# 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. đ [OK. Poor joke. Acknowledged.]

Just to recap the essentials, I part ways with mainstream physicists in regard to theÂ interpretationÂ of the wavefunction. For mainstream physicists, the wavefunction is just some mathematical construct. NothingÂ real. Of course, I acknowledge mainstream physicists have very good reasons for that, but… Well… I believe that, if there is interference, or diffraction, thenÂ somethingÂ must be interfering, or something must be diffracting. I won’t dwell on this because… Well… I have done that too many times already. MyÂ hypothesisÂ is that the wavefunction is, in effect, aÂ rotatingÂ field vector, so itâ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, theÂ (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 forceÂ per unit massÂ (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/s2), so thatâ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 doÂ notÂ involve any mass: theyâre just an oscillatingÂ field. Nothing more. Nothing less. Why would I believe there must still be some pointlike particle involved? Well…Â As Feynman puts it: âWhen you do find the electron some place, the entire charge is there.â (FeynmanâsÂ Lectures, III-21-4) So… Well… That’s why.

The third difference is one that I thought of only recently: theÂ planeÂ of the oscillation cannotÂ be perpendicular to the direction of motion of our electron, because then we canâ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. đ

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 toÂ imagineÂ it. That’s great mental exercise, so… Well… Just try it. đ

Let’s now think about rotating reference frames and transformations. If theÂ z-direction is the direction along which we measure the angular momentum (or the magnetic moment), then theÂ up-direction will be theÂ positiveÂ z-direction. We’ll also assume theÂ y-direction is the direction of travel of our elementary particleâand let’s just consider an electron here so we’re moreÂ real. đ 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 apparatusâwhich I’ll refer to as a beam splitterâillustrates this geometry.

So I think the magnetic momentâor the angular momentum, reallyâcomes from an oscillatory motion in the x– and y-directions. One is theÂ realÂ component (the cosine function) and the other is the imaginary component (the sine function), as illustrated below.Â

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 isÂ notÂ in theÂ xz-plane, but in theÂ 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-directionâso that’s the direction in which our particle is movingâ, then we might imagine how it would look like whenÂ weÂ would make a 180Â°Â 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.Â 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 theÂ z-axis as it is (pointing upwards), and we will also want to define theÂ x– andÂ y-axis using the familiar right-hand rule for defining a coordinate frame. So our newÂ x-axis and our newÂ y-axis will the same as the oldÂ x- andÂ y-axes but with the sign reversed. In short, we’ll have the following mini-transformation: (1)Â z‘ =Â z, (2) x’ = âx, and (3) y’ =Â ây.

So… Well… If we’re effectively looking at somethingÂ realÂ that was moving along theÂ y-axis, then it will now still be moving along the y’-axis, butÂ in theÂ negativeÂ direction. Hence, our elementary wavefunctionÂ eiÎ¸Â = cosÎ¸ +Â iÂˇsinÎ¸ willÂ transformÂ intoÂ âcosÎ¸ âÂ iÂˇsinÎ¸ =Â âcosÎ¸ âÂ iÂˇsinÎ¸ =Â cosÎ¸ âÂ iÂˇsinÎ¸.Â 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 theÂ z-axis:

Now, ifÂ Ď is equal to 180Â° (so that’s Ď in radians), then theseÂ eiĎ/2Â andÂ eâiĎ/2/â2Â factors areÂ equal toÂ eiĎ/2Â =Â +iÂ andÂ eâiĎ/2Â = âiÂ respectively. Hence, ourÂ eiÎ¸Â = cosÎ¸ +Â iÂˇsinÎ¸ becomes…

Hey ! Wait a minute ! We’re talking about twoÂ veryÂ different things here, right? TheÂ eiÎ¸Â = cosÎ¸ +Â iÂˇsinÎ¸ is anÂ elementaryÂ wavefunction which, we presume, describes some real-life particleâwe talked about an electron with its spin in theÂ up-directionâwhile these transformation matrices are to be applied to amplitudes describing… Well… Either anÂ up– or a down-state, right?

