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. đ I added a footnote, though, which is interesting to reflect upon. Also, I know many of my readers aren’t interested in the mathâeven 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 360/720Â° symmetry in quantum physics orâto be preciseâwe 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 aÂ reference frameÂ and a representational baseÂ (or representationÂ tout court). I won’t repeat that explanation, but think of the following.

If we just rotate the reference frame over 360Â°, we’re just using the same reference frame and so we see the same thing: some object which we, vaguely, describe by someÂ eiÂˇÎ¸Â 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 spinningâas 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 someÂ eâiÂˇÎ¸Â function (again, you should read: aÂ FourierÂ sum of such functions). Now, the weird thing is is the following: if we rotate the object itself, over the sameÂ 360Â°, we get aÂ differentÂ object: ourÂ eiÂˇÎ¸Â andÂ eâiÂˇÎ¸Â function (again: think of aÂ FourierÂ sum, so that’s a waveÂ packet, really) becomes aÂ âeÂąiÂˇÎ¸Â thing. We get aÂ minusÂ sign in front of it.Â 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 fullÂ 360Â° (so that’s a rotation of your reference frame over 360Â°), then you’ll end up where you were when you started and, importantly, you’ll see the same thing: me. đ I mean… You’ll seeÂ exactlyÂ the same thing: if I was anÂ e+iÂˇÎ¸Â wave packet, I am still anÂ anÂ e+iÂˇÎ¸Â wave packet now. OrÂ if I was an eâiÂˇÎ¸Â wave packet, then I am still anÂ an eâiÂˇÎ¸Â wave packet now. Easy. Logical. Obvious, right?

But so now we try something different:Â IÂ turn around, over a fullÂ 360Â° turn, and youÂ stay where you are and watch meÂ 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 wasâlooking at you again, so to speakâthen… Well… I am not quite the same any more. Or… Well… Perhaps I am but youÂ seeÂ me differently. If I wasÂ e+iÂˇÎ¸Â wave packet, then I’ve become aÂ âe+iÂˇÎ¸Â wave packet now.

Not hugely different but… Well… ThatÂ minusÂ sign matters, right? OrÂ If I wasÂ wave packet built up from elementaryÂ aÂˇeâiÂˇÎ¸Â waves, then I’ve become aÂ âeâiÂˇÎ¸Â wave packet now. What happened?

It makes me think of the twin paradox in special relativity. We know it’s aÂ paradoxâso that’s anÂ apparentÂ 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 someÂ force.

Can we relate this to the twin paradox? Maybe. Note that aÂ minusÂ sign in front of theÂ eâÂąiÂˇÎ¸Â 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 180Â°: âcosÎ¸ =Â cos(Î¸ Âą Ď) andÂ âsinÎ¸ =Â sin(Î¸ Âą Ď). Now, adding or subtracting aÂ commonÂ phase factor to/from the argument of the wavefunction amounts toÂ changingÂ the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360Â° and 720Â° symmetries are, effectively, related. đ

Post scriptum:Â GoogleÂ honors Max Born’s 135th birthday today. đ I think that’s a great coincidence in light of the stuff I’ve been writing about lately (possible interpretations of the wavefunction). đ

Wavefunctions, perspectives, reference frames, representations and symmetries

Ouff ! This title is quite a mouthful, isn’t it? đ 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 (orÂ physical) interpretations of the wavefunction is the fact that a wavefunction does not seem to obey the classical 360Â° 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 (orÂ representation, tout court)Â 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 frameâand so that explains the weird stuff (like that 720Â° 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 paperÂ first.

The reality of directions

Huh? TheÂ realityÂ of directions? Yes. I warned you. This post may cause brain damage. đÂ The whole argument revolves around a thoughtÂ experimentâbut one whose results have been verified in zillions of experiments in university student labs so… Well… We do notÂ doubt the results and, therefore, we do not doubt the basic mathematical results: we just want to try to understandÂ 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 âimprovedâ Stern-Gerlach apparatuses. The apparatus basically splits and then re-combines the two new beams along theÂ z-axis. It is also possible to block one of the beams, so we filter out only particles with their spinÂ upÂ or, alternatively, with their spinÂ down. Spin (or angular momentum or the magnetic moment) as measured along theÂ z-axis, of courseâI should immediately add: we’re talking theÂ z-axis of the apparatus here.

The two situations involve a different relative orientation of the apparatuses: in (a), the angle is 0Â°, while in (b) we have a (right-handed) rotation of 90Â° about the z-axis. He then provesâusing geometry and logic onlyâthat the probabilities and, therefore, the magnitudes of the amplitudes (denoted byÂ C+ and Câ and Câ+ and Cââ in the S and T representation respectively) must be the same, but the amplitudes must have different phases, notingâin his typical style, mixing academic and colloquial languageâthat âthere must be some way for a particle to tell that it has turned a corner in (b).â

The various interpretations of what actually happens here may shed some light on the heated discussions on the reality of the wavefunctionâand of quantum states. In fact, I should note that Feynman’s argument revolves around quantum states. To be precise, the analysis is focused on two-state systems only, and the wavefunctionâwhich captures a continuum of possible states, so to speakâis introduced only later. However, we may look at the amplitude for a particle to be in theÂ up– or down-state as a wavefunction and, therefore (but do note that’s my humble opinion once more), the analysis is actuallyÂ notÂ 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 (90Â°), we know that the amplitudes are given by Câ+ = eiâĎ/2âC+ = e iâĎ/4âC+ and Cââ = eâiâĎ/2âC+ = eâ iâĎ/4âCâ respectively (the amplitude to go from the down to the up state, or vice versa, is zero). Hence, yes, weânotÂ the particle, Mr. Feynman!âknowÂ that, in (b), the electron has, effectively, turned a corner.

The more subtle question here is the following: is the reality of the particle in the two setups the same? Feynman, of course, stays away from such philosophical question. He just notes that, while â(a) and (b) are differentâ, âthe probabilities are the sameâ. 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 turnâso it is just going in some other direction. Thatâs all.

However, common sense is seldom a good guide when thinking about quantum-mechanical realities. Also, from a more philosophical point of view, one may argue that the reality of the particle is not the same: something mightâor mustâhave happened to the electron because, when everything is said and done, the particle did take a turn in (b). It did not in (a). [Note that the difference between âmightâ and âmustâ 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…Â Wait a minute…Â Nothing is obvious in quantum mechanics, right? How can weÂ experimentally confirmÂ thatÂ they are different?

Huh?Â I must be joking, right? You canÂ seeÂ they are different, right? No.Â I am not joking. In physics, two things are different if we get differentÂ measurementÂ 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 thingâsame probabilities, remember?âwhy are they different? Think of this: if we look at the two beam splitters as one singleÂ tube (anÂ 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 sameÂ even when it takes a turn, we could say the tube is still the same, despite us having wrenched it over a 90Â° corner.

Now, I am sure you think I’ve just gone nuts, but just tryÂ to stick with me a little bit longer. Feynman actually acknowledges the same: we need to experimentallyÂ proveÂ (a) and (b) are different. He does so by getting aÂ thirdÂ apparatus in (U), as shown below, whose relative orientation to T is the same in both (a) and (b), so there is no difference there.

Now, the axis ofÂ UÂ is not theÂ z-axis: it is theÂ x-axis in (a), and theÂ y-axis in (b). So what? Well… I will quote Feynman hereânot (only) because his words are more important than mine but also because every word matters here:

“The two apparatuses in (a) and (b) are, in fact, different, as we can see in the following way. Suppose that we put an apparatus in front ofÂ SÂ which produces a pure +xÂ state. Such particles would be split into +z andÂ âz intoÂ beams inÂ S,Â but the two beams would be recombined to give aÂ +xÂ state again at P1âthe exit ofÂ S.Â The same thing happens again inÂ T.Â If we followÂ TÂ by a third apparatusÂ U,Â whose axis is in the +xÂ direction and, as shown in (a), all the particles would go into the +Â beam ofÂ U.Â Now imagine what happens ifÂ TÂ and UÂ are swung aroundÂ togetherÂ by 90Â°Â to the positions shown in (b).Â Again, theÂ TÂ apparatus puts out just what it takes in, so the particles that enterÂ UÂ are in a +xÂ stateÂ with respect toÂ S,Â which is different. By symmetry, we would now expect only one-half of the particles to get through.”

I should note that (b) shows theÂ UÂ apparatus wide open so… Well… I must assume that’s a mistake (and should alert the current editors of the LecturesÂ to it): Feynman’s narrative tells us we should also imagine it with theÂ minus channel shut. InÂ thatÂ case, it should, effectively, filter approximately half of the particles out, while they all get through in (a). So that’s aÂ measurementÂ result which shows the direction, as weÂ seeÂ it, makes a difference.

Now, Feynman would be very angry with meâbecause, as mentioned, he hates philosophersâbut I’d say: this experiment proves that a direction is something real. Of course, the next philosophical question then is: whatÂ isÂ a direction? I could answer this by pointing to the experiment above: a direction is something that alters the probabilities between the STU tube as set up in (a) versus the STU tube in (b). In factâbut, I admit, that would be pretty ridiculousâwe could use the varying probabilities as we wrench this tube over varying angles toÂ 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. đ 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 720Â° 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 frameâorÂ representation, as it’s referred to in quantum mechanicsâto another?

Well… I should immediately correct myself here: a reference frame and a representation are two different things. They areÂ relatedÂ but… Well… Different… Quite different. Not same-same but different. đ I’ll explain why later. Let’s go for it.

Before talking representations, let us first think about what we reallyÂ 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 ourÂ 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 theÂ shape, if that sounds easier to you) of the measurement apparatusÂ (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 usâ230 years agoâthat the mathematical idea of a three-dimensional reference frame is grounded in our intuitive notions of up and down, and left and right. [If you doubt this, think about the necessity of the various right-hand rules and conventions that we cannot do without in math, and in physics.] But so we do not want to sound academic. Let us be practical. Just think about the following.Â The apparatus gives us two directions:

(1) TheÂ upÂ direction, whichÂ weÂ associate with theÂ positive direction of theÂ z-axis, and

(2) the direction of travel of our particle, whichÂ we associateÂ with the positive direction of theÂ y-axis.

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

Of course, now you’ll say: if you think that ârelativeâ and âabsoluteâ are ambiguous terms and that we, therefore, may want to avoid them a bit more, then ârealâ and its opposite (unreal?) are ambiguous terms too, right? WellâŚ Maybe. What language would youÂ suggest? đ Just stick to the story for a while. I am not done yet. So… Yes… WhatÂ isÂ theirÂ reality?Â Let’s think about that in the next section.

Perspectives, reference frames and symmetries

You’ve done some mental exercises already as you’ve been working your way through the previous section, but you’ll need to do plenty more. In fact, they may become physical exercise too: when I first thought about these things (symmetries and, more importantly, asymmetries in space), I found myself walking around the table with some asymmetrical everyday objects and papers with arrows and clocks and other stuff on itâeffectively analyzing what right-hand screw, thumb or grip rules actuallyÂ mean. đ

So… Well… I want you to distinguishâjust for a whileâbetween the notion of a reference frame (think of the xyz reference frame that comes with the apparatus) and yourÂ 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 sideâwhich, if you think about it, you can only defineÂ in terms of the various positive and negative directions of the various axes. đÂ If you think this is getting ridiculous… Well… Don’t. Feynman himselfÂ 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 theÂ axesÂ (i.e. the reference frame) that comes with it. You don’t believe me? This is theÂ very first illustration of hisÂ LectureÂ 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 positiveÂ y-directionâso thatâs the direction in which our particle is movingâthen we might imagine how it would look like whenÂ weÂ would make a 180Â°Â turn and look at the situation from the other side, so to speak. 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 comingÂ towardsÂ us, out of the apparatus.

What happensâbut that’s not scientific language, of courseâis that left becomes right, and right becomes left. Top still is top, and bottom is bottom. We are looking now in theÂ negativeÂ y-direction, and the positive direction of the x-axisâwhich pointed right when we were looking in the positiveÂ y-directionânow points left. I see you nodding your head nowâbecause you’ve heard about parity inversions, mirror symmetries and what have youâand I hear you say: “That’s the mirror world, right?”

No. It is not. I wrote about this in another post: the world in the mirror is theÂ 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 toÂ this story, which is about transformations of amplitudes (or wavefunctions). [If you really want to knowâbut I know this sounds counterintuitiveâthe mirror world doesn’t really switch left for right. Your reflection doesn’t do a 180 degree turn: it is just reversed front to back, with no rotation at all. It’s only your brain which mentallyÂ 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 onlyÂ apparent. It’s a common misconception, and… Well… I’ll let you figure this out yourself. I need to move on.]Â Just note the following:

1. TheÂ xyzÂ 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. OurÂ perspectiveÂ on it has changed.
2. 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)âa cosine and sine function respectivelyâthen our change in perspectiveÂ might, effectively, mess up our convention for measuring angles.

I am not saying itÂ does. Not now, at least. I am just saying it might. It depends on the plane of the oscillation, as I’ll explain in a few moments. Think of this: we measure angles counterclockwise, right? As shown below… But… Well… If the thing below would be some funny clock going backwardsâyou’ve surely seen them in a bar or so, right?âthen… Well… If they’d be transparent, and you’d go around them, you’d see them as going… Yes… Clockwise. đ [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, ifÂ we wouldÂ assume this clock represents something realâand, of course, I am thinking of theÂ elementary wavefunctionÂ eiÎ¸Â =Â cosÎ¸ +Â iÂˇsinÎ¸ nowâthen… Well… Then it will look different when we go around it. When going around our backwards clock above and looking at it from… Well… The back, we’d describe it, naively, as… Well…Â Think! What’s your answer? Give me the formula!Â đ

[…]

We’d see it asÂ eâiÎ¸Â =Â cos(âÎ¸) +Â iÂˇsin(âÎ¸) =Â cosÎ¸ âÂ iÂˇsinÎ¸, right? The hand of our clock now goes clockwise, so that’s theÂ oppositeÂ direction of our convention for measuring angles. Hence, instead ofÂ eiÎ¸, we writeÂ eâiÎ¸, right? So that’s the complex conjugate. So we’ve got a differentÂ imageÂ of the same thing here. Not good. Not good at all.

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

You are right. WeÂ couldÂ use the right-hand rule more consistently. We could, in fact, use it as anÂ alternativeÂ convention for measuring angles: we could, effectively, measure them clockwise or counterclockwise depending on the direction of our particle.Â But… Well… The fact is:Â we don’t. We do not use that alternative convention when we talk about the wavefunction. Physicists do use theÂ counterclockwiseÂ convention all of the time and just jot down these complex exponential functions and don’t realize that,Â if they are to represent something real, ourÂ perspectiveÂ on the reference frame matters. To put it differently, theÂ directionÂ in which we are looking at things matters! Hence, the direction is not…Â Well… I am tempted to say… NotÂ relative at all but then… Well… We wanted to avoid that term, right? đ

[…]

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 symmetryâorÂ asymmetry, I should say.

