A lot of the Uncertainty in quantum mechanics is suspiciously certain. For example, we know an electron will always have its spin up or down, in any direction along which we choose to measure it, and the value of the angular momentum will, accordingly, be measured as plus or minus ħ/2. That doesn’t sound uncertain to me. In fact, it sounds remarkably certain, doesn’t it? We know – we are sure, in fact, because of countless experiments – that the electron will be in either of those two states, and we also know that these two states are separated by ħ, Planck’s quantum of action, exactly.
When modeling electromagnetic waves, the notion of left versus right circular polarization is quite clear and fully integrated in the mathematical treatment. In contrast, quantum math sticks to the very conventional idea that the imaginary unit (i) is – always! – a counter-clockwise rotation by 90 degrees. We all know that –i would do just as an imaginary unit as i, because the definition of the imaginary unit says the only requirement is that its square has to be equal to –1, and (–i)2 is also equal to –1.
So we actually have two imaginary units: i and –i. However, physicists stubbornly think there is only one direction for measuring angles, and that is counter-clockwise. That’s a mathematical convention, Professor: it’s something in your head only. It is not real. Nature doesn’t care about our conventions and, therefore, I feel the spin ‘up’ versus spin ‘down’ should correspond to the two mathematical possibilities: if the ‘up’ state is represented by some complex function, then the ‘down’ state should be represented by its complex conjugate.
This ‘additional’ rule wouldn’t change the basic quantum-mechanical rules – which are written in terms of state vectors in a Hilbert space (and, yes, a Hilbert space is an unreal as it sounds: its rules just say you should separate cats and dogs while adding them – which is very sensible advice, of course). However, they would, most probably (just my intuition – I need to prove it), avoid these crazy 720 degree symmetries which inspire the likes of Penrose to say there is no physical interpretation on the wavefunction.
Oh… As for the title of my post… I think it would be a great title for a book – because I’ll need some space to work it all 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 (albeit on my other site, where I had posted the same). 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.
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/meff)·∇2ψ(x, t)
Now, the key difference with the diffusion equation – let me write it for you once again: ∂φ(x, t)/∂t = D·∇2φ(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/meff)·Im(∇2ψ)
Im(∂ψ/∂t) = (1/2)·(ħ/meff)·Re(∇2ψ)
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/meff)∙∇2ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i2 = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i2∙d = − d + i∙c. [Now that we’re getting a bit technical, let me note that the meff 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 meff = 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 = c2∇×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 – something with a magnitude and a direction, in a real vector space – as opposed to these state vectors from a 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)∙ex×E, with ex 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)∙ex× 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)∙ex×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 (not state vectors – objects from an abstract Hilbert space – but real 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? 🙂
In one of my previous posts, I developed a flywheel model for an electron – or for charged matter-particles in general. You can review the basics in the mentioned post or, else, download a more technical paper from this link. Here I just want to present some more general remarks about the conceptual difficulties. Let’s recall the basics of the model. We think of an electron as a point charge moving about some center, so that’s the green dotin the animation below. We can then analyze its movement in terms of two perpendicular oscillations, i.e. the sine (blue) and cosine (red) functions.
The idea is that space itself – by some mechanism we’d need to elaborate – provides a restoring force when our charge moves away from the center. So it wants to go back to the r= (x, y) = 0, but it never actually does because of these restoring forces, which are perpendicular to each other and just make it spin around. This may sound crazy but the original inspiration for this – the mathematical similarity between the E = m·c2 and E = m·a2·ω2 formulas – is intriguing enough to pursue the idea. Let me show you.
Think of an electron with energy E (the equivalent mass of this electron is, of course, m = E/c2). Now look at the following. If we substitute ω for E/ħ (the frequency of the matter-wave) and a for ħ/m·c (i.e. the Compton scattering radius of an electron), we can write:
a·ω = [ħ/(m·c)]·[E/ħ] = E/(m·c) = m·c2/(m·c) = c
We get the same when equating the E = m·a2·ω2 and E = m·c2 formulas:
m·a2·ω2 = m·c2 ⇔ c = a·ω
Wow! Did we just prove something? No. Not really. We only showed that our E = m·a2·ω2 = m·c2 equation might make sense. Note that the a·ω = c product is the tangential velocity vtangential = r·ω = a·ω = c.
