}^{2}·sin^{2}(θ + Δ) = m·ω_{0}^{2}·sin^{2}(ω_{0}·t + Δ)
- P.E. = U = k·x^{2}/2 = k·cos^{2}(θ + Δ)
The coefficient k in the potential energy formula characterizes the force: F = −k·x. The minus sign reminds us our oscillator wants to return to the center point, so the force pulls back. From the dynamics involved, it is obvious that k must be equal to m·ω_{0}^{2}., so that gives us the famous T + U = m·ω_{0}^{2}/2 formula or, including a once again, T + U = m·a^{2}·ω_{0}^{2}/2.
Now, if we normalize our functions by equating k to one (k = 1), then the motion of our first oscillator is given by the cosθ function, and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:
d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dt = 2∙sinθ∙cosθ
Let’s look at the second oscillator now. Just think of the second piston going up and down in our V-twin engine. Its motion is given by the sinθ function which, as mentioned above, is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin^{2}(θ−π /2), and how it changes – as a function of θ – will be equal to:
2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ
We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the rotating shaft moves at constant speed. Linear motion becomes circular motion, and vice versa, in a frictionless Universe. We have the metaphor we were looking for!
Somehow, in this beautiful interplay between linear and circular motion, energy is being borrowed from one place to another, and then returned. From what place to what place? I am not sure. We may call it the real and imaginary energy space respectively, but what does that mean? One thing is for sure, however: the interplay between the real and imaginary part of the wavefunction describes how energy propagates through space!
How exactly? Again, I am not sure. Energy is, obviously, mass in motion – as evidenced by the E = m·c^{2} equation, and it may not have any direction (when everything is said and done, it’s a scalar quantity without direction), but the energy in a linear motion is surely different from that in a circular motion, and our metaphor suggests we need to think somewhat more along those lines. Perhaps we will, one day, able to square this circle. 🙂
Schrödinger’s equation
Let’s analyze the interplay between the real and imaginary part of the wavefunction through an analysis of Schrödinger’s equation, which we write as:
i·ħ∙∂ψ/∂t = –(ħ^{2}/2m)∙∇^{2}ψ + V·ψ
We can do a quick dimensional analysis of both sides:
- [i·ħ∙∂ψ/∂t] = N∙m∙s/s = N∙m
- [–(ħ^{2}/2m)∙∇^{2}ψ] = N∙m^{3}/m^{2} = N∙m
- [V·ψ] = N∙m
Note the dimension of the ‘diffusion’ constant ħ^{2}/2m: [ħ^{2}/2m] = N^{2}∙m^{2}∙s^{2}/kg = N^{2}∙m^{2}∙s^{2}/(N·s^{2}/m) = N∙m^{3}. Also note that, in order for the dimensions to come out alright, the dimension of V – the potential – must be that of energy. Hence, Feynman’s description of it as the potential energy – rather than the potential tout court – is somewhat confusing but correct: V must equal the potential energy of the electron. Hence, V is not the conventional (potential) energy of the unit charge (1 coulomb). Instead, the natural unit of charge is used here, i.e. the charge of the electron itself.
