Of course, it’s not because I am done with Feynman’s Lectures, that I am done with physics. I want to move on to more advanced – or funnier š – topics now. Check my new blog. Onwards! š

# Looking back…

Well… I think this is it, folks ! With my last posts on superconductivity, I think I am done. I’ve gone through all of theĀ *Lectures*Ā and it’s been a amazing adventure.

Looking back at it, I’d say: there is really no substitute for buying theseĀ *Lectures*Ā yourself, and just grind through it. The only thing this blog really does is, perhaps, raise a question here and there – or help with figuring something out. But then… Well… If I can do it, you can do it. Don’t go for other sources if you can go for the original writings ! Read a classic rather than yet another second-hand or half-cooked thing !

I should also note that I started off using the print copy of Feynman’s *Lectures* but, at this point, I realize I should really acknowledge the incredible effort of two extraordinary people: Michael Gottlieb and Rudolf Pfeiffer, who have worked for decades to get thoseĀ *Lectures online*. I borrowed a lot of stuff from it. In fact, in the coming weeks and months, I want to make sure I duly acknowledge that for all of the illustrations and quotes I’ve used, and if I haven’t been paraphrasing a bit too much, but… Well… That will be quite an effort. These two extraordinary guys alsoĀ created a website for theseĀ *Lectures*Ā which offers many more resources. That makes it accessible to all and everyone.

However, let me repeat: there is no substitute for buying the *Lectures*Ā yourself, and grinding through it yourself. I wish you all the best on this journey. It’s been a nice journey for me, and I am therefore pretty sure you’ll enjoy it at least as much as I did.

Jean Louis Van Belle, 26 February 2018

**Post scriptum**: The material I have copied and republished from this wonderful online edition of Gottlieb and Pfeiffer is under copyright. The site mentions that, without explicit permission, only some limited copying is permitted under Fair UseĀ laws, for non-commercial publications (which this blog surely is), and with proper attribution. I realize that, despite my best efforts to provide hyperlinks to theĀ *LecturesĀ *themselves whenever I’d borrow from them, I should probably go through it all to make sure that’s effectively the case. If I have been lacking in this regard, it was surely not intentional.

# Superconductivity and flux quantization

This post continues my mini-series on Feynman’sĀ *Seminar on Superconductivity*. Superconductivity is a state which produces many wondrous phenomena, but… Well… The flux quantization phenomenon may not be part of your regular YouTube feed but, as far as I am concerned, itĀ may well beĀ *theĀ *most amazing manifestation of aĀ quantum-mechanical phenomenon at aĀ *macroscopicĀ *scale. I mean… Super currents that keep going, with zero resistance, are weirdāthey explain how we can *trapĀ *a magnetic flux in the first placeābut the fact that such fluxes are *quantized* is even weirder.

The key idea is the following. When we cool a ring-shaped piece of superconducting material in a magnetic field, all the way down to the critical temperature that causes the electrons to condense into a superconducting fluid, then a super current will emergeāthink of an eddy current, here, but with zero resistanceāthat will force the magnetic fieldĀ out*Ā *of the material, as shown below. This current will permanently trap some of the magnetic field, even when the external field is being removed.Ā As said, that’s weird enough by itself but… Well… If we think of the super current as an eddy current encountering zero resistance, then the idea of a permanently trapped magnetic field makes sense, right? In case you’d doubt the effect… Well… Just watch one of the many videos on the effect on YouTube. š The amazing thing here is *notĀ *the permanently trapped magnetic field, but the fact that it’s *quantized*.

To be precise, the trapped flux will always be an integer times 2ĻÄ§/q. In other words, the magnetic field which Feynman denotes by Ī¦ (the *capitalized*Ā Greek letter*Ā *phi), will always be equal to:

Ī¦ =Ā *nĀ·*2ĻÄ§/q, withĀ *nĀ *= 0, 1, 2, 3,…

Hence, the flux can be 0,Ā 2ĻÄ§/q, 4ĻÄ§/q, 6ĻÄ§/q , and so on. The fact that it’s a multiple of 2Ļ shows us it’s got to do with the fact that our piece of material is, effectively, a ring. The nice thing about this phenomenon is that the mathematical analysis is, in fact, fairly easy to followāor… Well… Much easier than what we discussed before. š Let’s quickly go through it.

We have a formula for the magnetic flux. It must be equal to the line integral of the vector potential (**A**) around a closed loopĀ Ī¤, so we write:

Now, we can *choose* the loopĀ Ī¤ to be well inside the body of the ring, so that it never gets near the surface, as illustrated below. So we know that the currentĀ ** J** is zero there. [In case you doubt this, see my previous post.]

One of the equations we introduced in our previous post,Ā Ä§**ā***Īø*Ā = mĀ·** v**Ā + qĀ·

**A**, will then reduce to:

Ä§**ā***Īø*Ā = qĀ·**A**

Why? TheĀ ** v **in theĀ mĀ·

**Ā term (the velocity of the superconducting fluid, really), is zero. Remember the analysis is for this particular loop (well inside the ring) only. So… Well… If we integrate the expression above, we get:**

*v*Combining the two expressions with the integrals, we get:

Now, the lineĀ integral of a gradient from one point to another (say from point 1 to point 2)Ā is the difference of the values of the function at the two points, so we can write:

Now what *constraints*Ā are there on the values of Īø_{1}Ā andĀ Īø_{2}? Well… You might think that, if they’re associated with the same point (we’re talking a closed loop, right?), then the two values should be the same, but… Well… No. All we can say is that theĀ *wavefunctionĀ *must have the same value. We wrote that wavefunction as:

Ļ =Ā Ļ(** r**)

^{1/2}

*e*

^{Īø(r)}

TheĀ *valueĀ *of this function at some pointĀ * rĀ *is the same ifĀ ĪøĀ

*changesĀ*by

*nĀ·*2Ļ. Hence, when doing one

*complete*turn around the ring, theĀ ā«ā

*Īø*Ā·d

**Ā integral in the integral formulas we wrote down must be equal toĀ**

*s**nĀ·*2Ļ. Therefore, the second integral expression above can be re-written as:

That’s the result we wanted to explain so… Well… We’re done. Let me wrap up by quoting Feynman’s account of the 1961 experiment which confirmed London’s *prediction* of the effect, which goes back to 1950! It’s interesting, because… Well… It shows howĀ *up to dateĀ *Feynman’s Lectures really areāor *were*, back in 1963, at least!

# Feynman’s Seminar on Superconductivity (2)

We didn’t get very far in our first post on Feynman’s Seminar on Superconductivity, and then I shifted my attention to other subjects over the past few months. So… Well… Let me re-visit the topic here.

One of the difficulties one encounters when trying to read this so-called *seminar*āwhich, according to Feynman, is ‘for entertainment only’ and, therefore, not really part of the *LecturesĀ *themselvesāis that Feynman throws in a lot of stuff that is not all thatĀ relevant to the topic itself but… Well… He apparently didn’t manage to throw all that he wanted to throw into his (other)Ā *LecturesĀ on Quantum Mechanics*Ā and so he inserted a lot of stuff which he could, perhaps, have discussed elsewhere. So let us try to re-construct the main lines of reasoning here.

The first equation isĀ SchrĆ¶dinger’s equation for some particle with chargeĀ *q*Ā that is moving in an electromagnetic field that is characterized not only by the (scalar) potentialĀ Ī¦ but also by a vector potentialĀ *A*:

This closely resemblesĀ SchrĆ¶dinger’s equation for an electron that is moving in anĀ *electricĀ *field only, which we used to find the energy states of electrons in a hydrogen atom:Ā *i*Ā·Ä§Ā·āĻ/āt = ā(1/2)Ā·(Ä§^{2}/m)ā^{2}Ļ + VĀ·Ļ. We just need to note the following:

- On the left-hand side, we can, obviously, replaceĀ ā1/
*i*by*i*. - On the right-hand side, we can replace V by qĀ·Ī¦, because the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (Ī¦).
- As for the other term on the right-hand sideāso that’s the ā(1/2)Ā·(Ä§
^{2}/m)ā^{2}Ļ termāwe can re-write āÄ§^{2}Ā·ā^{2}Ļ as [(Ä§/*i*)Ā·ā]Ā·[(Ä§/*i*)Ā·ā]Ļ because (1/*i*)Ā·(1/*i*) = 1/*i*^{2}Ā = 1/(ā1) = ā1. š - So all that’s left now, is that additional āqĀ·
Ā term in the (Ä§/*A**i*)ā ā qĀ·expression. In our post, we showed that’s easily explained because we’re talking magneto*A*Ā*dynamics*: we’ve got to allow for the possibility ofĀ*changingĀ*magnetic fields, and so that’s what theĀ āqĀ·Ā term captures.*A*

Now, the latter point isĀ *notĀ *so easy to grasp but… Well… I’ll refer you that first post of mine, in which I show that some charge in a *changing* magnetic field will effectively gather someĀ *extraĀ *momentum, whose magnitude will be equal to p = mĀ·*v*Ā =Ā āqĀ·A. So that’s why we need to introduce anotherĀ *momentumĀ *operator here, which we write as:

OK. Next. But… Then… Well… All of what follows are either digressionsālike the section on the local conservation of probabilitiesāor, else, quite intuitive arguments. Indeed, Feynman doesĀ *not *give us the nitty-gritty of the Bardeen-Cooper-Schrieffer theory, nor is the rest of the argument nearly as rigorous as the derivation of the electron orbitals from SchrĆ¶dinger’s equation in an electrostatic field. So let us closely stick to what heĀ *doesĀ *write, and try our best to follow the arguments.

### Cooper pairs

The key assumption is that there is some *attraction* between electrons which, at low enough temperatures, can overcome the Coulomb repulsion. Where does this attraction come from? Feynman does not give us any clues here. He just makes a reference to the BCS theory but notes this theory is “not the subject of this seminar”, and that we should just “accept the idea that the electrons do, in some manner or other, work in pairs”, and that “we can think of thosāe pairs as behaving more or less like particles”, and that “we can, therefore, talk about the wavefunction for a pair.”

So we have a new particle, so to speak, which consists of two electrons who move through the conductor as one. To be precise, the electron pair behaves as a boson. Now, bosons have integer spin. According to the spin addition rule, we have four possibilities here but only three possible values:ā 1/2 + 1/2 = 1;Ā ā1/2 + 1/2 = 0; +1/2 ā 1/2 = 0; ā1/2 ā 1/2 = ā 1. Of course, it is tempting to think these Cooper pairs are just like the electron pairs in the atomic orbitals, whose spin is always opposite because of the Fermi exclusion principle. Feynman doesn’t say anything about this, but the Wikipedia article on the BCS theory notes that the two electrons in a Cooper pair are, effectively, correlated because of their opposite spin. Hence, we must assume the Cooper pairs effectively behave like spin-zero particles.

Now, unlike fermions, bosons can collectively share the same energy state. In fact, they areĀ *likelyĀ *to share the same state into what is referred to as a Bose-Einstein condensate. As Feynman puts it: “Since electron pairs are bosons, when there are a lot of them in a given state there is an especially large amplitude for other pairs to go to the same state. So nearly all of the pairs will be locked down at the lowest energy inĀ *exactly the same state*āit wonāt be easy to get one of them into another state. Thereās more amplitude to go into the same state than into an unoccupied state by the famous factorĀ ān, where nā1 is theĀ occupancy of the lowest state. So we would expect all the pairs to be moving in the same state.”

Of course, this only happens at very low temperatures, because even if the thermal energy is very low, it will give the electrons sufficient energy to ensure the attractive force is overcome and all pairs are broken up. It is only atĀ *veryĀ *low temperature that they will pair up and go into a Bose-Einstein condensate. Now, Feynman derives thisĀ ānĀ factor in a rather abstruse introductoryĀ *LectureĀ *in the third volume, and I’d advise you to *google* other material on Bose-Einstein statistics because… Well… The mentioned *Lecture* is not among Feynman’s finest. OK. Next step.

### Cooper pairs and wavefunctions

We know the *probability* of finding a Cooper pair is equal to the absolute square of its wavefunction. Now, it isĀ *veryĀ *reasonable to assume that this probability will be proportional to the charge density (Ļ), so we can write:

|Ļ|^{2Ā }=Ā ĻĻ*Ā ā¼Ā Ļ(** r**)

The argument here (* r*)

*Ā*is just the position vector. The next step, then, is to write Ļ as the

*square rootĀ*ofĀ Ļ(

**) times some phase factorĀ Īø. Abstracting away from time, this phase factor will also depend onĀ**

*r***, of course. So this is what Feynman writes:**

*r*Ļ =Ā Ļ(** r**)

^{1/2}

*e*

^{Īø(r)}

As Feynman notes, we can write any complex function ofĀ * rĀ *like this but… Well… The charge density is, obviously, somethingĀ

*real*. Something we canĀ

*measure*, so we’re

*notĀ*writing the obvious here. The next step is even less obvious.

In our first post, we spent quite some time on Feynman’s digression on the local conservation of probability and… Well… I wrote above I didn’t think this digression was very useful. It now turns out it’s a central piece in the puzzle that Feynman is trying to solve for us here. The key formula here is the one for the so-calledĀ **probabilityĀ *** current*, whichāas Feynman showsāwe write as:

This current ** J**Ā can also be written as:

Now, Feynman skips all of the math here (he notes “it’s just a change of variables” but so he doesn’t want to go through all of the algebra), and so I’ll just believe him when he says that, when substitutingĀ Ļ for our wavefunction Ļ =Ā Ļ(** r**)

^{1/2}

*e*

^{Īø(r)}, then we can express this ‘current’ (

**) in terms ofĀ Ļ and Īø. To be precise, he writesĀ**

*J**as:Ā So what? Well… It’s really fascinating to see what happens next. WhileĀ*

**J**Ā*was some rather abstract concept so farāwhat’s a probability current,*

**J**Ā*really*?āFeynman now suggests we may want to think of it as a very classical

*electricĀ*currentāthe charge density times the velocity of the

*fluidĀ*of electrons. Hence, we equateĀ

*toĀ*

**J**Ā*= Ā ĻĀ·*

**J**Ā**. Now, if the equation above holds true, butĀ**

*v**Ā is also equal toĀ*

**J***= ĻĀ·*

**J**Ā**, then the equation above is equivalent to:**

*v*Now, that gives us a formula forĀ Ä§**ā**Īø. We write:

Ä§**ā***Īø* = mĀ·** v**Ā + qĀ·

**A**

Now, in my previous post on this *Seminar*, I noted that Feynman attaches a lot of importance to this mĀ·** v**Ā + qĀ·

**A**Ā quantity because… Well… It’s actually an

*invariantĀ*quantity. The argument can be, very briefly, summarized as follows. During the

*build-up*Ā of (or a change in) a magnetic flux, a charge will pick up some (classical)Ā

*momentum*Ā that is equal toĀ

**p**Ā =Ā mĀ·

**Ā =Ā āqĀ·**

*v***A**. Hence, theĀ mĀ·

**Ā + qĀ·**

*v***A**Ā sum is zero, and so… Well… That’s it, really: it’s some quantity that… Well… It has a significance in quantum mechanics. What significance? Well… Think of what we’ve been writing here. TheĀ

*and the*

**v**Ā**A**have aĀ

*physicalĀ*significance, obviously. Therefore, that phase factorĀ

*Īø*(

**) must also have a physical significance.**

*r*But the question remains: * what physical significance, exactly?* Well… Let me quote Feynman here:

“The phase is just as observable as the charge density Ļ. It is aĀ piece of the current densityĀ ** J**. TheĀ

*absolute*Ā phase (

*Īø*) is not observable, but if the gradient of the phase (

**ā**

*Īø*) is known everywhere, then the phase is known except for a constant. You can define the phase at one point, and then the phase everywhere is determined.”

That makes sense, doesn’t it? But it still doesn’t quite answer the question: whatĀ *isĀ *the physical significance ofĀ *Īø*(** r**). WhatĀ

*isĀ*it,

*really*? We may be able to answer that question after exploring the equations above a bit more, so let’s do that now.

### Superconductivity

The phenomenon of superconductivity itself is easily explained by the mentionedĀ *condensationĀ *of the Cooper pairs: they all go into the sameĀ *energyĀ *state. They form, effectively, a superconductingĀ *fluid*. Feynman’s description of this is as follows:

“There is no electrical resistance. Thereās no resistance because all the electrons are collectively in the same state. In the ordinary flow of current you knock one electron or the other out of the regular flow, gradually deteriorating the general momentum. But here to get one electron away from what all the others are doing is very hard because of the tendency of all Bose particles to go in the same state. A current once started, just keeps on going forever.”

Frankly, I’ve re-read this a couple of times, but I don’t think it’s the best description of what we think is going on here. I’d rather compare the situation to… Well… Electrons moving around in an electron orbital. That’s doesn’t involve any radiation or energy transfer either. There’s just movement. Flow. The kind of flow we have in the wavefunction itself. Here I think the video on Bose-Einstein condensates on the FrenchĀ *Tout est quantiqueĀ *site is quite instructive: all of the Cooper pairs join to become one giant wavefunctionāone superconductingĀ *fluid*, really. š

OK… Next.

### The Meissner effect

Feynman describes the Meissner effect as follows:

“If you have a piece of metal in the superconducting state and turn on a magnetic field which isnāt too strong (we wonāt go into the details of how strong), the magnetic field canāt penetrate the metal. If, as you build up the magnetic field, any of it were to build up inside the metal, there would be a rate of change of flux which would produce an electric field, and an electric field would immediately generate a current which, by Lenzās law, would oppose the flux. Since all the electrons will move together, an infinitesimal electric field will generate enough current to oppose completely any applied magnetic field. So if you turn the field on after youāve cooled a metal to the superconducting state, it will be excluded.

Even more interesting is a related phenomenon discovered experimentally by Meissner.Ā If you have a piece of the metal at a high temperature (so that it is a normal conductor) and establish a magnetic field through it, and then you lower the temperature below the critical temperature (where the metal becomes a superconductor),Ā *the field is expelled*. In other words, it starts up its own currentāand in just the right amount to push the field out.”

The math here is interesting. Feynman first notes that, in any lump of superconducting metal, the *divergence* of the current must be zero, so we write:Ā Ā āĀ·** J**Ā = 0. At any point? Yes. The current that goes in must go out. No point is a sink or a source. Now the divergence operator (āĀ·

**) is a linear operator. Hence, that means that, when applying the divergence operator to theĀ**

*J**=Ā (Ä§/m)Ā·[*

**J**Ā**ā**

*Īø*ā (q/Ä§)Ā·

**A**]Ā·Ļ equation, we’ll need to figure out whatĀ āĀ·

**ā**

*Īø*=Ā Ā =Ā ā

^{2}

*Īø*andĀ āĀ·

**A**Ā are. Now, as explained in my post on gauges, we canĀ

*chooseĀ*to make āĀ·

**A**Ā equal to zero so… Well… We’ll make that choice and, hence, the term withĀ āĀ·

**A**Ā in it vanishes. So… Well… IfĀ āĀ·

**Ā equals zero, then the term withĀ ā**

*J*^{2}

*Īø*Ā has to be zero as well, soĀ ā

^{2}

*Īø*Ā has to be zero. That, in turn, impliesĀ

**ā**

*Īø*has to be some constant (vector).

Now, there is a pretty big error in Feynman’s *Lecture*Ā here, as it notes: “Now the only way that ā^{2}*Īø*Ā can be zero everywhere inside the lump of metal is forĀ *ĪøĀ *to be a constant.” It should read:Ā ā^{2}*Īø*Ā can only be zero everywhere if **ā***Īø *is a constant (vector). So now we need to remind ourselves of theĀ *realityĀ *ofĀ *Īø*, as described by Feynman (quoted above): “TheĀ *absolute*Ā phase (*Īø*) is not observable, but if the gradient of the phase (**ā***Īø*) is known everywhere, then the phase is known except for a constant. You can define the phase at one point, and then the phase everywhere is determined.” So we can define, or *choose*, our constant (vector)Ā **ā***Īø *to beĀ **0**.

Hmm… We re-set not one but *twoĀ *gauges here: **A** andĀ **ā***Īø*. Tricky business, but let’s go along with it. [If we want to understand Feynman’s argument, then we*Ā *actuallyĀ have no choice than to go long with his argument, right?] The point is: theĀ (Ä§/m)Ā·**ā***Īø *term in theĀ * JĀ *=Ā (Ä§/m)Ā·[

**ā**

*Īø*ā (q/Ä§)Ā·

**A**]Ā·Ļ vanishes, so the equation we’re left with tells us theĀ currentāso that’s an actual as well as a

*probability*current!āis proportional to the vector potential:

Now, we’ve neglected any possible variation in the charge densityĀ Ļ so far because… Well… The charge density in a superconducting fluid must be uniform, right?Ā Why? When the metal is superconducting, an accumulation of electrons in one region would be immediately neutralized by a current, right? [Note that Feynman’s language is more careful here. He writes: the charge density isĀ *almostĀ *perfectly uniform.]

So what’s next? Well… We have a more general equation from the equations of electromagnetism:

[In case you’d want to know how we get this equation out of Maxwell’s equations, you can look it up online in one of the many standard textbooks on electromagnetism.] You recognize this as a Poisson equation… Well… Three Poisson equations: one for each component of **A** andĀ ** J**. We can now combine the two equations above by substitutingĀ

*JĀ*in that Poisson equation, so we get the following differential equation, which we need to solve forĀ

**:**

*A*TheĀ Ī»^{2}Ā in this equation is, of course, a shorthand for the following constant:

Now, it’s very easy to see that both *e*^{āĪ»r}Ā as well asĀ *e*^{āĪ»r}Ā are solutions for that Poisson equation. But what do they mean? In one dimension,Ā * rĀ *becomes the one-dimensional position variableĀ

*x*. You can check the shapes of these solutions with a graphing tool.