The flywheel model of an electron

In our previous posts, we offered a model that interprets the real and the imaginary part of the wavefunction as oscillations which each carry half of the total energy of the particle. These oscillations are perpendicular to each other, and the interplay between both is how energy propagates through spacetime. Let us recap the fundamental premises:

1. The dimension of the matter-wave field vector is forceÂ 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/s2), which is the dimension of the gravitational field.
2. We assume this gravitational disturbance causes our electron (or a charged massÂ 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, âwhen you do find the electron some place, the entire charge is there.â Of course, we cannot prove anything here, but our elegant yet simple derivation of the Compton radius of an electron is… Well… Just nice. đ
3. 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 compriseÂ the direction of its linear motion. Hence, unlike an electromagnetic wave, theÂ planeÂ of the two-dimensional oscillation (so that’s the polarization plane, really) cannotÂ 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.Â The direction of the angular momentum (and the magnetic moment) of an electronâor, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is travelingâcannotÂ be parallel to the direction of motion. On the contrary, it must be perpendicularÂ 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. theÂ planeÂ of the polarization)Â has toÂ compriseÂ 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 explanationâcombined with the quantization hypothesisâgoes a long way in explaining this: an object with an angular momentumÂ JÂ and a magnetic momentÂ ÎźÂ that is not exactly parallel to some magnetic fieldÂ B, willÂ notÂ line up: it willÂ precessâand, as mentioned, the quantization of angular momentum may well explain the rest.Â [Well… Maybe… We haveÂ detailed our attempts in this regard in various posts on this (just search for spinÂ orÂ angular momentumÂ 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 theÂ up-direction only, then it will be spinningÂ clockwise if its angular momentum is down (so itsÂ magnetic moment isÂ up). Conversely, it will be spinningÂ counterclockwise if its angular momentum isÂ up. Let us take theÂ 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. đ 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 asâI am not ashamed to admit thisâI did when thinking about this. So what do we get when we change the perspective? Let us walk around it, counterclockwise, let’s say, so we’re measuring our angle of rotation as someÂ positiveÂ angle.Â Walking around itâin whatever direction, clockwise or counterclockwiseâdoesn’t change the counterclockwise direction of our… Well… That weird object that mightâjust mightâ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 wrenchedÂ eiÎ¸Â =Â cosÎ¸ +Â iÂˇsinÎ¸ function, right? The x- andÂ y-axesÂ 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 180Â° angle so we’re looking in theÂ negativeÂ 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 theÂ 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 theÂ y-axis to point away, rather than towards us. In short, we have a transformation of the reference frame here:Â z’ =Â z,Â y’ = âÂ y, andÂ x’ =Â âÂ x. Mind you, this is still a regular right-handed reference frame. [That’s the difference with aÂ mirrorÂ image: aÂ mirroredÂ right-hand reference frame is no longer right-handed.]Â So, in our new reference frame, that we choose to coincide with ourÂ perspective,Â we will now describe the same thing as someÂ âcosÎ¸ âÂ iÂˇsinÎ¸ =Â âeiÎ¸Â function. Of course,Â âcosÎ¸ =Â cos(Î¸ +Â Ď) andÂ âsinÎ¸ =Â sin(Î¸ +Â Ď) so we can write this as:

âcosÎ¸ âÂ iÂˇsinÎ¸ =Â cos(Î¸ +Â Ď) +Â iÂˇsinÎ¸ =Â eiÂˇ(Î¸+Ď)Â =Â eiĎÂˇeiÎ¸Â = âeiÎ¸.

Sweet ! But… Well… First note this isÂ notÂ the complex conjugate:Â eâiÎ¸Â =Â cosÎ¸ âÂ iÂˇsinÎ¸Â â Â âcosÎ¸ âÂ iÂˇsinÎ¸ =Â âeiÎ¸. 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. đ And, yes, let me lighten up the discussion with that painting here. đ We need to haveÂ someÂ fun while torturing our brain, right?

So, because we assume the plane of polarization is different, we get anÂ âeiÎ¸Â function instead of aÂ eâiÎ¸Â function.

Let us now think about the eiÂˇ(Î¸+Ď)Â function. It’s the same asÂ âeiÎ¸Â but… Well… We walked around theÂ z-axis taking a full 180Â° turn, right? So that’s Ď in radians. So that’s the phase shiftÂ here. Hey!Â Try the following now. Go back and walk around the apparatus once more, but letÂ 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 theÂ z-axis (which points out of this page, toward you, as you are looking at this), let’s say by some small angleÂ Îą. So we rotate the reference frame about theÂ z-axis byÂ Îą and… Well… Of course, ourÂ eiÂˇÎ¸Â now becomes anÂ ourÂ eiÂˇ(Î¸+Îą)Â function, right? We’ve just derived the transformation coefficient for a rotation about theÂ z-axis, didn’t we? It’s equal toÂ eiÂˇÎą, right? We get the transformed wavefunction in the new reference frame by multiplying the old one byÂ eiÂˇÎą, right? It’s equal toÂ eiÂˇÎąÂˇeiÂˇÎ¸Â =Â eiÂˇ(Î¸+Îą), right?

Well…

[…]

No. The answer is: no. TheÂ transformation coefficient is notÂ eiÂˇÎąÂ butÂ eiÂˇÎą/2. So we get an additional 1/2 factor in theÂ phase shift.

Huh?Â Yes.Â That’s what it is: when we change the representation, by rotating our apparatus over some angle Îą 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 onlyÂ half ofÂ the rotation angle only.

Huh?Â Yes. It’s even weirder than that. For a spin downÂ electron, the transformation coefficient is eâiÂˇÎą/2, so we get an additional minus sign in the argument.

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

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:

OurÂ eiÂˇÎąÂ coefficient describes a rotation of the reference frame. In contrast, theÂ eiÂˇÎą/2Â andÂ eâiÂˇÎą/2Â coefficients describe what happens when we rotate the T apparatus! Now thatÂ is a very different proposition.Â

Right! You got it! RepresentationsÂ and reference frames are different things.Â QuiteÂ different, I’d say: representations areÂ real, reference frames aren’tâbut then you don’t like philosophical language, do you? đÂ But think of it. When we just go about theÂ z-axis, a full 180Â°, but we don’t touch thatÂ T-apparatus, we don’t changeÂ 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 theÂ z-axis, a full 180Â°, 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 frameâfrom xyz to x’y’z’ to be precise: we doÂ not changeÂ the representation.

In contrast, when we rotate theÂ TÂ apparatus over a full 180Â°, 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 throughÂ S, and now it goes in the opposite directionârelative to the direction it was going in S, that is.

So what happens,Â really, when weÂ change the representation, rather than the reference frame? Well… Let’s think about that. đ

Quantum-mechanical weirdness?

The transformation matrix for the amplitude of a system to be in anÂ upÂ orÂ downÂ state (and, hence, presumably, for a wavefunction) for a rotation about theÂ z-axis is the following one:

Feynman derives this matrix in a rather remarkable intellectualÂ tour de forceÂ in the 6th of hisÂ 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 howÂ IÂ 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? đ So… Well… Because heÂ knowsâfrom experimentâthat the coefficient isÂ eiÂˇÎą/2Â instead of eiÂˇÎą, he just says the phase shiftâwhich he denotes by Îťâmust be someÂ proportionalÂ to the angle of rotationâwhich he denotes byÂ Ď rather than Îą (so as to avoid confusion with the EulerÂ angleÂ Îą). So he writes:

Îť =Â mÂˇĎ

Initially, he also tries the obvious thing: m should be one, right? SoÂ Îť = Ď, 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:

“SupposeÂ TÂ is rotated byÂ 360Â°; then, clearly, it is right back at zero degrees, and we should haveÂ Câ+ = C+Â andÂ Cââ =Â CâÂ or,Â what is the same thing,Â eiÂˇmÂˇ2ĎÂ = 1. We get m =Â 1. [But no!]Â This argument is wrong!Â To see that it is, consider thatÂ TÂ is rotated byÂ 180Â°. If mÂ were equal to 1, we would have Câ+ =Â eiÂˇĎC+Â = âC+Â and Cââ =Â eâiÂˇĎCâÂ =Â âCâ. [Feynman works with statesÂ here, instead of the wavefunction of the particle as a whole. I’ll come back to this.] However, this is just theÂ originalÂ state all over again.Â BothÂ amplitudes are just multiplied byÂ â1Â which gives back the original physical system. (It is again a case of a common phase change.) This means that if the angle betweenÂ TÂ andÂ SÂ is increased to 180Â°, the system would be indistinguishable from the zero-degree situation, and the particles would again go through the (+)Â state of theÂ UÂ apparatus. AtÂ 180Â°, though, the (+)Â state of theÂ UÂ apparatus is theÂ (âx)Â state of the originalÂ SÂ apparatus. So a (+x)Â state would become aÂ (âx)Â state. But we have done nothing toÂ changeÂ the original state; the answer is wrong. We cannot haveÂ m = 1.Â We must have the situation that a rotation byÂ 360Â°, andÂ no smaller angleÂ reproduces the same physical state. This will happen ifÂ m = 1/2.”

The result, of course, is this weird 720Â° symmetry. While we get the same physics after a 360Â° rotation of the T apparatus, we doÂ notÂ get the same amplitudes. We get the opposite (complex) number:Â Câ+ =Â eiÂˇ2Ď/2C+Â = âC+Â and Cââ =Â eâiÂˇ2Ď/2CâÂ =Â âCâ. That’s OK, because… Well… It’s aÂ commonÂ phase shift, so it’s just like changing the origin of time. Nothing more. Nothing less. Same physics. Same reality. But… Well…Â Câ+ â Â âC+Â andÂ Cââ â Â âCâ, right? We only get our original amplitudes back if we rotate theÂ T apparatus two times, so that’s by a full 720 degreesâas opposed to the 360Â° we’d expect.

Now, space is isotropic, right? So this 720Â° 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 orÂ âx, or t and ât.Â I’ve said this a couple of times already again, and I’ll keep saying it many times more:Â NatureÂ surely can’t be bothered by how we measure stuff, right? In the positive or the negative directionâthat’s just our choice, right?Â OurÂ convention. So… Well… It’s just like thatÂ âeiÎ¸Â function we got when looking at theÂ same experimental set-up from the other side: ourÂ eiÎ¸Â and âeiÎ¸Â functions didÂ notÂ describe a different reality. We just changed our perspective. TheÂ reference frame. As such, the reference frame isn’tÂ real. The experimental set-up is. AndâI know I will anger mainstream physicists with thisâtheÂ representationÂ is. Yes. Let me say it loud and clear here:

A different representation describes a different reality.

In contrast, a different perspectiveâor a different reference frameâdoes not.

Conventions

While you might have had a lot of trouble going through all of the weird stuff above, the point is: it isÂ notÂ all that weird. WeÂ canÂ 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 formulaâeiÎ¸Â =Â cosÎ¸ +Â iÂˇsinÎ¸âwould, one day, be used to representÂ something real: an electron, or any elementary particle, really. If he wouldÂ have known, I am sure he would have noted what I am noting here:Â NatureÂ can’t be bothered by our conventions. Hence, ifÂ eiÎ¸Â represents something real, thenÂ eâiÎ¸Â 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 notedâand, if he would have known about circularly polarized waves, probably agreed toâthatÂ alternative convention for measuring angles: we could, effectively, measure angles clockwise or counterclockwise depending on the direction of our particleâas opposed to Euler’s ‘one-size-fits-all’ counterclockwise convention. But so we didÂ notÂ adopt that alternative convention because… Well… We want to keep honoring Euler, I guess. đ

So… Well… If we’re going to keep honoring Euler by sticking to that ‘one-size-fits-all’ counterclockwise convention, then I doÂ believe thatÂ eiÎ¸Â and eâiÎ¸Â represent twoÂ differentÂ 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 seeÂ something different when they go through a Stern-Gerlach apparatus. So it’s not just some convention toÂ countÂ things like 0, 1, 2, etcetera versus 0,Â â1,Â â2 etcetera. It’s the same story again: different but relatedÂ mathematicalÂ notions are (often) related to different but relatedÂ physicalÂ possibilities. So… Well… I think that’s what we’ve got here.Â Think of it. Mainstream quantum math treats all wavefunctions as right-handed but… Well…Â A particle with up spin is a different particle than one withÂ downÂ spin, right? And, again,Â NatureÂ surely cannotÂ be bothered about our convention of measuring phase angles clockwise or counterclockwise, right? So… Well… Kinda obvious, right? đ

Let me spell out my conclusions here:

1. The angular momentum can be positive or, alternatively, negative: J = +Ä§/2 orÂ âÄ§/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 particleâthink of an electron, or whatever other particle you’d think ofâcomes in twoÂ variants:Â right-handed and left-handed. They will, therefore,Â either consist of (elementary) right-handed waves or,Â else, (elementary) left-handed waves. An elementary right-handed wave would be written as: Ď(Î¸i)Â = eiÎ¸iÂ = aiÂˇ(cosÎ¸i + iÂˇsinÎ¸i). In contrast,Â an elementary left-handed wave would be written as: Ď(Î¸i)Â =Â eâiÎ¸iÂ = aiÂˇ(cosÎ¸i â iÂˇsinÎ¸i).Â So that’s the complex conjugate.

So… Well… Yes, I think complex conjugates are not just someÂ mathematicalÂ notion: I believe they represent something real. It’s the usual thing:Â NatureÂ has shown us that (most) mathematical possibilities correspond to realÂ physical situations so… Well… Here you go. It is reallyÂ 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 differencesâdifferent polarization plane and dimensions and what have youâI’ve already summed those up, so I won’t repeat myself here.]Â The point is: ifÂ we have two differentÂ physicalÂ situations, we’ll want to have two different functions to describe it. Think of it like this: why would we haveÂ twoâyes, I admit, two relatedâamplitudes to describe the upÂ or downÂ state of the same system, but only one wavefunction for it?Â 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 (orÂ physical) interpretation of the wavefunction might be, it won’t be compatible with theÂ isotropyÂ of space. You cannot imagineÂ an object with a 720Â° symmetry. That’sÂ geometrically impossible.”

Well… Almost three years ago, I wrote the following on this blog: “As strange as it sounds, aÂ spin-1/2 particle needsÂ twoÂ full rotations (2Ă360Â°=720Â°) until it is again in the same state. Now, in regard to that particularity, youâll often read something like: âThere isÂ nothingÂ in our macroscopic world which has a symmetry like that.â Or, worse, âCommon sense tells us that something like that cannot exist, that it simply is impossible.â [I wonât quote the site from which I took this quotes, because it is, in fact, the site of a very respectable Â research center!]Â Bollocks!Â TheÂ Wikipedia article on spinÂ has this wonderful animation: look at how the spirals flip between clockwise and counterclockwise orientations, and note that itâs only after spinning a full 720 degrees that this âpointâ 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, weÂ canÂ 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 goodÂ image, I should say, because… Well… A representation is something real, remember? đ

Post scriptum (10 December 2017):Â Our flywheel model of an electron makes sense, but also leaves many unanswered questions. The most obvious one question, perhaps, is: why theÂ upÂ andÂ downÂ state only?

I am not so worried about that question, even if I can’t answer it right away because… Well… Our apparatusâthe way weÂ measureÂ realityâis set up to measure the angular momentum (or the magnetic moment, to be precise) in one direction only. If our electron isÂ capturedÂ by someÂ harmonicÂ (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 spinÂ upÂ situation (magnetic moment down) is quite peculiar: if our electron is aÂ mini-magnet, why would itÂ notÂ line up with the magnetic field? We understand the precession of a spinning top in a gravitational field, but…Â Hey… It’s actually not that different. Try to imagine some spinning top on the ceiling. đ I am sure we can work out the math. đ 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. đ […] At least, that’s why my instinct tells me. I admit I need to work out the math to convince you. đ

The second question is more important. If we just rotate the reference frame over 360Â°, we see the same thing: some rotating object which we, vaguely, describe by someÂ e+iÂˇÎ¸Â functionâto be precise, I should say: by some Fourier sum of such functionsâor, if the rotation is in the other direction, by someÂ eâiÂˇÎ¸Â function (again, you should read: aÂ FourierÂ 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 sameÂ 360Â°, we get aÂ differentÂ object: ourÂ eiÂˇÎ¸Â andÂ eâiÂˇÎ¸Â function (again: think of aÂ FourierÂ sum, so that’s a waveÂ packet, really) becomes aÂ âeÂąiÂˇÎ¸Â thing. We get aÂ minusÂ sign in front of it.Â 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 fullÂ 360Â°, then you’ll end up where you were when you started and, importantly, you’ll see the same thing.Â ExactlyÂ the same thing: if I was anÂ e+iÂˇÎ¸Â wave packet, I am still anÂ anÂ e+iÂˇÎ¸Â wave packet now. OrÂ if I was an eâiÂˇÎ¸Â wave packet, then I am still anÂ an eâiÂˇÎ¸Â wave packet now. Easy. Logical. Obvious, right?

But so now we try something different:Â IÂ turn around, over a fullÂ 360Â° turn, and youÂ stay where you are. When I am back where I wasâlooking at you again, so to speakâthen… Well… I am not quite the same any more. Or… Well… Perhaps I am but youÂ seeÂ me differently. If I wasÂ e+iÂˇÎ¸Â wave packet, then I’ve become aÂ âe+iÂˇÎ¸Â wave packet now. Not hugely different but… Well… ThatÂ minusÂ sign matters, right? OrÂ If I wasÂ wave packet built up from elementaryÂ aÂˇeâiÂˇÎ¸Â waves, then I’ve become aÂ âeâiÂˇÎ¸Â wave packet now. What happened?

It makes me think of the twin paradox in special relativity. We know it’s aÂ paradoxâso that’s anÂ apparentÂ 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 someÂ force.