Our rotating charge should also have some angular momentum, right? In fact, the Stern-Gerlach experiment tells us the angular momentum should be equal to ± ħ/2. Is it? We will need a formula for the angular mass, aka the moment of inertia. Let’s try the one for a rotating disk: I = (1/2)·m·a2. The formula for the angular momentum itself is just L = I·ω. Combining both gives us the following result:
L = ħ/2 = I·ω = (1/2)·m·a2·ω = (1/2)·(E/c2)·a2·(E/ħ) ⇔ a = ħ/(m·c)
Did we just get the formula for the Compton scattering radius out of our model? Well… Yes. It is a rather Grand Result, I think. So… Well… It might make sense, but… Well… Does it, really?
First, we should note that the math is not so easy as it might seem at first: we would need to use relativistically correct formulas everywhere. I made a start at that, and it’s quite encouraging. For example, I showed in one of my previous posts, that the classical E = m·v2/2 formula for the kinetic energy of a mass on a spring effectively morphs into the relativistically correct E = mv·c2 = m0·γ·c2 formula. So it looks like the annoying 1/2 factors in the classical formulas effectively disappear in a relativistically correct analysis. However, there is a lot more to be checked than just the math for one oscillator, and I haven’t done that yet.
The key issue is the interpretation of the mass factor. Our pointlike charge can only travel at light speed if its rest mass is equal to zero. But that is a problem for the model, because the idea of accelerations or decelerations as a result of a restoring force is no longer consistent: if there is norest mass, even the tiniest force on it will give it an infinite acceleration or deceleration. Hence, it is tempting to assume the charge must have some mass, and the concepts of the electromagnetic and/or the effective mass spring to mind. However, I don’t immediately see how the specifics of this could be worked out. So we’re a bit stuck here. Still, it is tempting to try to think of it like a photon, for which we can effectively state that all of its energy – and the equivalent mass – is in the oscillation: E = ħ·ω and m = E/c2 = ħ·ω/c2.
Let’s think of a few other questions. What if the energy of our electron increases? Our a·ω = c tells that a = c/ω = c·ħ/E should decrease. Does that make any sense? We may think of it like this. The E = m·c2 equation tells us that the ratio between the energy and the mass of a particle is constant: E/m = c2. So we can write:
E/m = c2 = a2·ω2 = a2·E2/ħ2 ⇔ a·E/ħ = c
This tells us the same: a is inversely proportional to E, and the constant of inverse proportionality is c·ħ. Hence, our disk becomes smaller but spins faster. What happens to the angular mass and the angular momentum? The angular momentum must L = ħ/2 = I·ω must remain the same, and you can check it does. Hence, the angular mass I effectively diminishes as ω = E/ħ goes up.
Hence, it all makes sense – intuitively, that is. In fact, our model is basically an adaptation of the more general idea of electromagnetic mass: we’re saying the energy in the (two-dimensional) oscillation gives our electron the equivalent mass that we measure in real-life experiments. Having said that, the idea of a pointlike charge with zero rest mass – accelerating and decelerating in two perpendicular directions – remains a strange one.
Let us think about the nature of the restoring force. The oscillator model is, effectively, based on the assumption we have a linear force here. To be precise, the restoring force is thought of as being directly proportional with the distance from the equilibrium: F = −k·x. The proportionality constant k is the stiffness of our spring. It just comes with the system. So what’s the equivalent of k in our model here? For a non-relativistic spring, it is easy to show that – while a constant – k will always be equal to k = m·ω2. Hence, if we put a larger mass on the spring, then the frequency of the oscillation must go down, and vice versa. But what mass concept should we apply here? Before we further explore this question, let us look at those relativistic corrections. Why? Well… I think it’s about time I show you how they affect the analysis, right? Let me gratefully acknowledge an obviously brilliant B.A. student who did an excellent paper on relativistic springs (link and references here), from which I copy the illustration below.
The graphs show what happens to the trajectory and the velocity when speeds go up all the way to the relativistic limit. The light grey is the classical (non-relativistic) case: we recognize the sinusoidal motion and velocity function. See how these shapes change when velocities go to the relativistic limit (see the grey and (for velocities reaching the speed of light) the black curves). The sinusoidal shape of the functions disappears: the velocity very rapidly changes between −c and +c. In fact, the velocity function starts resembling a square wave function.