Note that only one half of each graph counts: the vector potential mustĀ *decreaseĀ *when we go from the surface into the material, and there is a cut-off at the surface of the material itself, of course. So all depends on the size ofĀ Ī», as compared to the size of our piece of superconducting metal (or whatever other substance our piece is made of). In fact, if we look atĀ *e*^{āĪ»x}Ā as as an exponential decay function, thenĀ Ļ = 1/Ī» is the so-called scaling constant (it’s the inverse of the decay constant, which is Ī» itself). [You can work this out yourself. Note that forĀ *xĀ *=Ā Ļ = 1/Ī», the value of our functionĀ *e*^{āĪ»x}Ā will be equal toĀ *e*^{āĪ»(1/Ī»)}Ā =Ā *e*^{ā1}Ā ā 0.368, so it means the value of our function is reduced to about 36.8% of its initial value. For all practical purposes, we may sayāas Feynman notesāthat the field will, effectively, only penetrate to a thin layer at the surface: a layer of about 1/1/Ī» in thickness. He illustrates this as follows:

Moreover, he calculates the 1/Ī» distance forĀ *lead*. Let me copy him here:

Well… That says it all, right? We’re talking twoĀ *millionthsĀ *of a centimeter here… š

So what’s left? A lot, like flux quantization, or theĀ *equations of motionĀ *for the superconducting electron fluid. But we’ll leave that for the next posts. š

# Wavefunctions and the twin paradox

My previous post was *awfully* long, so I must assume many of my readers may have started to read it, but… Well… Gave up halfway or even sooner. š I added a footnote, though, which is interesting to reflect upon. Also, I know many of my readers aren’t interested in the mathāeven if they understand one cannot really appreciate quantum theory without the math. But… Yes. I may have left some readers behind. Let me, therefore, pick up the most interesting bit of all of the stories in my last posts in as easy a language as I can find.

We have that weird 360/720Ā° symmetry in quantum physics orāto be preciseāwe have it for elementary matter-particles (think of electrons, for example). In order to, hopefully, help you understand what it’s all about, I had to explain the often-confused but substantially different concepts of aĀ *reference frameĀ *and a *representational baseĀ *(or representationĀ *tout court*). I won’t repeat that explanation, but think of the following.

If we just rotate the *reference frame* over 360Ā°, we’re just using the same reference frame and so we see the same thing: some object which we, vaguely, describe by someĀ *e*^{i}^{Ā·Īø}Ā function. Think of some spinning object. In its own reference frame, it will just spin around some center or, in ours, it will spin while moving along some axis in its own reference frame or, seen from ours, as moving in some direction while it’s spinningāas illustrated below.

To be precise, I should say that we describe it by some *Fourier* sum of such functions. Now, if its spin direction is… Well… In the other direction, then we’ll describe it by by someĀ *e*^{āi}^{Ā·Īø}Ā function (again, you should read: aĀ *FourierĀ *sum of such functions). Now, the weird thing is is the following: if we rotate *the object itself*, over the sameĀ 360Ā°, we get aĀ *differentĀ *object: ourĀ *e*^{i}^{Ā·Īø}Ā andĀ *e*^{āi}^{Ā·Īø}Ā function (again: think of aĀ *FourierĀ *sum, so that’s a waveĀ *packet*, really) becomes aĀ ā*e*^{Ā±i}^{Ā·Īø}Ā thing. We get aĀ *minusĀ *sign in front of it.Ā So what happened here? What’s the difference, *really*?

Well… I don’t know. It’s very deep. Think of you and me as two electrons who are watching each other. If I do nothing, and you keep watching me *while turning around me*, for a fullĀ 360Ā° (so that’s a rotation of your reference frame over 360Ā°), then you’ll end up where you were when you started and, importantly, you’ll see the same thing: *me*. š I mean… You’ll seeĀ *exactlyĀ *the same thing: if I was anĀ *e*^{+i}^{Ā·Īø}Ā wave packet, I am still anĀ anĀ *e*^{+i}^{Ā·Īø}Ā wave packet now. OrĀ if I was an *e*^{āi}^{Ā·Īø}Ā wave packet, then I am still anĀ an *e*^{āi}^{Ā·Īø}Ā wave packet now. Easy. Logical. *Obvious*, right?

But so now we try something different:Ā *IĀ *turn around, over a fullĀ 360Ā° turn, and *youĀ *stay where you are and watch *meĀ *while I am turning around. What happens? Classically, nothing should happen but… Well… This is the weird world of quantum mechanics: when I am back where I wasālooking at you again, so to speakāthen… Well… I am not quite the same any more. Or… Well… Perhaps I am but youĀ *seeĀ *me differently. If I wasĀ *e*^{+i}^{Ā·Īø}Ā wave packet, then I’ve become aĀ ā*e*^{+i}^{Ā·Īø}Ā wave packet now.

Not *hugely* different but… Well… ThatĀ *minusĀ *sign matters, right? OrĀ If I wasĀ wave packet built up from elementaryĀ *a*Ā·*e*^{āi}^{Ā·Īø}Ā waves, then I’ve become aĀ ā*e*^{āi}^{Ā·Īø}Ā wave packet now. What happened?

It makes me think of the twin paradox in special relativity. We know it’s aĀ *paradox*āso that’s anĀ *apparentĀ *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 someĀ *force*.

Can we relate this to the twin paradox? Maybe. Note that aĀ *minusĀ *sign in front of theĀ *e*^{āĀ±i}^{Ā·Īø}Ā 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 180Ā°: ā*cos*Īø =Ā *cos*(Īø Ā± Ļ) andĀ ā*sin*Īø =Ā *sin*(Īø Ā± Ļ). Now, adding or subtracting aĀ *commonĀ *phase factor to/from the argument of the wavefunction amounts toĀ *changingĀ *the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360Ā° and 720Ā° symmetries are, effectively, related. š

**Post scriptum**:Ā *GoogleĀ *honors Max Born’s 135th birthday today. š I think that’s a great coincidence in light of the stuff I’ve been writing about lately (possible interpretations of the wavefunction). š

# Wavefunctions, perspectives, reference frames, representations and symmetries

Ouff ! This title is quite a mouthful, isn’t it? š 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 (orĀ *physical*) interpretations of the wavefunction is the fact that a wavefunction does not seem to obey the classical 360Ā° 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 (orĀ representation, *tout court*)Ā 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 frameāand so that explains the weird stuff (like that 720Ā° 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 paperĀ first.

### The reality of directions

*Huh? *TheĀ *realityĀ *of directions? Yes. I warned you. This post may cause brain damage. šĀ The whole argument revolves around a *thoughtĀ *experimentābut one whose results have been verified in zillions of experiments in university student labs so… Well… We do *notĀ *doubt the results and, therefore, we do not doubt the basic mathematical results: we just want to try to *understandĀ *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 āimprovedā Stern-Gerlach apparatuses. The apparatus basically splits and then re-combines the two new beams along theĀ *z*-axis. It is also possible to block one of the beams, so we filter out only particles with their spinĀ *up*Ā or, alternatively, with their spinĀ *down*. Spin (or angular momentum or the magnetic moment) as measured along theĀ *z*-axis, of courseāI should immediately add: we’re talking **theĀ z-axis of the apparatus** here.

The two situations involve a different *relative *orientation of the apparatuses: in (a), the angle is 0**Ā°**, while in (b) we have a (right-handed) rotation of 90Ā° about the *z*-axis. He then provesāusing geometry and logic onlyāthat the probabilities and, therefore, **the magnitudes of the amplitudes** (denoted byĀ

*C*

_{+}and

*C*

_{ā}and

*Cā*

_{+}and

*Cā*

_{ā}in the

*S*and

*T*representation respectively)

**must be the same, but the amplitudes**, notingāin his typical style, mixing academic and colloquial languageāthat āthere must be some way for a particle to tell that it has turned a corner in (b).ā

*must*have different phasesThe various interpretations of what actually *happens* here may shed some light on the heated discussions on the *reality *of the wavefunctionāand 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 wavefunctionāwhich captures a continuum of possible states, so to speakāis introduced only later. However, we may look at the amplitude for a particle to be in theĀ *up*– or *down*-state as a wavefunction and, therefore (but do note that’s my humble opinion once more), the analysis is actuallyĀ *notĀ *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 (90Ā°), we *know* that the amplitudes are given by *Cā*_{+} = *e ^{i}*

^{āĻ/2}ā

*C*

_{+}=

*e*

^{ i}^{āĻ/4}ā

*C*

_{+}and

*Cā*

_{ā}=

*e*

^{āiāĻ/2}ā

*C*

_{+}=

*e*

^{ā iāĻ/4}ā

*C*

_{ā}respectively (the amplitude to go from the down to the up state, or vice versa, is zero). Hence, yes,

**ā**

*we**notĀ*the particle, Mr. Feynman!ā

*that, in (b), the electron has, effectively, turned a corner.*

**know**ĀThe 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 ā(a) and (b) are differentā, āthe probabilities are the sameā. 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 turnāso it is just going in some other direction. Thatās 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 mightāor *must*āhave *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 āmightā and āmustā 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…Ā *Wait a minute…*Ā Nothing is obvious in quantum mechanics, right? How can weĀ *experimentally confirmĀ *thatĀ they are different?

* Huh?Ā *I must be joking, right? You canĀ

*seeĀ*they are different, right? No.Ā I am not joking. In physics, two things are different if we get differentĀ

*measurement*Ā 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 thingāsame

*probabilities*, remember?āwhy are they different? Think of this: if we look at the two beam splitters as one singleĀ tube (anĀ

*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 sameĀ

*even when it takes a turn*, we could say the tube is still the same, despite us having wrenched it over a 90Ā° corner.

Now, I am sure you think I’ve just gone nuts, but just try*Ā *to stick with me a little bit longer. Feynman actually acknowledges the same: we need to *experimentallyĀ **proveĀ *(a) and (b) are different. He does so by getting **aĀ thirdĀ apparatus **in

**(**, as shown below,

*U*)**whose**, so there is no difference there.

*relative*orientation to*T*is the same in both (a) and (b)Now, the axis ofĀ *UĀ *is not theĀ *z*-axis: it is theĀ *x*-axis in (a), and theĀ *y*-axis in (b). So what? Well… I will quote Feynman hereānot (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 ofĀ *SĀ *which produces a pure +*x*Ā state. Such particles would be split into +*z* andĀ ā*z* intoĀ beams inĀ *S*,Ā but the two beams would be recombined to give aĀ +*x*Ā state again at P_{1}āthe exit ofĀ *S*.Ā The same thing happens again inĀ *T*.Ā If we followĀ *TĀ *by a third apparatusĀ *U*,Ā whose axis is in the +*x*Ā direction and, as shown in (a), all the particles would go into the +Ā beam ofĀ *U*.Ā Now imagine what happens ifĀ *TĀ *and *UĀ *are swung aroundĀ *together*Ā by 90Ā°Ā to the positions shown in (b).Ā Again, theĀ *TĀ *apparatus puts out just what it takes in, so the particles that enterĀ *UĀ *are in a +*xĀ *stateĀ ** with respect toĀ S**,Ā which is different. By symmetry, we would now expect only one-half of the particles to get through.”

I should note that (b) shows theĀ *UĀ *apparatus wide open so… Well… I must assume that’s a mistake (and should alert the current editors of the *LecturesĀ *to it): Feynman’s narrative tells us we should also imagine it with theĀ *minus *channel shut. InĀ *thatĀ *case, it should, effectively, filter approximately half of the particles out, while they all get through in (a). So that’s aĀ *measurementĀ *result which shows the direction, as weĀ *seeĀ *it, makes a difference.

Now, Feynman would be very angry with meābecause, as mentioned, he hates philosophersābut I’d say: this experiment proves that a direction is something real. Of course, the next philosophical question then is: whatĀ *isĀ *a direction? I could answer this by pointing to the experiment above: a direction is something that alters the probabilities between the *S**T**U* tube as set up in (a) versus the *S**T**U* tube in (b). In factābut, I admit, that would be pretty ridiculousāwe could use the varying probabilities as we wrench this tube over varying angles toĀ *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. š 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 720Ā° 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 frameāorĀ *representation*, as it’s referred to in quantum mechanicsāto another?

Well… I should immediately correct myself here: a reference frame and a representation are two different things. They areĀ *relatedĀ *but… Well… Different… *Quite* different. Not same-same but different. š I’ll explain why later. Let’s go for it.

Before talking representations, let us first think about what we reallyĀ *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 *ourĀ *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 theĀ *shape*, if that sounds easier to you)** of the measurement apparatus**Ā (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 usā230 years agoāthat 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.Ā The apparatus gives us two *directions*:

(1) TheĀ *upĀ *direction, whichĀ *weĀ associate* with theĀ positive direction of theĀ *z*-axis, and

(2) the direction of travel of our particle, whichĀ *we associate*Ā with the positive direction of theĀ *y*-axis.

Now, if we have two axes, then the third axis (theĀ *x*-axis) will be given by the right-hand rule, right? So we may say the apparatus gives us the reference frame. Full stop.Ā So… Well… Everything is relative? Is this reference frame relative? Are directions relative? That’s what you’ve been told, but think about this:Ā relativeĀ *to what?*Ā Here is where the object meets the subject. What’s relative? What’s absolute?Ā Frankly, I’ve started to think that, in this particular situation, we should, perhaps, not use these two terms. I am *notĀ *saying thatĀ 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 theĀ *realityĀ *that we are observing, right? And the third one… Well… That’s given by our perspectiveāby our right-hand rule, which is… Well… *OurĀ *right-hand rule.

Of course, now you’ll say: if you think that ārelativeā and āabsoluteā are ambiguous terms and that we, therefore, may want to avoid them a bit more, then ārealā and its opposite (unreal?) are ambiguous terms too, right? Wellā¦ Maybe. What language would *youĀ *suggest? š Just stick to the story for a while. I am not done yet. So… Yes… WhatĀ *isĀ *theirĀ *reality*?Ā 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, *a*symmetries in space), I found myself walking around the table with some asymmetrical everyday objects and papers with arrows and clocks and other stuff on itāeffectively analyzing what right-hand screw, thumb or grip rules actuallyĀ *mean*. š

So… Well… **I want you to distinguishājust for a whileābetween the notion of a reference frame (think of the x–y–z reference frame that comes with the apparatus) and yourĀ 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 sideāwhich, if you think about it, you can only

*defineĀ*in terms of the various positive and negative directions of the various axes. šĀ If you think this is getting ridiculous… Well… Don’t. Feynman himselfĀ 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 theĀ

*axesĀ*(i.e. the reference frame) that comes with it. You don’t believe me? This is theĀ

*very*first illustration of hisĀ

*LectureĀ*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*Ā *y*-directionāso thatās the direction in which our particle is movingāthen we might imagine how it would look like whenĀ *weĀ *would make a 180Ā°Ā 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 comingĀ *towardsĀ *us, out of the apparatus.

What happensābut that’s not scientific language, of courseāis that left becomes right, and right becomes left. Top still is top, and bottom is bottom. We are looking now in theĀ *negativeĀ y*-direction, and the positive direction of the *x*-axisāwhich pointed right when we were looking in the positiveĀ *y*-directionānow points left. I see you nodding your head nowābecause you’ve heard about parity inversions, mirror symmetries and what have youāand 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 theĀ 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 toĀ *this *story, which is about transformations of amplitudes (or wavefunctions). [If you really want to knowābut I know this sounds counterintuitiveāthe 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*Ā 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 onlyĀ *apparent*. It’s a common misconception, and… Well… I’ll let you figure this out yourself. I need to move on.]Ā Just note the following:

- TheĀ
*xyz*Ā 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. OurĀ*perspectiveĀ*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)āa cosine and sine function respectivelyāthen our change in perspectiveĀ
*might*, effectively, mess up our convention for measuring angles.

I am not saying itĀ *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 *counter*clockwise, right? As shown below… But… Well… If the thing below would be some funny clock going backwardsāyou’ve surely seen them in a bar or so, right?āthen… Well… If they’d be transparent, and you’d go around them, you’d see them as going… Yes… Clockwise. š [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, *ifĀ *we wouldĀ assume this clock represents something realāand, of course, **I am thinking of theĀ elementary wavefunctionĀ e^{i}^{Īø}Ā =Ā cosĪø +Ā iĀ·sinĪø now**āthen… 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…Ā

*Think! What’s your answer? Give me the formula!Ā*š

[…]

We’d see it asĀ *e*^{āi}^{Īø}Ā =Ā *cos*(āĪø) +Ā *i*Ā·*sin*(āĪø) =Ā *cos*Īø āĀ *i*Ā·*sin*Īø, right? The hand of our clock now goes clockwise, so that’s theĀ *oppositeĀ *direction of our convention for measuring angles. Hence, instead ofĀ *e*^{i}^{Īø}, we writeĀ *e*^{āi}^{Īø}, right? So that’s the complex conjugate. So we’ve got a differentĀ *imageĀ *of the same thing here. *Not* good. *Not good at all.*

You’ll say: *so what? *We can fix this thing easily, right?Ā YouĀ don’t need the convention for measuring angles or for the imaginary unit (*i*) here.Ā 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).Ā *To hell with the counterclockwise convention for measuring angles!*

You are right. WeĀ *couldĀ *use the right-hand rule more consistently. We could, in fact, use it as anĀ *alternativeĀ *convention for measuring angles: we could, effectively, measure them clockwise *or* counterclockwise depending on the direction of our particle.Ā But… Well… The fact is:Ā *we don’t*. We do *not* use that alternative convention when we talk about the wavefunction. Physicists do use theĀ *counterclockwise*Ā convention ** all of the time** and just jot down these complex exponential functions and don’t realize that,Ā

*if they are to represent something real*, ourĀ

*perspective*Ā on the reference frame matters. To put it differently, theĀ

*directionĀ*in which we are looking at things matters! Hence, the direction is

*not…Ā*Well… I am tempted to say…

*NotĀ*relative at all but then… Well… We wanted to avoid that term, right? š

[…]

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 symmetryāorĀ *a*symmetry, 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 forceĀ 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*Ā 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, āwhen you do find the electron some place, the entire charge is there.ā Of course, we cannot prove anything here, but our elegant yet simple derivation of the Compton radius of an electron is… Well… Just nice. š - 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**Hence, unlike an electromagnetic wave, theĀ*compriseĀ*the direction of its linear motion.*planeĀ*of the two-dimensional oscillation (so that’s the polarization plane, really) can*notĀ*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.Ā The direction of the angular momentum (and the magnetic moment) of an electronāor, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is travelingācan*not*Ā be parallel to the direction of motion. On the contrary, it must be *perpendicular*Ā 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. theĀ *planeĀ *of the polarization)Ā has toĀ *compriseĀ *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 explanationācombined with the quantization hypothesisāgoes a long way in explaining this: an object with an angular momentumĀ ** J**Ā and a magnetic momentĀ

**Ā that is**

*Ī¼**not exactly*parallel to some magnetic fieldĀ

**B**, willĀ

*notĀ*line up: it willĀ

*precess*āand, as mentioned, the quantization of angular momentum may well explain the rest.Ā [Well… Maybe… We haveĀ detailed our attempts in this regard in various posts on this (just search for

*spinĀ*orĀ

*angular momentumĀ*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 theĀ *up*-direction *only*, then it will be spinningĀ *clockwise *if its angular momentum is down (so itsĀ *magnetic moment *isĀ *up*). Conversely, it will be spinningĀ *counter*clockwise if its angular momentum isĀ *up*. Let us take theĀ *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. š 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 asāI am not ashamed to admit thisāI 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 someĀ *positiveĀ *angle.Ā Walking around itāin whatever direction, clockwise or counterclockwiseādoesn’t change the counterclockwise direction of our… Well… That weird object that mightājust *mightā*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 wrenchedĀ *e ^{i}*

^{Īø}Ā =Ā

*cos*Īø +Ā

*i*Ā·

*sin*Īø function, right? The

*x-*andĀ

*y*-axesĀ

*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 180Ā° angle so we’re looking in theĀ *negativeĀ *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 theĀ *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 theĀ *y*-axis to point away, rather than towards us. In short, we have a transformation of the reference frame here:Ā *z’* =Ā *z*,Ā *y’* = āĀ *y*, andĀ *x’* =Ā āĀ *x*. Mind you, this is still a regular right-handed reference frame. [That’s the difference with aĀ *mirrorĀ *image: aĀ *mirroredĀ *right-hand reference frame is no longer right-handed.]Ā So, in our new reference frame, that we choose to coincide with ourĀ *perspective*,Ā we will now describe the same thing as someĀ ā*cos*Īø āĀ *i*Ā·*sin*Īø =Ā ā*e ^{i}*

^{Īø}Ā function. Of course,Ā ā

*cos*Īø =Ā

*cos*(Īø +Ā Ļ) andĀ ā

*sin*Īø =Ā

*sin*(Īø +Ā Ļ) so we can write this as:

ā*cos*Īø āĀ *i*Ā·*sin*Īø =Ā *cos*(Īø +Ā Ļ) +Ā *i*Ā·*sin*Īø =Ā *e ^{i}*

^{Ā·(}

^{Īø+Ļ)}Ā =Ā

*e*

^{i}^{Ļ}Ā·

*e*

^{i}^{Īø}Ā = ā

*e*

^{i}^{Īø}.

Sweet ! But… Well… First note this isĀ *notĀ *the complex conjugate:Ā *e*^{āi}^{Īø}Ā =Ā *cos*Īø āĀ *i*Ā·*sin*ĪøĀ ā Ā ā*cos*Īø āĀ *i*Ā·*sin*Īø =Ā ā*e ^{i}*

^{Īø}. 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. š And, yes, let me lighten up the discussion with that painting here. š We need to haveĀ

*someĀ*fun while torturing our brain, right?

So, because we assume the plane of polarization is different, we get anĀ ā*e ^{i}*

^{Īø}Ā function instead of aĀ

*e*

^{āi}

^{Īø}Ā function.

Let us now think about the *e ^{i}*

^{Ā·(}

^{Īø+Ļ)}Ā function. It’s the same asĀ ā

*e*

^{i}^{Īø}Ā but… Well… We walked around theĀ

*z*-axis taking a full 180Ā° turn, right? So that’s Ļ in radians. So that’s the

*phase shiftĀ*here.

*Hey!Ā*Try the following now. Go back and walk around the apparatus once more, but letĀ 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 theĀ

*z*-axis (which points out of this page,

*toward*you, as you are looking at this), let’s say by some small angleĀ Ī±. So we rotate the reference frame about theĀ

*z*-axis byĀ Ī± and… Well… Of course, ourĀ

*e*

^{i}^{Ā·}

^{Īø}Ā now becomes anĀ ourĀ

*e*

^{i}^{Ā·(}

^{Īø+Ī±)}Ā function, right? We’ve just derived the transformation coefficient for a rotation about theĀ

*z*-axis, didn’t we? It’s equal toĀ

*e*

^{i}^{Ā·}

^{Ī±}, right? We get the transformed wavefunction in the new reference frame by multiplying the old one byĀ

*e*

^{i}^{Ā·}

^{Ī±}, right? It’s equal toĀ

*e*

^{i}^{Ā·}

^{Ī±}Ā·

*e*

^{i}^{Ā·}

^{Īø}Ā =Ā

*e*

^{i}^{Ā·(}

^{Īø+Ī±)}, right?