Right. But… Well… Is itÂ so different, really? Suppose ourÂ eiÎ¸Â = cosÎ¸ +Â iÂˇsinÎ¸ wavefunction describes anÂ up-electron, then we still have to apply thatÂ eiĎ/2Â =Â eiĎ/2Â =Â +iÂ factor, right? So we get a new wavefunction that will be equal toÂ eiĎ/2ÂˇeiÎ¸Â =Â eiĎ/2ÂˇeiÎ¸Â =Â +iÂˇeiÎ¸Â =Â iÂˇcosÎ¸ +Â i2ÂˇsinÎ¸ =Â sinÎ¸ âÂ iÂˇcosÎ¸, right? So how can we reconcile that with the cosÎ¸ âÂ iÂˇsinÎ¸ 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 180Â°, so it is going in the other direction now !Â 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 theÂ sinÎ¸ âÂ iÂˇcosÎ¸ andÂ cosÎ¸ âÂ iÂˇsinÎ¸ 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: cosÎ¸ =Â sin(Î¸ +Â Ď/2) andÂ âsinÎ¸ =Â cos(Î¸ +Â Ď/2). Let’s see what we can do with that. We can write the following, for example:

sinÎ¸ âÂ iÂˇcosÎ¸ =Â âcos(Î¸ +Â Ď/2) âÂ iÂˇsin(Î¸ +Â Ď/2) =Â â[cos(Î¸ +Â Ď/2) +Â iÂˇsin(Î¸ +Â Ď/2)] =Â âeiÂˇ(Î¸ +Â Ď/2)

Well… I guess that’s something at least ! The eiÂˇÎ¸Â and âeiÂˇ(Î¸ +Â Ď/2)Â functions differ by a phase shiftÂ andÂ a minus sign so… Well… That’s what it takes to reverse the direction of an electron. đ Let us mull over that in the coming days. As I mentioned, these more philosophical topics are not easily exhausted. đ

# 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 particleâfor an electron, at least: for the proton, we only got the order of magnitude rightâbut then a proton is not an elementary particle.Â We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between theÂ E =Â mÂˇa2ÂˇĎ2Â andÂ E =Â mÂˇc2Â relations that we… Well… We need to be moreÂ specific about it.

Indeed, I’ve been ambiguous here and thereâoscillatingÂ between various interpretations, so to speak. đ In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the form factorÂ problem.Â So… Well… That explains the title of my post. But so… Well… I do want to be somewhat moreÂ conclusiveÂ in this post. So let’s go and see where we end up. đ

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. đ

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 ourÂ 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 =Â ĎÂˇI. Of course,Â the moment of inertia (aka the angular mass) will depend on theÂ formÂ (orÂ shape) of our flywheel:

1. I = mÂˇa2Â for a rotating pointÂ mass m or, what amounts to the same, for a circular hoop of mass m and radiusÂ rÂ =Â a.
2. For a rotating (uniformly solid)Â disk, we must add a 1/2 factor: IÂ =Â mÂˇa2/2.

How can we relate those formulas to the E =Â mÂˇa2ÂˇĎ2Â formula? TheÂ kinetic energy that is being stored in a flywheel is equal EkineticÂ = IÂˇĎ2/2, so that is only halfÂ of theÂ E =Â mÂˇa2ÂˇĎ2Â product if we substitute I forÂ I = mÂˇa2. [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, theÂ E =Â mÂˇa2ÂˇĎ2Â 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 justÂ transfersÂ energy from one oscillator to the other, but so… Well… We don’tÂ includeÂ it in our energy calculations. The essence of our model is thatÂ two-dimensional oscillation whichÂ drivesÂ the electron, and which is reflected in Einstein’sÂ E =Â mÂˇc2Â formula.Â That two-dimensional oscillationâtheÂ a2ÂˇĎ2Â = c2Â equation, reallyâtells us that theÂ resonantÂ (orÂ natural) frequencyÂ of the fabric of spacetime is given by theÂ speed of lightâbut measured in units ofÂ a. [If you don’t quite get this, re-write theÂ a2ÂˇĎ2Â = c2Â equation asÂ Ď = c/a: the radius of our electron appears as a naturalÂ 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 exampleâ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 toÂ vÂ =Â aÂˇĎÂ =Â 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 radiusÂ 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 asÂ aÂ =Â Ä§Âˇ/(mÂˇc) = 3.8616Ă10â13Â 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:

cÂ = aÂˇĎÂ =Â aÂˇE/Ä§ =Â aÂˇmÂˇc2/Ä§Â Â âÂ aÂ =Â Ä§/(mÂˇc)

The question is: whatÂ is that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’tÂ proveÂ anything in this regard. But myÂ hypothesisÂ is that it is, in effect, aÂ rotatingÂ 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:

1. 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 massÂ (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/s2), so that’s the dimension of a gravitational field.
2. 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 doÂ notÂ 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)
3. The third difference is one that I thought of only recently: theÂ planeÂ of the oscillation cannotÂ 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.Â The basic idea here is illustrated below (credit for this illustration goes toÂ another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to aÂ YouTube videoÂ from theÂ Quantum Made SimpleÂ site.