Now, Schrödinger’s equation – without the V·ψ term – can be written as the following pair of equations:
- Re(∂ψ/∂t) = −(1/2)∙(ħ/m)∙Im(∇^{2}ψ)
- Im(∂ψ/∂t) = (1/2)∙(ħ/m)∙Re(∇^{2}ψ)
This closely resembles the propagation mechanism of an electromagnetic wave as described by Maxwell’s equation for free space (i.e. a space with no charges), but E and B are vectors, not scalars. How do we get this result. Well… ψ is a complex function, which we can write as a + i∙b. Likewise, ∂ψ/∂t is a complex function, which we can write as c + i∙d, and ∇^{2}ψ can then be written as e + i∙f. If we temporarily forget about the coefficients (ħ, ħ^{2}/m and V), then Schrödinger’s equation – including V·ψ term – amounts to writing something like this:
i∙(c + i∙d) = –(e + i∙f) + (a + i∙b) ⇔ a + i∙b = i∙c − d + e+ i∙f ⇔ a = −d + e and b = c + f
Hence, we can now write:
- V∙Re(ψ) = −ħ∙Im(∂ψ/∂t) + (1/2)∙( ħ^{2}/m)∙Re(∇^{2}ψ)
- V∙Im(ψ) = ħ∙Re(∂ψ/∂t) + (1/2)∙( ħ^{2}/m)∙Im(∇^{2}ψ)
This simplifies to the two equations above for V = 0, i.e. when there is no potential (electron in free space). Now we can bring the Re and Im operators into the brackets to get:
- V∙Re(ψ) = −ħ∙∂Im (ψ)/∂t + (1/2)∙( ħ^{2}/m)∙∇^{2}Re(ψ)
- V∙Im(ψ) = ħ∙∂Re(ψ)/∂t + (1/2)∙( ħ^{2}/m)∙∇^{2}Im(ψ)
This is very interesting, because we can re-write this using the quantum-mechanical energy operator H = –(ħ^{2}/2m)∙∇^{2} + V· (note the multiplication sign after the V, which we do not have – for obvious reasons – for the –(ħ^{2}/2m)∙∇^{2} expression):
- H[Re (ψ)] = −ħ∙∂Im(ψ)/∂t
- H[Im(ψ)] = ħ∙∂Re(ψ)/∂t
A dimensional analysis shows us both sides are, once again, expressed in N∙m. It’s a beautiful expression because – if we write the real and imaginary part of ψ as r∙cosθ and r∙sinθ, we get:
- H[cosθ] = −ħ∙∂sinθ/∂t = E∙cosθ
- H[sinθ] = ħ∙∂cosθ/∂t = E∙sinθ
Indeed, θ = (E∙t − p∙x)/ħ and, hence, −ħ∙∂sinθ/∂t = ħ∙cosθ∙E/ħ = E∙cosθ and ħ∙∂cosθ/∂t = ħ∙sinθ∙E/ħ = E∙sinθ. Now we can combine the two equations in one equation again and write:
H[r∙(cosθ + i∙sinθ)] = r∙(E∙cosθ + i∙sinθ) ⇔ H[ψ] = E∙ψ
The operator H – applied to the wavefunction – gives us the (scalar) product of the energy E and the wavefunction itself. Isn’t this strange?
Hmm… I need to further verify and explain this result… I’ll probably do so in yet another post on the same topic… 🙂
Post scriptum: The symmetry of our V-2 engine – or perpetuum mobile – is interesting: its cross-section has only one axis of symmetry. Hence, we may associate some angle with it, so as to define its orientation in the two-dimensional cross-sectional plane. Of course, the cross-sectional plane itself is at right angles to the crankshaft axis, which we may also associate with some angle in three-dimensional space. Hence, its geometry defines two orthogonal directions which, in turn, define a spherical coordinate system, as shown below.
We may, therefore, say that three-dimensional space is actually being implied by the geometry of our V-2 engine. Now that is interesting, isn’t it? 🙂
Re-visiting uncertainty…
I re-visited the Uncertainty Principle a couple of times already, but here I really want to get at the bottom of the thing? What’s uncertain? The energy? The time? The wavefunction itself? These questions are not easily answered, and I need to warn you: you won’t get too much wiser when you’re finished reading this. I just felt like freewheeling a bit. [Note that the first part of this post repeats what you’ll find on the Occam page, or my post on Occam’s Razor. But these post do not analyze uncertainty, which is what I will be trying to do here.]
Let’s first think about the wavefunction itself. It’s tempting to think it actually is the particle, somehow. But it isn’t. So what is it then? Well… Nobody knows. In my previous post, I said I like to think it travels with the particle, but then doesn’t make much sense either. It’s like a fundamental property of the particle. Like the color of an apple. But where is that color? In the apple, in the light it reflects, in the retina of our eye, or is it in our brain? If you know a thing or two about how perception actually works, you’ll tend to agree the quality of color is not in the apple. When everything is said and done, the wavefunction is a mental construct: when learning physics, we start to think of a particle as a wavefunction, but they are two separate things: the particle is reality, the wavefunction is imaginary.
But that’s not what I want to talk about here. It’s about that uncertainty. Where is the uncertainty? You’ll say: you just said it was in our brain. No. I didn’t say that. It’s not that simple. Let’s look at the basic assumptions of quantum physics:
- Quantum physics assumes there’s always some randomness in Nature and, hence, we can measure probabilities only. We’ve got randomness in classical mechanics too, but this is different. This is an assumption about how Nature works: we don’t really know what’s happening. We don’t know the internal wheels and gears, so to speak, or the ‘hidden variables’, as one interpretation of quantum mechanics would say. In fact, the most commonly accepted interpretation of quantum mechanics says there are no ‘hidden variables’.