Well…

[…]

No. The answer is: no. TheĀ transformation coefficient is notĀ *e ^{i}*

^{Ā·}

^{Ī±}Ā butĀ

*e*

^{i}^{Ā·}

^{Ī±/2}. So we get an additional 1/2 factor in theĀ

*phase shift*.

* Huh?Ā *Yes.Ā That’s what it is: when we change the representation, by rotating our apparatus over some angle Ī± 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 onlyĀ

*half*ofĀ the rotation angle only.

** Huh?Ā **Yes. It’s even weirder than that. For a spin

*downĀ*electron, the transformation coefficient is

*e*

^{āiĀ·}

^{Ī±/2}, so we get an additional minus sign in the argument.

* Huh?Ā *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…Ā *Hey! Wait a minute! That’s it, right?Ā *

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:

**OurĀ e^{i}^{Ā·}^{Ī±}Ā coefficient describes a rotation of the reference frame. In contrast, theĀ e^{i}^{Ā·}^{Ī±/2}Ā andĀ e^{āiĀ·}^{Ī±/2}Ā coefficients describe what happens when we rotate the T apparatus! Now thatĀ is a very different proposition.Ā **

Right! You got it! *Representations*Ā and reference frames are different things.Ā *QuiteĀ *different, I’d say: representations areĀ *real*, reference frames aren’tābut then you don’t like philosophical language, do you? šĀ But think of it. When we just go about theĀ *z*-axis, a full 180Ā°, but we don’t touch thatĀ *T*-apparatus, we don’t changeĀ *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 theĀ *z*-axis, a full 180Ā°, 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 frameāfrom *xyz* to *x’y’z’* to be precise: we doĀ *not *changeĀ the representation.

In contrast, **when we rotate theĀ TĀ apparatus over a full 180Ā°, 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 throughĀ

*S*, and now it goes in the opposite directionā

*relative to the direction it was going in S*, that is.

So what happens,Ā *really*, when weĀ change the *representation*, rather than the reference frame? Well… Let’s think about that. š

### Quantum-mechanical weirdness?

The transformation matrix for the amplitude of a system to be in anĀ *upĀ *orĀ *downĀ *state (and, hence, presumably, for a wavefunction) for a rotation about theĀ *z*-axis is the following one:

Feynman derives this matrix in a rather remarkable intellectualĀ *tour de forceĀ *in the 6th of hisĀ *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 howĀ *IĀ *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? š So… Well… Because heĀ *knows*āfrom experimentāthat the coefficient isĀ *e ^{i}*

^{Ā·}

^{Ī±/2}Ā instead of

*e*

^{i}^{Ā·}

^{Ī±}, he just says the phase shiftāwhich he denotes by Ī»āmust be someĀ

*proportionalĀ*to the angle of rotationāwhich he denotes byĀ Ļ rather than Ī± (so as to avoid confusion with the

*EulerĀ*angleĀ Ī±). So he writes:

Ī» =Ā mĀ·Ļ

Initially, he also tries the obvious thing: m should be one, right? SoĀ Ī» = Ļ, 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:

“SupposeĀ *T*Ā is rotated byĀ 360Ā°; then, clearly, it is right back at zero degrees, and we should haveĀ *Cā*_{+} = *C*_{+}Ā andĀ *Cā*_{ā} =Ā *C*_{ā}Ā or,Ā what is the same thing,Ā *e ^{i}*

^{Ā·mĀ·2Ļ}Ā = 1. We get m =Ā 1. [But no!]Ā

*This argument is wrong!*Ā To see that it is, consider thatĀ

*TĀ*is rotated byĀ 180Ā°. If mĀ were equal to 1, we would have

*Cā*

_{+}=Ā

*e*

^{i}^{Ā·Ļ}

*C*

_{+}Ā = ā

*C*

_{+}Ā and

*Cā*

_{ā}=Ā

*e*

^{ā}

^{i}^{Ā·Ļ}

*C*

_{ā}Ā =Ā ā

*C*

_{ā}. [Feynman works with

*statesĀ*here, instead of the wavefunction of the particle as a whole. I’ll come back to this.] However, this is just theĀ

*original*Ā state all over again.Ā

**Ā amplitudes are just multiplied byĀ ā1Ā which gives back the original physical system. (It is again a case of a**

*Both***phase change.) This means that if the angle betweenĀ**

*common**TĀ*andĀ

*SĀ*is increased to 180Ā°, the system would be indistinguishable from the zero-degree situation, and the particles would again go through the (+)Ā state of theĀ

*UĀ*apparatus. AtĀ 180Ā°, though, the (+)Ā state of theĀ

*UĀ*apparatus is theĀ (ā

*x*)Ā state of the originalĀ

*S*Ā apparatus. So a (+

*x*)Ā state would become aĀ (ā

*x*)Ā state. But we have done nothing toĀ

*change*Ā the original state; the answer is wrong. We cannot haveĀ m = 1.Ā We must have the situation that a rotation byĀ 360Ā°, andĀ

*no smaller angle*Ā reproduces the same physical state. This will happen ifĀ m = 1/2.”

The result, of course, is this weird 720Ā° symmetry. While we get the same *physics* after a 360Ā° rotation of the *T* apparatus, we doĀ *notĀ *get the same amplitudes. We get the opposite (complex) number:Ā *Cā*_{+} =Ā *e ^{i}*

^{Ā·2Ļ/2}

*C*

_{+}Ā = ā

*C*

_{+}Ā and

*Cā*

_{ā}=Ā

*e*

^{ā}

^{i}^{Ā·2Ļ/2}

*C*

_{ā}Ā =Ā ā

*C*

_{ā}. That’s OK, because… Well… It’s aĀ

*commonĀ*phase shift, so it’s just like changing the origin of time. Nothing more. Nothing less. Same physics. Same

*reality.*But… Well…Ā

*Cā*

_{+}ā Ā ā

*C*

_{+}Ā andĀ

*Cā*

_{ā}ā Ā ā

*C*

_{ā}, right? We only get our original amplitudes back if we rotate theĀ

*T*apparatus two times, so that’s by a full 720 degreesāas opposed to the 360Ā° we’d expect.

Now, space is isotropic, right? So this 720Ā° 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* orĀ ā*x*, or *t* and ā*t.Ā *I’ve said this a couple of times already again, and I’ll keep saying it many times more:Ā *NatureĀ *surely can’t be bothered by how we measure stuff, right? In the positive or the negative directionāthat’s just our choice, right?Ā *OurĀ *convention. So… Well… It’s just like thatĀ ā*e ^{i}*

^{Īø}Ā function we got when looking at theĀ

*same*experimental set-up from the other side: ourĀ

*e*

^{i}^{Īø}Ā and ā

*e*

^{i}^{Īø}Ā functions didĀ

*notĀ*describe a different reality. We just changed our perspective. TheĀ

*reference frame*. As such, the reference frame isn’tĀ

*real*. The experimental set-up is. AndāI know I will anger mainstream physicists with thisātheĀ

*representationĀ*is. Yes. Let me say it loud and clear here:

**A different representation describes a different reality. **

In contrast, a different perspectiveāor a different reference frameādoes not.

### Conventions

While you might have had a lot of trouble going through all of the weird stuff above, the point is: it isĀ *notĀ *all that weird. WeĀ *canĀ *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 formulaā*e*^{i}^{Īø}Ā =Ā *cos*Īø +Ā *i*Ā·*sin*Īøāwould, one day, be used to representĀ *something real*: an electron, or any elementary particle, really. If he *wouldĀ *have known, I am sure he would have noted what I am noting here:Ā *NatureĀ *can’t be bothered by our conventions. Hence, ifĀ *e*^{i}^{Īø}Ā represents something real, thenĀ *e*^{āi}^{Īø}Ā 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*āand, if he would have known about circularly polarized waves, probably *agreed* toāthatĀ *alternative *convention for measuring angles: we could, effectively, measure angles clockwise *or* counterclockwise depending on the direction of our particleāas opposed to Euler’s ‘one-size-fits-all’ counterclockwise convention. But so we didĀ *notĀ *adopt that alternative convention because… Well… We want to keep honoring Euler, I guess. š

So… Well… If we’re going to keep honoring Euler by sticking to that ‘one-size-fits-all’ counterclockwise convention, then **I doĀ believe thatĀ e^{i}^{Īø}Ā and e^{āi}^{Īø}Ā represent twoĀ differentĀ 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 *seeĀ *something different when they go through a Stern-Gerlach apparatus. So it’s *not* just some convention toĀ *countĀ *things like 0, 1, 2, etcetera versus 0,Ā ā1,Ā ā2 etcetera. It’s the same story again: different but relatedĀ *mathematicalĀ *notions are (often) related to different but relatedĀ *physicalĀ *possibilities. So… Well… I think that’s what we’ve got here.Ā Think of it. Mainstream quantum math treats all wavefunctions as right-handed but… Well…Ā A particle with *up *spin is a different particle than one withĀ *downĀ *spin, right? And, again,Ā *Nature*Ā surely can*not*Ā be bothered about our convention of measuring phase angles clockwise or counterclockwise, right? So… Well… Kinda obvious, right? š

Let me spell out my conclusions here:

**1.** The angular momentum can be positive or, alternatively, negative: *J* = +Ä§/2 orĀ āÄ§/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* particleāthink of an electron, or whatever other particle you’d think ofācomes in twoĀ *variants*:Ā right-handed and left-handed. They will, therefore,Ā *either* consist of (elementary) right-handed waves or,Ā *else*, (elementary) left-handed waves. An elementary right-handed wave would be written as: Ļ(Īø* _{i}*)Ā

*=*

*e*^{i}^{Īøi}

*Ā = a*Ā·(

_{i}*cos*Īø

*+*

_{i}*iĀ·sin*Īø

*). In contrast,Ā an elementary left-handed wave would be written as: Ļ(Īø*

_{i}*)Ā*

_{i}*=Ā*

*e*^{āi}^{Īøi}

*Ā·(*

*Ā =*a_{i}*cos*Īø

*ā*

_{i}*iĀ·sin*Īø

*).Ā So that’s the complex conjugate.*

_{i}So… Well… Yes, I think complex conjugates are not just someĀ *mathematicalĀ *notion: I believe they represent something real. It’s the usual thing:Ā *NatureĀ *has shown us that (most) mathematical possibilities correspond to *realĀ *physical situations so… Well… Here you go. It is reallyĀ 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 differencesādifferent polarization plane and dimensions and what have youāI’ve already summed those up, so I won’t repeat myself here.]Ā The point is: ifĀ we have two differentĀ *physicalĀ *situations, we’ll want to have two different functions to describe it. Think of it like this: why would we haveĀ *two*āyes, I admit, two *relatedā*amplitudes to describe the *upĀ *or *downĀ *state of the same system, but only one wavefunction for it?Ā 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 (orĀ *physical*) interpretation of the wavefunction might be, it won’t be compatible with theĀ *isotropyĀ *of space. You cannot *imagineĀ *an object with a 720Ā° symmetry. That’sĀ *geometrically *impossible.”

Well… Almost three years ago, I wrote the following on this blog: “As strange as it sounds, aĀ spin-1/2 particle needsĀ *twoĀ *full rotations (2Ć360Ā°=720Ā°) until it is again in the same state. Now, in regard to that particularity, youāll often read something like: ā*There isĀ **nothing**Ā in our macroscopic world which has a symmetry like that.*ā Or, worse, ā*Common sense tells us that something like that cannot exist, that it simply is impossible.*ā [I wonāt quote the site from which I took this quotes, because it is, in fact, the site of a very respectable Ā research center!]*Ā Bollocks!*Ā TheĀ Wikipedia article on spinĀ has this wonderful animation: look at how the spirals flip between clockwise and counterclockwise orientations, and note that itās only after spinning a full 720 degrees that this āpointā 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, weĀ canĀ 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 goodĀ *image*, I should say, because… Well… A representation is something real, remember? š

**Post scriptum** (10 December 2017):Ā Our flywheel model of an electron makes sense, but also leaves many unanswered questions. The most obvious one question, perhaps, is: why theĀ *upĀ *andĀ *downĀ *state only?

I am not so worried about that question, even if I can’t answer it right away because… Well… Our apparatusāthe way weĀ *measureĀ *realityāis set up to measure the angular momentum (or the *magnetic moment*, to be precise) in one direction only. If our electron isĀ *captured*Ā by someĀ *harmonicĀ *(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 spinĀ *upĀ *situation (magnetic moment *down*) is quite peculiar: if our electron is aĀ *mini*-magnet, why would itĀ *notĀ *line up with the magnetic field? We understand the precession of a spinning top in a gravitational field, but…Ā *Hey**… It’s actually not that different*. Try to imagine some spinning top on the ceiling. š I am sure we can work out the math. š 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*. š […] At least, that’s why my instinct tells me. I admit I need to work out the math to convince you. š

The second question is more important. If we just rotate the reference frame over 360Ā°, we see the same thing: some rotating object which we, vaguely, describe by someĀ *e*^{+i}^{Ā·Īø}Ā functionāto be precise, I should say: by some *Fourier* sum of such functionsāor, if the rotation is in the other direction, by someĀ *e*^{āi}^{Ā·Īø}Ā function (again, you should read: aĀ *FourierĀ *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 sameĀ 360Ā°, we get aĀ *differentĀ *object: ourĀ *e*^{i}^{Ā·Īø}Ā andĀ *e*^{āi}^{Ā·Īø}Ā function (again: think of aĀ *FourierĀ *sum, so that’s a waveĀ *packet*, really) becomes aĀ ā*e*^{Ā±i}^{Ā·Īø}Ā thing. We get aĀ *minusĀ *sign in front of it.Ā 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 fullĀ 360Ā°, then you’ll end up where you were when you started and, importantly, you’ll see the same thing.Ā *ExactlyĀ *the same thing: if I was anĀ *e*^{+i}^{Ā·Īø}Ā wave packet, I am still anĀ anĀ *e*^{+i}^{Ā·Īø}Ā wave packet now. OrĀ if I was an *e*^{āi}^{Ā·Īø}Ā wave packet, then I am still anĀ an *e*^{āi}^{Ā·Īø}Ā wave packet now. Easy. Logical. *Obvious*, right?

But so now we try something different:Ā *IĀ *turn around, over a fullĀ 360Ā° turn, and *youĀ *stay where you are. When I am back where I wasālooking at you again, so to speakāthen… Well… I am not quite the same any more. Or… Well… Perhaps I am but youĀ *seeĀ *me differently. If I wasĀ *e*^{+i}^{Ā·Īø}Ā wave packet, then I’ve become aĀ ā*e*^{+i}^{Ā·Īø}Ā wave packet now. Not *hugely* different but… Well… ThatĀ *minusĀ *sign matters, right? OrĀ If I wasĀ wave packet built up from elementaryĀ *a*Ā·*e*^{āi}^{Ā·Īø}Ā waves, then I’ve become aĀ ā*e*^{āi}^{Ā·Īø}Ā wave packet now. What happened?

It makes me think of the twin paradox in special relativity. We know it’s aĀ *paradox*āso that’s anĀ *apparentĀ *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 someĀ *force*.

Can we relate this to the twin paradox? Maybe. Note that aĀ *minusĀ *sign in front of theĀ *e*^{āĀ±i}^{Ā·Īø}Ā 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 180Ā°: ā*cos*Īø =Ā *cos*(Īø Ā± Ļ) andĀ ā*sin*Īø =Ā *sin*(Īø Ā± Ļ). Now, adding or subtracting aĀ *commonĀ *phase factor to/from the argument of the wavefunction amounts toĀ *changingĀ *the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360Ā° and 720Ā° symmetries are, effectively, related. š

# 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. š [OK. Poor joke. Acknowledged.]

Just to recap the essentials, I part ways with mainstream physicists in regard to theĀ *interpretationĀ *of the wavefunction. For mainstream physicists, the wavefunction is just some mathematical construct. NothingĀ *real*. Of course, I acknowledge mainstream physicists have very good reasons for that, but… Well… I believe that, if there is interference, or diffraction, thenĀ *somethingĀ *must be interfering, or something must be diffracting. I won’t dwell on this because… Well… I have done that too many times already. MyĀ *hypothesisĀ *is that the wavefunction is, in effect, aĀ *rotatingĀ **field vector*, so itās just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).

Of course, it must be different, and it is. First, theĀ (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 forceĀ *per unit massĀ *(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 thatās 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 doĀ *notĀ *involve any mass: theyāre just an oscillatingĀ *field*. Nothing more. Nothing less. Why would I believe there must still be some pointlike particle involved? Well…Ā As Feynman puts it: āWhen you do find the electron some place, the entire charge is there.ā (FeynmanāsĀ *Lectures*, III-21-4) So… Well… That’s why.

The third difference is one that I thought of only recently: theĀ *planeĀ *of the oscillation can*notĀ *be perpendicular to the direction of motion of our electron, because then we canāt 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. š

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 toĀ *imagineĀ *it. That’s great mental exercise, so… Well… Just try it. š

Let’s now think about rotating reference frames and transformations. If theĀ *z*-direction is the direction along which we measure the angular momentum (or the magnetic moment), then theĀ *up*-direction will be theĀ *positiveĀ *z-direction. We’ll also assume theĀ *y*-direction is the direction of travel of our elementary particleāand let’s just consider an electron here so we’re moreĀ real. š 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 apparatusāwhich I’ll refer to as a beam splitterāillustrates this *geometry*.

So I think the magnetic momentāor the angular momentum, reallyācomes from an oscillatory motion in the *x*– and *y*-directions. One is theĀ *realĀ *component (the cosine function) and the other is the imaginary component (the sine function), as illustrated below.Ā

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 isĀ *notĀ *in theĀ *xz*-plane, but in theĀ *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*-directionāso that’s the direction in which our particle is movingā, then we might imagine how it would look like whenĀ *weĀ *would make a 180Ā°Ā 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.Ā 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 theĀ *z*-axis as it is (pointing upwards), and we will also want to define theĀ *x*– andĀ *y-*axis using the familiar right-hand rule for defining a coordinate frame. So our newĀ *x*-axis and our newĀ *y-*axis will the same as the oldĀ *x-* andĀ *y-*axes but with the sign reversed. In short, we’ll have the following mini-transformation: (1)Ā *z*‘ =Ā *z*, (2) *x’* = ā*x*, and (3) *y’* =Ā ā*y*.

So… Well… If we’re effectively looking at somethingĀ *realĀ *that was moving along theĀ *y*-axis, then it will now still be moving along the *y’*-axis, butĀ in theĀ *negativeĀ *direction. Hence, our elementary wavefunctionĀ *e ^{i}*

^{Īø}Ā =

*cos*Īø +Ā

*i*Ā·

*sin*Īø willĀ

*transformĀ*intoĀ ā

*cos*Īø āĀ

*i*Ā·

*sin*Īø =Ā ā

*cos*Īø āĀ

*i*Ā·

*sin*Īø =Ā

*cos*Īø āĀ

*i*Ā·

*sin*Īø.Ā 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 theĀ *z*-axis:

Now, ifĀ Ļ is equal to 180Ā° (so that’s Ļ in radians), then theseĀ *e*^{i}^{Ļ/2}Ā andĀ *e*^{āi}^{Ļ/2}/ā2Ā factors areĀ equal toĀ *e*^{i}^{Ļ/2}Ā =Ā *+i*Ā andĀ *e*^{āi}^{Ļ/2}Ā = ā*i*Ā respectively. Hence, ourĀ *e ^{i}*

^{Īø}Ā =

*cos*Īø +Ā

*i*Ā·

*sin*Īø becomes…

** Hey !** Wait a minute ! We’re talking about twoĀ

*veryĀ*different things here, right? TheĀ

*e*

^{i}^{Īø}Ā =

*cos*Īø +Ā

*i*Ā·

*sin*Īø is anĀ

*elementaryĀ*wavefunction which, we presume, describes some real-life particleāwe talked about an electron with its spin in theĀ

*up*-directionāwhile these transformation matrices are to be applied to amplitudes describing… Well… Either anĀ

*up*– or a

*down*-state, right?

Right. But… Well… Is itĀ so different, really? Suppose ourĀ *e ^{i}*

^{Īø}Ā =

*cos*Īø +Ā

*i*Ā·

*sin*Īø wavefunction describes anĀ

*up*-electron, then we still have to apply thatĀ

*e*

^{i}^{Ļ/2}Ā =Ā

*e*

^{i}^{Ļ/2}Ā =Ā

*+i*Ā factor, right? So we get a new wavefunction that will be equal toĀ

*e*

^{i}^{Ļ/2}Ā·

*e*

^{i}^{Īø}Ā =Ā

*e*

^{i}^{Ļ/2}Ā·

*e*

^{i}^{Īø}Ā =Ā

*+i*Ā·

*e*

^{i}^{Īø}Ā =Ā

*i*Ā·

*cos*Īø +Ā

*i*

^{2}Ā·

*sin*Īø =Ā

*sin*Īø āĀ

*i*Ā·

*cos*Īø, right? So how can we reconcile that with the

*cos*Īø āĀ

*i*Ā·

*sin*Īø 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 180Ā°, so it is going in the other direction now !Ā *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 theĀ *sin*Īø āĀ *i*Ā·*cos*Īø andĀ *cos*Īø āĀ *i*Ā·*sin*Īø 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: *cos*Īø =Ā *sin*(Īø +Ā Ļ/2) andĀ ā*sin*Īø =Ā *cos*(Īø +Ā Ļ/2). Let’s see what we can do with that. We can write the following, for example:

*sin*Īø āĀ *i*Ā·*cos*Īø =Ā ā*cos*(Īø +Ā Ļ/2) āĀ *i*Ā·*sin*(Īø +Ā Ļ/2) =Ā ā[*cos*(Īø +Ā Ļ/2) +Ā *i*Ā·*sin*(Īø +Ā Ļ/2)] =Ā ā*e ^{i}*

^{Ā·(Īø +Ā Ļ/2)}

Well… I guess that’s something at least ! The *e ^{i}*

^{Ā·Īø}Ā and ā

*e*

^{i}^{Ā·(Īø +Ā Ļ/2)}Ā functions differ by a phase shiftĀ

*andĀ*a minus sign so… Well… That’s what it takes to reverse the direction of an electron. š Let us mull over that in the coming days. As I mentioned, these more philosophical topics are not easily exhausted. š

# Transforming amplitudes for spin-1/2 particles

Some say it is not possibleĀ to *fully*Ā *understand*Ā quantum-mechanical spin. Now, I do agree it is difficult, but I do *notĀ *believe it is impossible. That’s why I wrote so many posts on it. Most of these focused on elaborating how the classical view of how a rotating charge precesses in a magnetic field might translate into the weird world of quantum mechanics. Others were more focused on the corollary of theĀ *quantizationĀ *of the angular momentum, which is that, in the quantum-mechanical world, the angular momentum is never quite all in *one* direction onlyāso that explains some of the seemingly inexplicable randomness in particle behavior.