Can we relate this to the twin paradox? Maybe. Note that aÂ minusÂ sign in front of theÂ eâÂąiÂˇÎ¸Â 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 180Â°: âcosÎ¸ =Â cos(Î¸ Âą Ď) andÂ âsinÎ¸ =Â sin(Î¸ Âą Ď). Now, adding or subtracting aÂ commonÂ phase factor to/from the argument of the wavefunction amounts toÂ changingÂ the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360Â° and 720Â° symmetries are, effectively, related. đ

The reality of the wavefunction

If you haven’t read any of my previous posts on the geometry of the wavefunction (this link goes to the most recent one of them), then don’t attempt to read this one. It brings too much stuff together to be comprehensible. In fact, I am not even sure if I am going to understand what I write myself. đ [OK. Poor joke. Acknowledged.]

Just to recap the essentials, I part ways with mainstream physicists in regard to theÂ interpretationÂ of the wavefunction. For mainstream physicists, the wavefunction is just some mathematical construct. NothingÂ real. Of course, I acknowledge mainstream physicists have very good reasons for that, but… Well… I believe that, if there is interference, or diffraction, thenÂ somethingÂ must be interfering, or something must be diffracting. I won’t dwell on this because… Well… I have done that too many times already. MyÂ hypothesisÂ is that the wavefunction is, in effect, aÂ rotatingÂ field vector, so itâs just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).

Of course, it must be different, and it is. First, theÂ (physical) dimension of the field vector of the matter-wave must be different. So what is it? Well… I am tempted to associate the real and imaginary component of the wavefunction with a forceÂ per unit massÂ (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s2), so thatâs the dimension of a gravitational field.

Second, I also am tempted to think that this gravitational disturbance causes an electron (or any matter-particle) to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves doÂ notÂ involve any mass: theyâre just an oscillatingÂ field. Nothing more. Nothing less. Why would I believe there must still be some pointlike particle involved? Well…Â As Feynman puts it: âWhen you do find the electron some place, the entire charge is there.â (FeynmanâsÂ Lectures, III-21-4) So… Well… That’s why.

The third difference is one that I thought of only recently: theÂ planeÂ of the oscillation cannotÂ be perpendicular to the direction of motion of our electron, because then we canât explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus. I am more explicit on that in the mentioned post, so you may want to check there. đ

I wish I mastered the software to make animations such as the one above (for which I have to credit Wikipedia), but so I don’t. You’ll just have toÂ imagineÂ it. That’s great mental exercise, so… Well… Just try it. đ

Let’s now think about rotating reference frames and transformations. If theÂ z-direction is the direction along which we measure the angular momentum (or the magnetic moment), then theÂ up-direction will be theÂ positiveÂ z-direction. We’ll also assume theÂ y-direction is the direction of travel of our elementary particleâand let’s just consider an electron here so we’re moreÂ real. đ So we’re in the reference frame that Feynman used to derive the transformation matrices for spin-1/2 particles (or for two-state systems in general). His ‘improved’ Stern-Gerlach apparatusâwhich I’ll refer to as a beam splitterâillustrates this geometry.

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

So the crucial difference with the animations above (which illustrate left- and a right-handed polarization respectively) is that we, somehow, need to imagine the circular motion isÂ notÂ in theÂ xz-plane, but in theÂ yz-plane. Now what happens if we change the reference frame?

Well… That depends on what you mean by changing the reference frame. Suppose we’re looking in the positive y-directionâso that’s the direction in which our particle is movingâ, then we might imagine how it would look like whenÂ weÂ would make a 180Â°Â turn and look at the situation from the other side, so to speak. Now, I did a post on that earlier this year, which you may want to re-read.Â When we’re looking at the same thing from the other side (from the back side, so to speak), we will want to use our familiar reference frame. So we will want to keep theÂ z-axis as it is (pointing upwards), and we will also want to define theÂ x– andÂ y-axis using the familiar right-hand rule for defining a coordinate frame. So our newÂ x-axis and our newÂ y-axis will the same as the oldÂ x- andÂ y-axes but with the sign reversed. In short, we’ll have the following mini-transformation: (1)Â z‘ =Â z, (2) x’ = âx, and (3) y’ =Â ây.

So… Well… If we’re effectively looking at somethingÂ realÂ that was moving along theÂ y-axis, then it will now still be moving along the y’-axis, butÂ in theÂ negativeÂ direction. Hence, our elementary wavefunctionÂ eiÎ¸Â = cosÎ¸ +Â iÂˇsinÎ¸ willÂ transformÂ intoÂ âcosÎ¸ âÂ iÂˇsinÎ¸ =Â âcosÎ¸ âÂ iÂˇsinÎ¸ =Â cosÎ¸ âÂ iÂˇsinÎ¸.Â It’s the same wavefunction. We just… Well… We just changed our reference frame. We didn’t change reality.

Now you’ll cry wolf, of course, because we just went through all that transformational stuff in our last post. To be specific, we presented the following transformation matrix for a rotation along theÂ z-axis:

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

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

Right. But… Well… Is itÂ so different, really? Suppose ourÂ eiÎ¸Â = cosÎ¸ +Â iÂˇsinÎ¸ wavefunction describes anÂ up-electron, then we still have to apply thatÂ eiĎ/2Â =Â eiĎ/2Â =Â +iÂ factor, right? So we get a new wavefunction that will be equal toÂ eiĎ/2ÂˇeiÎ¸Â =Â eiĎ/2ÂˇeiÎ¸Â =Â +iÂˇeiÎ¸Â =Â iÂˇcosÎ¸ +Â i2ÂˇsinÎ¸ =Â sinÎ¸ âÂ iÂˇcosÎ¸, right? So how can we reconcile that with the cosÎ¸ âÂ iÂˇsinÎ¸ function we thought we’d find?

We can’t. So… Well… Either my theory is wrong or… Well… Feynman can’t be wrong, can he? I mean… It’s not only Feynman here. We’re talking all mainstream physicists here, right?

Right. But think of it. Our electron in that thought experiment does, effectively, make a turn of 180Â°, so it is going in the other direction now !Â That’s more than just… Well… Going around the apparatus and looking at stuff from the other side.

Hmm… Interesting. Let’s think about the difference between theÂ sinÎ¸ âÂ iÂˇcosÎ¸ andÂ cosÎ¸ âÂ iÂˇsinÎ¸ functions. First, note that they will give us the same probabilities: the square of the absolute value of both complex numbers is the same. [It’s equal to 1 because we didn’t bother to put a coefficient in front.] Secondly, we should note that the sine and cosine functions are essentially the same. They just differ by a phase factor: cosÎ¸ =Â sin(Î¸ +Â Ď/2) andÂ âsinÎ¸ =Â cos(Î¸ +Â Ď/2). Let’s see what we can do with that. We can write the following, for example:

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

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

Transforming amplitudes for spin-1/2 particles

Some say it is not possibleÂ to fullyÂ understandÂ quantum-mechanical spin. Now, I do agree it is difficult, but I do notÂ 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 theÂ quantizationÂ of the angular momentum, which is that, in the quantum-mechanical world, the angular momentum is never quite all in one direction onlyâso 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 aÂ geometric (orÂ physical)Â interpretation of the wavefunction: the extra mile. đ

Now, in one of these many posts on spin and angular momentum, I advise my readers –Â you, that isÂ – 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 theÂ physicalÂ (orÂ geometric) interpretation of the wavefunction that we presented in previous posts is, somehow,Â true, then we need to relate it to the abstract math of these so-calledÂ transformationsÂ between representations.Â 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. đ

The thought experiment that Feynman uses throughout his LectureÂ 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 analyzeÂ 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 itselfâor, to be precise, ourÂ perspectiveÂ of itâgives us two directions: (1) theÂ upÂ 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 theÂ 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 +SÂ state,Â what is the probability (or theÂ amplitude) that it will get through aÂ second apparatus TÂ if that was set to filter out the +TÂ 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 argumentâwho, remember, represents mainstream physicsâsomehow 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 justÂ perspective: somethingÂ happenedÂ to the electron. That something does not onlyÂ changeÂ 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. đ Let’s focus on the math here. The philosophy will follow later. đÂ Next remark.

Second, we assume theÂ (a) and (b) illustrations above represent the sameÂ physicalÂ reality because the relative orientation between the two apparatuses, as measured by the angle Îą, is the same. NowÂ thatÂ isÂ 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 noÂ aetherÂ allowing us to establish some sense of absoluteÂ direction. Directions areÂ relativeârelative to the observer, that is… But… Well… Again, in my next post, I’ll argue that it’sÂ notÂ because directions areÂ relativeÂ that they are, somehow,Â notÂ real. Indeed, in my humble opinion, it does matter whether an electron goes here or, alternatively, there. These twoÂ differentÂ 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, FeynmanÂ also assumes the two set-ups below are, somehow,Â equivalent.

You’ll say: Huh?Â If not, say it!Â Huh? đÂ Yes. Good.Â Huh? Feynman writesÂ equivalentânotÂ the same because… Well… They’re not the same, obviously:

1. In the first set-up (a), TÂ is wide open, so the apparatus is not supposed to do anything with the beam: it just splits and re-combines it.
2. In set-up (b) theÂ TÂ apparatus is, quite simply,Â not there, so… Well… Again. Nothing is supposed to happen with our particles as they come out ofÂ S and travel toÂ U.

TheÂ 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!Â Now that is aÂ very tricky assumption, because… Well… While the netÂ turn of our electron is the same, it is quite obvious it has to takeÂ twoÂ turns to get to U in (a), while it only takesÂ oneÂ turn in (b). And so… Well… You can probably think of other differences too.Â So… Yes. And no.Â Same-same but different, right? đ

Right. That isÂ 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 TÂ in set-up (a), time goes by and, therefore, the amplitude would be different by someÂ phase factorÂ Î´. [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:Â Î¸ = (Eât âÂ pâx)/Ä§).]Â Now, if we believe that the amplitude is just some mathematical constructâso that’s what mainstream physicists (not me!) believeâthen weÂ couldÂ effectively say that the physics of (a) and (b) are the same, as Feynman does. In fact, let me quote him here:

“TheÂ physicsÂ 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 theÂ physicsÂ really be the same? I mean…Â ExactlyÂ the same? It all boils down to that unfathomable belief that, somehow, the particle is real but the wavefunction thatÂ describesÂ it, is not.Â 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 beÂ notÂ real. But… Well… The time and distance traveled in theÂ TÂ apparatus is the time and distance traveled in theÂ TÂ apparatus, right?

Bon…Â I have to stay away from these questions as for nowâwe need to move on with the math hereâbut 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 toÂ derive in this post, but… Well… Let’s assume we know them already. đ 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 RST,Â  RTU,Â  andÂ RSUÂ respectively. Adding the subscripts for theÂ baseÂ states in each representation, theÂ equivalenceÂ 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. đ Now, Feynman develops a beautiful mathematical argument to show that theÂ eiÎ´Â factor effectively disappears if weÂ convertÂ our rotation matrices to some rather specialÂ 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 beÂ very intriguing.] Indeed, the calculation below shows that theÂ determinantÂ of theseÂ specialÂ 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, theÂ eiÎ´Â factor vanishes and the formula above reduces to:

So… Yes. End of excursion.Â 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 +SÂ state,Â what is the probability (or theÂ amplitude) that it will get through aÂ second apparatus (T) if that was set to filter out the +TÂ state?

We said the result would depend on the angles between the two apparatuses S and T. I wrote: anglesâplural. Why? Because a rotation will generally be described by the three so-calledÂ Euler angles:Â  Îą, Î˛ and Îł. Now, it is easy to make a mistake here, because there is a sequence to these so-calledÂ elemental rotationsâand right-hand rules, of courseâbut I will let you figure that out. đ

The basic idea is the following: if we can work out the transformation matrices for each of theseÂ elementalÂ rotations, then we can combine them and find the transformation matrix forÂ anyÂ rotation. So… Well… That fills most of Feynman’sÂ LectureÂ 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. đ

So… TheÂ 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 theÂ up-axis of the apparatus. In the illustration below, that direction pointsÂ 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 thatâdespite the angleâtheÂ probabilityÂ for a particle to beÂ upÂ orÂ downÂ with regard toÂ SÂ andÂ T respectively should be the same. Well… Let’s be fair. He does not onlyÂ sayÂ that: experimentÂ showsÂ it to be true. [Again, I am tempted to interject here that it isÂ notÂ because the probabilities for (a) and (b) are the same, that theÂ realityÂ 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.]Â The probability is, of course, the square of theÂ absolute valueÂ of the amplitude, which we will denote asÂ C+,Â Câ, C’+, andÂ C’âÂ respectively. Hence, we can write the following:

Now, theÂ absolute values (or the magnitudes)Â are the same, but theÂ amplitudes may differ. In fact, theyÂ mustÂ 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 himselfâjokingly or seriously: “There must be some way for a particle to know that it has turned the corner at P1.” [P1Â is the midwayÂ pointÂ betweenÂ SÂ andÂ TÂ in the illustration, of courseânot some probability.]

So… Well… We write:

C’+Â =Â eiÎťÂ ÂˇC+Â andÂ C’âÂ =Â eiÎźÂ ÂˇCâ

Now, Feynman notes that anÂ equal phase change in all amplitudes has no physical consequence (think of re-defining our t0Â = 0 point), so we can add some arbitrary amount to bothÂ Îť and Îź without changing any of the physics. So then we canÂ chooseÂ this amount asÂ â(Îť + Îź)/2. We write:

Now, it shouldn’t you too long to figure out thatÂ Îť’ is equal toÂ Îť’ =Â Îť/2 + Îź/2 =Â âÎź’. So… Well… Then we can just adopt the convention thatÂ Îť = âÎź. So ourÂ C’+Â =Â eiÎťÂ ÂˇC+Â andÂ C’âÂ =Â eiÎźÂ ÂˇCâÂ equations can now be written as:

C’+Â =Â eiÎťÂ ÂˇC+Â andÂ C’âÂ =Â eâiÎťÂˇCâ

The absolute values are the same, but theÂ phasesÂ are different. Right. OK. Good move. What’s next?

Well… The next assumption is that the phase shiftÂ Îť is proportional to the angle (Îą) between the two apparatuses. Hence,Â Îť is equal to Îť =Â mÂˇÎą, and we can re-write the above as:

C’+Â =Â eimÎąÂˇC+Â andÂ C’âÂ =Â eâimÎąÂˇCâ

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 weÂ canÂ writeÂ Îť as Îť =Â mÂˇÎą. The next question then is:Â what is the value for m? Now, we obviously do get exactly the same physicsÂ if we rotateÂ TÂ by 360Â°, or 2Ď radians. So weÂ mightÂ conclude that the amplitudes should be the same and, therefore, that eimÎąÂ =Â eimÂˇ2ĎÂ has to be equal to one, soÂ C’+Â =Â C+Â andÂ C’âÂ =Â Câ . That’s the case if m is equal to 1. But… Well… No. It’s the same thing again: theÂ probabilities (or theÂ magnitudes)Â have to be the same, but the amplitudes may be different because of some phase factor. In fact, theyÂ should be different. If m = 1/2, then we also get the same physics, even if the amplitudes areÂ notÂ the same. They will be each other’s opposite:

Huh?Â Yes. Think of it. The coefficient of proportionality (m) cannot be equal to 1. If it would be equal to 1, and we’d rotateÂ TÂ by 180Â° only, then we’d also get thoseÂ C’+Â =Â âC+Â andÂ C’âÂ =Â âCâÂ equations, and so these coefficients would, therefore,Â also describeÂ the same physical situation. Now, you will understand,Â intuitively, that a rotation of theÂ TÂ apparatusÂ byÂ 180Â° willÂ notÂ give us the sameÂ physicalÂ situation… So… Well… In case you’d want a more formal argument proving a rotation by 180Â° does not give us the same physical situation, Feynman has one for you. đ

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 â I quote Feynman here â is the following:

If we know the amplitudesÂ C+Â andÂ CâÂ 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 frameÂ TÂ which is obtained from S byÂ a rotationÂ Îą around theÂ z-axis, the new amplitudes are given in terms of the old by the following formulas:

[Feynman denotes our angleÂ Îą byÂ phi (Ď) 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 theÂ CâÂ to theÂ C’+Â state, and from theÂ C+Â to the C’âÂ 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:Â “It is very curious to say that if you turn the apparatus 360Â° 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 toÂ defineÂ the idea that it has been rotatedÂ 360Â°âas distinct from zero net rotationâif 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 720Â° does give us the same amplitudes and, therefore, cannot 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 720Â° rotations but… Well… I’ll let you think about this. đ