- However, as Shakespeare has one of his characters say: there is a method in the madness, and the pioneers– I mean Werner Heisenberg, Louis de Broglie, Niels Bohr, Paul Dirac, etcetera – discovered that method: all probabilities can be found by taking the square of the absolute value of a complex-valued wavefunction (often denoted by Ψ), whose argument, or phase (θ), is given by the de Broglie relations ω = E/ħ and k = p/ħ. The generic functional form of that wavefunction is:
Ψ = Ψ(x, t) = a·e^{−iθ }= a·e^{−i(ω·t − k ∙x)} = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]}
That should be obvious by now, as I’ve written more than a dozens of posts on this. 🙂 I still have trouble interpreting this, however—and I am not ashamed, because the Great Ones I just mentioned have trouble with that too. It’s not that complex exponential. That e^{−iφ} is a very simple periodic function, consisting of two sine waves rather than just one, as illustrated below. [It’s a sine and a cosine, but they’re the same function: there’s just a phase difference of 90 degrees.]
No. To understand the wavefunction, we need to understand those de Broglie relations, ω = E/ħ and k = p/ħ, and then, as mentioned, we need to understand the Uncertainty Principle. We need to understand where it comes from. Let’s try to go as far as we can by making a few remarks:
- Adding or subtracting two terms in math, (E/ħ)·t − (p/ħ)∙x, implies the two terms should have the same dimension: we can only add apples to apples, and oranges to oranges. We shouldn’t mix them. Now, the (E/ħ)·t and (p/ħ)·x terms are actually dimensionless: they are pure numbers. So that’s even better. Just check it: energy is expressed in newton·meter (energy, or work, is force over distance, remember?) or electronvolts (1 eV = 1.6×10^{−19 }J = 1.6×10^{−19 }N·m); Planck’s constant, as the quantum of action, is expressed in J·s or eV·s; and the unit of (linear) momentum is 1 N·s = 1 kg·m/s = 1 N·s. E/ħ gives a number expressed per second, and p/ħ a number expressed per meter. Therefore, multiplying E/ħ and p/ħ by t and x respectively gives us a dimensionless number indeed.
- It’s also an invariant number, which means we’ll always get the same value for it, regardless of our frame of reference. As mentioned above, that’s because the four-vector product p_{μ}x_{μ }= E·t − p∙x is invariant: it doesn’t change when analyzing a phenomenon in one reference frame (e.g. our inertial reference frame) or another (i.e. in a moving frame).
- Now, Planck’s quantum of action h, or ħ – h and ħ only differ in their dimension: h is measured in cycles per second, while ħ is measured in radians per second: both assume we can at least measure one cycle – is the quantum of energy really. Indeed, if “energy is the currency of the Universe”, and it’s real and/or virtual photons who are exchanging it, then it’s good to know the currency unit is h, i.e. the energy that’s associated with one cycle of a photon. [In case you want to see the logic of this, see my post on the physical constants c, h and α.]
- It’s not only time and space that are related, as evidenced by the fact that t − x itself is an invariant four-vector, E and p are related too, of course! They are related through the classical velocity of the particle that we’re looking at: E/p = c^{2}/v and, therefore, we can write: E·β = p·c, with β = v/c, i.e. the relative velocity of our particle, as measured as a ratio of the speed of light. Now, I should add that the t − x four-vector is invariant only if we measure time and space in equivalent units. Otherwise, we have to write c·t − x. If we do that, so our unit of distance becomes c meter, rather than one meter, or our unit of time becomes the time that is needed for light to travel one meter, then c = 1, and the E·β = p·c becomes E·β = p, which we also write as β = p/E: the ratio of the energy and the momentum of our particle is its (relative) velocity.