Frankly, I think those explanations help us quite a bit already but… Well… We need to go the extra mile, right? In fact, that’s drives my search for aĀ *geometric *(orĀ *physical*)*Ā *interpretation of the wavefunction: the extra mile. š

Now, in one of these many posts on spin and angular momentum, I advise my readers –Ā *you*, that is*Ā *– to try to work yourself through Feynman’s 6th Lecture on quantum mechanics, which is highly abstract and, therefore, usually skipped. [Feynman himself told his students to skip it, so I am sure that’s what they did.] However, if we believe theĀ *physicalĀ *(orĀ *geometric*) interpretation of the wavefunction that we presented in previous posts is, somehow,Ā *true*, then we need to relate it to the abstract math of these so-calledĀ *transformationsĀ *between *representations*.Ā That’s what we’re going to try to do here. It’s going to be just a start, and I will probably end up doing several posts on this but… Well… We do have to start *somewhere*, right? So let’s see where we get today. š

The thought experiment that Feynman uses throughout his LectureĀ makes use of what Feynman’s refers to as modified or improved Stern-Gerlach apparatuses. They allow us to prepare a pure state or, alternatively, as Feynman puts it, to *analyze*Ā a state. In theory, that is. The illustration below present a side and top view of such apparatus. We may already note that the apparatus itselfāor, to be precise, ourĀ *perspectiveĀ *of itāgives us two directions: (1) theĀ *upĀ *direction, so that’s the positive direction of the *z*-axis, and (2) the direction of travel of our particle, which coincides with the positive direction of theĀ *y*-axis. [This is obvious and, at the same time, not so obvious, but I’ll talk about that in my next post. In this one, we basically need to work ourselves through the math, so we don’t want to think too much about philosophical stuff.]

The kind of questions we want to answer in this post are variants of the following basic one: if a spin-1/2 particle (let’s think of an electron here, even if the Stern-Gerlach experiment is usually done with an atomic beam) was prepared in a given condition by one apparatus *S*, say the +*S*Ā state,Ā what is the probability (or theĀ *amplitude*) that it will get through aĀ second apparatus *T*Ā if that was set to filter out the +*T*Ā state?

The result will, of course, depend on the angles between the two apparatuses *S* and *T*, as illustrated below. [Just to respect copyright, I should explicitly note here that all illustrations are taken from the mentioned *Lecture*, and that the line of reasoning sticks close to Feynman’s treatment of the matter too.]

We should make a few remarks here. First, this thought experiment assumes our particle doesn’t get lost. That’s obvious but… Well… If you haven’t thought about this possibility, I suspect you will at some point in time. So we do assume that, somehow, this particle makes a turn. It’s an important point because… Well… Feynman’s argumentāwho, remember, represents mainstream physicsāsomehow assumes that doesn’t really matter. It’s the same particle, right? It just took a turn, so it’s going in some other direction. That’s all, right? Hmm… That’s where I part ways with mainstream physics: the transformation matrices for the amplitudes that we’ll find here describe something real, I think. It’s not justĀ *perspective*: somethingĀ *happenedĀ *to the electron. That something does not onlyĀ *changeĀ *the amplitudes but… Well… It describes a different electron. It describes *an electron that goes in a different direction* now. But… Well… As said, these are reflections I will further develop in my next post. š Let’s focus on the math here. The philosophy will follow later. šĀ Next remark.

Second, we assume theĀ (a) and (b) illustrations above represent the sameĀ *physicalĀ *reality because the *relative* orientation between the two apparatuses, as measured by the angle Ī±, is the same. NowĀ *thatĀ *isĀ obvious, you’ll say, but, as Feynman notes, we can only make that assumption because experiments effectively confirm that spacetime is, effectively, isotropic. In other words, there is noĀ *aether*Ā allowing us to establish some sense of *absoluteĀ *direction. Directions areĀ *relative**ā*relative to the observer, that is… But… Well… Again, in my next post, I’ll argue that it’sĀ *notĀ *because directions areĀ *relativeĀ *that they are, somehow,Ā *notĀ *real. Indeed, in my humble opinion, it does matter whether an electron goes *here* or, alternatively, *there*. These twoĀ *differentĀ *directions are not just two different coordinate frames. But… Well… Again. The philosophy will follow later. We need to stay focused on the math here.

Third and final remark. This one is actually *very* tricky. In his argument, FeynmanĀ also assumes** the two set-ups below are, somehow,Ā equivalent.**

You’ll say:* Huh?Ā *If not, say it!Ā

**šĀ Yes. Good.Ā**

*Huh?***Feynman writesĀ**

*Huh?***because… Well… They’re not the same, obviously:**

*equivalent*ā*notĀ*the*same*- In the first set-up (a),
*T*Ā is wide open, so the apparatus is not supposed to do anything with the beam: it just splits and re-combines it. - In set-up (b) theĀ
*T*Ā apparatus is, quite simply,Ā not there, so… Well… Again. Nothing is supposed to happen with our particles as they come out ofĀ*S*and travel toĀ*U*.

**TheĀ fundamental idea here is that our spin-1/2 particle **(again, think of an electron here)** enters apparatus U in the same state as it left apparatus S. In both set-ups, that is!Ā **Now that is aĀ

*very*tricky assumption, because… Well… While the

*netĀ*turn of our electron is the same, it is quite obvious it has to takeĀ

*twoĀ*turns to get to

*U*in (a), while it only takesĀ

*oneĀ*turn in (b). And so… Well… You can probably think of other differences too.Ā So… Yes. And no.Ā

**š**

*Same-same but different*, right?Right. That isĀ why Feynman goes out of his way to explain the nitty-gritty behind: he actually devotes a full page in small print on this, which I’ll try to summarize in just a few paragraphs here. [And, yes, you should check my summary against Feynman’s actual writing on this.] It’s like this. While traveling through apparatus *T*Ā in set-up (a), time goes by and, therefore, the amplitude would be different by someĀ *phase factorĀ *Ī“. [Feynman doesn’t say anything about this, but… Well… In the particle’s own frame of reference, this phase factor depend on the energy, the momentum and the time and distance traveled. Think of the argument of the elementary wavefunction here:Ā Īø = (Eāt āĀ **p**ā**x**)/Ä§).]Ā Now, *if* we *believe* that the amplitude is just some mathematical constructāso that’s what mainstream physicists (*not* *me!*) believeāthen weĀ *couldĀ *effectively say that the physics of (a) and (b) are the same, as Feynman does. In fact, let me quote him here:

“TheĀ *physics*Ā of set-up (a) and (b) should be the same but the amplitudes could be different by some phase factor without changing the result of any calculation about the real world.”

Hmm… It’s one of those mysterious short passages where we’d all like geniuses like Feynman (or Einstein, or whomever) to be more explicit on their *world view*: if the *amplitudes* are different, can theĀ *physicsĀ *really be the same? I mean…Ā *ExactlyĀ *the same? It all boils down to that unfathomable belief that, somehow, the particle is real but the wavefunction thatĀ *describesĀ *it, is not.Ā Of course, I admit that it’s true that choosing another zero point for the time variable would also change all amplitudes by a common phase factor and… Well… That’s something that I consider to beĀ *notĀ *real. But… Well… The time and distance traveled in theĀ *TĀ *apparatus is the time and distance traveled in theĀ *TĀ *apparatus, right?

*Bon…Ā *I have to stay away from these questions as for nowāwe need to move on with the math hereābut I will come back to it later. But… Well… Talking math, I should note a very interesting *mathematical* point here. We have these transformation matrices for amplitudes, right? Well… Not yet. In fact, the coefficient of these matrices are exactly what we’re going to try toĀ *derive *in this post, but… Well… Let’s assume we know them already. š So we have a 2-by-2 matrix to go from *S* to *T*, from *T* to *U*, and then one to go from *S* to *U* without going through *T*, which we can write as *R ^{ST}*,Ā

*R*,Ā andĀ

^{TU}*R*Ā respectively. Adding the subscripts for theĀ

^{SU}*baseĀ*states in each representation, theĀ

*equivalenceĀ*between the (a) and (b) situations can then be captured by the following formula:

So we have that phase factor here: the left- and right-hand side of this equation is, effectively, *same-same but different*, as they would say in Asia. š Now, Feynman develops a beautiful mathematical argument to show that theĀ *e*^{i}^{Ī“}Ā factor effectively disappears if weĀ *convertĀ *our rotation matrices to some rather specialĀ form that is defined as follows:

I won’t copy his argument here, but I’d recommend you go over it because it is wonderfully easy to follow and very intriguing at the same time. [Yes. Simple things can beĀ very intriguing.] Indeed, the calculation below shows that theĀ *determinantĀ *of theseĀ specialĀ rotation matrices will be equal to 1.

So… Well… So what? You’re right. I am being sidetracked here. The point is that, if we put all of our rotation matrices in this special form, theĀ *e*^{i}^{Ī“}Ā factor vanishes and the formula above reduces to:

So… Yes. End of excursion.Ā Let us remind ourselves of what it is that we are trying to do here. As mentioned above, the kind of questions we want to answer will be variants of the following basic one: if a spin-1/2 particle was prepared in a given condition by one apparatus (*S*), say the +*S*Ā state,Ā what is the probability (or theĀ *amplitude*) that it will get through aĀ second apparatus (*T*) if that was set to filter out the +*T*Ā state?

We said the result would depend on the angles between the two apparatuses *S* and *T*. I wrote: angle** s**āplural. Why? Because a rotation will generally be described by the three so-calledĀ

*Euler angles*:Ā Ī±, Ī² and Ī³. Now, it is easy to make a mistake here, because there is a sequence to these so-calledĀ

*elemental*rotationsāand right-hand rules, of courseābut I will let you figure that out. š

The basic idea is the following: **if we can work out the transformation matrices for each of theseĀ elementalĀ rotations, then we can combine them and find the transformation matrix forĀ anyĀ rotation.** So… Well… That fills most of Feynman’sĀ

*LectureĀ*on this, so we don’t want to copy all that. We’ll limit ourselves to the logic for a rotation about the

*z-*axis, and then… Well… You’ll see. š

So… TheĀ *z*-axis… We take that to be the direction along which we are measuring the angular momentum of our electron, so that’s the direction of the (magnetic) field gradient, so that’s theĀ *up*-axis of the apparatus. In the illustration below, that direction pointsĀ *out of the page*, so to speak, because it is perpendicular to the direction of the *x*– and the *y*-axis that are shown. Note that the *y*-axis is the initial direction of our beam.

Now, because the (physical) orientation of the fields and the field gradients of *S* and *T* is the same, Feynman says thatādespite the angleātheĀ *probabilityĀ *for a particle to beĀ *upĀ *orĀ *downĀ *with regard toĀ *SĀ *andĀ *T *respectively should be the same. Well… Let’s be fair. He does not onlyĀ *sayĀ *that: experimentĀ *showsĀ *it to be true. [Again, I am tempted to interject here that it isĀ *notĀ *because the probabilities for (a) and (b) are the same, that theĀ *reality*Ā of (a) and (b) is the same, but… Well… You get me. That’s for the next post. Let’s get back to the lesson here.]*Ā *The probability is, of course, the square of theĀ *absolute value*Ā of the* *amplitude, which we will denote asĀ *C*_{+},Ā *C*_{ā}, *C’*_{+}, andĀ *C’*_{ā}Ā respectively. Hence, we can write the following:

Now, theĀ *absolute values *(or the* magnitudes*)*Ā *are the same, but theĀ *amplitudes *may differ. In fact, theyĀ *mustĀ *be different by some phase factor because, otherwise, we would not be able to distinguish the two situations, which are obviously different. As Feynman, finally, admits himselfājokingly or seriously: “There must be some way for a particle to know that it has turned the corner at P_{1}.” [P_{1}Ā is the midwayĀ *pointĀ *betweenĀ *SĀ *andĀ *TĀ *in the illustration, of courseānot some probability.]

So… Well… We write:

*C’*_{+}Ā =Ā *e*^{i}^{Ī»}Ā Ā·*C*_{+}Ā andĀ *C’*_{ā}Ā =Ā *e*^{i}^{Ī¼}Ā Ā·*C*_{ā}

*all*amplitudes has no physical consequence (think of re-defining our t

_{0}Ā = 0 point), so we can add some arbitrary amount to bothĀ Ī» and Ī¼ without changing any of the

*physics*. So then we canĀ

*chooseĀ*this amount asĀ ā(Ī» + Ī¼)/2. We write:

Now, it shouldn’t you too long to figure out thatĀ Ī»’ is equal toĀ Ī»’ =Ā Ī»/2 + Ī¼/2 =Ā āĪ¼’. So… Well… Then we can just adopt the convention thatĀ Ī» = āĪ¼. So ourĀ *C’*_{+}Ā =Ā *e*^{i}^{Ī»}Ā Ā·*C*_{+}Ā andĀ *C’*_{ā}Ā =Ā *e*^{i}^{Ī¼}Ā Ā·*C*_{ā}Ā equations can now be written as:

*C’*_{+}Ā =Ā *e*^{i}^{Ī»}Ā Ā·*C*_{+}Ā andĀ *C’*_{ā}Ā =Ā *e*^{āiĪ»}Ā·*C*_{ā}

The absolute values are the same, but theĀ *phasesĀ *are different. Right. OK. Good move. What’s next?

Well… The next assumption is that the phase shiftĀ Ī» is proportional to the angle (Ī±) between the two apparatuses. Hence,Ā Ī» is equal to Ī» =Ā mĀ·Ī±, and we can re-write the above as:

*C’*_{+}Ā =Ā *e*^{i}^{mĪ±}Ā·*C*_{+}Ā andĀ *C’*_{ā}Ā =Ā *e*^{āimĪ±}Ā·*C*_{ā}

Now, this assumption may or may not seem reasonable. Feynman justifies it with a continuity argument, arguing any rotation can be built up as a sequence of infinitesimal rotations and… Well… Let’s not get into the nitty-gritty here. [If you want it, check Feynman’s Lecture itself.] Back to the main line of reasoning. So we’ll assume weĀ *canĀ *writeĀ Ī» as Ī» =Ā mĀ·Ī±. The next question then is:Ā what is the value for m? Now, we obviously do get *exactly* *the same* * physicsĀ *if we rotateĀ

*TĀ*by 360Ā°, or 2Ļ radians. So weĀ

*mightĀ*conclude that the amplitudes should be the same and, therefore, that

*e*

^{i}^{mĪ±}Ā =Ā

*e*

^{i}^{mĀ·2Ļ}Ā has to be equal to one, soĀ

*C’*

_{+}Ā =Ā

*C*

_{+}Ā andĀ

*C’*

_{ā}Ā =Ā

*C*

_{ā}. That’s the case if m is equal to 1. But… Well… No. It’s the same thing again: theĀ

*probabilities*(or theĀ

*magnitudes*)Ā have to be the same, but the amplitudes may be different because of some phase factor. In fact, theyĀ

*should*be different. If m = 1/2, then we also get the same physics, even if the amplitudes areĀ

*notĀ*the same. They will be each other’s opposite:

* Huh?Ā *Yes. Think of it. The coefficient of proportionality (m) cannot be equal to 1. If it

*would*be equal to 1, and we’d rotateĀ

*TĀ*by 180Ā° only, then we’d also get thoseĀ

*C’*

_{+}Ā =Ā ā

*C*

_{+}Ā andĀ

*C’*

_{ā}Ā =Ā ā

*C*

_{ā}Ā equations, and so these coefficients would, therefore,Ā

*also*describeĀ the same

*physical*situation. Now, you will understand,Ā

*intuitively*, that a rotation of theĀ

*TĀ*apparatusĀ byĀ 180Ā° willĀ

*notĀ*give us the sameĀ

*physicalĀ*situation… So… Well… In case you’d want a more formal argument proving a rotation by 180Ā° does

*not*give us the same physical situation, Feynman has one for you. š

I know that, by now, you’re totally tired and bored, and so you only want the grand conclusion at this point. Well… All of what I wrote above should, hopefully, help you to understand that conclusion, which ā I quote Feynman here ā is the following:

If we know the amplitudesĀ *C*_{+}Ā andĀ *C*_{ā}Ā of spin one-half particles with respect to a reference frame *S*, and we then use new base states, defined with respect to a reference frameĀ *T*Ā which is obtained from *S* byĀ a rotationĀ Ī± around theĀ *z*-axis, the new amplitudes are given in terms of the old by the following formulas:

[Feynman denotes our angleĀ Ī± byĀ *phi* (Ļ) because… He uses the Euler angles a bit differently. But don’t worry: it’s the same angle.]

What about the amplitude to go from theĀ *C*_{ā}Ā to theĀ *C’*_{+}Ā state, and from theĀ *C*_{+}Ā to the *C’*_{ā}Ā state? Well… That amplitude is zero. So the transformation matrix is this one:

Let’s take a moment and think about this. Feynman notes the following, among other things:Ā “It is very curious to say that if you turn the apparatus 360Ā° you get new amplitudes. [They aren’t really new, though, because the common change of sign doesn’t give any different physics.] But if something has been rotated by a sequence of small rotations whose net result is to return it to the original orientation, then it is possible toĀ define*Ā *the idea that it has been rotatedĀ 360Ā°āas distinct from zero net rotationāif you have kept track of the whole history.”

This is very deep. It connects space and time into one single geometric space, so to speak. But… Well… I’ll try to explain this rather sweeping statement later. Feynman also notes that a net rotation of 720Ā° does give us the same amplitudes and, therefore, can*not* be distinguished from the original orientation. Feynman finds that intriguing but… Well… I am not sure if it’s very significant. I do note some symmetries in quantum physics involve 720Ā° rotations but… Well… I’ll let you think about this. š

Note that the determinant of our matrix is equal to *a*Ā·*dĀ *ā *b*Ā·*c* =Ā *e*^{i}^{Ļ/2}Ā·*e*^{āi}^{Ļ/2}Ā = 1. So… Well… Our rotation matrix is, effectively, in that special form! How comes? Well… When equatingĀ Ī» = āĪ¼, we are effectively putting the transformation into that special form.Ā Let us also, just for fun, quickly check the normalization condition.Ā It requires that the *probabilities*, in any given representation*,*Ā add to up to one. So… Well… Do they? When they come out ofĀ *S*, our electrons are equally likely to be in the *upĀ *orĀ *downĀ *state. So theĀ *amplitudesĀ *are 1/ā2. [To be precise, they areĀ Ā±1/ā2 but… Well… It’s the phase factor story once again.] That’s normalized:Ā |1/ā2|^{2}Ā +Ā |1/ā2|^{2} = 1. The amplitudes to come out of theĀ *TĀ *apparatus in the *up* or *down* state areĀ *e*^{i}^{Ļ/2}/ā2 andĀ *e*^{i}^{Ļ/2}/ā2 respectively, so the probabilities add up toĀ |*e*^{i}^{Ļ/2}/ā2|^{2}Ā +Ā |*e*^{āi}^{Ļ/2}/ā2|^{2} = … Well… It’s 1. Check it. š

Let me add an extra remark here. The normalization condition will result in matrices whose determinant will be equal to some pure imaginary exponential, likeĀ *e*^{i}^{Ī±}. So is that what we have here? Yes. We can re-write 1 as 1 =Ā *e*^{i}^{Ā·0}Ā = *e*^{0}, soĀ Ī± = 0. š *Capito?* Probably not, but… Well… Don’t worry about it. Just think about the grand results. As Feynman puts it, this Lecture is really “a sort of *cultural* excursion.” š

Let’s do a practical calculation here. Let’s suppose the angle is, effectively, 180Ā°. So theĀ *e*^{i}^{Ļ/2}Ā and *e*^{āi}^{Ļ/2}/ā2Ā factors areĀ equal toĀ *e*^{i}^{Ļ/2}Ā =Ā *+i* andĀ *e*^{āi}^{Ļ/2}Ā = ā*i*, so… Well… What does thatĀ *mean*āin terms of theĀ *geometryĀ *of the wavefunction?Ā Hmm… We need to do some more thinking about the implications of all this transformation business for ourĀ *geometricĀ *interpretation of he wavefunction, but so we’ll do that in our next post. Let us first work our way out of this rather hellish transformation logic. š [See? I do admit it is all quite difficult and abstruse, but… Well… We can do this, right?]

So what’s next? Well… Feynman develops a similar argument (I should sayĀ *same-same but different*Ā once more) to derive the coefficients for a rotation ofĀ Ā±90Ā° around theĀ *y*-axis. Why 90Ā° only? Well… Let me quote Feynman here, as I can’t sum it up more succinctly than he does: “With just two transformationsā90Ā°Ā about theĀ *y*-axis,Ā and an arbitrary angle about theĀ *z*-axis [which we described above]āwe can generate any rotation at all.”

So how does that work? Check the illustration below. In Feynman’s words again: “Suppose that we want the angleĀ Ī± around *x*. We know how to deal with the angleĀ Ī±Ā Ī±Ā aroundĀ *z*, but now we want it aroundĀ *x*.Ā How do we get it? First, we turn the axisĀ *zĀ *down ontoĀ *x*āwhich is a rotation ofĀ +90Ā°.Ā Then we turn through the angleĀ Ī±Ā aroundĀ *xĀ *=Ā *z’*. Then we rotateĀ ā90Ā°Ā aboutĀ *y”*. The net result of the three rotations is the same as turning aroundĀ *x*Ā by the angleĀ Ī±. It is a property of space.”

Besides helping us greatly to derive the transformation matrix forĀ *anyĀ *rotation, the mentioned property of space is rather mysterious and deep. It sort of reduces theĀ *degrees of freedom*, so to speak. FeynmanĀ writes the following about this:

“These facts of the combinations of rotations, and what they produce, are hard to grasp intuitively. It is rather strange, because we live in three dimensions, but it is hard for us to appreciate what happens if we turn this way and then that way. Perhaps, if we were fish or birds and had a real appreciation of what happens when we turn somersaults in space, we could more easily appreciate such things.”

In any case, I should limit the number of philosophical interjections. If you go through the motions, then you’ll find the following elemental rotation matrices:

What about the determinants of the *R _{x}*(Ļ) andĀ

*R*(Ļ) matrices? They’re also equal toĀ

_{y}*one*, so… Yes.Ā A pure imaginary exponential, right? 1 =Ā

*e*

^{i}^{Ā·0}Ā =

*e*

^{0}. š

What’s next? Well… We’re done. We can now combine theĀ *elementalĀ *transformations above in a more general format, using the standardized Euler angles. Again, just go through the motions. The Grand Result is:

Does it give us normalized amplitudes? It should, but it looks like our determinant is going to be a much more complicated complex exponential. š Hmm… Let’s take some time to mull over this. As promised, I’ll be back with more reflections in my next post.