Note that the determinant of our matrix is equal to aÂˇdÂ â bÂˇc =Â eiĎ/2ÂˇeâiĎ/2Â = 1. So… Well… Our rotation matrix is, effectively, in that special form! How comes? Well… When equatingÂ Îť = âÎź, we are effectively putting the transformation into that special form.Â  Let us also, just for fun, quickly check the normalization condition.Â It requires that the probabilities, in any given representation,Â add to up to one. So… Well… Do they? When they come out ofÂ S, our electrons are equally likely to be in the upÂ orÂ downÂ state. So theÂ amplitudesÂ are 1/â2. [To be precise, they areÂ Âą1/â2 but… Well… It’s the phase factor story once again.] That’s normalized:Â |1/â2|2Â +Â |1/â2|2 = 1. The amplitudes to come out of theÂ TÂ apparatus in the up or down state areÂ eiĎ/2/â2 andÂ eiĎ/2/â2 respectively, so the probabilities add up toÂ |eiĎ/2/â2|2Â +Â |eâiĎ/2/â2|2 = … Well… It’s 1. Check it. đ

Let me add an extra remark here. The normalization condition will result in matrices whose determinant will be equal to some pure imaginary exponential, likeÂ eiÎą. So is that what we have here? Yes. We can re-write 1 as 1 =Â eiÂˇ0Â = e0, soÂ Îą = 0. đ 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.” đ

Let’s do a practical calculation here. Let’s suppose the angle is, effectively, 180Â°. So theÂ eiĎ/2Â and eâiĎ/2/â2Â factors areÂ equal toÂ eiĎ/2Â =Â +i andÂ eâiĎ/2Â = âi, so… Well… What does thatÂ meanâin terms of theÂ geometryÂ of the wavefunction?Â Hmm… We need to do some more thinking about the implications of all this transformation business for ourÂ geometricÂ 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. đ [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 sayÂ same-same but differentÂ once more) to derive the coefficients for a rotation ofÂ Âą90Â° around theÂ y-axis. Why 90Â° only? Well… Let me quote Feynman here, as I can’t sum it up more succinctly than he does: “With just two transformationsâ90Â°Â about theÂ y-axis,Â and an arbitrary angle about theÂ z-axis [which we described above]âwe 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 angleÂ Îą around x. We know how to deal with the angleÂ ÎąÂ ÎąÂ aroundÂ z, but now we want it aroundÂ x.Â How do we get it? First, we turn the axisÂ zÂ down ontoÂ xâwhich is a rotation ofÂ +90Â°.Â Then we turn through the angleÂ ÎąÂ aroundÂ xÂ =Â z’. Then we rotateÂ â90Â°Â aboutÂ y”. The net result of the three rotations is the same as turning aroundÂ xÂ by the angleÂ Îą. It is a property of space.”

Besides helping us greatly to derive the transformation matrix forÂ anyÂ rotation, the mentioned property of space is rather mysterious and deep. It sort of reduces theÂ degrees of freedom, so to speak. FeynmanÂ 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 Rx(Ď) andÂ Ry(Ď) matrices? They’re also equal toÂ one, so… Yes.Â A pure imaginary exponential, right? 1 =Â eiÂˇ0Â = e0. đ

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

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

To help focus our mind, let us recall the metaphor of the V-2 perpetuum mobile, as illustrated below. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down. It provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring: it is described by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs. Of course, instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft, but then that’s not fancy enough for me. đ

At first sight, the analogy between our flywheel model of an electron and the V-twin engine seems to be complete: the 90 degree angle of ourÂ V-2 engine makes it possible to perfectly balance the pistons and we may, therefore, think of the flywheel as a (symmetric) rotating mass, whose angular momentum is given by the product of the angular frequency and the moment of inertia: L =Â ĎÂˇI. Of course,Â the moment of inertia (aka the angular mass) will depend on theÂ formÂ (orÂ shape) of our flywheel:

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

How can we relate those formulas to the E =Â mÂˇa2ÂˇĎ2Â formula? TheÂ kinetic energy that is being stored in a flywheel is equal EkineticÂ = IÂˇĎ2/2, so that is only halfÂ of theÂ E =Â mÂˇa2ÂˇĎ2Â product if we substitute I forÂ I = mÂˇa2. [For a disk, we get a factor 1/4, so that’s even worse!] However, our flywheel model of an electron incorporates potential energy too. In fact, theÂ E =Â mÂˇa2ÂˇĎ2Â formula just adds the (kinetic and potential) energy of two oscillators: we do not really consider the energy in the flywheel itself because… Well… The essence of our flywheel model of an electron is not the flywheel: the flywheel justÂ transfersÂ energy from one oscillator to the other, but so… Well… We don’tÂ includeÂ it in our energy calculations. The essence of our model is thatÂ two-dimensional oscillation whichÂ drivesÂ the electron, and which is reflected in Einstein’sÂ E =Â mÂˇc2Â formula.Â That two-dimensional oscillationâtheÂ a2ÂˇĎ2Â = c2Â equation, reallyâtells us that theÂ resonantÂ (orÂ natural) frequencyÂ of the fabric of spacetime is given by theÂ speed of lightâbut measured in units ofÂ a. [If you don’t quite get this, re-write theÂ a2ÂˇĎ2Â = c2Â equation asÂ Ď = c/a: the radius of our electron appears as a naturalÂ distance unit here.]

Now, we were extremely happy with this interpretation not only because of the key results mentioned above, but also because it has lots of other nice consequences. Think of our probabilities as being proportional to energy densities, for exampleâand all of the other stuff I describe in my published paper on this. But there is even more on the horizon: a follower of this blog (a reader with an actual PhD in physics, for a change) sent me an article analyzing elementary particles as tiny black holes because… Well… If our electron is effectively spinning around, then its tangential velocity is equal toÂ vÂ =Â aÂˇĎÂ =Â c. Now, recent research suggest black holes are also spinning at (nearly) the speed of light. Interesting, right? However, in order to understand what she’s trying to tell me, I’ll first need to get a better grasp of general relativity, so I can relate what I’ve been writing here and in previous posts to the Schwarzschild radiusÂ and other stuff.

Let me get back to the lesson here. In the reference frame of our particle, the wavefunction really looks like the animation below: it has two components, and the amplitude of the two-dimensional oscillation is equal to a, which we calculated asÂ aÂ =Â Ä§Âˇ/(mÂˇc) = 3.8616Ă10â13Â m, so that’s the (reduced) Compton scattering radius of an electron.

In my original article on this, I used a more complicated argument involving the angular momentum formula, but I now prefer a more straightforward calculation:

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

The question is: whatÂ is that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’tÂ proveÂ anything in this regard. But myÂ hypothesisÂ is that it is, in effect, aÂ rotatingÂ field vector, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).

There are a number of crucial differences though:

1. The (physical) dimension of the field vector of the matter-wave is different: I associate the real and imaginary component of the wavefunction with a force per unit massÂ (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s2), so that’s the dimension of a gravitational field.
2. I do believe this gravitational disturbance, so to speak, does cause an electron to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves doÂ notÂ involve any mass: they’re just an oscillating field. Nothing more. Nothing less. In contrast, as Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s Lectures, III-21-4)
3. The third difference is one that I thought of only recently: theÂ planeÂ of the oscillation cannotÂ be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus.

I mentioned that in my previous post but, for your convenience, I’ll repeat what I wrote there.Â The basic idea here is illustrated below (credit for this illustration goes toÂ another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to aÂ YouTube videoÂ from theÂ Quantum Made SimpleÂ site.

The point is: the direction of the angular momentum (and the magnetic moment) of an electronâor, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron isÂ travelingâcannotÂ be parallel to the direction of motion. On the contrary, it isÂ perpendicularÂ to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will compriseÂ the direction of motion.

However, we need to add an interesting detail here. As you know, we don’t really have a precise direction of angular momentum in quantum physics. [If you don’t know this… Well… Just look at one of my many posts on spin and angular momentum in quantum physics.] Now, we’ve explored a number of hypotheses but, when everything is said and done, a rather classical explanation turns out to be the best: an object with an angular momentum JÂ and a magnetic momentÂ ÎźÂ (I used bold-face because these areÂ vector quantities) that is parallel to some magnetic field B, will notÂ line up, as you’d expect a tiny magnet to do in a magnetic fieldâor not completely, at least: it willÂ precess. I explained that in another post on quantum-mechanical spin, which I advise you to re-read if you want to appreciate the point that I am trying to make here. That post integrates some interesting formulas, and so one of the things on my ‘to do’ list is to prove that these formulas are, effectively, compatible with the electron model we’ve presented in this and previous posts.

Indeed, when one advances a hypothesis like this, it’s not enough to just sort ofÂ showÂ that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever weÂ thinkÂ our electronâor its wavefunctionâmight be, it needs to be compatible with stuff like the observedÂ precession frequencyÂ of an electron in a magnetic field.

Our model also needs to be compatible with the transformation formulas for amplitudes. I’ve been talking about this for quite a while now, and so it’s about time I get going on that.

Last but not least, those articles that relate matter-particles to (quantum) gravityâsuch as the one I mentioned aboveâare intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? đ Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years aheadâI hope. đ

Post scriptum: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But itÂ isÂ that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillatorsÂ driveÂ the flywheel but, without the flywheel, nothing is happening. It is really theÂ transferÂ of energyâthrough the flywheelâwhich explains why our flywheel goes round and round.”

It may or may not be useful to remind ourselves of the math in this regard.Â The motionÂ ofÂ our first oscillator is given by the cos(ĎÂˇt) = cosÎ¸ function (Î¸ = ĎÂˇt), and its kinetic energy will be equal toÂ sin2Î¸. Hence, the (instantaneous)Â changeÂ in kinetic energy at any point in time (as a function of the angle Î¸) isÂ equal to:Â d(sin2Î¸)/dÎ¸ = 2âsinÎ¸âd(sinÎ¸)/dÎ¸ = 2âsinÎ¸âcosÎ¸. Now, the motion of theÂ second oscillator (just look at that second piston going up and down in the V-2 engine) is given by theÂ sinÎ¸ function, which is equal to cos(Î¸ â Ď /2). Hence, its kinetic energy is equal toÂ sin2(Î¸ â Ď /2), and how itÂ changesÂ (as a function of Î¸ again) is equal toÂ 2âsin(Î¸ â Ď /2)âcos(Î¸ â Ď /2) =Â = â2âcosÎ¸âsinÎ¸ = â2âsinÎ¸âcosÎ¸. So here we have our energy transfer: the flywheel organizes the borrowing and returning of energy, so to speak. That’s the crux of the matter.

So… Well… WhatÂ if the relevant energy formula isÂ E =Â mÂˇa2ÂˇĎ2/2 instead ofÂ E =Â mÂˇa2ÂˇĎ2? What are the implications? Well… We get aÂ â2 factor in our formula for the radiusÂ a, as shown below.

Now that isÂ notÂ so nice. For the tangential velocity, we getÂ vÂ =Â aÂˇĎ =Â â2Âˇc. This is alsoÂ notÂ so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precessionâtheÂ wobbling of our flywheel in a magnetic field. Remember we may think of Jzâthe angular momentum or, to be precise, its component in theÂ z-direction (the direction in which weÂ measureÂ itâas the projection of theÂ realÂ angular momentumÂ J. Let me insert Feynman’s illustration here again (Feynman’s Lectures, II-34-3), so you get what I am talking about.

Now, all depends on the angle (Î¸) betweenÂ JzÂ andÂ J, of course. We did a rather obscure post on these angles, but the formulas there come in handy now. Just click the link and review it if and when you’d want to understand the following formulas for theÂ magnitudeÂ of theÂ presumedÂ actualÂ momentum:In this particular case (spin-1/2 particles),Â j is equal to 1/2 (in units ofÂ Ä§, of course). Hence,Â JÂ is equal toÂ â0.75Â â 0.866. Elementary geometry then tells us cos(Î¸) =Â (1/2)/â(3/4) =Â  = 1/â3. Hence,Â Î¸Â â 54.73561Â°. That’s a big angleâlarger than the 45Â° angle we had secretly expected because… Well… The 45Â° angle has thatÂ â2 factor in it:Â cos(45Â°) =Â sin(45Â°) = 1/â2.

Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? đ We’ll solve the puzzle. One day… But not today, I am afraid. I’ll call it the form factor problem… Because… Well… It sounds better than the 1/2 orÂ â2 problem, right? đ

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

To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of twoÂ circularÂ oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation â around any axis â will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensionalÂ oscillation as an oscillation of a polar and azimuthal angle. It’s just a thought but… Well… I am sure it’s going to keep me busy for a while. đThey are oscillations, still, so I am not thinking ofÂ twoÂ flywheels that keep going around in the same direction. No. More like a wobbling object on a spring. Something like the movement of a bobblehead on a spring perhaps. đ

Electron spin and the geometry of the wavefunction

In our previous posts, we interpreted the elementary wavefunction Ď = aÂˇeâiâÎ¸Â = aÂˇcosÎ¸ âÂ iÂˇaÂˇsinÎ¸Â 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 minusÂ Ä§/2 respectively.

This allowed us to derive the Compton scattering radius of an elementary particle. Indeed, we interpreted the rotating vector as aÂ resultant vector, which we get byÂ addingÂ the sine and cosine motions, which represent the real and imaginary components of our wavefunction.Â The energy of this two-dimensional oscillation isÂ twiceÂ the energy of aÂ one-dimensional oscillator and, therefore, equal toÂ E =Â mÂˇa2ÂˇĎ2. Now, the angular frequency is given byÂ Ď = E/Ä§ and E must, obviously, also be equal to E = mÂˇc2. Substitition and re-arranging the terms gives us the Compton scattering radius:

The value given above is the (reduced) Compton scattering radius for anÂ electron. For a proton, we get a value of aboutÂ 2.1Ă10â16Â m, which is about 1/4 of theÂ 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 magnitudeâwhich is fine because we know a proton is not an elementary particle and, hence, the motion of its constituent parts (quarks) is… Well… It complicates the picture hugely.

If we’d presume the electron charge would, effectively, be rotating about the center, then its tangential velocity is given byÂ vÂ =Â aÂˇĎ =Â [Ä§Âˇ/(mÂˇc)]Âˇ(E/Ä§) =Â c, which is yet another wonderful implication of our hypothesis. Finally, theÂ cÂ =Â aÂˇĎ formula allowed us to interpret the speed of light as theÂ resonant frequencyÂ 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 =Â mÂˇc2Âˇ 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 theÂ plane of our two-dimensional oscillation cannotÂ 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 theÂ directionÂ 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 theÂ Quantum Made SimpleÂ site.

The point is: the direction of the angular momentum (and the magnetic moment) of an electronâor, to be precise, its component as measured in the direction of the (inhomogenous) magnetic field through which our electron is travelingâcannotÂ 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 actuallyÂ compriseÂ the direction of motion.

What are the implications? I am not sure. I will definitely need to review whatever I wrote about theÂ de BroglieÂ 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 = mÂˇr2/2), and the other is the one for a rotating mass (I = mÂˇr2). 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 = mÂˇvÂˇr. [See also Feynman, Vol. II-34-2, in this regard)] Hence, for our model, we get L =Â mÂˇvÂˇrÂ =Â mÂˇcÂˇa =Â mÂˇcÂˇÄ§/(mÂˇc) =Â Ä§. Now, weÂ knowÂ it’s equal toÂ ÂąÄ§/2, so we need that 1/2 factor in the formula.

Can we relate this 1/2 factor to theÂ 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? đ

I would just like to signal another interesting consequence of our model. If we would interpret the radius of our disk (a)âso that’s the Compton radius of our electron, as opposed to the Thomson radiusâas theÂ uncertainty in the position of our electron, then ourÂ L =Â mÂˇvÂˇrÂ =Â mÂˇcÂˇaÂ = pÂˇa = Ä§/2 formula as a very particular expression of the Uncertainty Principle:Â pÂˇÎx= Ä§/2. Isn’t that just plainÂ nice? đ

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.Â 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 motionâbut then our particle doesn’t move in its own reference frame, obviously. Hence, we could be looking at our particle from any direction and we should, presumably, see a similar two-dimensional oscillation. That is interesting because… Well… If we rotate this circle around its center (in whatever direction we’d choose), we get a sphere, right? It’s only when it starts moving, that it loses its symmetry. Now, that isÂ very intriguing, butÂ let’s think about that later.