Combining all of the above, we may want to assume that we are measuring energy and momentum in terms of the Planck constant, i.e. the ‘natural’ unit for both. In addition, we may also want to assume that we’re measuring time and distance in equivalent units. Then the equation for the phase of our wavefunctions reduces to:
θ = (ω·t − k ∙x) = E·t − p·x
Now, θ is the argument of a wavefunction, and we can always re-scale such argument by multiplying or dividing it by some constant. It’s just like writing the argument of a wavefunction as v·t–x or (v·t–x)/v = t –x/v with v the velocity of the waveform that we happen to be looking at. [In case you have trouble following this argument, please check the post I did for my kids on waves and wavefunctions.] Now, the energy conservation principle tells us the energy of a free particle won’t change. [Just to remind you, a ‘free particle’ means it’s in a ‘field-free’ space, so our particle is in a region of uniform potential.] So we can, in this case, treat E as a constant, and divide E·t − p·x by E, so we get a re-scaled phase for our wavefunction, which I’ll write as:
φ = (E·t − p·x)/E = t − (p/E)·x = t − β·x
Alternatively, we could also look at p as some constant, as there is no variation in potential energy that will cause a change in momentum, and the related kinetic energy. We’d then divide by p and we’d get (E·t − p·x)/p = (E/p)·t − x) = t/β − x, which amounts to the same, as we can always re-scale by multiplying it with β, which would again yield the same t − β·x argument.
The point is, if we measure energy and momentum in terms of the Planck unit (I mean: in terms of the Planck constant, i.e. the quantum of energy), and if we measure time and distance in ‘natural’ units too, i.e. we take the speed of light to be unity, then our Platonic wavefunction becomes as simple as:
Φ(φ) = a·e^{−iφ }= a·e^{−i(t − β·x)}
This is a wonderful formula, but let me first answer your most likely question: why would we use a relative velocity?Well… Just think of it: when everything is said and done, the whole theory of relativity and, hence, the whole of physics, is based on one fundamental and experimentally verified fact: the speed of light is absolute. In whatever reference frame, we will always measure it as 299,792,458 m/s. That’s obvious, you’ll say, but it’s actually the weirdest thing ever if you start thinking about it, and it explains why those Lorentz transformations look so damn complicated. In any case, this fact legitimately establishes c as some kind of absolute measure against which all speeds can be measured. Therefore, it is only natural indeed to express a velocity as some number between 0 and 1. Now that amounts to expressing it as the β = v/c ratio.
Let’s now go back to that Φ(φ) = a·e^{−iφ }= a·e^{−i(t − β·x) }wavefunction. Its temporal frequency ω is equal to one, and its spatial frequency k is equal to β = v/c. It couldn’t be simpler but, of course, we’ve got this remarkably simple result because we re-scaled the argument of our wavefunction using the energy and momentum itself as the scale factor. So, yes, we can re-write the wavefunction of our particle in a particular elegant and simple form using the only information that we have when looking at quantum-mechanical stuff: energy and momentum, because that’s what everything reduces to at that level.
So… Well… We’ve pretty much explained what quantum physics is all about here. You just need to get used to that complex exponential: e^{−iφ} = cos(−φ) + i·sin(−φ) = cos(φ) −i·sin(φ). It would have been nice if Nature would have given us a simple sine or cosine function. [Remember the sine and cosine function are actually the same, except for a phase difference of 90 degrees: sin(φ) = cos(π/2−φ) = cos(φ+π/2). So we can go always from one to the other by shifting the origin of our axis.] But… Well… As we’ve shown so many times already, a real-valued wavefunction doesn’t explain the interference we observe, be it interference of electrons or whatever other particles or, for that matter, the interference of electromagnetic waves itself, which, as you know, we also need to look at as a stream of photons , i.e. light quanta, rather than as some kind of infinitely flexible aether that’s undulating, like water or air.
However, the analysis above does not include uncertainty. That’s as fundamental to quantum physics as de Broglie‘s equations, so let’s think about that now.
Introducing uncertainty
Our information on the energy and the momentum of our particle will be incomplete: we’ll write E = E_{0 }± σ_{E}, and p = p_{0 }± σ_{p}. Huh? No ΔE or ΔE? Well… It’s the same, really, but I am a bit tired of using the Δ symbol, so I am using the σ symbol here, which denotes a standard deviation of some density function. It underlines the probabilistic, or statistical, nature of our approach.
The simplest model is that of a two-state system, because it involves two energy levels only: E = E_{0 }± A, with A some constant. Large or small, it doesn’t matter. All is relative anyway. 🙂 We explained the basics of the two-state system using the example of an ammonia molecule, i.e. an NH_{3 }molecule, so it consists on one nitrogen and three hydrogen atoms. We had two base states in this system: ‘up’ or ‘down’, which we denoted as base state | 1 〉 and base state | 2 〉 respectively. This ‘up’ and ‘down’ had nothing to do with the classical or quantum-mechanical notion of spin, which is related to the magnetic moment. No. It’s much simpler than that: the nitrogen atom could be either beneath or, else, above the plane of the hydrogens, as shown below, with ‘beneath’ and ‘above’ being defined in regard to the molecule’s direction of rotation around its axis of symmetry.