This is quite interesting, but… Well… It doesn’t make us much wiser, and it surely does not answer our question: how should we interpret the stiffness coefficient k here? The relativistically correct force law is given below, but that doesn’t tell us anything new: k tells us that the restoring force is going to be directly proportional to the distance from the equilibrium (zero) point:
F = dp/dt = d(mv)/dt = mvdv/dt = mva = −k·x
However, we may think of this, perhaps: when we analyze a mechanical (or any other typical) oscillator, we think of k being given. But for our flywheel model, we find that the value of k is going to depend on the energy and the amplitude of the oscillation. To be precise, if we just take the E = m·a2·ω2 = m·c2 and k = m·ω2 formulas for granted, we can write k as a function of the energy:
E = k·a2 ⇔ k = E/a2 = (E·E2)/(c2ħ2) ⇔ k = E3/(c2ħ2)
It is a funny formula: it tells us the stiffness – or whatever the equivalent is in this model of ours – increases exponentially with the energy. It is a weird formula, but it is consistent with the other conclusion above: if the energy goes up, the radius a becomes smaller. It’s because space becomes stickier. 🙂
But space is space, right? And its defining characteristic – the constant speed of light c – is… Well… It is what it is: it just tells us that the ratio between the energy and mass for any particle is always the same:
What else does it tell us? Well… If our model is correct, it also tells us the tangential velocity of our charge will always be equal to c. In that sense, then, it tells us space is space. There is one more relation that we might mention here:
c = a·ω ⇔ ω = c/a
The energy defines the radius in our model (remember the a = c·ħ/E relation). Now, the relationship above tells us that, in turn, defines the frequency, and in a rather particular way: the frequency is inversely proportional to the radius, and the proportionality coefficient is, once again, c. 🙂
One of the fundamental questions for any freshman studying physics is the following: if an electron orbits around a nucleus, the various accelerations and decelerations should cause it to emit electromagnetic radiation. Hence, it should gradually lose energy and, therefore, the orbital cannot be stable. Therefore, the atom cannot be stable. So we have a problem. 🙂
A related but different question is: why don’t the electron and the nucleus simply crash into each other? They attract each other, very strongly, right? Well… Yes. But, as I said, that’s a related but different question. Let me first try to handle the first one – as good as I can. 🙂
So… Well… It’s a simple question but – as you know by now – the science of physics seldom gives us simple answers to simple questions. Worse, I’ve studied physics for many years now – admittedly, in my own stubborn, critical and generally limited way 🙂 – and I feel the answer I am getting is not only complicated but also not very real. So… Well… We might want to think we probably do not quite understand what is going on really.
This lack of understanding is nothing to be ashamed of, as great physicists such as Richard Feynman (and others) acknowledge: “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, there you have it: physicists don’t understand their own science, it seems. 🙂 But let’s go beyond that. Let’s talk about the wavefunction because… Well… You know it’s supposed to describe the electron, right? So what is it?
Well… Unfortunately, physics textbooks won’t tell you what the wavefunction is. They’ll tell you it’s a mathematical construct. A solution to some differential equation (Schrödinger’s equation), to be precise. 😦 However, they will you – from time to time, at least – tell you what it isn’t. For example, Feynman’s most precise description of the model of an electron – or an electron orbital, I should say – might be the one he offers when, while deriving the electron orbitals from Schrödinger’s equation, he says what the wavefunction is surely not:
“The wave function Ψ(r) for an electron in an atom does not describe a smeared-out electron with a smooth charge density. The electron is either here, or there, or somewhere else, but wherever it is, it is a point charge.” (Feynman’s Lectures, Vol. III, p. 21-6)
So… Well… That’s not too bad as an explanation. 🙂 But… Well… While fairly precise, I’d think we can improve on Feynman’s language. For starters, we should distinguish the concept of an electron and the concept of its charge. When the electron is in some stable configuration – i.e. in an orbital as described by its wavefunction Ψ(r) – the idea of the electron combines both the orbital and the point charge. Let’s be precise here:
The charge is what, when probing, we’ll effectively find “here, there, or somewhere else” in the space that is being described by our wavefunction Ψ(r).
As for the electron… Well… We know that – by applying operators to the wavefunction – we’ll not only get information about its position, but also about its linear or angular momentum, its energy, and whatever other characteristic of the electron that we’re describing. In that sense, we might say that the wavefunction completely describes the electron and that, therefore, the electron is not the point charge itself, but the orbital, as described by the wavefunction, with its point charge somewhere.