Let’s assume we’re looking at it from some specificÂ direction. ThenÂ 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 respectivelyâas we observe it, so to speak. [Of course, you’ve been told you can’t observe wavefunctions so… Well… You should probably stop reading this. :-)] We write:

Ď = =Â aÂˇeâiâÎ¸Â =Â aÂˇeâiâEÂˇt/Ä§ = aÂˇcos(âEât/Ä§) + iÂˇaÂˇsin(âEât/Ä§) = aÂˇcos(Eât/Ä§) â iÂˇaÂˇsin(Eât/Ä§)Â

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 pÂˇx term to the argument (Î¸ = EÂˇt âÂ pâx). It is easy to show this term doesn’t change the argument (Î¸), because we also get a different value for the energy in the new reference frame:Â EvÂ =Â ÎłÂˇE0Â 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 EvÂ and pvÂ and, importantly,Â the coordinates xÂ and tÂ relativisticallyÂ transform ensures the invariance.

In fact, I’ve always wanted to readÂ de Broglie‘sÂ 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 doesÂ notÂ imply there is no need for a relativistic waveÂ equation: the wavefunction is aÂ solutionÂ for the wave equation and, yes, I am the first to note the SchrĂśdinger equation has some obvious issues, which I briefly touch upon in one of my other postsâand which is why SchrĂśdinger 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 SchrĂśdinger published his). In my humble opinion, the key issue is notÂ that SchrĂśdinger’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 waveÂ equationÂ for what it is and goÂ back to our wavefunction.

You’ll note the argument (orÂ phase) of our wavefunction moves clockwiseâorÂ counterclockwise, depending on whether you’re standing in front of behind the clock. Of course,Â NatureÂ doesn’t care about where we stand orâto put it differentlyâwhether we measure time clockwise, counterclockwise, in the positive, the negative or whatever direction. Hence, I’ve argued we can have both left- as well as right-handed wavefunctions, as illustrated below (for pÂ â  0). Our hypothesis is that these two physical possibilities correspond to the angular momentum of our electron being eitherÂ positive or negative: JzÂ =Â +Ä§/2 or, else, JzÂ =Â âÄ§/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 definedÂ 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 toÂ calculateÂ the amplitudeÂ a. We got a result that tentatively confirms we’re on the right track with our interpretation: we found thatÂ aÂ =Â Ä§/meÂˇc, so that’s theÂ Compton scattering radiusÂ of our electron. All good ! But we were still a bit stuckâorÂ ambiguous, I should sayâon what the components of our wavefunction actuallyÂ are. Are we really imagining the tip of that rotating arrow is a pointlike electric chargeÂ spinning around the center? [Pointlike or… Well… Perhaps we should think of theÂ ThomsonÂ radius of the electron here, i.e. the so-calledÂ classical electron radius, which isÂ equal to the Compton radius times the fine-structure constant:Â rThomsonÂ =Â ÎąÂˇrComptonÂ â 3.86Ă10â13/137.]

So that would be the flywheel model.

In contrast, we may also think the whole arrow is some rotatingÂ field vectorâsomething like the electric field vector, with the same or some other physicalÂ dimension, like newton per charge unit, or newton per mass unit? So that’s the fieldÂ model. Now, theseÂ interpretations may or may not be compatibleâorÂ complementary, I should say. I sure hopeÂ 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 theÂ interactionÂ between a photon and an electron, as I demonstrated in my previous post: the electromagnetic energy of the photon willÂ driveÂ the circulatory motion of our electron… So… Well… That’s a nice physicalÂ explanation for the transfer of energy.Â 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 flywheelÂ metaphorÂ to its logical limits. Let me remind you of what triggered it all: it was theÂ mathematicalÂ equivalence of the energy equation for an oscillator (E =Â mÂˇa2ÂˇĎ2) and Einstein’s formula (E =Â mÂˇc2), which tells us energy and mass areÂ equivalentÂ but… Well… They’re not the same. So whatÂ areÂ they then? WhatÂ isÂ energy, and whatÂ isÂ massâin the context of these matter-waves that we’re looking at. To be precise, theÂ E =Â mÂˇa2ÂˇĎ2Â formula gives us the energy ofÂ twoÂ oscillators, so we need aÂ two-spring model whichâbecause I love motorbikesâI referred to as my V-twin engine model, but it’s not anÂ engine, really: it’s two frictionless pistons (or springs) whose direction of motion is perpendicular to each other, so they are in a 90Â° degree angle and, therefore, their motion is, effectively, independent. In other words: they will not interfereÂ 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 = aÂˇcos(ĎÂˇt + Î).Â Needless to say, Î is just a phase factor which defines our t = 0 point, and ĎÂ is theÂ naturalÂ angular frequency of our oscillator. Because of the 90Â° angle between the two cylinders, Î would be 0 for one oscillator, and âĎ/2 for the other. Hence, the motion of one piston is given by x = aÂˇcos(ĎÂˇt), while the motion of the other is given by x = aÂˇcos(ĎÂˇtâĎ/2) = aÂˇsin(ĎÂˇt). TheÂ kinetic and potential energy of one oscillator â think of one piston or one spring only â can then be calculated as:

1. K.E. = T = mÂˇv2/2 =Â (1/2)ÂˇmÂˇĎ2Âˇa2Âˇsin2(ĎÂˇt + Î)
2. P.E. = U = kÂˇx2/2 = (1/2)ÂˇkÂˇa2Âˇcos2(ĎÂˇt + Î)

The coefficient k in the potential energy formula characterizes the restoring force: F = âkÂˇx. From the dynamics involved, it is obvious that k must be equal to mÂˇĎ2. Hence, the total energyâforÂ oneÂ piston, or one springâis equal to:

E = T + U = (1/2)Âˇ mÂˇĎ2Âˇa2Âˇ[sin2(ĎÂˇt + Î) + cos2(ĎÂˇt + Î)] = mÂˇa2ÂˇĎ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 = mÂˇa2ÂˇĎ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, butÂ 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 theÂ E =Â mÂˇa2ÂˇĎ2Â and E =Â mÂˇc2Â formulas, we need to explain why we could, potentially, writeÂ cÂ asÂ cÂ =Â aÂˇĎÂ =Â aÂˇâ(k/m). We’ve done that alreadyâto some extent at least. TheÂ tangentialÂ velocity of a pointlike particle spinning around some axis is given byÂ vÂ =Â rÂˇĎ. Now, the radiusÂ rÂ is given byÂ aÂ =Â Ä§/(mÂˇc), andÂ Ď = E/Ä§ =Â mÂˇc2/Ä§, soÂ vÂ is equal to toÂ v = [Ä§/(mÂˇc)]Âˇ[mÂˇc2/Ä§] =Â c. Another beautiful result, but what does itÂ mean? We need to think about theÂ meaning of theÂ Ď =Â â(k/m) formula here. In the mentioned article, we boldly wrote that the speed of light is to be interpreted as theÂ resonantÂ frequency of spacetime, but so… Well… What do we reallyÂ meanÂ by that? Think of the following.

Einsteinâs E = mc2 equation implies the ratio between the energy and the mass of any particle is always the same:

This effectively reminds us of theÂ Ď2 = C1/L or Ď2 = k/mÂ formula for harmonic oscillators.Â The key difference is that the Ď2= C1/L and Ď2 = k/m formulas introduce two (or more) degrees of freedom. In contrast, c2= 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 inÂ oneÂ physical space only:Â ourÂ spacetime. Hence, the speed of light (c) emerges here as the defining property ofÂ 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 Ď2= 1/LC formula here. It’s basically the same concept:Â the Ď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 Ď2= Câ1/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring, so… Well… You get it, right? The Ď2= C1/L and Ď2 = k/m sort of describe the same thing: harmonic oscillation. It’s just… Well… Unlike theÂ Ď2= C1/L, theÂ Ď2 = k/m isÂ directlyÂ compatible with our V-twin engine metaphor, because it also involves physical distances, as I’ll show you here.] TheÂ kÂ in theÂ Ď2 = k/m is, effectively, the stiffness of the spring. It isÂ definedÂ by Hooke’s Law, which states thatÂ the force that is needed to extend or compress a springÂ by some distanceÂ xÂ Â is linearly proportional to that distance, so we write: F = kÂˇx.

NowÂ that is interesting, isn’t it? We’re talkingÂ exactlyÂ the same thing here: spacetime is, presumably,Â isotropic, so it should oscillate the same in any directionâI am talking those sine and cosine oscillations now, but inÂ physicalÂ spaceâso there is nothing imaginary here: all is realÂ or… Well… As real as we can imagine it to be. đ

We can elaborate the point as follows. TheÂ F = kÂˇxÂ equation implies k is a forceÂ per unit distance: k = F/x. Hence, its physical dimension isÂ newton per meterÂ (N/m). Now, theÂ xÂ in this equation may be equated to theÂ maximumÂ extension of our spring, or theÂ amplitudeÂ of the oscillation, so that’s the radiusÂ rÂ =Â aÂ in the metaphor we’re analyzing here. NowÂ look at how we can re-write theÂ cÂ =Â aÂˇĎÂ =Â aÂˇâ(k/m) equation:

In case you wonder about the E =Â FÂˇa substitution: just remember thatÂ energyÂ is force times distance. [Just do a dimensional analysis: you’ll see it works out.] So we have aÂ spectacular result here, for several reasons. The first, and perhaps most obvious reason, is that we can actuallyÂ deriveÂ Einstein’s E = mÂˇc2Â formula from ourÂ flywheel model. Now, thatÂ isÂ truly glorious, I think. However, even more importantly, this equation suggests we doÂ not necessarilyÂ 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 aÂ forceÂ acting over some distance, regardless of whether or not it is actually acting on a particle.Â Now,Â thatÂ energy will have anÂ equivalentÂ mass which isâor should be, I’d say… Well… The mass of our electron or, generalizing, the mass of the particle we’re looking at.

Huh?Â Yes. In case you wonder what I am trying to get at, I am trying to convey the idea that theÂ two interpretationsâthe field versus the flywheel modelâare actually fullyÂ equivalent, orÂ compatible, if you prefer that term. In Asia, they would say: they are the “same-same but different” đ 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. đ 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 aboutÂ equivalence. đ So it’s just like Einstein’s equation. đ

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 theÂ RydbergÂ energy, even Feynman casually writes that “we need not trust our answer [to questions like this] within factors like 2,Â Ď, etcetera.” Frankly, that’s disappointing. Factors like 2, 1/2, Ď or 2Ď 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 doesÂ not come into play: think of the Cooper pairs we use to explain superconductivity (so that’s the BCS theory). The 1/2 factorÂ in SchrĂśdinger’s equation also has weird consequences (when you plug in the elementary wavefunction and do the derivatives, you get a weird energy concept: E = mÂˇv2, to be precise). This problem may also be solved when assuming we’re actually calculating orbitals for aÂ pairÂ of electrons, rather than orbitals for just one electron only. [We’d getÂ twiceÂ the mass (and, presumably, the charge, so… Well… It might workâbut I haven’t done it yet. It’s on my agendaâas so many other things, but I’ll get there… One day. :-)]

So… Well… Let’s get back to the lesson here. In this particular context (i.e. in the context of trying to find some reasonable physicalÂ interpretation of the wavefunction), you may or may not remember (if not, check my post on it) ‘ll remember I had to use theÂ I = mÂˇr2/2 formula for the angular momentum, as opposed to the I = mÂˇr2Â formula.Â I = mÂˇr2/2 (with the 1/2 factor) gives us the angular momentum of aÂ diskÂ with radiusÂ r, as opposed to aÂ pointÂ mass going around some circle with radiusÂ r. I noted that “the addition of this 1/2 factor may seem arbitrary”âand it totallyÂ is, of courseâbut so it gave us the result we wanted: theÂ exactÂ (Compton scattering)Â radius of our electron.

Now, the arbitraryÂ 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 toÂ xÂ =Â a, we need to calculate it as the following integral:

So… Well… That will give you some food for thought, I’d guess. đ If it racks your brain too muchâor if you’re too exhausted by this point (which is OK, because it racks my brain too!)âjust note we’ve also shown that the energy is proportional to theÂ squareÂ of the amplitude here, so that’s a nice result as well… đ

Talking food for thought, let me make one final point here. TheÂ c2Â = a2Âˇk/m relation implies a value for k which is equal to k = mÂˇc2/a = E/a. What does this tell us? In one of our previous posts, we wrote that the radius of our electron appeared as aÂ naturalÂ distance unit. We wrote that because of another reason: the remark was triggered by the fact that we can write theÂ c/Ď ratioÂ asÂ c/Ď =Â aÂˇĎ/Ď =Â a. This implies the tangential and angular velocity in our flywheel model of an electron would be the same if weâd measure distance in units ofÂ a. Now, the E = aÂˇk =Â aÂˇF/xÂ (just re-writing…) implies that the force is proportional to the energyâ F = (x/a)ÂˇE â and the proportionality coefficient is… Well…Â x/a. So that’s the distance measured in units ofÂ a.Â So… Well… Isn’t that great? The radius of our atom appearing as aÂ naturalÂ distance unit does fit in nicely with ourÂ geometricÂ interpretation of the wavefunction, doesn’t it? I mean…Â 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 youÂ cÂ appears as a resonant frequency of spacetime and, in this post, I tried to explain what that reallyÂ means. I’d appreciate if you could let me know if you got it. If not, I’ll try again. đ When everything is said and done, one only truly understands stuff when one is able to explain it to someone else, right? đ Please do think of more innovative or creative ways if you can! đ

OK. That’s it but… Well…Â 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 someÂ specificÂ direction. It could be anyÂ direction but… Well… It’sÂ someÂ direction. We have noÂ 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 beÂ anyÂ direction, our analysis is valid for any direction. Hence, ifÂ our interpretation would happen to be some trueâand that’s a bigÂ if, of courseâthenÂ our particle has to be spherical, right? Why? Well… Because we see this circular thing from any direction, so itÂ hasÂ to be a sphere, right?

Well… Yes. But then… Well… While that logic seems to beÂ 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. đŚ

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), theÂ aÂ =Â Ä§/(mÂˇc) formula gives a muchÂ smaller radius: 1835 timesÂ smaller, to be precise, so that’s around 2.1Ă10â16Â m, which is about 1/4 of the so-calledÂ chargeÂ 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 isÂ notÂ 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 evenÂ 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 wasÂ veryÂ short:Â “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 aÂ field. We may call it aÂ matterÂ field (yes, think of the scalar Higgs field here), but we could also think of it as an oscillation of theÂ spacetime fabric itself: aÂ tiny gravitational wave, in effect. All we need to do here is to associate the sine and cosine component with aÂ physicalÂ dimension. The analogy here is the electromagnetic field vector, whose dimension isÂ forceÂ per unitÂ chargeÂ (newton/coulomb). So we may associate the sine and cosine components of the wavefunction with, say, theÂ force per unitÂ massÂ dimension (newton/kg) which, using Newton’s Law (F = mÂˇa) reduces to the dimension ofÂ accelerationÂ (m/s2), which is the dimension of gravitational fields.Â I’ll refer to this interpretation as theÂ fieldÂ interpretation of the matter wave (or wavefunction).

(2) The other interpretation is what I refer to as theÂ flywheelÂ interpretation of the electron. If you google this, you won’t find anything. However, you will probably stumble upon the so-calledÂ ZitterbewegungÂ interpretation of quantum mechanics, which is a more elaborate theory based on the same basic intuition. TheÂ ZitterbewegungÂ (a term which was coined by Erwin SchrĂśdinger 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’sÂ spin and magnetic moment. All that I am doing, is… Well… I think I do push the envelope of this interpretation quite a bit. đ

The first interpretation implies our rotating arrow is, effectively, someÂ 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

The flywheel interpretation has an advantage over the field interpretation, because it also gives us a wonderfully simple physicalÂ interpretation of the interactionÂ between electrons and photonsâor, further speculating, between matter-particles (fermions) and force-carrier particles (bosons) in general. In fact,Â FeynmanÂ shows how this might workâbut in a rather theoreticalÂ LectureÂ on symmetries and conservation principles, and heÂ doesn’t elaborate much, so let me do that for him.Â The argument goes as follows.