In any case, for the details, I’ll refer you to the post(s) on it. Here I just want to mention the result. We wrote the amplitude to find the molecule in either one of these two states as:
- C_{1 }= 〈 1 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E0 − A)·t }+ (1/2)·e^{−(i/ħ)·(E0 + A)·t}
- C_{2 }= 〈 2 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E0 − A)·t }– (1/2)·e^{−(i/ħ)·(E0 + A)·t}
That gave us the following probabilities:
If our molecule can be in two states only, and it starts off in one, then the probability that it will remain in that state will gradually decline, while the probability that it flips into the other state will gradually increase.
Now, the point you should note is that we get these time-dependent probabilities only because we’re introducing two different energy levels: E_{0 }+ A and E_{0 }− A. [Note they separated by an amount equal to 2·A, as I’ll use that information later.] If we’d have one energy level only – which amounts to saying that we know it, and that it’s something definite – then we’d just have one wavefunction, which we’d write as:
a·e^{−iθ }= a·e^{−(}^{i/ħ)·(E0·t − p·x)}^{ }= a·e^{−(}^{i}^{/ħ)·(E}^{0·t)}·e^{(}^{i/ħ)·(p·x)}
Note that we can always split our wavefunction in a ‘time’ and a ‘space’ part, which is quite convenient. In fact, because our ammonia molecule stays where it is, it has no momentum: p = 0. Therefore, its wavefunction reduces to:
a·e^{−iθ }= a·e^{−(}^{i/ħ)·(E0·t)}
As simple as it can be. 🙂 The point is that a wavefunction like this, i.e. a wavefunction that’s defined by a definite energy, will always yield a constant and equal probability, both in time as well as in space. That’s just the math of it: |a·e^{−iθ}|^{2 }= a^{2}. Always! If you want to know why, you should think of Euler’s formula and Pythagoras’ Theorem: cos^{2}θ +sin^{2}θ = 1. Always! 🙂
That constant probability is annoying, because our nitrogen atom never ‘flips’, and we know it actually does, thereby overcoming a energy barrier: it’s a phenomenon that’s referred to as ‘tunneling’, and it’s real! The probabilities in that graph above are real! Also, if our wavefunction would represent some moving particle, it would imply that the probability to find it somewhere in space is the same all over space, which implies our particle is everywhere and nowhere at the same time, really.
So, in quantum physics, this problem is solved by introducing uncertainty. Introducing some uncertainty about the energy, or about the momentum, is mathematically equivalent to saying that we’re actually looking at a composite wave, i.e. the sum of a finite or potentially infinite set of component waves. So we have the same ω = E/ħ and k = p/ħ relations, but we apply them to n energy levels, or to some continuous range of energy levels ΔE. It amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ. In our two-state system, n = 2, obviously! So we’ve two energy levels only and so our composite wave consists of two component waves only.
We know what that does: it ensures our wavefunction is being ‘contained’ in some ‘envelope’. It becomes a wavetrain, or a kind of beat note, as illustrated below:
[The animation comes from Wikipedia, and shows the difference between the group and phase velocity: the green dot shows the group velocity, while the red dot travels at the phase velocity.]