In short, for all practical purposes, we might say that the electron is the wavefunction, and vice versa. 🙂 Indeed, when studying quantum mechanics, one effectively does end up equating the particle with its wavefunction, not with its charge. And rightly so ! An elementary particle – be it an electron or a quark – is more than just its charge: it has energy, momentum (linear or angular), occupies some space and – in the case of quarks – has a color too ! 🙂
But that still doesn’t answer the simple question I started out with: the electrons – or the point charges in those orbitals – don’t emit radiation. Why not? Well… If I’d be your professor, and you’d be sitting for an exam in front of me, then I’d expect you to start talking about the Uncertainty Principle, wavefunctions, energy states and what have you. But I am not your professor (I am not a professor at all, in fact), and so I don’t want hear that answer. To be precise, I don’t like that answer because, just like Feynman, I don’t quite understand it the way I would like to understand it! So… What other answer can we think of? Can we think of something that is, perhaps, more intuitive?
“Physical objects are not in space, but these objects are spatially extended. In this way, the concept “empty space” loses its meaning.”
In fact, I’d go one step further and say: objects create their own space.
Huh? Yes. Think of the planets – including Earth – going around the Sun. Einstein’s general relativity theory tells us they are in their own space. Indeed, Einstein told us we should not think of gravitation as a force: the masses involved just curve the space-time fabric around them in such a way that these planets just keep going and going around and around. They are basically moving in free space: their path just happens to be curved from our perspective. If we wouldn’t impose this abstract (or mathematical, I should say) rectangular Cartesian coordinate space on our description of them, but accept this system of large objects creates its own space, we’d understand why it’s stable: it doesn’t need any energy from the outside to keep going. Nor does it radiate anything out.
Let me emphasize this point: they are in their own space because they don’t radiate anything out. And, I should add, nor do they absorb any energy from the outside. Of course, you’ve heard about gravitational waves – and most notably the one detected by the LIGO Lab last year – but note that gravitational wave was created when two black holes spectacularly merged. That’s because black holes do emit radiation, as a result of which do lose mass and, therefore, this system of large objects became unstable. Of course, if we’d detonate all of the atomic bombs we’ve built, we might also cause our planetary system to become unstable, but you’ll understand that’s a different discussion altogether. 🙂
So… Well… I like to think a wavefunction for an orbital represents the same: we’re looking at a charge that moves around in its own space. In our Cartesian reference frame, this looks like a terribly complicated oscillation. In fact, the oscillation is not only complicated but also – literally – complex, because we’re keeping track of two dimensions simultaneously: the real and imaginary component of the wavefunction. Both are equally real, of course, in a physical sense (and we can argue about what that means, exactly, but not about the statement itself). But so… Well… It’s just a spacetime blob. The charge itself just moves around along a geodesic in its spacetime, and that’s why it doesn’t emit or absorb any energy from the outside. 🙂
Of course, the question now becomes: if an electron orbital is nothing but a weird blob of curved spacetime – in which our charge moves around like a planet moves around in a planetary system – then what’s causing the curvature of space? For our planetary system, we know it’s mass.
So… Well… What can I say? Well… What’s mass? Energy has an equivalent mass, and mass has an equivalent energy. In my previous posts, I look at mass as an oscillation itself and, as I show in one of my papers, that might allow us to interpret Schrödinger’s wave equation as an energy diffusion equation, and the wavefunction itself as a self-contained and self-sustaining gravitational wave. So… Well… If the wavefunction represents a blob of energy – some two-dimensional oscillation – then… Well… Then it could create its own space, right? Just like our Sun and the planets create their own space, in which they move without absorbing or radiating any energy away. In other words, they move in a straight line in their own space. I am tempted to think our pointlike charge must also be moving in a straight line in its own space because… Well… It would, effectively, be emitting radiation otherwise. 🙂
So what’s the nature of Nature, then? Well… All is movement, it seems. Panta rhei ! 🙂 And… Well… I’ll let you do the philosophy now. For example, if objects create their own space, how should we think of their interactions? 🙂
This post is actually not about Einstein or anything he wrote. It is just for fun. It really is because… Well… While it’s about one of the most fundamental questions in physics (what is mass?), I am not going to talk about the Higgs field or other terribly complicated stuff. In fact, I’ll be talking about relativity using formulas that I shouldn’t be using. 🙂 But you should find this very straightforward and thought-provoking. Or so I hope, at least. So see if you like it and – if you do or you don’t, whatever – please let me know by posting a comment.