A light beamâan electromagnetic waveâconsists of a large number of photons. These photons are thought of as being circularly polarized: look at those animations above again. The Planck-Einstein equation tells us the energy of each photon is equal to E =Â Ä§ÂˇĎ = hÂˇf. [I should, perhaps, quickly note that the frequencyÂ fÂ is, obviously, the frequency of the electromagnetic wave. It, therefore, is notÂ to be associated with aÂ matterÂ wave: theÂ de BroglieÂ 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 ofÂ NÂ photons, the total energy of our beam will be equal to W =Â NÂˇE =Â NÂˇÄ§ÂˇĎ. It is crucially important to note that this energy is to be interpreted as the energy that is carried by the beamÂ 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 carryÂ 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 canÂ see there is angular momentum there. Photons are spin-1 particles, so the angular momentum will be equal toÂ Âą Ä§. Hence,Â thenÂ theÂ totalÂ angular momentum JzÂ (the direction of propagation is supposed to be theÂ z-axis here) will be equal toÂ Jz =Â NÂˇÄ§. [This, of course, assumesÂ 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 theÂ W =Â NÂˇÄ§ÂˇĎ andÂ Jz =Â NÂˇÄ§ equations, we get:

Jz =Â NÂˇÄ§ = W/Ď

For a photon, we do accept the field interpretation, as illustrated below. As mentioned above, theÂ 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 asÂ Îľ (epsilon) so as to not cause any confusion with the Î 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, aÂ relativisticÂ effect which vanishes in the reference frame of the charge itself.] TheÂ phaseÂ of the electric field vector isÂ Ď =Â ĎÂˇt.

Now, a chargeÂ (so that’s our electron now) will experience a force which is equal to F = qÂˇÎľ. We use bold letters here because F andÂ Îľ 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, weÂ calculatedÂ the radiusÂ rÂ based on a similar argument as the one Feynman used to get thatÂ Jz =Â NÂˇÄ§ = W/Ď equation. I’ll refer you those posts and just mention the result here:Â 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 velocityÂ vÂ was equal to v =Â rÂˇĎ =Â [Ä§Âˇ/(mÂˇc)]Âˇ(E/Ä§) =Â c. Hence, in our flywheel model of an electron, it is effectively spinning around at the speed of light. Note that the angular frequency (Ď) in theÂ v =Â rÂˇĎ equation isÂ not the angular frequency of our photon: it’s the frequency of our electron. So we use the same Planck-Einstein equation (Ď = 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 areÂ veryÂ 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 asÂ Ď0.

Just to be clear here, at this point, our analysis here diverges from Feynman’s. Feynman had no intention whatsoever to talk about SchrĂśdinger’sÂ ZitterbewegungÂ hypothesis when he wrote what he wrote back in the 1960s. In fact, Feynman is very reluctant to venture intoÂ physicalÂ interpretations of the wavefunction in all hisÂ Lectures on quantum mechanicsâwhich is surprising. Because he comes so tantalizing close at many occasionsâas he does here: he describes the motion of the electron here as that of a harmonic oscillator which can be driven by an external electric field. Now thatÂ isÂ a physical interpretation, and it is totally consistent with the one I’ve advanced in my recent posts.Â Indeed, Feynman also describes it as an oscillation in two dimensionsâperpendicular to each other and to the direction of motion, as we doâ 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. Weâll suppose that the atom is isotropic, so that it can oscillate equally well in theÂ x– orÂ y-Â directions. Then in the circularly polarized light, theÂ xÂ displacement and theÂ yÂ displacement are the same, but one is 90Â°Â behind the other. The net result is that the electron moves in a circle.”

Right on! But so what happens really? As our light beamâthe photons, reallyâare being absorbed by our electron (or our atom), it absorbsÂ angular momentum. In other words, there is aÂ torqueÂ about the central axis. Let me remind you of the formulas for the angular momentum and for torqueÂ respectively: L = rĂp andÂ Ď =Â rĂF. Needless to say, we have twoÂ vector cross-products here. Hence, if we use theÂ Ď =Â rĂFÂ formula, we need to find theÂ tangentialÂ component of the force (Ft), whose magnitude will be equal to Ft = qÂˇÎľt.Â Now, energy is force over some distance so… Well… You may need to think about it for a while but, if you’ve understood all of the above, you should also be able to understand the following formula:

dW/dt =Â qÂˇÎľtÂˇv

[If you have trouble, rememberÂ vÂ is equal to ds/dt =Â Îs/Ît forÂ ÎtÂ â 0, and re-write the equation above asÂ dW =Â qÂˇÎľtÂˇvÂˇdt =Â qÂˇÎľtÂˇds =Â FtÂˇds. Capito?]

Now, you may or may not remember thatÂ the time rate of change of angular momentumÂ must be equal to the torqueÂ that is being applied. Now, the torque is equal toÂ Ď = FtÂˇrÂ =Â qÂˇÎľtÂˇr, so we get:

dJz/dt =Â qÂˇÎľtÂˇv

TheÂ ratioÂ ofÂ dW/dt andÂ dJz/dt gives us the following interesting equation:

Now, Feynman tries to relate this to theÂ Jz =Â NÂˇÄ§ = W/Ď formula but… Well… We should remind ourselves that the angular frequency of these photons isÂ not the angular frequency of our electron. So… Well… WhatÂ canÂ we say about this equation? Feynman suggests to integrateÂ dJzÂ andÂ dW over some time interval, which makes sense: as mentioned, we interpreted W as the energy that is carried by the beam inÂ a certain time. So if we integrateÂ dW over this time interval, we get W. Likewise, if we integrateÂ dJzÂ over theÂ sameÂ time interval, we should get the totalÂ angular momentum that our electron isÂ absorbingÂ from the light beam. Now, becauseÂ dJzÂ =Â dW/Ď, we do concur withÂ 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/Ď.

It’s just… Well… TheÂ Ď here is the angular frequency of the electron. It’sÂ notÂ the angular frequency of the beam. Not in our flywheel model of the electron which, admittedly, isÂ notÂ 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:Â how and where is the absorbed energy being stored?Â 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 aroundÂ afterÂ 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,Â Ď =Â E/Ä§ must increase, right? Right. At the same time, the velocityÂ vÂ =Â rÂˇĎ must still be equal toÂ v =Â rÂˇĎ =Â [Ä§Âˇ/(mÂˇc)]Âˇ(E/Ä§) =Â c, right? Right. So… IfÂ Ď increases, butÂ rÂˇĎ must equal the speed of light, thenÂ rÂ must actuallyÂ decreaseÂ somewhat, right?

Right. It’s a weird but inevitable conclusion, it seems. I’ll let you think about it. đ

To conclude this postâwhich, I hope, the reader who triggered it will find interestingâI would like to quote Feynman on an issue on which most textbooks remain silent: the two-state nature of photons. I will just quote him without trying to comment or alter what he writes, because what he writes is clear enough, I think:

“Now letâs ask the following question: If light is linearly polarized in the x-direction, what is its angular momentum? Light polarized in the x-direction can be represented as the superposition of RHC and LHC polarized light. […] The interference of these two amplitudes produces the linear polarization, but it hasÂ equalÂ 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 momentumâthe average angular momentum is zero.]

Now, we have said that any spin-one particle can have three values of Jz, namelyÂ +1,Â 0,Â â1Â (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 spinÂ jÂ which is standing still, there must be theÂ 2j+1Â possible states with values of JzÂ going in steps ofÂ 1Â fromÂ âjÂ toÂ +j. But it turns out that for something of spinÂ jÂ with zero mass only the states with the componentsÂ +jÂ andÂ âjÂ along the direction of motion exist. For example, light does not have three states, but only twoâalthough a photon is still an object of spin one.”

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

“How is this consistent with our earlier proofsâbased on what happens under rotations in spaceâthat for spin-one particles three states are necessary? For a particle at rest, rotations can be made about any axis without changing the momentum state. Particles with zero rest mass (like photons and neutrinos) cannot be at rest; only rotations about the axis along the direction of motion do not change the momentum state. Arguments about rotations around one axis only are insufficient to prove that three states are required. We have tried to find at least a proof that the component of angular momentum along the direction of motion must for a zero mass particle be an integral multiple ofÂ Ä§/2âand not something likeÂ Ä§/3.Â Even using all sorts of properties of the Lorentz transformation and what not, we failed. Maybe itâs not true. Weâll have to talk about it with Prof. Wigner, who knows all about such things.”

The reference to Eugene Wigner is historically interesting. Feynman probably knew him very wellâif only because they had both worked together on the Manhattan Projectâand it’s true Wigner was not only a great physicist but a mathematical genius as well. However, Feynman probably quotes him here for the 1963 Nobel Prize he got for… Well… Wigner’s “contributions to the theory of the atomic nucleus and elementary particles,Â particularly through the discovery and application of fundamental symmetry principles.” đ I’ll let you figure out how what I write about in this post, and symmetry arguments, might be related. đ

That’s it for today, folks! I hope you enjoyed this. đ

Post scriptum: The mainÂ disadvantage of the flywheel interpretation is that it doesn’t explain interference: waves interfereâsome rotating mass doesn’t. Ultimately, the wave and flywheel interpretation must, somehow, be compatible. One way to think about it is that the electron can only move as it doesâin a “local circulatory motion”âif there is aÂ forceÂ 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 hereâsome negative charge orbiting around a positive nucleus. We’re just considering the electron itself.] So we just need to prove that our rotating arrow willÂ alsoÂ 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. đ

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 modelÂ (illustrated below), but then whatÂ isÂ the mass here? One hypothesis that comes to mind is that we’re constantly accelerating and decelerating an electric charge (the electron charge)âagainst all other charges in the Universe, so to speak. So that’s a force over a distanceâenergy. And energy has an equivalent mass.

The question which remains open, then, is the following: what is the nature of this force? In previous posts, I suggested it might be gravitational, but so here we’re back to the drawing board: we’re talking an electrical force, but applied to someÂ massÂ which acquires mass because of… Well… Because of the forceâbecause 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 asÂ kinetic energy. It is entirely compatible with ourÂ physicalÂ interpretation of the wavefunction of an electron as… Well… We’ve basically been looking at the electron as a little flywheel, right? đ 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 withÂ threeÂ possible values for the angular momentum (JzÂ = +Ä§, 0 orÂ âÄ§), 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. đ

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 âintroductoryâ 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 Feynmanâs âintroductoryâ 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 1960âs or 1980âs. 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, whoÂ 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.Â I respect the views in the mentioned article mentioned aboveâif only because the author,Â 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.Â Frankly, I must agree I struggled with Feynman’s LecturesÂ 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 willÂ notÂ find in any other textbook. Feynman tries to give us aÂ physicalÂ 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 veryÂ disappointing. I must admit thatÂ Feynman has racked my brainâbut 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 tooâand 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 wavefunctionâhalf of my life, in a wayâand 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 â very surprisingly â 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.”Â 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-studyâsome productive, in the classical sense of the word (moving ahead) and… Some… Well… Much of what I read didâand still doesâkeep me awake at night. đ

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. đŚ But I’ll do my best to try to explain what I am thinking of. Remember the formula (orÂ definition) of theÂ elementary wavefunction:

Ď =Â aÂˇeâi[EÂˇt â pâx]/Ä§ =Â aÂˇcos(pâx/Ä§ â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ âÂ Eât/Ä§)

How should we interpret this? We know an actual particle will be represented by aÂ wave packet: a sum of wavefunctions, each with its own amplitude ak and its own argument Î¸k = (Ekât â pkâx)/Ä§. But… Well… Let’s see how far we get when analyzing theÂ elementaryÂ wavefunction itself only.

According to mathematicalÂ convention, the imaginary unit (i) is a 90Â°Â angle in theÂ counterclockwise direction. However, NatureÂ surely cannot be bothered about our convention of measuring phase angles – orÂ timeÂ itself – clockwiseÂ 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:

Ď =Â aÂˇei[EÂˇt â pâx]/Ä§ =Â aÂˇcos(pâx/Ä§ â Eât/Ä§)Â âÂ iÂˇaÂˇsin(pâx/Ä§ âÂ Eât/Ä§)

In my previous posts, I hypothesized that the two physical possibilities correspond to the angular momentum of our particle – say, an electron – being eitherÂ positive or negative: J = +Ä§/2 or, else,Â J = âÄ§/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 theÂ directionÂ as well as theÂ magnitudeÂ of the (linear) momentum (p) are relative: they depend on the orientation and relative velocity of our reference frame – which are, in effect, relative to the reference frame of our object. As such, the wavefunction itself is relative: another observer will obtain a different value for both the momentum (p) as well as for the energy (E). Of course, this makes us think of the relativity of the electric and magnetic field vectors (E and B) but… Well… It’s not quite the same because – as I will explain in a moment – the argument of the wavefunction, considered as a whole, is actually invariant under a Lorentz transformation.

Let me elaborate this point.Â 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:Â p = 0. Let us now look at howÂ the argument of the wavefunction transforms. The E and p in the argument of the wavefunction (Î¸ = Ďât â kâx = (E/Ä§)ât â (p/Ä§)âx =Â (Eât â pâ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 = Ev and p = pv = pvâv. If we then use natural time and distanceÂ units (hence, the numerical value of c is equal to 1 and, hence, the (relative) velocity is then measured as a fraction ofÂ 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:

EvÂ =Â ÎłÂˇE0Â and pvÂ = ÎłÂˇm0âvÂ =Â ÎłÂˇE0âv/c2

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

Î¸ = [ÎłÂˇE0/Ä§]ât â [(ÎłÂˇE0âv/c2)/Ä§]âx = (E0/Ä§)Âˇ(t â vâx/c2)ÂˇÎł =Â (E0/Ä§)ât’

The Îł in these formulas is, of course, the Lorentz factor, and t’ is theÂ properÂ time: t’Â = (t â vâx/c2)/â(1âv2/c2). 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 primedÂ Î¸ (Î¸’):Â Î¸ = (Ev/Ä§)Âˇt â (pv/Ä§)Âˇx =Â (E0/Ä§)ât’.

2.Â TheÂ E0/Ä§ coefficient pops up as an angular frequency:Â E0/Ä§ =Â Ď0. We may refer to it asÂ theÂ frequency of the elementary wavefunction.

Now, if you don’t like the concept ofÂ angular frequency, we can also write:Â f0Â =Â Ď0/2Ď = (E0/Ä§)/2Ď = E0/h.Â Alternatively, and perhaps more elucidating, we get the following formula for theÂ periodÂ of the oscillation:

T0Â = 1/f0Â =Â h/E0

This is interesting, because we can look at the period as aÂ naturalÂ unit of time for our particle. This period is inverselyÂ proportional to the (rest) energy of the particle, and the constant of proportionality is h. Substituting E0Â for m0Âˇc2, we may also say it’s inversely proportional to the (rest) mass of the particle, with the constant of proportionality equal to h/c2. The period of an electron, for example, would be equal to about 8Ă10â21Â s. That’sÂ veryÂ small, and it only gets smaller for larger objects ! But what does all of this really tellÂ us? What does it actuallyÂ mean?

We can look at the sine and cosine components of the wavefunction as an oscillation inÂ twoÂ dimensions, as illustrated below.

Look at the little green dot going around. Imagine it is someÂ 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 hasÂ angular momentum, which we can write as the vector cross-product L = rĂp or, alternatively, as the product of an angular velocity (Ď) and rotational inertia (I), aka as theÂ moment of inertia or the angular mass:Â L = IÂˇĎ. [Note we writeÂ L andÂ Ď in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = IÂˇĎ (no boldface).]

We can now do some calculations. We already know the angular velocity (Ď) is equal toÂ E0/Ä§. Now, theÂ magnitude ofÂ rÂ in the L =Â rĂpÂ vector cross-product should equal theÂ magnitudeÂ ofÂ Ď =Â aÂˇeâiâEÂˇt/Ä§, so we write:Â r = a. What’s next? Well… The momentum (p) is the product of a linear velocity (v) – in this case, theÂ tangentialÂ velocity –Â and some mass (m): p = mÂˇv. If we switch to scalarÂ instead ofÂ vector quantities, then the (tangential) velocity is given by v = rÂˇĎ.

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 pointÂ mass going around some center, then the formula to use isÂ I = mÂˇr2. 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 becomesÂ I = mÂˇr2/2. 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 = IÂˇĎ = (mÂˇr2/2)Âˇ(E/Ä§) = (1/2)Âˇa2Âˇ(E/c2)Âˇ(E/Ä§) =Â a2ÂˇE2/(2ÂˇÄ§Âˇc2)

Note that our frame of reference is that of the particle itself, so we should actually write Ď0, m0Â and E0Â instead ofÂ Ď, m and E. The value of the rest energy of an electron is about 0.510 MeV, or 8.1871Ă10â14 Nâm. Now, this momentum should equal J = ÂąÄ§/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 3.8616Ă10â13 m, which is the (reduced) Compton scattering radius of an electron.Â The (tangential) velocity (v) can now be calculated as being equal toÂ v = rÂˇĎ = aÂˇĎ = [Ä§Âˇ/(mÂˇc)]Âˇ(E/Ä§) =Â c. This is an amazing result. Let us think about it.