# 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 particleāfor an electron, at least: for the proton, we only got the order of magnitude rightābut then a proton is not an elementary particle.Ā We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between theĀ E =Ā mĀ·*a*^{2}Ā·Ļ^{2}Ā andĀ E =Ā mĀ·*c*^{2}Ā relations that we… Well… We need to be moreĀ *specific *about it.

Indeed, I’ve been ambiguous here and thereā*oscillatingĀ *between various interpretations, so to speak. š In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the *form factorĀ *problem.Ā So… Well… That explains the title of my post. But so… Well… I do want to be somewhat moreĀ *conclusiveĀ *in this post. So let’s go and see where we end up. š

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. š

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 ourĀ 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 =Ā ĻĀ·I. Of course,Ā the moment of inertia (aka the angular mass) will depend on theĀ *formĀ *(orĀ *shape*) of our flywheel:

- I = mĀ·
*a*^{2}Ā for a rotating*pointĀ*mass m or, what amounts to the same, for a circular*hoop*of mass m and radiusĀ*rĀ*=Ā*a*. - For a rotating (uniformly solid)Ā
*disk*, we must add a 1/2 factor: I*Ā*=Ā mĀ·*a*^{2}/2.

How can we relate those formulas to the E =Ā mĀ·*a*^{2}Ā·Ļ^{2}Ā formula? TheĀ *kinetic *energy that is being stored in a flywheel is equal E* _{kinetic}*Ā = IĀ·Ļ

*/2, so that is only*

^{2}*halfĀ*of theĀ E =Ā mĀ·

*a*

^{2}Ā·Ļ

^{2}Ā product if we substitute I forĀ I = mĀ·

*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, theĀ E =Ā mĀ·

*a*

^{2}Ā·Ļ

^{2}Ā 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 justĀ

*transfersĀ*energy from one oscillator to the other, but so… Well… We don’tĀ

*includeĀ*it in our energy calculations.

**The essence of our model is thatĀ two-dimensional oscillation whichĀ**That two-dimensional oscillationātheĀ

*drivesĀ*the electron, and which is reflected in Einstein’sĀ E =Ā mĀ·*c*^{2}Ā formula.Ā*a*

^{2}Ā·Ļ

^{2}Ā =

*c*

^{2}Ā equation, reallyātells us that

**theĀ**ābut measured in units ofĀ

*resonant*Ā (orĀ*natural*)*frequencyĀ*of the fabric of spacetime is given by theĀ speed of light*a*. [If you don’t quite get this, re-write theĀ

*a*

^{2}Ā·Ļ

^{2}Ā =

*c*

^{2}Ā equation asĀ Ļ =

*c*/

*a*: the radius of our electron appears as a

*naturalĀ*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 exampleāand 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 toĀ *vĀ *=Ā *aĀ·*ĻĀ =Ā *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 radiusĀ 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 asĀ *aĀ *=Ā Ä§Ā·/(mĀ·*c*) = 3.8616Ć10^{ā13}Ā 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*Ā = *a*Ā·ĻĀ =Ā *a*Ā·E/Ä§ =Ā *a*Ā·mĀ·*c*^{2}/Ä§Ā Ā āĀ *aĀ *=Ā Ä§/(mĀ·*c*)

The question is: whatĀ *is *that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’tĀ *proveĀ *anything in this regard. But myĀ *hypothesisĀ *is that it is, in effect, aĀ *rotatingĀ **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 massĀ*(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 doĀ
*notĀ*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: theĀ
*planeĀ*of the oscillation can*notĀ*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.Ā The basic idea here is illustrated below (credit for this illustration goes toĀ another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to aĀ YouTube videoĀ from theĀ *Quantum Made SimpleĀ *site.

The point is: the direction of the angular momentum (and the magnetic moment) of an electronāor, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron isĀ *traveling*ācan*not*Ā be parallel to the direction of motion. On the contrary, it isĀ *perpendicular*Ā to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will *compriseĀ *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**Ā and a magnetic momentĀ

**Ā (I used bold-face because these areĀ**

*Ī¼**vector*quantities) that is parallel to some magnetic field

**B**, will

*notĀ*line up, as you’d expect a tiny magnet to do in a magnetic fieldāor not

*completely*, at least: it willĀ

*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 ofĀ show*Ā *that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever weĀ *thinkĀ *our electronāor its wavefunctionāmight be, it needs to be compatible with stuff like the *observed*Ā precession frequency*Ā *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) gravityāsuch as the one I mentioned aboveāare intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? š Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years aheadāI hope. š

**Post scriptum**: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But itĀ *isĀ *that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillatorsĀ *driveĀ *the flywheel but, without the flywheel, nothing is happening. It is really theĀ *transferĀ *of energyāthrough the flywheelāwhich explains why our flywheel goes round and round.”

It may or may not be useful to remind ourselves of the math in this regard.Ā The *motionĀ *ofĀ our first oscillator is given by the cos(ĻĀ·t) = cosĪø function (Īø = ĻĀ·t), and its kinetic energy will be equal toĀ sin^{2}Īø. Hence, the (instantaneous)Ā *changeĀ *in kinetic energy at any point in time (as a function of the angle Īø) isĀ equal to:Ā d(sin^{2}Īø)/dĪø = 2āsinĪøād(sinĪø)/dĪø = 2āsinĪøācosĪø. Now, the motion of theĀ second oscillator (just look at that second piston going up and down in the V-2 engine) is given by theĀ sinĪø function, which is equal to cos(Īø ā Ļ /2). Hence, its kinetic energy is equal toĀ sin^{2}(Īø ā Ļ /2), and how itĀ *changesĀ *(as a function of Īø again) is equal toĀ 2āsin(Īø ā Ļ /2)ācos(Īø ā Ļ /2) =Ā = ā2ācosĪøāsinĪø = ā2āsinĪøācosĪø. 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… WhatĀ *if *the relevant energy formula isĀ E =Ā mĀ·*a*^{2}Ā·Ļ^{2}/2 instead ofĀ E =Ā mĀ·*a*^{2}Ā·Ļ^{2}? What are the implications? Well… We get aĀ ā2 factor in our formula for the radiusĀ *a*, as shown below.

Now that isĀ *notĀ *so nice. For the tangential velocity, we getĀ *vĀ *=Ā *a*Ā·Ļ =Ā ā2Ā·*c*. This is alsoĀ *notĀ *so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precessionātheĀ *wobbling *of our flywheel in a magnetic field. Remember we may think of * J_{z}*āthe angular momentum or, to be precise, its component in theĀ

*z*-direction (the direction in which weĀ

*measureĀ*itāas the projection of theĀ

*realĀ*angular momentumĀ

*. Let me insert Feynman’s illustration here again (Feynman’s*

**J***Lectures*, II-34-3), so you get what I am talking about.

Now, all depends on the angle (Īø) betweenĀ * J_{z}*Ā andĀ

**, 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 theĀ**

*J**magnitudeĀ*of theĀ presumedĀ

*actualĀ*momentum:In this particular case (spin-1/2 particles),Ā

*j*is equal to 1/2 (in units ofĀ Ä§, of course). Hence,Ā

*JĀ*is equal toĀ ā0.75Ā ā 0.866. Elementary geometry then tells us cos(Īø) =Ā (1/2)/ā(3/4) =Ā = 1/ā3. Hence,Ā ĪøĀ ā 54.73561Ā°. That’s a big angleālarger than the 45Ā° angle we had secretly expected because… Well… The 45Ā° angle has thatĀ ā2 factor in it:Ā cos(45Ā°) =Ā sin(45Ā°) = 1/ā2.

Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? š 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 orĀ ā2 problem, right?* *š

**Note**: If you’re into quantum math, you’ll noteĀ *aĀ *=Ā *Ä§*/(mĀ·*c*) is theĀ *reducedĀ *Compton scattering radius. The standard Compton scattering radius is equal toĀ *aĀ·*2Ļ*Ā *= (2ĻĀ·*Ä§*)/(mĀ·*c*) =Ā *h*/(mĀ·*c*) = *h*/(mĀ·*c*). It doesn’t solve theĀ ā2 problem. Sorry. The form factor problem. š

To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of twoĀ *circularĀ *oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation ā around any axis ā will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensionalĀ 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. šThey are oscillations, still, so I am not thinking ofĀ *twoĀ *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. š

# Electron spin and the geometry of the wavefunction

In our previous posts, we interpreted the elementary wavefunction Ļ = *aĀ·e*^{āiāĪø}Ā = *a*Ā·*cos*Īø ā*Ā i*Ā·*a*Ā·*sin*ĪøĀ 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 minusĀ Ä§/2 respectively.

This allowed us to derive the Compton scattering radius of an elementary particle. Indeed, we interpreted the rotating vector as aĀ *resultant* vector, which we get byĀ *addingĀ *the sine and cosine motions, which represent the real and imaginary components of our wavefunction.Ā The energy of this *two*-dimensional oscillation isĀ *twiceĀ *the energy of aĀ *one*-dimensional oscillator and, therefore, equal toĀ E =Ā mĀ·*a*^{2}Ā·Ļ^{2}. Now, the angular frequency is given byĀ Ļ = E/Ä§ and E must, obviously, also be equal to E = mĀ·*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 anĀ *electron*. For a proton, we get a value of aboutĀ 2.1Ć10^{ā16}Ā m, which is about 1/4 of theĀ 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 magnitudeāwhich 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 byĀ *v*Ā =Ā *a*Ā·Ļ =Ā [Ä§Ā·/(mĀ·*c*)]Ā·(E/Ä§) =Ā *c*, which is yet another wonderful implication of our hypothesis. Finally, theĀ *cĀ *=Ā *a*Ā·Ļ formula allowed us to interpret the speed of light as theĀ *resonant frequencyĀ *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 =Ā mĀ·*c*^{2}Ā· 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 theĀ *plane *of our two-dimensional oscillation can*notĀ *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 theĀ *directionĀ *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 theĀ *Quantum Made SimpleĀ *site.

The point is: the direction of the angular momentum (and the magnetic moment) of an electronāor, to be precise, its component as measured in the direction of the (inhomogenous) magnetic field through which our electron is *traveling*ācan*not*Ā 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 actuallyĀ *compriseĀ *the direction of motion.

What are the implications? I am not sure. I will definitely need to review whatever I wrote about theĀ *de BroglieĀ *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 = mĀ·*r*^{2}/2), and the other is the one for a rotating mass (I = mĀ·*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 = mĀ·*v**Ā·r*. [See also Feynman, Vol. II-34-2, in this regard)] Hence, for our model, we get L =Ā mĀ·*v**Ā·r*Ā =Ā mĀ·*c**Ā·a** =Ā mĀ·cĀ·*Ä§/(mĀ·*c*) =Ā Ä§. Now, weĀ *knowĀ *it’s equal toĀ Ā±Ä§/2, so we need that 1/2 factor in the formula.

Can we relate this 1/2 factor to theĀ *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? š

I would just like to signal another interesting consequence of our model. If we would interpret the radius of our disk (*a*)āso that’s the Compton radius of our electron, as opposed to the Thomson radiusāas theĀ *uncertainty in the position of our electron*, then ourĀ L =Ā mĀ·*v**Ā·r*Ā =Ā mĀ·*c**Ā·aĀ *= p*Ā·a** =* Ä§/2 formula as a very particular expression of the Uncertainty Principle:Ā pĀ·Ī*x=* Ä§/2. Isn’t that just plainĀ *nice*? š

# 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.Ā 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 motionābut 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 isĀ *very *intriguing, butĀ let’s think about that later.

Let’s assume we’re looking at it from *some *specificĀ direction. ThenĀ 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 respectivelyāas 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:

Ļ = =Ā *aĀ·e*^{āiāĪø}Ā =Ā *aĀ·e*^{āiāEĀ·t/Ä§} = *a*Ā·cos(āEāt/Ä§)* + i*Ā·aĀ·sin(āEāt/Ä§)* = a*Ā·cos(Eāt/Ä§) *ā** i*Ā·aĀ·sin(Eāt/Ä§)*Ā *

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*Ā·

*term to the argument (Īø = EĀ·t āĀ*

**x****p**ā

**x**). It is easy to show this term doesn’t change the argument (Īø), because we also get a different value for the energy in the new reference frame:Ā E

*=Ā Ī³Ā·E*

_{vĀ }_{0}Ā 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

*Ā and p*

_{v}*Ā and, importantly,Ā the coordinates*

_{v}*xĀ*and

*t*Ā relativistically

*Ā transform*ensures the invariance.

In fact, I’ve always wanted to readĀ *de Broglie*‘sĀ 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 doesĀ * notĀ *imply there is no need for a relativistic waveĀ

*equation*: the wavefunction is aĀ

*solutionĀ*for the wave equation and, yes, I am the first to note the SchrĆ¶dinger equation has some obvious issues, which I briefly touch upon in one of my other postsāand which is why SchrĆ¶dinger 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 SchrĆ¶dinger published his). In my humble opinion, the key issue is

*notĀ*that SchrĆ¶dinger’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 waveĀ

*equationĀ*for what it is and goĀ back to our wave

*function*.

You’ll note the argument (orĀ *phase*) of our wavefunction moves clockwiseāorĀ *counter*clockwise, depending on whether you’re standing in front of behind the clock. Of course,Ā *NatureĀ *doesn’t care about where we stand orāto put it differentlyāwhether 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**Ā ā **0**). Our hypothesis is that these two *physical* possibilities correspond to the angular momentum of our electron being eitherĀ positive or negative: *J _{z}*Ā =Ā +Ä§/2 or, else,

*J*Ā =Ā āÄ§/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

_{z}*defined*Ā 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 toĀ *calculateĀ *the amplitudeĀ *a*. We got a result that tentatively confirms we’re on the right track with our interpretation: we found thatĀ *aĀ *=Ā Ä§/m_{e}Ā·*c*, so that’s theĀ *Compton scattering radius*Ā of our electron. All good ! But we were still a bit stuckāorĀ *ambiguous*, I should sayāon what the components of our wavefunction actuallyĀ *are*. Are we really imagining the tip of that rotating arrow is a pointlike electric chargeĀ spinning around the center? [Pointlike or… Well… Perhaps we should think of theĀ *ThomsonĀ *radius of the electron here, i.e. the so-calledĀ *classical *electron radius, which isĀ equal to the Compton radius times the fine-structure constant:*Ā r _{Thomson}Ā =Ā Ī±Ā·r_{Compton}*Ā ā 3.86Ć10

^{ā13}/137.]

So that would be the flywheel model.

In contrast, we may also think the whole arrow is some rotatingĀ *field vector*āsomething like the electric field vector, with the same or some other *physicalĀ *dimension, like newton per charge unit, or newton per mass unit? So that’s the *fieldĀ *model. Now, theseĀ interpretations may or may not be compatibleāorĀ *complementary*, I should say. I sure *hopeĀ *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 theĀ *interactionĀ *between a photon and an electron, as I demonstrated in my previous post: the electromagnetic energy of the photon willĀ *driveĀ *the circulatory motion of our electron… So… Well… That’s a nice *physicalĀ *explanation for the transfer of energy.Ā 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 flywheelĀ *metaphor*Ā to its logical limits. Let me remind you of what triggered it all: it was theĀ *mathematicalĀ *equivalence of the energy equation for an oscillator (E =Ā mĀ·*a*^{2}Ā·Ļ^{2}) and Einstein’s formula (E =Ā mĀ·*c*^{2}), which tells us energy and mass areĀ *equivalentĀ *but… Well… They’re not the same. So whatĀ *areĀ *they then? WhatĀ *isĀ *energy, and whatĀ *isĀ *massāin the context of these matter-waves that we’re looking at. To be precise, theĀ E =Ā mĀ·*a*^{2}Ā·Ļ^{2}Ā formula gives us the energy ofĀ *twoĀ *oscillators, so we need aĀ *two*-spring model whichābecause I love motorbikesāI referred to as my V-twin engine model, but it’s not anĀ *engine*, really: it’s two frictionless pistons (or springs) whose direction of motion is perpendicular to each other, so they are in a 90Ā° degree angle and, therefore, their motion is, effectively, independent. In other words: they will not interfereĀ *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*Ā·cos(ĻĀ·t + Ī).Ā Needless to say, Ī is just a phase factor which defines our *t* = 0 point, and ĻĀ is theĀ *naturalĀ angular *frequency of our oscillator. Because of the 90Ā° angle between the two cylinders, Ī would be 0 for one oscillator, and āĻ/2 for the other. Hence, the motion of one piston is given by *x* = *a*Ā·cos(ĻĀ·t), while the motion of the other is given by *x* = *a*Ā·cos(ĻĀ·tāĻ/2) = *a*Ā·sin(ĻĀ·t). TheĀ kinetic and potential energy of *one *oscillator ā think of one piston or one spring only ā can then be calculated as:

- K.E. = T = mĀ·
*v*^{2}/2 =Ā (1/2)Ā·mĀ·Ļ^{2}Ā·*a*^{2}Ā·sin^{2}(ĻĀ·t + Ī) - P.E. = U = kĀ·x
^{2}/2 = (1/2)Ā·kĀ·*a*^{2}Ā·cos^{2}(ĻĀ·t + Ī)

The coefficient k in the potential energy formula characterizes the restoring force: F = ākĀ·x. From the dynamics involved, it is obvious that k must be equal to mĀ·Ļ^{2}. Hence, the total energyāforĀ *oneĀ *piston, or one springāis equal to:

E = T + U = (1/2)Ā· mĀ·Ļ^{2}Ā·*a*^{2}Ā·[sin^{2}(ĻĀ·t + Ī) + cos^{2}(ĻĀ·t + Ī)] = mĀ·*a*^{2}Ā·Ļ^{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 = mĀ·*a*^{2}Ā·Ļ^{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, butĀ 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 theĀ E =Ā mĀ·*a*^{2}Ā·Ļ^{2}Ā and E =Ā mĀ·*c*^{2}Ā formulas, we need to explain why we could, potentially, writeĀ *cĀ *asĀ *cĀ *=Ā *a*Ā·ĻĀ =Ā *a*Ā·ā(k/m). We’ve done that alreadyāto some extent at least. TheĀ *tangentialĀ *velocity of a pointlike particle spinning around some axis is given byĀ *v*Ā =Ā *r*Ā·Ļ. Now, the radiusĀ *rĀ *is given byĀ *aĀ *=Ā Ä§/(mĀ·*c*), andĀ Ļ = E/Ä§ =Ā mĀ·*c*^{2}/Ä§, soĀ *vĀ *is equal to toĀ *v *= [Ä§/(mĀ·*c*)]Ā·[mĀ·*c*^{2}/Ä§] =Ā *c*. Another beautiful result, but what does itĀ *mean*? We need to think about theĀ *meaning *of theĀ Ļ =Ā ā(k/m) formula here. In the mentioned article, we boldly wrote that the speed of light is to be interpreted as theĀ *resonantĀ *frequency of spacetime, but so… Well… What do we reallyĀ *meanĀ *by that? Think of the following.

Einsteinās E = m*c*^{2} equation implies the *ratio* between the energy and the mass of *any *particle is always the same:

This effectively reminds us of theĀ Ļ^{2} = *C*^{–}^{1}/*L* or Ļ^{2} = k/mĀ formula for harmonic oscillators.Ā The key difference is that the Ļ^{2}= *C*^{–}^{1}/*L* and Ļ^{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 inĀ *oneĀ *physical space only:Ā *ourĀ *spacetime. Hence, the speed of light (*c*) emerges here as *the* defining property ofĀ 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 Ļ^{2}= 1/*LC* formula here. It’s basically the same concept:Ā the Ļ^{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 Ļ^{2}= *C*^{ā1}/*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 Ļ^{2}= *C*^{–}^{1}/*L* and Ļ^{2} = k/m sort of describe the same thing: harmonic oscillation. It’s just… Well… Unlike theĀ Ļ^{2}= *C*^{–}^{1}/*L*, theĀ Ļ^{2} = k/m isĀ *directlyĀ *compatible with our V-twin engine metaphor, because it also involves *physical distances*, as I’ll show you here.] TheĀ *kĀ *in theĀ Ļ^{2} = k/m is, effectively, the stiffness of the spring. It isĀ *definedĀ *by Hooke’s Law, which states thatĀ the force that is needed to extend or compress a springĀ by some distanceĀ *xĀ *Ā is linearly proportional to that distance, so we write: F = kĀ·*x*.

NowĀ *that *is interesting, isn’t it? We’re talkingĀ *exactlyĀ *the same thing here: spacetime is, presumably,Ā *isotropic*, so it should oscillate the same in any directionāI am talking those sine and cosine oscillations now, but inĀ *physicalĀ *spaceāso there is nothing imaginary here: all is *realĀ *or… Well… As real as we can imagine it to be. š

We can elaborate the point as follows. TheĀ F = kĀ·*x*Ā equation implies k is a forceĀ *per unit distance*: k = F/*x*. Hence, its physical dimension isĀ *newton per meter*Ā (N/m). Now, theĀ *xĀ *in this equation may be equated to theĀ *maximumĀ *extension of our spring, or theĀ *amplitudeĀ *of the oscillation, so that’s the radiusĀ *rĀ *=Ā *aĀ *in the metaphor we’re analyzing here. NowĀ look at how we can re-write theĀ *cĀ *=Ā *a*Ā·ĻĀ =Ā *a*Ā·ā(k/m) equation:

In case you wonder about the E =Ā FĀ·*a* substitution: just remember thatĀ *energyĀ is force times distance*. [Just do a dimensional analysis: you’ll see it works out.] So we have aĀ spectacular result here, for several reasons. The first, and perhaps most obvious reason, is that we can actuallyĀ *deriveĀ *Einstein’s E = mĀ·*c*^{2}Ā formula from ourĀ flywheel model. Now, thatĀ *isĀ *truly glorious, I think. However, even more importantly, this equation suggests we doĀ *not necessarilyĀ *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 aĀ forceĀ acting over some distance**

**, regardless of whether or not it is**

*actually*acting*Now,Ā*

**on a particle.Ā***thatĀ*energy will have anĀ

*equivalentĀ*mass which isāor

*should*be, I’d say… Well… The mass of our electron or, generalizing, the mass of the particle we’re looking at.