So… OK. That should be clear enough. Let’s now apply these thoughts to our ‘reduced’ wavefunction
Φ(φ) = a·e^{−iφ }= a·e^{−i(t − β·x)}
Thinking about uncertainty
Frankly, I tried to fool you above. If the functional form of the wavefunction is a·e^{−(}^{i/ħ)·(E·t − p·x)}, then we can measure E and p in whatever unit we want, including h or ħ, but we cannot re-scale the argument of the function, i.e. the phase θ, without changing the functional form itself. I explained that in that post for my kids on wavefunctions:, in which I explained we may represent the same electromagnetic wave by two different functional forms:
F(ct−x) = G(t−x/c)
So F and G represent the same wave, but they are different wavefunctions. In this regard, you should note that the argument of F is expressed in distance units, as we multiply t with the speed of light (so it’s like our time unit is 299,792,458 m now), while the argument of G is expressed in time units, as we divide x by the distance traveled in one second). But F and G are different functional forms. Just do an example and take a simple sine function: you’ll agree that sin(θ) ≠ sin(θ/c) for all values of θ, except 0. Re-scaling changes the frequency, or the wavelength, and it does so quite drastically in this case. 🙂 Likewise, you can see that a·e^{−i(φ/E) }= [a·e^{−iφ}]^{1/E}, so that’s a very different function. In short, we were a bit too adventurous above. Now, while we can drop the 1/ħ in the a·e^{−(}^{i/ħ)·(E·t − p·x)}^{ }function when measuring energy and momentum in units that are numerically equal to ħ, we’ll just revert to our original wavefunction for the time being, which equals
Ψ(θ) = a·e^{−iθ }= a·e^{−}^{i·[(E/ħ)·t − (p/ħ)·x]}
Let’s now introduce uncertainty once again. The simplest situation is that we have two closely spaced energy levels. In theory, the difference between the two can be as small as ħ, so we’d write: E = E_{0 }± ħ/2. [Remember what I said about the ± A: it means the difference is 2A.] However, we can generalize this and write: E = E_{0 }± n·ħ/2, with n = 1, 2, 3,… This does not imply any greater uncertainty – we still have two states only – but just a larger difference between the two energy levels.
Let’s also simplify by looking at the ‘time part’ of our equation only, i.e. a·e^{−}^{i·(E/ħ)·t}. It doesn’t mean we don’t care about the ‘space part’: it just means that we’re only looking at how our function varies in time and so we just ‘fix’ or ‘freeze’ x. Now, the uncertainty is in the energy really but, from a mathematical point of view, we’ve got an uncertainty in the argument of our wavefunction, really. This uncertainty in the argument is, obviously, equal to:
(E/ħ)·t = [(E_{0 }± n·ħ/2)/ħ]·t = (E_{0}/ħ ± n/2)·t = (E_{0}/ħ)·t ± (n/2)·t
So we can write:
a·e^{−}^{i·(E/ħ)·t} = a·e^{−}^{i·[(E0/ħ)·t ± (1/2)·t]} = a·e^{−}^{i·[(E0/ħ)·t]}·e^{i·[±(n/2)·t]}
This is valid for any value of t. What the expression says is that, from a mathematical point of view, introducing uncertainty about the energy is equivalent to introducing uncertainty about the wavefunction itself. It may be equal to a·e^{−}^{i·[(E0/ħ)·t]}·e^{i·(n/2)·t}, but it may also be equal to a·e^{−}^{i·[(E0/ħ)·t]}·e^{−i·(n/2)·t}. The phases of the e^{−i·t/2} and e^{i·t/2 }factors are separated by a distance equal to t.
So… Well…
[…]
Hmm… I am stuck. How is this going to lead me to the ΔE·Δt = ħ/2 principle? To anyone out there: can you help? 🙂
[…]
The thing is: you won’t get the Uncertainty Principle by staring at that formula above. It’s a bit more complicated. The idea is that we have some distribution of the observables, like energy and momentum, and that implies some distribution of the associated frequencies, i.e. ω for E, and k for p. The Wikipedia article on the Uncertainty Principle gives you a formal derivation of the Uncertainty Principle, using the so-called Kennard formulation of it. You can have a look, but it involves a lot of formalism—which is what I wanted to avoid here!
I hope you get the idea though. It’s like statistics. First, we assume we know the population, and then we describe that population using all kinds of summary statistics. But then we reverse the situation: we don’t know the population but we do have sample information, which we also describe using all kinds of summary statistics. Then, based on what we find for the sample, we calculate the estimated statistics for the population itself, like the mean value and the standard deviation, to name the most important ones. So it’s a bit the same here, except that, in quantum mechanics, there may not be any real value underneath: the mean and the standard deviation represent something fuzzy, rather than something precise.
Hmm… I’ll leave you with these thoughts. We’ll develop them further as we will be digging into all much deeper over the coming weeks. 🙂
Post scriptum: I know you expect something more from me, so… Well… Think about the following. If we have some uncertainty about the energy E, we’ll have some uncertainty about the momentum p according to that β = p/E. [By the way, please think about this relationship: it says, all other things being equal (such as the inertia, i.e. the mass, of our particle), that more energy will all go into more momentum. More specifically, note that ∂p/∂p = β according to this equation. In fact, if we include the mass of our particle, i.e. its inertia, as potential energy, then we might say that (1−β)·E is the potential energy of our particle, as opposed to its kinetic energy.] So let’s try to think about that.