In our previous posts, we introduced the metaphor of twoÂ springsÂ or oscillators, whose energy was equal to E =Â mÂˇĎ2. Is this compatible with Einstein’s E =Â mÂˇc2Â mass-energy equivalence relation? It is. TheÂ E =Â mÂˇc2Â impliesÂ E/m =Â c2. We, therefore, can write the following:

Ď = E/Ä§ =Â mÂˇc2/Ä§ = mÂˇ(E/m)Âˇ/Ä§Â â Ď =Â E/Ä§

Hence, we should actually have titled this and the previous post somewhat differently: the speed of light appears as aÂ tangentialÂ velocity. Think of the following: theÂ ratioÂ ofÂ c andÂ Ď is equal toÂ c/Ď =Â aÂˇĎ/Ď =Â a. Hence, the tangential and angular velocity would be the same if we’d measure distance in units ofÂ a. In other words,Â the radius of an electron appears as a natural distance unit here: if we’d measureÂ Ď inÂ units ofÂ aÂ 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 naturalÂ unit of velocity.Â Huh?Â Yes. Just re-write theÂ c/Ď =Â a asÂ Ď =Â c/a. What does it say? Exactly what I said, right? As such, the radius of an electron is not only aÂ normÂ for measuring distance but also for time.Â đ

If you don’t quite get this, think of the following. For an electron, we get an angular frequency that is equal toÂ Ď = E/Ä§ = (8.19Ă10â14Â NÂˇm)/(1.05Ă10â34Â NÂˇmÂˇs) â 7.76Ă1020Â radiansÂ per second. That’s an incredible velocity, because radians are expressed in distance unitsâso that’s inÂ meter. However, our mass is not moving along theÂ unitÂ circle, but along a much tinier orbit. TheÂ ratioÂ of the radius of the unit circle andÂ aÂ is equal to 1/a âÂ (1 m)/(3.86Ă10â13 m) â 2.59Ă1012. Now, if we divide theÂ above-mentionedÂ velocityÂ ofÂ 7.76Ă1020Â radiansÂ per second by this factor, we get… Right ! The speed of light: 2.998Ă1082Â m/s. đ

Post scriptum: I have no clear answer to the question as to why we should use the I = mÂˇr2/2 formula, as opposed to theÂ I = mÂˇr2Â 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 SchrĂśdinger’s equation, which is also quite enigmatic. In my view, the 1/2 factor should not be there in SchrĂśdinger’s equation. Electron orbitals tend to be occupied byÂ twoÂ electrons with opposite spin. That’s why their energy levels should beÂ 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. đ But then that’s just my personal opinion or… Well… I’ve always been intrigued by the difference between the originalÂ printedÂ 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:

“Donât forget thatÂ meff has nothing to do with the real mass of an electron. It may be quite differentâalthough in commonly used metals and semiconductors it often happens to turn out to be the same general order of magnitude, about 2 to 20 timesÂ 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 meffNEWÂ = 2âmeffOLDÂ – 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. đ

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):Â Einsteinâs view on aether theories probably still holds true: âWe may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an aether. According to the general theory of relativity, space without aether is unthinkable â 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.â

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: âIt is ironic that Einstein’s most creative work, the general theory of relativity, should boil down to conceptualizing space as a medium when his original premise [in special relativity] was that no such medium existed. [âŚ] 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. [âŚ]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.â

I really love this: a relativistic aether. MyÂ interpretation of the wavefunction is veryÂ consistent with that.

A physical explanation for relativistic length contraction?

My last posts were all about a possible physicalÂ 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 aÂ physicalÂ 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 unitÂ chargeÂ (newton perÂ coulomb), while the other gives us a force per unitÂ 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:

Ď =Â aÂˇeâi(EÂˇt â pâx)/Ä§ =Â aÂˇcos(pâx/Ä§ – Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§)

Nature should not care about our conventions for measuring the phase angle clockwise or counterclockwise and, therefore, the Ď =Â aÂˇei[EÂˇt â pâx]/Ä§ function may also be permitted. We know that cos(Î¸) = cos(Î¸) and sinÎ¸ = sin(Î¸), so we can write:Â  Â Â

Ď =Â aÂˇei(EÂˇt â pâx)/Ä§ =Â aÂˇcos(Eât/Ä§ – pâx/Ä§) + iÂˇaÂˇsin(Eât/Ä§ – pâx/Ä§)

= aÂˇcos(pâx/Ä§ – Eât/Ä§) iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§)

The vectors p and x are the the momentum and position vector respectively: p = (px, py, pz) and x = (x, y, z). However, if we assume there is no uncertainty about p â not about the direction nor the magnitude â 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 pâx/Ä§ reduces to pâx/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(Î¸) = cos(0) = 1 and sin(Î¸) = 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 âÄ§/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 Ď/2: sin(Î¸) = cos(Î¸ â Ď/2).

As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any wave F(kx – Ďt) is given by vp =Â Ď/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: vp =Â Ď/k = E/p.

The de Broglie relations

E/Ä§ = Ď 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 = Ď/2ĎÂ  and Îť = 2Ď/k, which gives us the two de Broglie relations:

1. E = Ä§âĎ = hâf
2. p = Ä§âk = h/Îť

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/Îť relation, the wavelength is inversely proportional to the momentum: Îť = h/p. The velocity of a photon, or a (theoretical) particle with zero rest mass (m0 = 0), is c and, therefore, we find that p = mvâv = mcâc = mâc (all of the energy is kinetic). Hence, we can write: pâc = mâc2 = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass, the wavelength can be written as:

Îť = h/p = hc/E = h/mc

However, this is a limiting situation â 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 pÂ â 0 then Îť âÂ â. How should we interpret this inverse proportionality between Îť and p? To answer this question, let us first see what this wavelength Îť actually represents.

If we look at the Ď = aÂˇcos(pâx/Ä§ – Eât/Ä§) – iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§) once more, and if we write pâx/Ä§ as Î, then we can look at pâx/Ä§ as a phase factor, and so we will be interested to know for what x this phase factor Î = pâx/Ä§ will be equal to 2Ď. So we write:

Î =pâx/Ä§ = 2Ď âÂ x = 2ĎâÄ§/p = h/p = Îť

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 Îť actually represents for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2ĎÂˇ(Ä§/E). Hence, we can now calculate the wave velocity:

v = Îť/T = (h/p)/[2ĎÂˇ(Ä§/E)] = E/p

Unsurprisingly, we just get the phase velocity that we had calculated already: v = vp = 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 = mâc2/mâvg Â = c/Î˛g, in which Î˛g is the relative classical velocity[3] of our particle Î˛g = vg/c) tells us that the phase velocities will effectively be superluminal (Î˛gÂ  < 1 so 1/ Î˛g > 1), but what if Î˛g approaches zero? The conclusion seems unavoidable: for a particle at rest, we only have a frequency in time, as the wavefunction reduces to:

Ď =Â aÂˇeâiÂˇEÂˇt/Ä§ =Â aÂˇcos(Eât/Ä§) – iÂˇaÂˇsin(Eât/Ä§)

How should we interpret this?

A physical interpretation of relativistic length contraction?

In my previous posts,Â 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 Îť decreases. Hence, the physical interpretation of the wavefunction that is offered here may explain relativistic length contraction.

đ

Yep. Think about it. đ

[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 (Îł), we can write the relativistically correct formula for the kinetic energy as KE = EÂ âÂ E0Â =Â mvc2Â â m0c2Â =Â m0Îłc2Â â m0c2Â =Â m0c2(Îł â 1). As v approaches c, Îł 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 vg.

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.Â The elementary wavefunction is written as:

Ď =Â aÂˇeâi[EÂˇtÂ â pâx]/Ä§ =Â aÂˇcos(pâx/Ä§Â â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§Â â Eât/Ä§)

Of course, NatureÂ (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 Ď =Â aÂˇei[EÂˇt â pâx]/Ä§ function is also permitted. We know that cos(Î¸) = cos(âÎ¸) and sinÎ¸ = âsin(âÎ¸), so we can write:Â  Â Â

Ď =Â aÂˇei[EÂˇt â pâx]/Ä§ =Â aÂˇcos(Eât/Ä§Â âÂ pâx/Ä§) + iÂˇaÂˇsin(Eât/Ä§Â âÂ pâx/Ä§)

= aÂˇcos(pâx/Ä§Â â Eât/Ä§) âÂ iÂˇaÂˇsin(pâx/Ä§Â â Eât/Ä§)

The vectors p and x are the momentum and position vector respectively: p = (px, py, pz) and x = (x, y, z). However, if we assume there is no uncertainty about p â not about the direction, and not about the magnitude â then the direction of p can be our x-axis. In this reference frame,Â x = (x, y, z) reduces to (x, 0, 0), and pâx/Ä§ reduces to pâx/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(Î¸) = cos(0) = 1 and sin(Î¸) = 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 âÄ§/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 Ď/2: sin(Î¸) = cos(Î¸ â Ď/2).

As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any waveform F(kx â Ďt) is given by vp =Â Ď/k. In our case, we find thatÂ vp =Â Ď/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 âp and, therefore, we would get a negative phase velocity: vp =Â Ď/k = (E/Ä§)/(âp/Ä§) = âE/p.

As you know, E/Ä§ = Ď 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 = Ď/2ĎÂ  and Îť = 2Ď/k, which gives us the two de Broglie relations:

1. E = Ä§âĎ = hâf
2. p = Ä§âk = h/Îť

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: Îť = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long, so we write:

If pÂ â 0 then Îť âÂ â.

For the limit situation, a particle with zero rest mass (m0 = 0), the velocity may be c and, therefore, we find that p = mvâv = mcâc = mâc (all of the energy is kinetic) and, therefore, pâc = mâc2 = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass (m0Â = 0), the wavelength can be written as:

Îť = h/p = hc/E = h/mc

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 neverÂ equalÂ c.[2] The question remains: how should we interpret the inverse proportionality between Îť and p?

Let us first see what this wavelength Îť actually represents. If we look at the Ď = aÂˇcos(pâx/Ä§ â Eât/Ä§) âÂ iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§) once more, and if we write pâx/Ä§ as Î, then we can look at pâx/Ä§ as a phase factor, and so we will be interested to know for what x this phase factor Î = pâx/Ä§ will be equal to 2Ď. So we write:

Î =pâx/Ä§ = 2Ď âÂ x = 2ĎâÄ§/p = h/p = Îť

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 Îť actually represent for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2ĎÂˇ(Ä§/E). Hence, we can now calculate the wave velocity:

v = Îť/T = (h/p)/[2ĎÂˇ(Ä§/E)] = E/p

Unsurprisingly, we just get the phase velocity that we had calculated already: v = vp = 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 â or the concept of a precise energy, and a precise momentum â does not make sense. This is where the Uncertainty Principle comes in: stating that p = 0, implies zero uncertainty. Hence, the Ďp factor in the ĎpâĎxÂ â¤ Ä§/2 would be zero and, therefore, ĎpâĎ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 Ďp refers to the standard deviation from the mean, as illustrated below. Of course, the distribution may be or may not be normal â we donât know â 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 thatâs about 0.726Ă10â17 m for the distance unit and 0.726Ă10â17 Nâs for the momentum unit, as the product of both gives us Ä§/2. To make this somewhat more real, we may note that 0.726Ă10â17 m is the attometer scale (1 am = 1Ă10â18 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 â which should correspond to the classical idea of the velocity of our particle â only makes sense in the context of wave packet. Indeed, the group velocity of a wave packet (vg) is calculated as follows:

vgÂ = âĎi/âkiÂ = â(Ei/Ä§)/â(pi/Ä§) = â(Ei)/â(pi)

This assumes the existence of a dispersion relation which gives us ĎiÂ as a function of ki â what amounts to the same â EiÂ as a function of pi. How do we get that?Â Well… There are a few ways to go about it but one interesting way of doing it is to re-write SchrĂśdinger’s equation as the following pairÂ of equations[4]:

1. Re(âĎ/ât) = â[Ä§/(2meff)]ÂˇIm(â2Ď)Â âÂ ĎÂˇcos(kx â Ďt) =Â k2Âˇ[Ä§/(2meff)]Âˇcos(kx â Ďt)
2. Im(âĎ/ât) = [Ä§/(2meff)]ÂˇRe(â2Ď)Â â ĎÂˇsin(kx â Ďt) = k2Âˇ[Ä§/(2meff)]Âˇsin(kx â Ďt)

These equations imply the following dispersion relation:

Ď =Â Ä§Âˇk2/(2m)

Of course, we need to think about the subscripts now: we have Ďi, ki, but… What about meff or, dropping the subscript, about m? Do we write it as mi? If so, what is it? Well… It is the equivalent mass of EiÂ obviously, and so we get it from the mass-energy equivalence relation: miÂ = Ei/c2. 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: Ďm = ĎE/c2. We are tempted to do a few substitutions here. Letâs first check what we get when doing the miÂ = Ei/c2 substitution:

Ďi =Â Ä§Âˇki2/(2mi) = (1/2)âÄ§Âˇki2âc2/Ei = (1/2)âÄ§Âˇki2âc2/(ĎiâÄ§)Â = (1/2)âÄ§Âˇki2âc2/Ďi

â Ďi2/ki2 = c2/2Â â Ďi/ki = vp = c/2 !?

We get a very interesting but nonsensical condition for the dispersion relation here. I wonder what mistake I made. đŚ

Let us try another substitution. The group velocity is what it is, right? It is the velocity of the group, so we can write: ki = p/Ä§ = mi Âˇvg. This gives us the following result:

Ďi =Â Ä§Âˇ(mi Âˇvg)2/(2mi) = Ä§ÂˇmiÂˇvg2/2

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 weÂ dropÂ it. Now you will object that SchrĂśdinger’s equation gives us the electron orbitals – and many other correct descriptions of quantum-mechanical stuff – so, surely,Â SchrĂśdinger’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 atÂ oneÂ of the two dimensions of the oscillation only and, therefore, it’s onlyÂ halfÂ 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(âĎ/ât) = â(Ä§/meff)ÂˇIm(â2Ď)Â âÂ ĎÂˇcos(kx â Ďt) =Â k2Âˇ(Ä§/meff)Âˇcos(kx â Ďt)
• Im(âĎ/ât) = (Ä§/meff)ÂˇRe(â2Ď)Â â ĎÂˇsin(kx â Ďt) = k2Âˇ(Ä§/meff)Âˇsin(kx â Ďt)

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

Ďi =Â Ä§Âˇki2/mi

TheÂ miÂ = Ei/c2 substitution then gives us the result we sort of expected to see:

Ďi =Â Ä§Âˇki2/miÂ = Ä§Âˇki2âc2/Ei = Ä§Âˇki2âc2/(ĎiâÄ§)Â â Ďi/ki = vp = c

Likewise, the other calculation also looks more meaningful now:

Ďi =Â Ä§Âˇ(mi Âˇvg)2/miÂ = Ä§ÂˇmiÂˇvg2

Sweet ! đ

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 â the speed with which those wave crests (or troughs) move â and (2) some kind of circular or tangential velocity â the velocity along the red contour lineÂ above. Weâll need the formula for a tangential velocity: vt = aâĎ.

Now, if Îť is zero, then vt = aâĎ = aâE/Ä§ is just all there is. We may double-check this as follows: the distance traveled in one period will be equal to 2Ďa, and the period of the oscillation is T = 2ĎÂˇ(Ä§/E). Therefore, vt will, effectively, be equal to vt = 2Ďa/(2ĎÄ§/E) = aâE/Ä§.Â However, if Îť is non-zero, then the distance traveled in one period will be equal to 2Ďa + Îť. The period remains the same: T = 2ĎÂˇ(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 theÂ real-life +Ä§/2 or âÄ§/2 values of its spin, and we got aÂ numericalÂ value forÂ a. It was the Compton radius: the scattering radius for an electron. Let us write it out:

Using the right numbers, youâll find the numerical value for a: 3.8616Ă10â13 m. But let us just substitute the formula itself here:Â

This is fascinating ! And we just calculated that vpÂ is equal toÂ c. For the elementary wavefunction, that is. Hence, we get this amazing result:

vt = 2c

ThisÂ tangentialÂ velocity isÂ twiceÂ the linearÂ velocity !

Of course, the question is: what is theÂ physicalÂ significance of this? I need to further look at this. Wave velocities are, essentially, mathematicalÂ concepts only: the wave propagates through space, butÂ nothing elseÂ 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 aÂ propertyÂ of the vacuum – or of the fabric of spacetime itself. đ

[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/c2. This mass combines the three known neutrino flavors.