* Huh?Ā *Yes. In case you wonder what I am trying to get at, I am trying to convey the idea that theĀ two interpretationsāthe field versus the flywheel modelāare actually fullyĀ

*equivalent*, orĀ

*compatible*, if you prefer that term. In Asia, they would say: they are the “same-same but different” š 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. š 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 aboutĀ *equivalence*. š So it’s just like Einstein’s equation. š

**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 theĀ *RydbergĀ *energy, even Feynman casually writes that “we need not trust our answer [to questions like this] within factors like 2,Ā Ļ, etcetera.” Frankly, that’s disappointing. Factors like 2, 1/2, Ļ or 2Ļ 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 doesĀ *not *come into play: think of the Cooper pairs we use to explain superconductivity (so that’s the BCS theory). The 1/2 factorĀ in SchrĆ¶dinger’s equation also has weird consequences (when you plug in the elementary wavefunction and do the derivatives, you get a weird energy concept: E = mĀ·*v*^{2}, to be precise). This problem may also be solved when assuming we’re actually calculating orbitals for aĀ *pairĀ *of electrons, rather than orbitals for just one electron only. [We’d getĀ *twiceĀ *the mass (and, presumably, the charge, so… Well… It might workābut I haven’t done it yet. It’s on my agendaāas 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 *physicalĀ *interpretation of the wavefunction), you may or may not remember (if not, check my post on it) ‘ll remember I had to use theĀ I = mĀ·*r*^{2}/2 formula for the angular momentum, as opposed to the I = mĀ·*r*^{2}Ā formula.Ā I = mĀ·*r*^{2}/2 (*with* the 1/2 factor) gives us the angular momentum of aĀ *diskĀ *with radiusĀ *r*, as opposed to aĀ *pointĀ *mass going around some circle with radiusĀ *r*. I noted that “the addition of this 1/2 factor may seem arbitrary”āand it totallyĀ *is*, of courseābut so it gave us the result we wanted: theĀ *exactĀ *(Compton scattering)Ā radius of our electron.

Now, the arbitraryĀ 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 toĀ *xĀ *=Ā *a*, we need to calculate it as the following integral:

So… Well… That will give you some food for thought, I’d guess. š If it racks your brain too muchāor if you’re too exhausted by this point (which is OK, because it racks my brain too!)ājust note we’ve also shown that the energy is proportional to theĀ *squareĀ *of the amplitude here, so that’s a nice result as well… š

Talking food for thought, let me make one final point here. TheĀ *c*^{2}*Ā *= *a*^{2}Ā·k/m relation implies a value for k which is equal to k = mĀ·*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 aĀ *natural*Ā distance unit. We wrote that because of another reason: the remark was triggered by the fact that we can write theĀ *c*/Ļ *ratioĀ *asĀ *c*/Ļ =Ā *a*Ā·Ļ/Ļ =Ā *a*. This implies the tangential and angular velocity in our flywheel model of an electron would be the same if weād measure distance in units ofĀ *a*. Now, the E = *a*Ā·k =Ā *a*Ā·F/*xĀ *(just re-writing…) implies that the force is proportional to the energyā F = (*x*/*a*)Ā·E ā and the proportionality coefficient is… Well…Ā *x*/*a*. So that’s the distance measured* in units ofĀ a.*Ā So… Well… Isn’t that great? The radius of our atom appearing as aĀ *naturalĀ *distance unit does fit in nicely with ourĀ *geometricĀ *interpretation of the wavefunction, doesn’t it? I mean…Ā 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 youĀ *cĀ *appears as a resonant frequency of spacetime and, in this post, I tried to explain what that reallyĀ *means*. I’d appreciate if you could let me know if you got it. If not, I’ll try again. š When everything is said and done, one only truly understands stuff when one is able to explain it to someone else, right? š Please do think of more innovative or creative ways if you can! š

OK. That’s it but… Well…Ā 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 someĀ *specificĀ *direction. It could be *anyĀ *direction but… Well… It’sĀ *someĀ *direction. We have noĀ *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 beĀ *anyĀ *direction, our analysis is valid for any direction. Hence, *if*Ā our interpretation would happen to be some *true*āand that’s a bigĀ *if*, of courseāthenĀ our particle has to be *spherical*, right? Why? Well… Because we see this circular thing from any direction, so itĀ *hasĀ *to be a sphere, right?

Well… Yes. But then… Well… While that logic seems to beĀ *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. š¦

**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), theĀ *aĀ *=Ā Ä§/(mĀ·*c*) formula gives a *muchĀ *smaller radius: 1835 timesĀ *smaller*, to be precise, so that’s around 2.1Ć10^{ā16}Ā m, which is about 1/4 of the so-calledĀ *chargeĀ *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 isĀ *notĀ *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 evenĀ *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 wasĀ *veryĀ *short:Ā “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 aĀ *field*. We may call it aĀ *matterĀ *field (yes, think of the scalar Higgs field here), but we could also think of it as an oscillation of theĀ spacetime fabric itself: aĀ tiny gravitational wave, in effect. All we need to do here is to associate the sine and cosine component with aĀ *physicalĀ *dimension. The analogy here is the electromagnetic field vector, whose dimension isĀ *forceĀ *per unitĀ *chargeĀ *(newton/coulomb). So we may associate the sine and cosine components of the wavefunction with, say, theĀ *force* per unitĀ *massĀ *dimension (newton/kg) which, using Newton’s Law (**F** = mĀ·** a**) reduces to the dimension ofĀ

*accelerationĀ*(m/s

^{2}), which is the dimension of

*gravitational*fields.Ā I’ll refer to this interpretation as theĀ

*fieldĀ*interpretation of the matter wave (or wavefunction).

**(2)** The other interpretation is what I refer to as theĀ *flywheelĀ *interpretation of the electron. If you *google* this, you won’t find anything. However, you will probably stumble upon the so-calledĀ *ZitterbewegungĀ *interpretation of quantum mechanics, which is a more elaborate theory based on the same basic intuition. TheĀ *Zitterbewegung*Ā (a term which was coined by Erwin SchrĆ¶dinger 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’sĀ spin and magnetic moment. All that I am doing, is… Well… I think I do push the envelope of this interpretation quite a bit. š

*field vector*. In contrast, the second interpretation implies it’s only the

*tip*of the rotating arrow that, literally,

*matters*: we should look at it as a pointlike

*charge*moving around a central axis, which is the direction of propagation. Let’s look at both.

### The flywheel interpretation

*physicalĀ*interpretation of the

*interactionĀ*between electrons and photonsāor, further speculating, between matter-particles (fermions) and force-carrier particles (bosons) in general. In fact,Ā FeynmanĀ shows how this might workābut in a rather theoreticalĀ

*LectureĀ*on symmetries and conservation principles, and heĀ doesn’t elaborate much, so let me do that for him.Ā The argument goes as follows.

A light beamāan electromagnetic waveāconsists 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 =Ā Ä§Ā·Ļ = hĀ·*f*. [I should, perhaps, quickly note that the frequencyĀ *fĀ *is, obviously, the frequency of the electromagnetic wave. It, therefore, is *notĀ *to be associated with aĀ *matterĀ *wave: theĀ *de BroglieĀ *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 ofĀ *NĀ *photons, the total energy of our beam will be equal to W =Ā *N*Ā·E =Ā *N*Ā·Ä§Ā·Ļ. It is crucially important to note that this energy is to be interpreted as the energy that is carried by the beamĀ *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 carryĀ *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 canĀ *see *there is angular momentum there. Photons are spin-1 particles, so the angular momentum will be equal toĀ Ā± Ä§. Hence,Ā thenĀ theĀ *totalĀ *angular momentum *J _{z}*Ā (the direction of propagation is supposed to be theĀ

*z*-axis here) will be equal toĀ

*J*=Ā

_{z}*N*Ā·Ä§. [This, of course, assumesĀ

*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 theĀ W =Ā

*N*Ā·Ä§Ā·Ļ andĀ

*J*=Ā

_{z}*N*Ā·Ä§ equations, we get:

*J _{z}* =Ā

*N*Ā·Ä§ = W/Ļ

For a photon, we do accept the field interpretation, as illustrated below. As mentioned above, theĀ *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 asĀ Īµ (*epsilon*) so as to not cause any confusion with the Ī 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, aĀ *relativisticĀ *effect which vanishes in the reference frame of the charge itself.] TheĀ *phaseĀ *of the electric field vector isĀ Ļ =Ā ĻĀ·t.

Now, a chargeĀ (so that’s our electron now) will experience a force which is equal to **F** = qĀ·**Īµ**. We use bold letters here because **F** andĀ **Īµ** 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, weĀ *calculatedĀ *the radiusĀ *rĀ *based on a similar argument as the one Feynman used to get thatĀ *J _{z}* =Ā

*N*Ā·Ä§ = W/Ļ equation. I’ll refer you those posts and just mention the result here:Ā

*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 *velocityĀ v*Ā was equal to *v* =Ā *r*Ā·Ļ =Ā [Ä§Ā·/(mĀ·*c*)]Ā·(E/Ä§) =Ā *c*. Hence, in our flywheel model of an electron, it is effectively spinning around at the speed of light. Note that the angular frequency (Ļ) in theĀ *v* =Ā *r*Ā·Ļ equation isĀ *not *the angular frequency of our photon: it’s the frequency of our electron. So we use the same Planck-Einstein equation (Ļ = E/Ä§) 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 areĀ *veryĀ *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 asĀ Ļ_{0}.

Just to be clear here, at this point, our analysis here diverges from Feynman’s. Feynman had no intention whatsoever to talk about SchrĆ¶dinger’sĀ *ZitterbewegungĀ *hypothesis when he wrote what he wrote back in the 1960s. In fact, Feynman is very reluctant to venture intoĀ *physicalĀ *interpretations of the wavefunction in all hisĀ *Lectures *on quantum mechanicsāwhich is surprising. Because he comes so tantalizing close at many occasionsāas 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 thatĀ

*isĀ*a physical interpretation, and it is totally consistent with the one I’ve advanced in my recent posts.Ā Indeed, Feynman also describes it as an oscillation in two dimensionsāperpendicular to each other and to the direction of motion, as we doā 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. Weāll suppose that the atom is isotropic, so that it can oscillate equally well in theĀ *x*– orĀ *y-Ā *directions. Then in the circularly polarized light, theĀ *x*Ā displacement and theĀ *yĀ *displacement are the same, but one is 90Ā°Ā behind the other. The net result is that the electron moves in a circle.”

Right on! But so what happens really? As our light beamāthe photons, reallyāare being absorbed by our electron (or our atom), it absorbsĀ *angular momentum*. In other words, there is aĀ *torqueĀ *about the central axis. Let me remind you of the formulas for the angular momentum and for torqueĀ respectively: **L** = ** r**Ć

**p**andĀ

**Ļ**=Ā

**Ć**

*r***F**. Needless to say, we have twoĀ

*vector*cross-products here. Hence, if we use theĀ

**Ļ**=Ā

**Ć**

*r***F**Ā formula, we need to find theĀ

*tangential*Ā component of the force (

**F**

_{t}), whose magnitude will be equal to F

_{t}= qĀ·Īµ

_{t}

*.Ā*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:

*d*W/*d*t =Ā qĀ·Īµ_{t}Ā·*v*

[If you have trouble, rememberĀ *vĀ *is equal to *d*s/*d*t =Ā Īs/Īt forĀ ĪtĀ ā 0, and re-write the equation above asĀ *d*W =Ā qĀ·Īµ_{t}Ā·*v*Ā·*d*t =Ā qĀ·Īµ_{t}Ā·*d*s =Ā F_{t}Ā·*d*s. *Capito?*]

Now, you may or may not remember thatĀ *the time rate of change of angular momentum*Ā *must be equal to the torqueĀ *that is being applied. Now, the torque is equal toĀ Ļ = F_{t}Ā·*r*Ā =Ā qĀ·Īµ_{t}Ā·*r*, so we get:

*d**J _{z}*/

*d*t =Ā qĀ·Īµ

_{t}Ā·

*v*

TheĀ *ratioĀ *ofĀ *d*W/*d*t andĀ *d**J _{z}*/

*d*t gives us the following interesting equation:

Now, Feynman tries to relate this to theĀ *J _{z}* =Ā

*N*Ā·Ä§ = W/Ļ formula but… Well… We should remind ourselves that the angular frequency of these photons isĀ

*not*the angular frequency of our electron. So… Well… WhatĀ

*canĀ*we say about this equation? Feynman suggests to integrateĀ

*d*

*J*Ā andĀ

_{z}*d*W over some time interval, which makes sense: as mentioned, we interpreted W as the energy that is carried by the beam

*in*

*Ā a certain time*. So if we integrateĀ

*d*W over this time interval, we get W. Likewise, if we integrateĀ

*d*

*J*Ā over theĀ

_{z}*sameĀ*time interval, we should get the

*totalĀ*angular momentum that our electron isĀ

*absorbingĀ*from the light beam. Now, becauseĀ

*d*

*J*Ā =Ā

_{z}*d*W/Ļ, we do concur withĀ 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/Ļ.

It’s just… Well… TheĀ Ļ here is the angular frequency of the electron. It’sĀ *notĀ *the angular frequency of the beam. Not in our flywheel model of the electron which, admittedly, isĀ *notĀ *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:Ā * how and where is the absorbed energy being stored?Ā *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 aroundĀ *afterĀ *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,Ā Ļ =Ā E/Ä§ must increase, right? Right. At the same time, the velocityĀ *v*Ā =Ā *r*Ā·Ļ must still be equal toĀ *v* =Ā *r*Ā·Ļ =Ā [Ä§Ā·/(mĀ·*c*)]Ā·(E/Ä§) =Ā *c*, right? Right. So… IfĀ Ļ increases, butĀ *r*Ā·Ļ must equal the speed of light, thenĀ *rĀ *must actuallyĀ *decreaseĀ *somewhat, right?

Right. It’s a weird but inevitable conclusion, it seems. I’ll let you think about it. š

To conclude this postāwhich, I hope, the reader who triggered it will find interestingāI 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 letās 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 hasĀ *equal*Ā 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 momentumāthe average angular momentum is zero.]

Now, we have said that any spin-one particle can have three values of *J _{z}*, namelyĀ +1,Ā 0,Ā ā1Ā (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 spinĀ

*jĀ*which is standing still, there must be theĀ 2

*j*+1Ā possible states with values of

*J*Ā going in steps ofĀ 1Ā fromĀ ā

_{z}*j*Ā toĀ +

*j*. But it turns out that for something of spinĀ

*j*Ā with zero mass only the states with the componentsĀ +

*j*Ā andĀ ā

*j*Ā along the direction of motion exist. For example, light does not have three states, but only twoāalthough a photon is still an object of spin one.”

In his typical style and franknessāfor which he is revered by some (like me) but disliked by othersāhe admits this is very puzzling, and not obvious at all! Let me quote him once more:

“How is this consistent with our earlier proofsābased on what happens under rotations in spaceāthat 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 ofĀ Ä§/2āand not something likeĀ Ä§/3.Ā Even using all sorts of properties of the Lorentz transformation and what not, we failed. Maybe itās not true. Weāll 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* wellāif only because they had both worked together on the Manhattan Projectāand 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,Ā particularly through the discovery and application of fundamental symmetry principles.” š I’ll let you figure out how what I write about in this post, and symmetry arguments, might be related. š

That’s it for today, folks! I hope you enjoyed this. š

**Post scriptum**: The mainĀ *dis*advantage of the flywheel interpretation is that it doesn’t explain interference: waves interfereāsome 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 doesāin a “local circulatory motion”āif there is *a*Ā *forceĀ *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 hereāsome negative charge orbiting around a positive nucleus. We’re just considering the electron itself.] So we just need to prove that our rotating arrow willĀ *alsoĀ *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. š

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 modelĀ (illustrated below), but then whatĀ *isĀ *the mass here? One hypothesis that comes to mind is that we’re constantly accelerating and decelerating an electric charge (the electron charge)āagainst all other charges in the Universe, so to speak. So that’s a force over a distanceāenergy. 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 someĀ *massĀ *which acquires mass because of… Well… Because of the forceābecause of the oscillation (the moving charge) itself. Hmm…. I need to think about this.

# Photons as spin-1 particles

After all of the lengthy and speculative excursions into the nature of the wavefunction for an electron, it is time to get back to Feynman’s Lectures and look at photon-electron *interactions*. So that’s chapter 17 and 18 of Volume III. Of all of the sections in those chapters – which are quite technical here and there – I find the one on the angular momentum of polarized light the most interesting.

Feynman provides an eminently readable explanation of how the electromagnetic energy of a photon may be absorbed by an electron asĀ *kinetic *energy. It is entirely compatible with ourĀ *physicalĀ *interpretation of the wavefunction of an electron as… Well… We’ve basically been looking at the electron as a little flywheel, right? š I won’t copy Feynman here, except the illustration, which speaks for itself.

However, I do recommend you explore these two Lectures for yourself. Among other interesting passages, Feynman notes that, while photons are spin-1 particles and, therefore, are supposed to be associated withĀ *threeĀ *possible values for the angular momentum (*J*_{z}Ā = +Ä§, 0 orĀ āÄ§), there are only two states: the zero case doesn’t exist. As Feynman notes: “This strange lack is related to the fact that light cannot stand still.” But I will let you explore this for yourself. š

# Feynman as the Great Teacher?

While browsing for something else, I stumbled on an article which derides Feynman’s qualities *as a teacher*, and the Caltech Feynman *Lectures* themselves. It is an interesting read. Let me quote (part of) the conclusion:

“Richard Feynman constructed an āintroductoryā physics course at Caltech suitable primarily for perhaps imaginary extreme physics prodigies like himself or how he pictured himself as an eighteen year old. It is an open question how well the actual eighteen year old Feynman would have done in the forty-three year old Feynmanās āintroductoryā physics course. Like many adults had Feynman lost touch with what it had been like to be eighteen? In any case, such extreme physics prodigies made up only a small fraction of the highly qualified undergraduate students at Caltech either in the 1960ās or 1980ās. An educational system designed by extreme prodigies for extreme prodigies, often from academic families, extremely wealthy families, or other unusual backgrounds rare even among most top students as conventionally defined, is a prescription for disaster for the vast majority of students and society at large.”

The article actually reacts to a blog post from Bill Gates, whoĀ extols Feynman’s virtues *as a teacher*. So… Was or wasn’t he a great teacher?

It all depends on your definition of a great teacher.Ā I respect the views in the mentioned article mentioned aboveāif only because the author,Ā *John F. McGowan*, is not just anyone: he is a B.S. from Caltech itself, and he has a Ph.D. in physics. I don’t, so… Well… He is an authority, obviously.Ā Frankly, I must agree I struggled with Feynman’s *Lectures*Ā too, and I will probably continue to do so as I read and re-read them time after time. On the other hand, below I copy one of those typical Feynman illustrations you willĀ *notĀ *find in any other textbook. Feynman tries to give us aĀ *physicalĀ *explanation of the photon-electron interaction here. Most introductory physics textbooks just don’t bother: they’ll give you the mathematical formalism and then some exercises, and that’s it. Worse, those textbooks will repeatedly tell you you can’t really ‘understand’ quantum math. Just go through the math and apply the rules. That’s the general message.

I find that *veryĀ *disappointing. I must admit thatĀ Feynman has racked my brainābut in a good way. I still feel I do not quite understand quantum physics “the way we would like to”. It is still “peculiar and mysterious”, but then that’s just how Richard Feynman feels about it tooāand he’s humble enough to admit that in the very first paragraph of his very first Lecture on QM.

I have spent a lot of my free time over the past years thinking about a physical or geometric interpretation of the wavefunctionāhalf of my life, in a wayāand I think I found it. The article I recently published on it got downloaded for the 100th time today, and *this* blog – as wordy, nerdy and pedantic as it is – attracted 5,000 visitors last *month* alone. People like me: people who want to understand physics beyond the equations.

So… Well… Feynman himself admits he was mainly interested in the “one or two dozen students who ā very surprisingly ā understood almost everything in all of the lectures, and who were quite active in working with the material and worrying about the many points in an excited and interested way.”Ā I think there are many people like those students. People like me: people who want to understand but can’t afford to study physics on a full-time basis.

For those, I think Feynman’s Lectures are truly inspirational. At the very least, they’ve provided me with many wonderful evenings of self-studyāsome productive, in the classical sense of the word (moving ahead) and… Some… Well… Much of what I read didāand still doesākeep me awake at night. š

# 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. š¦ But I’ll do my best to try to explain what I am thinking of. Remember the formula (orĀ *definition*) of theĀ *elementary* wavefunction:

Ļ =Ā *a*Ā·*e*^{āi[EĀ·t ā pāx]/Ä§} =Ā *a*Ā·cos(**p**ā**x**/Ä§ ā Eāt/Ä§) + *i*Ā·*a*Ā·sin(**p**ā**x**/Ä§ āĀ Eāt/Ä§)

How should we interpret this? We know an *actual* particle will be represented by aĀ wave *packet*: a sum of wavefunctions, each with its own amplitude *a*_{k} and its own argument Īø_{k} = (E_{k}āt ā **p**_{k}ā**x**)/Ä§. But… Well… Let’s see how far we get when analyzing theĀ *elementaryĀ *wavefunction itself only.

According to mathematical*Ā *convention, the imaginary unit (*i*) is a 90Ā°Ā angle in theĀ *counter*clockwise direction. However, *Nature*Ā surely cannot be bothered about our convention of measuring phase angles – orĀ *timeĀ *itself – clockwiseĀ 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:

Ļ =Ā *a*Ā·*e*^{i[EĀ·t ā pāx]/Ä§} =Ā *a*Ā·cos(**p**ā**x**/Ä§ ā Eāt/Ä§)Ā āĀ *i*Ā·*a*Ā·sin(**p**ā**x**/Ä§ āĀ Eāt/Ä§)

In my previous posts, I hypothesized that the two physical possibilities correspond to the angular momentum of our particle – say, an electron – being eitherĀ positive or negative: *J* = +Ä§/2 or, else,Ā *J* = āÄ§/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 theĀ *directionĀ *as well as theĀ *magnitudeĀ *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

**) but… Well… It’s not quite the same because – as I will explain in a moment – the argument of the wavefunction,**

*B**considered as a whole*, is actually invariant under a Lorentz transformation.