In my previous post, I noted the structural similarity between the E = mc2 and E = m·a2·ω2/2 formulas. The E = mc2 formula is very well known but also very mysterious. In contrast, the E = m·a2·ω2/2 formula is not so well known but not mysterious at all! It is the formula for the energy of a harmonic oscillator: think of an oscillating spring, for example. The only difference between the two formulas is the 1/2 factor and… Well… We would also need some interpretation of the c = a·ω identity that comes out of this, of course – but I will get there in a moment. 🙂
We can get rid of that 1/2 factor by combining two oscillators. Think of two frictionless pistons (or springs) in a 90° degree angle, as shown below. Because their motion is perpendicular to each other, their motion is independent, and so we can effectively add the power (and energy) of both. In fact, the 90° degree angle explains why a Ducati is more efficient than a Harley-Davidson, whose cylinders are at an angle of 45°. The 45° angle makes for great sound but… Well… Not so efficient. 🙂
The a in the E = m·a2·ω2/2 formula is the magnitude of the oscillation, and the motion of these pistons (or of a mass on a spring) will be described by an x = a·cos(ω·t + Δ) function. The x is just the displacement from the center, and the Δ is just a phase factor which defines our t = 0 point. The ω is the angular frequency of our oscillator: it is defined by the time that is needed for a complete cycle, which is referred to as the periodof the oscillation. We denote the period as t0, and it is easy to understand that t0 must be such that ω·t0 = 2π. Hence, t0 = 2π/ω. Note that, because of the 90° angle between the two cylinders, the phase factor Δ would be zero 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 animation below abstracts away from pistons, springs or whatever other physical oscillation we might think of – but it represents the same. We just denote the phase itself as θ = ω·t.
Now think of an electron as a charge moving about some center, so that’s the green dot in the animation above. We can then also analyze its movement in terms of two perpendicular oscillations, i.e. the sine and cosine functions shown above. Now, you may or may not know that an elementary wavefunction consists of the same: a sine and as cosine. Indeed, using Euler’s notation, we write:
Remembering that multiplication by the imaginary unit (i) amounts to a rotation by 90°. To be precise, because of the minus sign in front of the sine, we have a rotation by minus 90°here, but that doesn’t change the analysis: Nature doesn’t care about our convention for i, or for our convention for measuring angles clockwise or counter-clockwise, and it does allow the angular momentum to be either positive or negative (it is ± ħ/2 for an electron). But let’s further develop our analogy by getting back to our oscillators. The kinetic and potential energy of one oscillator – think of one piston or one spring only – can be calculated as:
K.E. = T = m·v2/2 = (1/2)·m·ω2·a2·sin2(ω·t + Δ)
P.E. = U = k·x2/2 = (1/2)·k·a2·cos2(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·ω2. Hence, the total energy—for one piston (or one spring) only—is equal to:
E = T + U = (1/2)· m·ω2·a2·[sin2(ω·t + Δ) + cos2(ω·t + Δ)] = m·a2·ω2/2
That is the formula we started out with and, yes, if we would add the energy of the two oscillators, we’d effectively have a perpetuum mobile storing an energy that is equal to twice this amount: E = m·a2·ω2.
Let us now think this through. If E and m are the energy and mass of an electron, then the E = m·a2·ω2 and E = m·c2 equations tell us that c = a·ω. What are a and ω here? Well… The de Broglie relations suggest we should equate ω to E/ħ. As for a, we could take the Compton scattering radius of the electron, which is equal to ħ/(m·c). So we write:
a·ω = [ħ/(m·c)]·[E/ħ] = E/(m·c) = m·c2/(m·c) = c
Did we prove anything here? No. We don’t prove anything in this post. We’re just having fun. We only showed that our E = m·a2·ω2 = m·c2 equation might (let me put in italics: might) make sense. 🙂
Let me show you something else. If this flywheel model makes sense, then we can, obviously, also calculate a tangential velocity for our charge. The tangential velocity is the product of the radius and the angular velocity: v = r·ω = a·ω = c. In our previous post, we wrote that we should think of the speed of light as the resonant frequency of the spacetime fabric, but we should probably take that back. The speed of light emerges as the speed of the charge in what I’ll now officially refer to as my flywheel model of an electron.