The point is: the direction of the angular momentum (and the magnetic moment) of an electronâor, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron isÂ travelingâcannotÂ be parallel to the direction of motion. On the contrary, it isÂ perpendicularÂ to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will compriseÂ 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 JÂ and a magnetic momentÂ ÎźÂ (I used bold-face because these areÂ vector quantities) that is parallel to some magnetic field B, will notÂ line up, as you’d expect a tiny magnet to do in a magnetic fieldâor not completely, at least: it willÂ 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 ofÂ showÂ that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever weÂ thinkÂ our electronâor its wavefunctionâmight be, it needs to be compatible with stuff like the observedÂ precession frequencyÂ 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) gravityâsuch as the one I mentioned aboveâare intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? đ Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years aheadâI hope. đ

Post scriptum: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But itÂ isÂ that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillatorsÂ driveÂ the flywheel but, without the flywheel, nothing is happening. It is really theÂ transferÂ of energyâthrough the flywheelâ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.Â The motionÂ ofÂ our first oscillator is given by the cos(ĎÂˇt) = cosÎ¸ function (Î¸ = ĎÂˇt), and its kinetic energy will be equal toÂ sin2Î¸. Hence, the (instantaneous)Â changeÂ in kinetic energy at any point in time (as a function of the angle Î¸) isÂ equal to:Â d(sin2Î¸)/dÎ¸ = 2âsinÎ¸âd(sinÎ¸)/dÎ¸ = 2âsinÎ¸âcosÎ¸. Now, the motion of theÂ second oscillator (just look at that second piston going up and down in the V-2 engine) is given by theÂ sinÎ¸ function, which is equal to cos(Î¸ â Ď /2). Hence, its kinetic energy is equal toÂ sin2(Î¸ â Ď /2), and how itÂ changesÂ (as a function of Î¸ again) is equal toÂ 2âsin(Î¸ â Ď /2)âcos(Î¸ â Ď /2) =Â = â2âcosÎ¸âsinÎ¸ = â2âsinÎ¸âcosÎ¸. 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… WhatÂ if the relevant energy formula isÂ E =Â mÂˇa2ÂˇĎ2/2 instead ofÂ E =Â mÂˇa2ÂˇĎ2? What are the implications? Well… We get aÂ â2 factor in our formula for the radiusÂ a, as shown below.

Now that isÂ notÂ so nice. For the tangential velocity, we getÂ vÂ =Â aÂˇĎ =Â â2Âˇc. This is alsoÂ notÂ so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precessionâtheÂ wobbling of our flywheel in a magnetic field. Remember we may think of Jzâthe angular momentum or, to be precise, its component in theÂ z-direction (the direction in which weÂ measureÂ itâas the projection of theÂ realÂ angular momentumÂ 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 (Î¸) betweenÂ JzÂ andÂ 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 theÂ magnitudeÂ of theÂ presumedÂ actualÂ momentum:In this particular case (spin-1/2 particles),Â j is equal to 1/2 (in units ofÂ Ä§, of course). Hence,Â JÂ is equal toÂ â0.75Â â 0.866. Elementary geometry then tells us cos(Î¸) =Â (1/2)/â(3/4) =Â  = 1/â3. Hence,Â Î¸Â â 54.73561Â°. That’s a big angleâlarger than the 45Â° angle we had secretly expected because… Well… The 45Â° angle has thatÂ â2 factor in it:Â cos(45Â°) =Â sin(45Â°) = 1/â2.

Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? đ 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 orÂ â2 problem, right? đ

Note: If you’re into quantum math, you’ll noteÂ aÂ =Â Ä§/(mÂˇc) is theÂ reducedÂ Compton scattering radius. The standard Compton scattering radius is equal toÂ aÂˇ2ĎÂ = (2ĎÂˇÄ§)/(mÂˇc) =Â  h/(mÂˇc) = h/(mÂˇc). It doesn’t solve theÂ â2 problem. Sorry. The form factor problem. đ

To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of twoÂ circularÂ oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation â around any axis â will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensionalÂ 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. đThey are oscillations, still, so I am not thinking ofÂ twoÂ 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. đ