Let me elaborate this point.Ā 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:Ā **p** = **0**. Let us now look at howĀ the argument of the wavefunction transforms. The E and **p** in the argument of the wavefunction (Īø = Ļāt ā **k**ā**x** = (E/Ä§)āt ā (**p**/Ä§)ā**x** =Ā (Eāt ā **p**ā**x**)/Ä§) 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

*= p*

_{v }*ā*

_{v}*v*. If we then use

*natural*time and distanceĀ units (hence, the

*numerical*value of

*c*is equal to 1 and, hence, the (relative) velocity is then measured as a fraction ofĀ

*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* _{vĀ }*=Ā Ī³Ā·E

_{0}Ā and p

*= Ī³Ā·m*

_{vĀ }_{0}ā

*v*Ā =Ā Ī³Ā·E

_{0}ā

*v*/

*c*

^{2}

The argument of the wavefunction can then be re-written as:

Īø = [Ī³Ā·E_{0}/Ä§]āt ā [(Ī³Ā·E_{0}ā*v*/*c*^{2})/Ä§]āx = (E_{0}/Ä§)Ā·(t ā *vāx*/*c*^{2})Ā·Ī³ =Ā (E_{0}/Ä§)āt’

The Ī³ in these formulas is, of course, the Lorentz factor, and t’ is theĀ *proper*Ā time: t’*Ā *= (t ā *vāx*/*c ^{2}*)/ā(1ā

*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 primedĀ Īø (Īø’):Ā Īø = (E* _{v}*/Ä§)Ā·t ā (p

*/Ä§)Ā·x =Ā (E*

_{v}_{0}/Ä§)āt’.

**2.**Ā **TheĀ E _{0}/Ä§ coefficient pops up as an angular**

**frequency:Ā E**. We may refer to it asĀ

_{0}/Ä§ =Ā Ļ_{0}*theĀ*frequency of the elementary wavefunction.

Now, if you don’t like the concept ofĀ *angular* frequency, we can also write:Ā *f*_{0}*Ā *=Ā Ļ_{0}/2Ļ = (E_{0}/Ä§)/2Ļ = E_{0}/h.Ā Alternatively, and perhaps more elucidating, we get the following formula for theĀ *periodĀ *of the oscillation:

T_{0}*Ā *= 1/*f*_{0}*Ā *=Ā h/E_{0}

This is interesting, because **we can look at the period as aĀ naturalĀ unit of time for our particle**. This period is

*inverselyĀ*proportional to the (rest) energy of the particle, and the constant of proportionality is h. Substituting E

_{0Ā }for m

_{0}Ā·

*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 8Ć10

^{ā21}Ā s. That’sĀ

*veryĀ*small, and it only gets smaller for larger objects ! But what does all of this really

*tellĀ*us? What does it actuallyĀ

*mean*?

We can look at the sine and cosine components of the wavefunction as an oscillation inĀ *twoĀ *dimensions, as illustrated below.

Look at the little green dot going around. Imagine it is someĀ *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 hasĀ *angular momentum*, which we can write as the vector cross-product **L** = ** r**Ć

**p**or, alternatively, as the product of an

*angular*velocity (

**Ļ**) and rotational inertia (I), aka as theĀ

*moment of inertia*or the

*angular mass*:Ā

**L**= IĀ·

**Ļ**. [Note we writeĀ

**L**andĀ

**Ļ**in

**boldface**here because they are (axial) vectors. If we consider their magnitudes only, we write L = IĀ·Ļ (no boldface).]

We can now do some calculations. We already know the angular velocity (Ļ) is equal toĀ E_{0}/Ä§. Now, theĀ magnitude ofĀ *r**Ā *in the **L** =Ā * r*Ć

**p**Ā vector cross-product should equal theĀ

*magnitudeĀ*ofĀ Ļ =Ā

*aĀ·e*

^{āiāEĀ·t/Ä§}, so we write:Ā

*r*=

*a*. What’s next? Well… The momentum (

**p**) is the product of a

*linear*velocity (

*) – in this case, theĀ*

**v***tangentialĀ*velocity –Ā and some mass (m):

**p**= mĀ·

*. If we switch to*

**v***scalarĀ*instead ofĀ vector quantities, then the (tangential) velocity is given by

*v*=

*r*Ā·Ļ.

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 *pointĀ *mass going around some center, then the formula to use isĀ I = mĀ·*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 becomesĀ I = mĀ·*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 = IĀ·Ļ = (mĀ·*r*^{2}/2)Ā·(E/Ä§) = (1/2)Ā·*a*^{2}Ā·(E/*c*^{2})Ā·(E/Ä§) =Ā *a*^{2}Ā·E^{2}/(2Ā·Ä§Ā·*c*^{2})

Note that our frame of reference is that of the particle itself, so we should actually write Ļ_{0}, m_{0}Ā and E_{0}Ā instead ofĀ Ļ, m and E. The value of the rest energy of an electron is about 0.510 MeV, or 8.1871Ć10^{ā14} Nām. Now, this momentum should equal *J* = Ā±Ä§/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 3.8616Ć10^{ā13} m, which is the (reduced) Compton scattering radius of an electron.Ā The (tangential) velocity (*v*) can now be calculated as being equal toĀ *v* = *r*Ā·Ļ = *a*Ā·Ļ = [Ä§Ā·/(mĀ·*c*)]Ā·(E/Ä§) =Ā *c*. This is an amazing result. Let us think about it.

In our previous posts, we introduced the metaphor of twoĀ *springsĀ *or oscillators, whose energy was equal to E =Ā mĀ·Ļ^{2}. Is this compatible with Einstein’s E =Ā mĀ·*c*^{2}Ā mass-energy equivalence relation? It is. TheĀ E =Ā mĀ·*c*^{2}Ā impliesĀ E/m =Ā *c*^{2}. We, therefore, can write the following:

Ļ = E/Ä§ =Ā mĀ·*c*^{2}/Ä§ = mĀ·(E/m)Ā·/Ä§Ā ā Ļ =Ā E/Ä§

Hence, we should actually have titled this and the previous post somewhat differently: the speed of light appears as aĀ *tangentialĀ *velocity. Think of the following: theĀ *ratioĀ *ofĀ *c *andĀ Ļ is equal toĀ *c*/Ļ =Ā *a*Ā·Ļ/Ļ =Ā *a*. Hence, the tangential and angular velocity would be the same if we’d measure distance in units ofĀ *a*. In other words,Ā the radius of an electron appears as a *natural* distance unit here: if we’d measureĀ Ļ inĀ *units of*Ā *aĀ *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*Ā *unit of* *velocity*.Ā * Huh?Ā *Yes. Just re-write theĀ

*c*/Ļ =Ā

*a*asĀ Ļ =Ā

*c*/

*a*. What does it say? Exactly what I said, right? As such, the radius of an electron is not only aĀ

*normĀ*for measuring distance but also for time.Ā š

If you don’t quite get this, think of the following. For an electron, we get an angular frequency that is equal toĀ Ļ = E/Ä§ = (8.19Ć10^{ā14}Ā NĀ·m)/(1.05Ć10^{ā34}Ā NĀ·mĀ·s) ā 7.76Ć10^{20}Ā *radiansĀ *per second. That’s an incredible *velocity*, because radians are expressed in distance unitsāso that’s inĀ *meter*. However, our mass is not moving along theĀ *unitĀ *circle, but along a much tinier orbit. TheĀ *ratioĀ *of the radius of the unit circle andĀ *aĀ *is equal to 1/*a ā*Ā (1 m)/(3.86Ć10^{ā13} m) ā 2.59Ć10^{12}. Now, if we divide theĀ above-mentionedĀ *velocityĀ *ofĀ 7.76Ć10^{20}Ā *radiansĀ *per second by this factor, we get… Right ! The speed of light: 2.998Ć10^{82}Ā m/s. š

**Post scriptum**: I have no clear answer to the question as to why we should use the I = mĀ·*r*^{2}/2 formula, as opposed to theĀ I = mĀ·*r*^{2}Ā 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 SchrĆ¶dinger’s equation, which is also quite enigmatic. In my view, the 1/2 factor should not be there in SchrĆ¶dinger’s equation. Electron orbitals tend to be occupied byĀ *twoĀ *electrons with opposite spin. That’s why their energy levels should beĀ *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. š But then that’s just my personal opinion or… Well… I’ve always been intrigued by the difference between the originalĀ *printedĀ *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:

“Donāt forget thatĀ m_{eff} has nothing to do with the real mass of an electron. It may be quite differentāalthough in commonly used metals and semiconductors it often happens to turn out to be the same general order of magnitude, about 2 to 20 timesĀ 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}Ā = 2ām

_{eff}

^{OLD}Ā – 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. š

# 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)**:Ā Einsteinās view on *aether* theories probably still holds true: āWe 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 ā 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.ā

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: āIt 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. [ā¦] 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. [ā¦]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.ā

I really love this: a *relativistic* aether. MyĀ interpretation of the wavefunction is *veryĀ *consistent with that.

# A physical explanation for relativistic length contraction?

My last posts were all about a possible *physicalĀ *interpretation of the quantum-mechanical wavefunction. To be precise, we have been interpreting the wavefunction as a gravitational wave. In this interpretation, the real and imaginary component of the wavefunction get aĀ *physicalĀ *dimension: force per unit mass (newton per kg). The inspiration here was the structural similarity between Coulomb’s and Newton’s force laws. They both look alike: it’s just that one gives us a force per unitĀ *chargeĀ *(newton perĀ *coulomb*), while the other gives us a force per unitĀ *mass*.

So… Well… Many nice things came out of this – and I wrote about that at length – but last night I was thinking this interpretation may also offer an explanation of relativistic length contraction. Before we get there, let us re-visit our hypothesis.

**The geometry of the wavefunction**

The elementary wavefunction is written as:

Ļ =Ā *aĀ·e*^{āi(EĀ·t ā pāx)/Ä§} =Ā *aĀ·cos*(**p**ā**x**/Ä§ – Eāt/Ä§) *+** iĀ·aĀ·sin*(**p**ā**x**/Ä§ – Eāt/Ä§)

Nature should not care about our conventions for measuring the phase angle clockwise or counterclockwise and, therefore, the Ļ =Ā *aĀ·e*^{i}^{[EĀ·t ā pāx]/Ä§} function may also be permitted. We know that *cos*(Īø) = *cos**(**–*Īø) and *sin*Īø = *–**sin**(**–*Īø), so we can write:Ā *Ā Ā *

Ļ =Ā *aĀ·e*^{i}^{(EĀ·t ā pāx)/Ä§} =Ā *aĀ·cos*(Eāt/Ä§ – **p**ā**x**/Ä§) *+** iĀ·aĀ·sin*(Eāt/Ä§ – **p**ā**x**/Ä§)

*= **aĀ·cos*(**p**ā**x**/Ä§ – Eāt/Ä§) *–** iĀ·aĀ·sin*(**p**ā**x**/Ä§ – Eāt/Ä§)

The vectors **p** and **x** are the the momentum and position vector respectively: **p** = (p_{x}, p_{y}, p_{z}) and **x** = (x, y, z). However, if we assume there is no uncertainty about **p** ā not about the direction nor the magnitude ā then we may choose an x-axis which reflects the direction of **p**. As such, **x** = (x, y, z) reduces to (x, 0, 0), and **p**ā**x**/Ä§ reduces to pāx/Ä§. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. Hence, the analysis is one-dimensional only.

The geometry of the elementary wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.

Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. You can check this as follows: if the origin is the (x, t) = (0, 0) point, then cos(Īø) = cos(0) = 1 and sin(Īø) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. We speculated this should correspond to the two possible values for the quantum-mechanical spin of the wave: +Ä§/2 or āÄ§/2. The cosine and sine components for the left-handed wave are shown below. Needless to say, the cosine and sine function are the same, except for a phase difference of Ļ/2: sin(Īø) = cos(Īø ā Ļ/2).

As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any wave F(kx – Ļt) is given by *v*_{p} =Ā Ļ/k = (E/Ä§)/(p/Ä§) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to -p and, therefore, we would get a negative phase velocity: *v*_{p} =Ā Ļ/k = *–*E/p.

**The de Broglie relations**

E/Ä§ = Ļ gives the frequency in time (expressed in radians per second), while p/Ä§ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = Ļ/2ĻĀ and Ī» = 2Ļ/k, which gives us the two de Broglie relations:

- E = Ä§āĻ = hāf
- p = Ä§āk = h/Ī»

The frequency in time is easy to interpret. The wavefunction of a particle with more energy, or more mass, will have a *higher density in time* than a particle with less energy.

In contrast, the second *de Broglie *relation is somewhat harder to interpret. According to the p = h/Ī» relation, the wavelength is *inversely *proportional to the momentum: Ī» = h/p. The velocity of a photon, or a (theoretical) particle with zero rest mass (m_{0} = 0), is *c* and, therefore, we find that p = m* _{v}*ā

*v*= m

*ā*

_{c}*c*= mā

*c*(all of the energy is kinetic). Hence, we can write: pā

*c*= mā

*c*

^{2}= E, which we may also write as: E/p =

*c*. Hence, for a particle with zero rest mass, the wavelength can be written as:

Ī» = h/p = hc/E = h/mc

However, this is a limiting situation ā applicable to photons only. Real-life *matter*-particles should have *some *mass[1] and, therefore, their velocity will never be *c*.[2]

Hence, if p goes to zero, then the wavelength becomes infinitely long: if pĀ ā 0 then Ī»* āĀ ā*. How should we interpret this inverse proportionality between Ī» and p? To answer this question, let us first see what this wavelength Ī» actually represents.

If we look at the Ļ = *a*Ā·cos(pāx/Ä§ – Eāt/Ä§) – *i*Ā·*a*Ā·sin(pāx/Ä§ – Eāt/Ä§) once more, and if we write pāx/Ä§ as Ī, then we can look at pāx/Ä§ as a phase factor, and so we will be interested to know for what x this phase factor Ī = pāx/Ä§ will be equal to 2Ļ. So we write:

Ī =pāx/Ä§ = 2Ļ āĀ x = 2ĻāÄ§/p = h/p = Ī»

So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: we measure crests and troughs against the y-axis here. Hence, our definition depend on the frame of reference.

Now we know what Ī» actually represents for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/*f *= 2ĻĀ·(Ä§/E). Hence, we can now calculate the wave velocity:

v = Ī»/T = (h/p)/[2ĻĀ·(Ä§/E)] = E/p

Unsurprisingly, we just get the phase velocity that we had calculated already: *v *= *v*_{p} = E/p. The question remains: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite wave velocity.

Now, re-writing the *v *= E/p as *v *= mā*c*^{2}/mā*v*_{g }*Ā *= *c*/Ī²_{g}, in which Ī²_{g} is the relative *classical *velocity[3] of our particle Ī²_{g} = *v*_{g}/*c*) tells us that the *phase *velocities will effectively be superluminal (Ī²_{g}Ā < 1 so 1/ Ī²_{g} > 1), but what if Ī²_{g} approaches zero? The conclusion seems unavoidable: for a particle at rest, we only have a frequency *in time*, as the wavefunction reduces to:

Ļ =Ā aĀ·e^{āiĀ·EĀ·t/Ä§} =Ā aĀ·cos(Eāt/Ä§) – iĀ·aĀ·sin(Eāt/Ä§)

How should we interpret this?

**A physical interpretation of relativistic length contraction?**

In my previous posts,Ā we argued that the oscillations of the wavefunction pack energy. Because the energy of our particle is finite, the wave train cannot be infinitely long. If we assume some *definite* number of oscillations, then the string of oscillations will be shorter as Ī» decreases. Hence, the physical interpretation of the wavefunction that is offered here may explain relativistic length contraction.

š

Yep. Think about it. š

[1] Even neutrinos have some (rest) mass. This was first confirmed by the US-Japan Super-Kamiokande collaboration in 1998. Neutrinos oscillate between three so-called flavors: electron neutrinos, muon neutrinos and *tau *neutrinos. Recent data suggests that the *sum *of their masses is less than a millionth of the rest mass of an electron. Hence, they propagate at speeds that are very near to the speed of light.

[2] Using the Lorentz factor (Ī³), we can write the relativistically correct formula for the kinetic energy as KE = EĀ āĀ E_{0}Ā =Ā m_{v}*c*^{2}Ā ā m_{0}*c*^{2}Ā =Ā m_{0}Ī³*c*^{2}Ā ā m_{0}*c*^{2}Ā =Ā m_{0}*c*^{2}(Ī³ ā 1). As *v *approaches *c*, Ī³ approaches infinity and, therefore, the kinetic energy would become infinite as well.

[3] Because our particle will be represented by a wave *packet*, i.e. a superimposition of elementary waves with different E and p, the classical velocity of the particle becomes the *group *velocity of the wave, which is why we denote it by *v*_{g}.

# The geometry of the wavefunction (2)

This post further builds on the rather remarkable results we got in our previous posts. Let us start with the basics once again.Ā The elementary wavefunction is written as:

Ļ =Ā *aĀ·e*^{āi[EĀ·tĀ ā pāx]/Ä§} =Ā *aĀ·cos*(**p**ā**x**/Ä§Ā ā Eāt/Ä§) *+** iĀ·aĀ·sin*(**p**ā**x**/Ä§Ā ā Eāt/Ä§)

Of course, *Nature*Ā (or God, as Einstein would put it) does not care about our conventions for measuring an angle (i.e. the phase of our wavefunction) clockwise or counterclockwise and, therefore, the Ļ =Ā *aĀ·e*^{i}^{[EĀ·t ā pāx]/Ä§} function is also permitted. We know that *cos*(Īø) = *cos*(āĪø) and *sin*Īø = ā*sin*(*ā*Īø), so we can write:Ā *Ā Ā *

Ļ =Ā *aĀ·e*^{i}^{[EĀ·t ā pāx]/Ä§} =Ā *aĀ·cos*(Eāt/Ä§Ā āĀ **p**ā**x**/Ä§) *+** iĀ·aĀ·sin*(Eāt/Ä§Ā āĀ **p**ā**x**/Ä§)

*= **aĀ·cos*(**p**ā**x**/Ä§Ā ā Eāt/Ä§) ā*Ā iĀ·aĀ·sin*(**p**ā**x**/Ä§Ā ā Eāt/Ä§)

The vectors **p** and **x** are the momentum and position vector respectively: **p** = (p_{x}, p_{y}, p_{z}) and **x** = (x, y, z). However, if we assume there is *no* uncertainty about **p** ā not about the direction, and not about the magnitude ā then the direction of **p** can be our x-axis. In this reference frame,Ā **x** = (x, y, z) reduces to (x, 0, 0), and **p**ā**x**/Ä§ reduces to pāx/Ä§. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. So we have an analysis in one dimension only then, which facilitates our calculations. The geometry of the wavefunction is then as illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.

Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. [You can check this as follows: if the origin is the (x, t) = (0, 0) point, then *cos*(Īø) = *cos*(0) = 1 and *sin*(Īø) = *sin*(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively.]

Now, you will remember that we speculated the two *polarizations* (left- versus right-handed) should correspond to the two possible values for the quantum-mechanical spin of the wave (+Ä§/2 or āÄ§/2). We will come back to this at the end of this post. Just focus on the essentials first: the cosine and sine components for the left-handed wave are shown below. Look at it carefully and try to understand. Needless to say, the cosine and sine function are the same, except for a phase difference of Ļ/2: *sin*(Īø) = *cos*(Īø ā Ļ/2).

As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any waveform F(kx ā Ļt) is given by *v*_{p} =Ā Ļ/k. In our case, we find thatĀ *v*_{p} =Ā Ļ/k = (E/Ä§)/(p/Ä§) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to āp and, therefore, we would get a negative phase velocity: *v*_{p} =Ā Ļ/k = (E/Ä§)/(āp/Ä§) = āE/p.

As you know, E/Ä§ = Ļ gives the *frequency in time* (expressed in radians per second), while p/Ä§ = k gives us the wavenumber, or the *frequency in space* (expressed in radians per meter). [If in doubt, check my post on essential wave math.] Now, you also know that *f* = Ļ/2ĻĀ and Ī» = 2Ļ/k, which gives us the two de Broglie relations:

- E = Ä§āĻ = hā
*f* - p = Ä§āk = h/Ī»

The frequency in time (oscillations or radians per second) is easy to interpret. A particle will always have some mass and, therefore, some energy, and it is easy to appreciate the fact that the wavefunction of a particle with more energy (or more mass) will have a *higher density in time* than a particle with less energy.

However, the second *de Broglie *relation is somewhat harder to interpret. Note that the wavelength is *inversely *proportional to the momentum: Ī» = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long, so we write:

If pĀ ā 0 then Ī»* āĀ ā.*

For the limit situation, a particle with zero *rest *mass (m_{0} = 0), the velocity may be *c* and, therefore, we find that p = m* _{v}*ā

*v*= m

*ā*

_{c}*c*= mā

*c*(all of the energy is kinetic) and, therefore, pā

*c*= mā

*c*

^{2}= E, which we may also write as: E/p =

*c*. Hence, for a particle with zero rest mass (m

_{0}Ā = 0), the wavelength can be written as:

Ī» = h/p = h*c*/E = h/m*c*

Of course, we are talking a *photon *here. We get the zero rest mass for a photon. In contrast, all *matter*-particles should have *some *mass[1] and, therefore, their velocity will *never*Ā equalĀ *c*.[2] The question remains: how should we interpret the inverse proportionality between *Ī»* and p?

Let us first see what this wavelength Ī» actually represents. If we look at the Ļ = aĀ·*cos*(pāx/Ä§ ā Eāt/Ä§) āĀ *iĀ·aĀ·sin*(pāx/Ä§ – Eāt/Ä§) once more, and if we write pāx/Ä§ as Ī, then we can look at pāx/Ä§ as a phase factor, and so we will be interested to know for what x this phase factor Ī = pāx/Ä§ will be equal to 2Ļ. So we write:

Ī =pāx/Ä§ = 2Ļ āĀ x = 2ĻāÄ§/p = h/p = Ī»

So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: they depend on your frame of reference.

So now we know what Ī» actually represent for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/*f *= 2ĻĀ·(Ä§/E). Hence, we can now calculate the wave velocity:

*v* = Ī»/T = (h/p)/[2ĻĀ·(Ä§/E)] = E/p

Unsurprisingly, we just get the phase velocity that we had calculated already: *v *= *v*_{p} = E/p. It does not answer the question: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite phase velocity. Now, we know *phase *velocities can be superluminal, but they should not be infinite. So what does the mathematical inconsistency tell us? Do these infinitely long wavelengths and infinite wave velocities tell us that our particle *has to* move? Do they tell us our notion of a particle at rest is mathematically inconsistent?