Let’s denote the uncertainty about the energy as ΔE. As should be obvious from the discussion above, it can be anything: it can mean two separate energy levels E = E_{0 }± A, or a potentially infinite set of values. However, even if the set is infinite, we know the various energy levels need to be separated by ħ, at least. So if the set is infinite, it’s going to be a countable infinite set, like the set of natural numbers, or the set of integers. But let’s stick to our example of two values E = E_{0 }± A only, with A = ħ so E + ΔE = E_{0 }± ħ and, therefore, ΔE = ± ħ. That implies Δp = Δ(β·E) = β·ΔE = ± β·ħ.
Hmm… This is a bit fishy, isn’t it? We said we’d measure the momentum in units of ħ, but so here we say the uncertainty in the momentum can actually be a fraction of ħ. […] Well… Yes. Now, the momentum is the product of the mass, as measured by the inertia of our particle to accelerations or decelerations, and its velocity. If we assume the inertia of our particle, or its mass, to be constant – so we say it’s a property of the object that is not subject to uncertainty, which, I admit, is a rather dicey assumption (if all other measurable properties of the particle are subject to uncertainty, then why not its mass?) – then we can also write: Δp = Δ(m·v) = Δ(m·β) = m·Δβ. [Note that we’re not only assuming that the mass is not subject to uncertainty, but also that the velocity is non-relativistic. If not, we couldn’t treat the particle’s mass as a constant.] But let’s be specific here: what we’re saying is that, if ΔE = ± ħ, then Δv = Δβ will be equal to Δβ = Δp/m = ± (β/m)·ħ. The point to note is that we’re no longer sure about the velocity of our particle. Its (relative) velocity is now:
β ± Δβ = β ± (β/m)·ħ
But, because velocity is the ratio of distance over time, this introduces an uncertainty about time and distance. Indeed, if its velocity is β ± (β/m)·ħ, then, over some time T, it will travel some distance X = [β ± (β/m)·ħ]·T. Likewise, it we have some distance X, then our particle will need a time equal to T = X/[β ± (β/m)·ħ].
You’ll wonder what I am trying to say because… Well… If we’d just measure X and T precisely, then all the uncertainty is gone and we know if the energy is E_{0 }+ ħ or E_{0 }− ħ. Well… Yes and no. The uncertainty is fundamental – at least that’s what’s quantum physicists believe – so our uncertainty about the time and the distance we’re measuring is equally fundamental: we can have either of the two values X = [β ± (β/m)·ħ] T = X/[β ± (β/m)·ħ], whenever or wherever we measure. So we have a ΔX and ΔT that are equal to ± [(β/m)·ħ]·T and X/[± (β/m)·ħ] respectively. We can relate this to ΔE and Δp:
- ΔX = (1/m)·T·Δp
- ΔT = X/[(β/m)·ΔE]
You’ll grumble: this still doesn’t give us the Uncertainty Principle in its canonical form. Not at all, really. I know… I need to do some more thinking here. But I feel I am getting somewhere. 🙂 Let me know if you see where, and if you think you can get any further. 🙂
The thing is: you’ll have to read a bit more about Fourier transforms and why and how variables like time and energy, or position and momentum, are so-called conjugate variables. As you can see, energy and time, and position and momentum, are obviously linked through the E·t and p·x products in the E_{0}·t − p·x sum. That says a lot, and it helps us to understand, in a more intuitive way, why the ΔE·Δt and Δp·Δx products should obey the relation they are obeying, i.e. the Uncertainty Principle, which we write as ΔE·Δt ≥ ħ/2 and Δp·Δx ≥ ħ/2. But so proving involves more than just staring at that Ψ(θ) = a·e^{−iθ }= a·e^{−}^{i·[(E/ħ)·t − (p/ħ)·x] }relation.
Having said, it helps to think about how that E·t − p·x sum works. For example, think about two particles, a and b, with different velocity and mass, but with the same momentum, so p_{a }= p_{b} ⇔ m_{a}·v_{a }= m_{a}·v_{a }⇔ m_{a}/v_{b }= m_{b}/v_{a}. The spatial frequency of the wavefunction would be the same for both but the temporal frequency would be different, because their energy incorporates the rest mass and, hence, because m_{a }≠ m_{b}, we also know that E_{a }≠ E_{b}. So… It all works out but, yes, I admit it’s all very strange, and it takes a long time and a lot of reflection to advance our understanding.