Is there a resonant frequency here? If so, how should we interpret it? Well… From our a·ω = c, we get that:
So the answer is: no. No resonant frequency of spacetime. The frequency is the frequency of our electron – not of the fabric of spacetime. However, we can, perhaps, think of another analogy. The natural frequency of a spring (ω) depends on (1) the mass on that spring (m) and (2) the restoring force, which is equal to F = −k·x. The k factorhere is the stiffness of the spring. Could we, likewise, talk about the stiffness of the spacetime fabric? We know that, for a spring, we can calculate k from m and ω. We wrote: k = m·ω2. Can we do anything with that? Probably not. The mass in our flywheel model is the equivalent mass of the energy in the (two-dimensional) oscillation. It is not some actual mass going up and down and back and forth. In fact, if the tangential velocity of our charge would be equal to c – which it is in our model – then the charge itself should have zero (rest) mass ! Hence, the stiffness would be equal to k = 0·ω2 = 0!
Let me offer another calculation instead. If this flywheel model makes sense, then the electron will have some angular momentum, right? The angular momentum is equal to L = ω·I, so that’s the product of the angular velocity (ω) and the moment of inertia (I), aka the angular mass. Now, from the Stern-Gerlach experiment, we know that the angular momentum of an electron is equal to ± ħ/2. So now we can calculate the moment of inertia as I = L/ω = (ħ/2)/(E/ħ) = ħ2/(2·E). Now, substituting E for E = m·c2 and remembering that a = ħ/(m·c), we can write this as:
I = ħ2/(2·m·c2) = (ħ2·m)/(2·m2·c2) = (1/2)·m·a2
Do we recognize that formula? Yes. It’s the formula for the angular mass of a solid disc, or a hoop about the diameter, as shown below. Which of the two makes most sense? I am not sure. I’ll let you think about that. 🙂
So… Well… That’s it! 🙂
Why am I smiling? Well… I hope this post makes you think about stuff yourself because… Well… That was my only objective: have fun by thinking about stuff yourself! 🙂
Do these calculations – and the analogy itself – make any sense? My truthful answer is: I am not sure. I really don’t know. Of course, I would very much like to think that this analogy may represent something real. Why? Because it would allow us to associate the wavefunction with something real and, therefore (see my paper on this), it would also allow us to think of Schrödinger’s equation as representing something real. To be precise, it would allow us to think of Schrödinger’s equation as an energy diffusion equation, but… Well… That is somewhat more difficult to explain than what I explained above. 😦
The essential question, of course, is: what gives that pointlike charge that circular motion? What is the origin of what Schrödinger himself referred to (I admit: he did so in a very different context) as a Zitterbewegung?
All that I’ve written above, assumes space is not just some abstract mathematical space. It is real, somehow, and perfectly elastic. So that’s why we can advance a model that assumes an an electron is nothing but a charge (with zero rest mass) that bounces around in it. All of its mass is the equivalent mass of the energy in the oscillation itself. It is, of course, a crazy hypothesis that we cannot prove but, as far as I can see, while crazy, the hypothesis is consistent with what we know about the weird wonderful world of quantum mechanics.
The main weakness in the argument is the following: if the charge itself has zero rest mass, then our E = m·a2·ω2/2 equation reduces to E = 0. Is the analogy still valid? And how can we possibly associate some angular mass with something that is going around but has zero rest mass? This is, effectively, a flywheel model without a flywheel. I may have explained matter as a two-dimensional oscillation, but I haven’t told you what is oscillating. Or… Well… I did. What is oscillating is the charge, with zero rest mass.
The analogy with a photon is obvious. A photon has zero rest mass too! Now, the wavefunction of a photon is the electromagnetic wave. Can we say what is oscillating there? Yes, we can! Of course: it is the electromagnetic field! But what’s the field? You will say the electromagnetic field has a physical dimension: newton per coulomb (N/C). This begs the next question: what’s the physical dimension of the wavefunction? Could it be the same?
My answer is: yes, or maybe. 🙂 Our model assumes it is, effectively, the charge of the electron that is oscillating. Hence, why wouldn’t it be the same?
In any case, I will let you think about that for yourself. As you can see, in physics, an answer to one question may trigger many more. 🙂 If you have any good feedback, please comment! 🙂
Post scriptum: The mentioned weakness in the argument should also be related to the fact that we are using classical non-relativistic formulas. If our charge is really moving at a speed at or near light speed around some center, we should probably have another look at our K.E. = T = m·v2/2 = (1/2)·m·ω2·a2·sin2(ω·t + Δ) formula, right? 🙂 The relativistically correct definition of kinetic energy is slightly different than the T = m·v2/2 formula. It may – or may not – make a difference. 🙂 I’ll talk about that in my next post.