Maybe. But maybe not. Perhaps the inconsistency just tells us our elementary wavefunction ā or the concept of a *precise *energy, and a *precise *momentum ā does not make sense. This is where the Uncertainty Principle comes in: stating that p = 0, implies zero uncertainty. Hence, the Ļ_{p} factor in the Ļ_{p}āĻ_{x}Ā ā¤ Ä§/2 would be zero and, therefore, Ļ_{p}āĻ_{x} would be zero which, according to the Uncertainty Principle, it cannot be: it can be very small, but it cannot be zero.

It is interesting to note here that Ļ_{p} refers to the standard deviation from the mean, as illustrated below. Of course, the distribution may be or may not be normal ā we donāt know ā but a normal distribution makes intuitive sense, of course. Also, if we assume the mean is zero, then the uncertainty is basically about the *direction *in which our particle is moving, as the momentum might then be positive *or *negative.

The question of natural units may pop up. The Uncertainty Principle suggests a *numerical *value of the natural unit for momentum and distance that is equal to the *square root *of Ä§/2, so thatās about 0.726Ć10^{ā17} m for the distance unit and 0.726Ć10^{ā17} Nās for the momentum unit, as the product of both gives us Ä§/2. To make this somewhat more real, we may note that 0.726Ć10^{ā17} m is the attometer scale (1 am = 1Ć10^{ā18} m), so that is very small but not unreasonably small.[3]

Hence, we need to superimpose a potentially infinite number of waves with energies and momenta centered on some mean value. It is only then that we get meaningful results. For example, the idea of a *group *velocity ā which should correspond to the classical idea of the velocity of our particle ā only makes sense in the context of wave *packet*. Indeed, the group velocity of a wave packet (*v*_{g}) is calculated as follows:

*v*_{g}Ā = āĻ* _{i}*/āk

*Ā = ā(E*

_{i}*/Ä§)/ā(p*

_{i}*/Ä§) = ā(E*

_{i}*)/ā(p*

_{i}*)*

_{i}This assumes the existence of a *dispersion relation* which gives us Ļ* _{i}*Ā as a function of k

*ā what amounts to the same ā E*

_{i}*Ā as a function of p*

_{i}*. How do we get that?Ā Well… There are a few ways to go about it but one interesting way of doing it is to re-write SchrĆ¶dinger’s equation as the following*

_{i}*pair*Ā of equations[4]:

*Re*(āĻ/āt) = ā[Ä§/(2m_{eff})]Ā·*Im*(ā^{2}Ļ)Ā āĀ ĻĀ·cos(kx ā Ļt) =Ā k^{2}Ā·[Ä§/(2m_{eff})]Ā·cos(kx ā Ļt)*Im*(āĻ/āt) = [Ä§/(2m_{eff})]Ā·*Re*(ā^{2}Ļ)Ā ā ĻĀ·sin(kx ā Ļt) = k^{2}Ā·[Ä§/(2m_{eff})]Ā·sin(kx ā Ļt)

These equations imply the following dispersion relation:

Ļ =Ā Ä§Ā·k^{2}/(2m)

Of course, we need to think about the subscripts now: we have Ļ* _{i}*, k

*, but… What about m*

_{i}_{eff}or, dropping the subscript, about m? Do we write it as m

*? If so, what is it? Well… It is the*

_{i}*equivalent*mass of E

*Ā obviously, and so we get it from the mass-energy equivalence relation: m*

_{i}*Ā = E*

_{i}*/*

_{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, and the two will, obviously, be related as follows: Ļ

_{m}= Ļ

_{E}/

*c*

^{2}. We are tempted to do a few substitutions here. Letās first check what we get when doing the m

*Ā = E*

_{i}*/*

_{i}*c*

^{2}substitution:

Ļ* _{i}* =Ā Ä§Ā·k

_{i}^{2}/(2m

*) = (1/2)āÄ§Ā·k*

_{i}

_{i}^{2}ā

*c*

^{2}/E

*= (1/2)āÄ§Ā·k*

_{i}

_{i}^{2}ā

*c*

^{2}/(Ļ

*āÄ§)Ā = (1/2)āÄ§Ā·k*

_{i}

_{i}^{2}ā

*c*

^{2}/Ļ

_{i}ā Ļ_{i}^{2}/k_{i}^{2} = *c*^{2}/2Ā ā Ļ* _{i}*/k

*=*

_{i}*v*

_{p}=

*c*/2 !?

We get a very interesting but nonsensical *condition* for the dispersion relation here. I wonder what mistake I made. š¦

Let us try another substitution. The group velocity is what it is, right? It is the velocity of the group, so we can write: k* _{i}* = p/Ä§ = m

_{i}Ā·

*v*

*. This gives us the following result:*

_{g}Ļ* _{i}* =Ā Ä§Ā·(m

*Ā·*

_{i}*v*

_{g})

^{2}/(2m

*) = Ä§Ā·m*

_{i}*Ā·*

_{i}*v*

_{g}

^{2}/2

It is yet another interesting *condition *for the dispersion relation. Does it make any more sense? I am not so sure. That factor 1/2 troubles us. It only makes sense when weĀ *dropĀ *it. Now you will object that SchrĆ¶dinger’s equation gives us the electron orbitals – and many other correct descriptions of quantum-mechanical stuff – so, surely,Ā SchrĆ¶dinger’s equation cannot be wrong. You’re right. It’s just that… Well… When we are splitting in up in two equations, as we are doing here, then we are looking atĀ *oneĀ *of the two dimensions of the oscillation only and, therefore, it’s onlyĀ *halfĀ *of the mass that counts. Complicated explanation but… Well… It should make sense, because the results that come out make sense. Think of it. So we write this:

*Re*(āĻ/āt) = ā(Ä§/m_{eff})Ā·*Im*(ā^{2}Ļ)Ā āĀ ĻĀ·cos(kx ā Ļt) =Ā k^{2}Ā·(Ä§/m_{eff})Ā·cos(kx ā Ļt)*Im*(āĻ/āt) = (Ä§/m_{eff})Ā·*Re*(ā^{2}Ļ)Ā ā ĻĀ·sin(kx ā Ļt) = k^{2}Ā·(Ä§/m_{eff})Ā·sin(kx ā Ļt)

We then get the dispersion relation *withoutĀ *that 1/2 factor:

Ļ* _{i}* =Ā Ä§Ā·k

_{i}^{2}/m

_{i}TheĀ m* _{i}*Ā = E

*/*

_{i}*c*

^{2}substitution then gives us the result we sort of expected to see:

Ļ* _{i}* =Ā Ä§Ā·k

_{i}^{2}/m

*Ā = Ä§Ā·k*

_{i}

_{i}^{2}ā

*c*

^{2}/E

*= Ä§Ā·k*

_{i}

_{i}^{2}ā

*c*

^{2}/(Ļ

*āÄ§)Ā ā Ļ*

_{i}*/k*

_{i}*=*

_{i}*v*

*=*

_{p}*c*

Likewise, the other calculation also looks more meaningful now:

Ļ* _{i}* =Ā Ä§Ā·(m

*Ā·*

_{i}*v*

_{g})

^{2}/m

*Ā = Ä§Ā·m*

_{i}*Ā·*

_{i}*v*

_{g}

^{2}

Sweet ! š

Let us put this aside for the moment and focus on something else. If you look at the illustrations above, you see we can sort of distinguish (1) a linear velocity ā the speed with which those wave crests (or troughs) move ā and (2) some kind of circular or tangential velocity ā the velocity along the red contour lineĀ above. Weāll need the formula for a tangential velocity: *v*_{t} = *a*āĻ.

Now, if Ī» is zero, then *v*_{t} = *a*āĻ = *a*āE/Ä§ is just all there is. We may double-check this as follows: the distance traveled in one period will be equal to 2Ļ*a*, and the period of the oscillation is T = 2ĻĀ·(Ä§/E). Therefore, *v*_{t} will, effectively, be equal to *v*_{t} = 2Ļ*a*/(2ĻÄ§/E) = *a*āE/Ä§.Ā However, if Ī» is non-zero, then the distance traveled in one period will be equal to 2Ļ*a *+ Ī». The period remains the same: T = 2ĻĀ·(Ä§/E). Hence, we can write:

For an electron, we did this weird calculation. We had an angular momentum formula (for an electron) which we equated with theĀ *real-life* +Ä§/2 or āÄ§/2 values of its spin, and we got aĀ *numericalĀ *value forĀ *a*. It was the Compton radius: the scattering radius for an electron. Let us write it out:

Using the right numbers, youāll find the *numerical* value for *a*: 3.8616Ć10^{ā13} m. But let us just substitute the formula itself here:Ā

This is fascinating ! And we just calculated that *v*_{p}Ā is equal toĀ *c*. For the elementary wavefunction, that is. Hence, we get this amazing result:

*v*_{t} = 2*c*

ThisĀ *tangentialĀ *velocity isĀ *twiceĀ *the *linearĀ *velocity !

Of course, the question is: what is theĀ *physicalĀ *significance of this? I need to further look at this. Wave velocities are, essentially, *mathematicalĀ *concepts only: the wave propagates through space, butĀ *nothing elseĀ *is really moving. However, the geometric implications are obviously quite interesting and, hence, need further exploration.

One conclusion stands out: all these results reinforce our interpretation of the speed of light as aĀ *propertyĀ *of the vacuum – or of the fabric of spacetime itself. š

[1] Even neutrinos should have some (rest) mass. In fact, the mass of the known neutrino flavors was estimated to be smaller than 0.12 eV/*c*^{2}. This mass combines the three known neutrino flavors.

[2] Using the Lorentz factor (Ī³), we can write the relativistically correct formula for the kinetic energy as KE = EĀ āĀ E_{0}Ā =Ā m_{v}*c*^{2}Ā ā m_{0}*c*^{2}Ā =Ā m_{0}Ī³*c*^{2}Ā ā m_{0}*c*^{2}Ā =Ā m_{0}*c*^{2}(Ī³ ā 1). As *v *approaches *c*, Ī³ approaches infinity and, therefore, the kinetic energy would become infinite as well.

[3] It is, of course, extremely small, but 1 am is the current sensitivity of the LIGO detector for gravitational waves. It is also thought of as the upper limit for the length of an electron, for quarks, and for fundamental strings in string theory. It is, in any case, 1,000,000,000,000,000,000 times larger than the order of magnitude of the Planck length (1.616229(38)Ć10^{ā35} m).

[4] 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. As for the equations, they are easily derived from noting that two complex numbers a +Ā *i*āb and c +Ā *i*ād are equal if, and only if, their real and imaginary parts are the same. Now, the āĻ/āt =Ā *i*ā(Ä§/m_{eff})āā^{2}Ļ equation amounts to writing something like this: a +Ā *i*āb =Ā *i*ā(c +Ā *i*ād). Now, remembering thatĀ *i*^{2}Ā = ā1, you can easily figure out thatĀ *i*ā(c +Ā *i*ād) =Ā *i*āc +Ā *i*^{2}ād = ā d +Ā *i*āc.

# The geometry of the wavefunction

My posts and article on the wavefunction as a gravitational wave are rather short on the exact geometry of the wavefunction, so let us explore that a bit here. By now, you know the formula for theĀ elementary wavefunction by heart:

Ļ =Ā *aĀ·e*^{āi[EĀ·t ā pāx]/Ä§} =Ā *aĀ·cos*(**p**ā**x**/Ä§ ā Eāt/Ä§) + *iĀ·aĀ·sin*(**p**ā**x**/Ä§ ā Eāt/Ä§)

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**ā**x**/Ä§ reduces to pāx/Ä§. This amounts to saying our particle is traveling along the x-axis. The geometry of the wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.

Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. If the origin is the (x, t) = (0, 0) point, then cos(Īø) = cos(0) = 1 and sin(Īø) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. I am convinced these correspond to the two possible values for the quantum-mechanical spin of the wave: +Ä§/2 or āÄ§/2. But… Well… Who am I? The *cosine *and *sine *components are shown below. Needless to say, the *cosine *and *sine *function are the same, except for a phase difference of Ļ/2: *sin*(Īø) = *cos*(Īø ā Ļ/2) Ā

Surely, Nature doesn’t care a hoot about our conventions for measuring the phase angle clockwise or counterclockwise and therefore, the Ļ =Ā *aĀ·e*^{i}^{[EĀ·t ā pāx]/Ä§} function should, effectively, also be permitted. We know that *cos*(Īø) = *cos**(**–*Īø) and *sin*Īø = *–**sin**(**–*Īø), so we can write: Ā *Ā Ā *

Ļ =Ā *aĀ·e*^{i}^{[EĀ·t ā pāx]/Ä§} =Ā *aĀ·cos*(Eāt/Ä§ ā pāx/Ä§) +* iĀ·aĀ·sin*(Eāt/Ä§ ā pāx/Ä§)

= *aĀ·cos*(pāx/Ä§ ā Eāt/Ä§) āĀ *iĀ·aĀ·sin*(pāx/Ä§ ā Eāt/Ä§)

E/Ä§ = Ļ gives the frequency in time (expressed in radians per second), while p/Ä§ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: *f* = Ļ/2ĻĀ and Ī» = 2Ļ/k, which gives us the two de Broglie relations:

- E = Ä§āĻ = hā
*f* - p = Ä§āk = h/Ī»

The frequency in time is easy to interpret (a particle will always have some mass and, therefore, some energy), but the wavelength is *inversely *proportional to the momentum: Ī» = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long: if pĀ ā 0, then Ī»* āĀ ā.Ā *For the limit situation, a particle with zero *rest *mass (m_{0} = 0), the velocity may be *c* and, therefore, we find that p = m* _{v}*ā

*v*= mā

*c*Ā and, therefore, pā

*c*= mā

*c*

^{2}= E, which we may also write as: E/p =

*c*. Hence, for a particle with zero rest mass, the wavelength can be written as:

Ī» = h/p = hc/E = h/mc

However, we argued that the physical dimension of the components of the wavefunction may be usefully expressed in N/kg units (force per unit *mass*), while the physical dimension of the electromagnetic wave are expressed in N/C (force per unit *charge*). This, in fact, explains the dichotomy between bosons (photons) and fermions (spin-1/2 particles). Hence, all matter-particles should have *some *mass.[1] But how we interpret the inverse proportionality between Ī» and p?

We should probably first ask ourselves what wavelength we are talking about. The wave only has a phase velocity here, which is equal to *v*_{p} =Ā Ļ/k = (E/Ä§)/(p/Ä§) = E/p. Of course, we know that, *classically*, the momentum will be equal to the *group *velocity times the mass: p = mĀ·*v*_{g}. However, when p is zero, we have a division by zero once more: if pĀ ā 0, then *v*_{p} = E/pĀ ā ā. Infinite wavelengths and infinite phase velocities probably tell us that our particle *has to* move: our notion of a particle at rest is mathematically inconsistent. If we associate this elementary wavefunction with some particle, and if we then imagine it to move, somehow, then we get an interesting relation between the group and the phase velocity:

*v*_{p}Ā =Ā Ļ/k = E/p = E/(mĀ·*v*_{g}) = (mĀ·*c*^{2})/(mĀ·*v*_{g}) = *c*^{2}/*v*_{g}

We can re-write this as *v*_{p}Ā·*v*_{g}Ā = *c*^{2}, which reminds us of the relationship between the electric and magnetic constant (1/Īµ_{0})Ā·(1/Ī¼_{0}) = *c*^{2}. But what is the group velocity of the *elementary *wavefunction? Is it a meaningful concept?

The phase velocity is just the ratio of Ļ/k. In contrast, the group velocity is the derivative of Ļ with respect to k. So we need to write Ļ as a function of k. Can we do that even if we have only one wave? We doĀ *notĀ *have a wave packet here, right? Just some hypotheticalĀ *building blockĀ *of a real-life wavefunction, right? Right. So we should introduce uncertainty about E and p and build up the wave packet, right? Well… Yes. But let’s wait with that, and see how far we can go in our interpretation of thisĀ *elementaryĀ *wavefunction. Let’s first get thatĀ Ļ = Ļ(k) relation. You’ll remember we can write SchrĆ¶dingerās equation – the equation that describes theĀ *propagationĀ *mechanism for matter-waves – asĀ the following *pair *of equations:

*Re*(āĻ/āt) = ā[Ä§/(2m)]Ā·*Im*(ā^{2}Ļ)Ā āĀ ĻĀ·cos(kx ā Ļt) =Ā k^{2}Ā·[Ä§/(2m)]Ā·cos(kx ā Ļt)*Im*(āĻ/āt) = [Ä§/(2m)]Ā·*Re*(ā^{2}Ļ)Ā ā ĻĀ·sin(kx ā Ļt) = k^{2}Ā·[Ä§/(2m)]Ā·sin(kx ā Ļt)

This tells us that Ļ =Ā Ä§Ā·k^{2}/(2m). Therefore, we can calculate āĻ/ākĀ as:

āĻ/āk = Ä§Ā·k/m = p/m = *v*_{g}

We learn nothing new. We are going round and round in circles here, and we always end up with a tautology: as soon as we have a non-zero momentum, we have a *mathematical*Ā formula for the group velocity – but we don’t know what it represents – and a finite wavelength. In fact, using the p = Ä§āk = h/Ī» relation, we can write one as a function of the other:

Ī» = h/p = h/m*v*_{g} āĀ *v*_{g} = h/mĪ»

What does this *mean*? ItĀ resembles the *c* = h/mĪ» relation we had for a particle with zero rest mass. Of course, it does: the Ī» = h/m*c* relation is, once again, a limit for *v*_{g} going to c. By the way, it is interesting to note that the *v*_{p}Ā·*v*_{g}Ā = *c*^{2} relation implies that the *phase *velocity is always superluminal. That’ easy to see when you re-write the equation in terms ofĀ *relativeĀ *velocities: (*v*_{p}/*c*)Ā·(*v*_{g}/*c*) =Ā Ī²* _{phase}*Ā·Ī²

*Ā = 1. Hence, ifĀ Ī²*

_{group}*Ā < 1, thenĀ Ī²*

_{group}*Ā > 1.*

_{phase}So whatĀ *isĀ *the geometry,Ā *really*? Letās look at the Ļ = *aĀ·cos*(pāx/Ä§ – Eāt/Ä§) *–** iĀ·aĀ·sin*(pāx/Ä§ – Eāt/Ä§) formula once more. If we write pāx/Ä§ as Ī, then we will be interested to know for what x this phase factor will be equal to 2Ļ. So we write:

Ī =pāx/Ä§ = 2Ļ āĀ x = 2ĻāÄ§/p = h/p = Ī»*Ā Ā *

So now we get a meaningful interpretation for that wavelength: it’s that distance between the crests of the wave, so to speak, as illustrated below.

Can we now find a *meaningful *(i.e.Ā geometric) interpretation for the group and phase velocity? If you look at the illustration above, you see we can sort of distinguish (1) a linear velocity (the speed with which those wave crests move) and (2) some kind of circular or tangential velocity (the velocity along the red contour lineĀ above). Weāll probably need the formula for the tangential velocity: *v* = *a*āĻ. If p = 0 (so we have that weird infinitesimally long wavelength), then we have two velocities:

- The tangential velocity around theĀ
*aĀ·e*^{i}^{Ā·EĀ·t}Ā circle, so to speak, and that will just be equal toĀ*v*=*a*āĻ =Ā*a*āE/Ä§. - The red contour line sort of gets stretched out, like infinitely long, and the velocity becomes… What does it do? Does it go toĀ ā , or toĀ
*c*?

Let’s think about this. For a particle at rest, we had this weird calculation. We had an angular momentum formula (for an electron) which we equated with theĀ *real-life* +Ä§/2 or āÄ§/2 values of its spin. And so we got aĀ *numericalĀ *value forĀ *a*. It was the Compton radius: the scattering radius for an electron. Let me copy it once again:

Just to bring this story a bit back to Earth, you should note the calculated value:Ā *aĀ *= 3.8616Ć10^{ā13} m.Ā We did then another weird calculation. We said all of the energy of the electron had to be packed in thisĀ *cylinderĀ *that might of might not be there. The point was: the energy is finite, so thatĀ *elementaryĀ *wavefunction can*notĀ *have an infinite length in space. Indeed, assuming that the energy was distributed uniformly, we jotted down this formula, which reflects the formula for theĀ *volume *of a cylinder:

E = ĻĀ·*a*^{2}Ā·*l* āĀ *lĀ *= E/(ĻĀ·*a*^{2})

Using the value we got for the Compton scattering radius (*aĀ *=Ā 3.8616Ć10^{ā13} m), we got an astronomical value forĀ *l*. Let me write it out:

*lĀ *=Ā (8.19Ć10^{ā14})/(ĻĀ·14.9Ć10^{ā26}) ā 0.175Ć10^{12}Ā m

It is,Ā *literally*, an astronomical value:Ā 0.175Ć10^{12}Ā m is 175 *millionĀ kilo*meter, so that’s like theĀ distance between the Sun and the Earth. We wrote, jokingly, that such space is too large to look for an electron and, hence, that we should really build a proper *packet *by making use of the Uncertainty Principle: allowing for uncertainty in the energy should, effectively, reduce the uncertainty in position.

But… Well… What if we use that value as the value forĀ Ī»? We’d get that linear velocity, right? Let’s try it. TheĀ *periodĀ *is equal to T =Ā T = 2ĻĀ·(Ä§/E) = h/E and Ī» =Ā E/(ĻĀ·*a*^{2}), so we write:We can write this as a function of m and theĀ *cĀ *andĀ Ä§ constants only:

A weird formula but not necessarily nonsensical: we get a finite speed. Now, if the wavelength becomes somewhat less astronomical, we’ll get different values of course. I have a strange feeling that, with these formula, we should, somehow, be able to explain relativistic length contraction. But I will let you think about that as for now. Here I just wanted toĀ *showĀ *the geometry of the wavefunction a bit more in detail.

[1]Ā The discussions on the mass of neutrinos are interesting in this regard. Scientists all felt the neutrinoĀ *had toĀ *haveĀ some (rest) mass, so my instinct on this is theirs. In fact, only recently experimental confirmation came in, and the mass of the known neutrino flavors was estimated to be something like 0.12 eV/*c*^{2}. This mass combines the three known neutrino flavors. To understand this number, you should note it is the same order of magnitude of the equivalent mass of low-energy photons, like infrared or microwave radiation.

# This year’s Nobel Prize for Physics…

One of my beloved brothers just sent me the news on this year’s Nobel Prize for Physics. Of course, it went to the MIT/Caltech LIGO scientists – who confirmed the reality of gravitational waves. That’s exactly the topic that I am exploring when trying to digest all this quantum math and stuff. Brilliant !

I actually sent the physicists a congratulatory message – and my paper ! I can’t believe I actually did that.

In the best case, I just made a fool of myself. In the worst case… Well… I just made a fool of myself. š