[2] Using the Lorentz factor (Îł), we can write the relativistically correct formula for the kinetic energy as KE = EÂ âÂ E0Â =Â mvc2Â â m0c2Â =Â m0Îłc2Â â m0c2Â =Â m0c2(Îł â 1). As v approaches c, Îł 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)Ă10â35 m).

[4] The meff is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write meff = m. As for the equations, they are easily derived from noting that two complex numbers a +Â iâb and c +Â iâd are equal if, and only if, their real and imaginary parts are the same. Now, the âĎ/ât =Â iâ(Ä§/meff)ââ2Ď equation amounts to writing something like this: a +Â iâb =Â iâ(c +Â iâd). Now, remembering thatÂ i2Â = â1, you can easily figure out thatÂ iâ(c +Â iâd) =Â iâc +Â i2âd = â d +Â iâc.

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 theÂ elementary wavefunction by heart:

Ď =Â aÂˇeâi[EÂˇt â pâx]/Ä§ =Â aÂˇcos(pâx/Ä§ â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ â Eât/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 pâx/Ä§ reduces to pâx/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(Î¸) = cos(0) = 1 and sin(Î¸) = 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 âÄ§/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 Ď/2: sin(Î¸) = cos(Î¸ â Ď/2) Â

Surely, Nature doesn’t care a hoot about our conventions for measuring the phase angle clockwise or counterclockwise and therefore, the Ď =Â aÂˇei[EÂˇt â pâx]/Ä§ function should, effectively, also be permitted. We know that cos(Î¸) = cos(Î¸) and sinÎ¸ = sin(Î¸), so we can write: Â Â Â

Ď =Â aÂˇei[EÂˇt â pâx]/Ä§ =Â aÂˇcos(Eât/Ä§ â pâx/Ä§) + iÂˇaÂˇsin(Eât/Ä§ â pâx/Ä§)

= aÂˇcos(pâx/Ä§ â Eât/Ä§) âÂ iÂˇaÂˇsin(pâx/Ä§ â Eât/Ä§)

E/Ä§ = Ď 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 = Ď/2ĎÂ  and Îť = 2Ď/k, which gives us the two de Broglie relations:

1. E = Ä§âĎ = hâf
2. p = Ä§âk = h/Îť

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: Îť = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long: if pÂ â 0, then Îť âÂ â.Â For the limit situation, a particle with zero rest mass (m0 = 0), the velocity may be c and, therefore, we find that p = mvâv = mâc Â and, therefore, pâc = mâc2 = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass, the wavelength can be written as:

Îť = 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 Îť 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 vp =Â Ď/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 = mÂˇvg. However, when p is zero, we have a division by zero once more: if pÂ â 0, then vp = E/pÂ â â. 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:

vpÂ =Â Ď/k = E/p = E/(mÂˇvg) = (mÂˇc2)/(mÂˇvg) = c2/vg

We can re-write this as vpÂˇvgÂ = c2, which reminds us of the relationship between the electric and magnetic constant (1/Îľ0)Âˇ(1/Îź0) = c2. But what is the group velocity of the elementary wavefunction? Is it a meaningful concept?

The phase velocity is just the ratio of Ď/k. In contrast, the group velocity is the derivative of Ď with respect to k. So we need to write Ď as a function of k. Can we do that even if we have only one wave? We doÂ notÂ have a wave packet here, right? Just some hypotheticalÂ building blockÂ 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 thisÂ elementaryÂ wavefunction. Let’s first get thatÂ Ď = Ď(k) relation. You’ll remember we can write SchrĂśdingerâs equation – the equation that describes theÂ propagationÂ mechanism for matter-waves – asÂ the following pair of equations:

1. Re(âĎ/ât) = â[Ä§/(2m)]ÂˇIm(â2Ď)Â âÂ ĎÂˇcos(kx â Ďt) =Â k2Âˇ[Ä§/(2m)]Âˇcos(kx â Ďt)
2. Im(âĎ/ât) = [Ä§/(2m)]ÂˇRe(â2Ď)Â â ĎÂˇsin(kx â Ďt) = k2Âˇ[Ä§/(2m)]Âˇsin(kx â Ďt)

This tells us that Ď =Â Ä§Âˇk2/(2m). Therefore, we can calculate âĎ/âkÂ as:

âĎ/âk = Ä§Âˇk/m = p/m = vg

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 mathematicalÂ formula for the group velocity – but we don’t know what it represents – and a finite wavelength. In fact, using the p = Ä§âk = h/Îť relation, we can write one as a function of the other:

Îť = h/p = h/mvg âÂ vg = h/mÎť

What does this mean? ItÂ resembles the c = h/mÎť relation we had for a particle with zero rest mass. Of course, it does: the Îť = h/mc relation is, once again, a limit for vg going to c. By the way, it is interesting to note that the vpÂˇvgÂ = c2 relation implies that the phase velocity is always superluminal. That’ easy to see when you re-write the equation in terms ofÂ relativeÂ velocities: (vp/c)Âˇ(vg/c) =Â Î˛phaseÂˇÎ˛groupÂ = 1. Hence, ifÂ Î˛groupÂ < 1, thenÂ Î˛phaseÂ > 1.

So whatÂ isÂ the geometry,Â really? Letâs look at the Ď = aÂˇcos(pâx/Ä§ – Eât/Ä§) iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§) formula once more. If we write pâx/Ä§ as Î, then we will be interested to know for what x this phase factor will be equal to 2Ď. So we write:

Î =pâx/Ä§ = 2Ď âÂ x = 2ĎâÄ§/p = h/p = ÎťÂ Â

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.Â 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 lineÂ above). Weâll probably need the formula for the tangential velocity: v = aâĎ. If p = 0 (so we have that weird infinitesimally long wavelength), then we have two velocities:

1. The tangential velocity around theÂ aÂˇeiÂˇEÂˇtÂ  circle, so to speak, and that will just be equal toÂ v = aâĎ =Â aâE/Ä§.
2. The red contour line sort of gets stretched out, like infinitely long, and the velocity becomes… What does it do? Does it go toÂ â , or toÂ 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 theÂ real-life +Ä§/2 or âÄ§/2 values of its spin. And so we got aÂ numericalÂ value forÂ 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:Â aÂ = 3.8616Ă10â13 m.Â We did then another weird calculation. We said all of the energy of the electron had to be packed in thisÂ cylinderÂ that might of might not be there. The point was: the energy is finite, so thatÂ elementaryÂ wavefunction cannotÂ have an infinite length in space. Indeed, assuming that the energy was distributed uniformly, we jotted down this formula, which reflects the formula for theÂ volume of a cylinder:

E = ĎÂˇa2Âˇl âÂ lÂ = E/(ĎÂˇa2)

Using the value we got for the Compton scattering radius (aÂ =Â 3.8616Ă10â13 m), we got an astronomical value forÂ l. Let me write it out:

lÂ =Â (8.19Ă10â14)/(ĎÂˇ14.9Ă10â26) â 0.175Ă1012Â m

It is,Â literally, an astronomical value:Â 0.175Ă1012Â m is 175 millionÂ kilometer, so that’s like theÂ 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 forÂ Îť? We’d get that linear velocity, right? Let’s try it. TheÂ periodÂ is equal to T =Â T = 2ĎÂˇ(Ä§/E) = h/E and Îť =Â E/(ĎÂˇa2), so we write:We can write this as a function of m and theÂ cÂ andÂ 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 toÂ showÂ the geometry of the wavefunction a bit more in detail.

[1]Â The discussions on the mass of neutrinos are interesting in this regard. Scientists all felt the neutrinoÂ had toÂ haveÂ 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/c2. 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. đ

Electron and photon strings

Note: I have published a paper that is very coherent and fully explains what the idea of a photon might be. There is nothing stringy. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

In my previous posts, I’ve been playing with… Well… At the very least, a new didactic approach to understanding the quantum-mechanical wavefunction. I just boldly assumed the matter-wave is a gravitational wave. I did so by associating its components with the dimension of gravitational field strength: newton per kg, which is the dimension of acceleration (N/kg = m/s2). Why? When you remember the physical dimension of the electromagnetic field is N/C (force per unitÂ charge), then that’s kinda logical, right? đÂ The math is beautiful. Key consequences include the following:

1. Schrodinger’s equation becomes an energy diffusion equation.
2. Energy densities give us probabilities.
3. The elementary wavefunction for the electron gives us the electron radius.
4. Spin angular momentum can be interpreted as reflecting the right- or left-handedness of the wavefunction.
5. Finally, the mysterious boson-fermion dichotomy is no longer “deep down in relativistic quantum mechanics”, as Feynman famously put it.

It’s all great. Every day brings something new. đ Today I want to focus on our weird electron model and how we get God’s number (aka the fine-structure constant) out of it. Let’s recall the basics of it.Â We had the elementary wavefunction:

Ď =Â aÂˇeâi[EÂˇt â pâx]/Ä§ =Â aÂˇeâi[EÂˇt â pâx]/Ä§ = aÂˇcos(pâx/Ä§ â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ â Eât/Ä§)

In one-dimensional space (think of a particle traveling along some line), the vectors (p and x) become scalars, and so we simply write:

Ď =Â aÂˇeâi[EÂˇt â pâx]/Ä§ =Â aÂˇeâi[EÂˇt â pâx]/Ä§ = aÂˇcos(pâx/Ä§ â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ â Eât/Ä§)

This wavefunction comes with constantÂ probabilities |Ď|2Â  = a2, so we need to define a space outside of whichÂ Ď = 0. Think of the particle-in-a-box model. This is obvious oscillations pack energy, and the energy of our particle is finite. Hence, each particle – be it a photon or an electron – will pack aÂ finiteÂ number of oscillations. It will, therefore, occupy a finite amount of space. Mathematically, this corresponds to the normalization condition:Â all probabilities have to add up to one, as illustrated below.Now, allÂ oscillations of the elementary wavefunction have the same amplitude:Â a. [Terminology is a bit confusing here because we use the term amplitude to refer to two very different things here: we may sayÂ a is the amplitude of the (probability) amplitudeÂ Ď. So how many oscillations do we have? What is theÂ sizeÂ of our box? Let us assume our particle is an electron, and we will reduce its motion to aÂ one-dimensionalÂ motion only: we’re thinking of it as traveling along the x-axis. We can then use the y- andÂ z-axes asÂ mathematical axes only: they will show us how the magnitude and direction of the real and imaginary component ofÂ Ď. The animation below (for which I have to credit Wikipedia) shows how it looks like.Of course, we can have right- as well as left-handed particle waves because, while timeÂ physicallyÂ goes by in one direction only (we can’t reverse time), we can countÂ it in two directions: 1, 2, 3, etcetera orÂ â1,Â â2,Â â3,Â etcetera. In the latter case, think of timeÂ tickingÂ away. đ Of course, in ourÂ physicalÂ interpretation of the wavefunction, this should explain the (spin) angular momentum of the electron, which is – for some mysterious reason that we now understand đ – always equal toÂ JÂ =Â Âą Ä§/2.

Now, becauseÂ a is some constant here, we may think of our box as a cylinder along the x-axis. Now, the rest mass of an electron is about 0.510 MeV, so that’s around 8.19Ă10â14 Nâm, so it will pack some 1.24Ă1020Â oscillations per second. So how long is our cylinder here? To answer that question, we need to calculate theÂ phaseÂ velocity of our wave. We’ll come back to that in a moment. Just note how this compares to a photon: the energy of a photon will typically be a few electronvoltÂ only (1 eVÂ â 1.6 Ă10â19Â NÂˇm) and, therefore, it will pack like 1015Â oscillations per second, so that’s a density (in time) that is about 100,000 timesÂ less.

Back to the angular momentum. The classical formula for it isÂ L = IÂˇĎ, so that’s angular frequency times angular mass. What’s the angular velocity here? That’s easy:Â Ď =Â E/Ä§. What’s the angular mass? If we think of our particle as a tiny cylinder,Â we may use the formula for its angular mass: I = mÂˇr2/2. We have m: that’s the electron mass, right? Right? So what is r? That should be the magnitude of the rotating vector, right? So that’sÂ a. Of course, the mass-energyÂ equivalence relation tells us that E = mc2, so we can write:

L = IÂˇĎ = (mÂˇr2/2)Âˇ(E/Ä§) = (1/2)Âˇa2ÂˇmÂˇ(mc2/Ä§) = (1/2)Âˇa2Âˇm2Âˇc2/Ä§

Does it make sense? Maybe. Maybe not. You can check the physical dimensions on both sides of the equation, and that works out: we do get something that is expressed in NÂˇmÂˇs, so that’s actionÂ orÂ angular momentumÂ units. Now, weÂ knowÂ L must be equal toÂ JÂ =Â Âą Ä§/2. [As mentioned above, the plus or minus sign depends on the left- or right-handedness of our wavefunction, so don’t worry about that.] How do we know that? Because of the Stern-Gerlach experiment, which has been repeated a zillion times, if not more. Now, if L =Â J, then we get the following equation for a:Â Â This is the formula for the radius of an electron. To be precise, it is theÂ Compton scattering radius, so that’s theÂ effectiveÂ radius of an electron as determined by scattering experiments. You can calculate it:Â it is about 3.8616Ă10â13 m, so that’s theÂ picometerÂ scale, as we would expect.

This isÂ a rather spectacular result. As far as I am concerned, it is spectacular enough for me to actuallyÂ believeÂ myÂ interpretation of the wavefunction makes sense.

Let us now try to think about theÂ lengthÂ of our cylinder once again. The period of our wave is equal to T = 1/f = 1/(Ď/2Ď) = 1/[(E/Ä§)Âˇ2Ď] =Â 1/(E/h) = h/E. Now, theÂ phaseÂ velocity (vp) will be given by:

vpÂ =Â ÎťÂˇfÂ = (2Ď/k)Âˇ(Ď/2Ď) =Â Ď/k =Â (E/Ä§)/(p/Ä§) = E/p = E/(mÂˇvg) = (mÂˇc2)/(mÂˇvg) = c2/vg

This isÂ veryÂ interesting, because it establishes anÂ inverseÂ proportionality betweenÂ the group and the phase velocity of our wave, withÂ c2Â as the coefficient ofÂ inverseÂ proportionality.Â In fact, this equation looks better if we write asÂ vpÂˇvgÂ =Â c2. Of course, theÂ groupÂ velocityÂ (vg) is theÂ classicalÂ velocity of our electron. This equation shows us the idea of an electron at rest doesn’t make sense: ifÂ vgÂ = 0, thenÂ vpÂ times zero must equalÂ c2, which cannot be the case: electronsÂ mustÂ move in space. More generally, speaking, matter-particles must move in space, with the photon as our limiting case: it moves at the speed of light. Hence, for a photon, we find that vpÂ =Â vgÂ = E/p =Â c.

How can we calculate theÂ lengthÂ of a photon or an electron? It is an interesting question. The mentioned orders or magnitude of the frequency (1015Â or 1020) gives us the number of oscillations per second. But how many do we have inÂ oneÂ photon, or inÂ one electron?

Let’s first think about photons, because we have more clues here. Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. We know how to calculate to calculate the Q of these atomic oscillators (see, for example, Feynman I-32-3):Â it is of the order of 108, which means the wave train will last about 10â8Â seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/e). Now, the frequency of sodium light, for example, is 0.5Ă1015Â oscillations per second, and the decay time is about 3.2Ă10â8Â seconds, so that makes for (0.5Ă1015)Âˇ(3.2Ă10â8) = 16 million oscillations. Now, the wavelength is 600 nanometer (600Ă10â9) m), so that gives us a wavetrain with a length of (600Ă10â9)Âˇ(16Ă106) = 9.6 m.

These oscillations may or may not have the same amplitude and, hence, each of these oscillations may pack a different amount of energies. However,Â if the total energy of our sodium light photon (i.e. about 2 eVÂ âÂ 3.3Ă10â19Â J) are to be packed in those oscillations, then each oscillation would pack about 2Ă10â26Â J, on average, that is. We speculated in other posts on how we might imagine the actual wave pulse that atoms emit when going from one energy state to another, so we don’t do that again here. However, the following illustration of the decay of a transient signal dies out may be useful.

This calculation is interesting. It also gives us an interesting paradox: if a photon is a pointlike particle, how can we say its length is like 10 meterÂ or more? Relativity theory saves us here. We need to distinguish the reference frame of the photon â riding along the wave as it is being emitted, so to speak â and our stationary reference frame, which is that of the emitting atom. Now, because the photon travels at the speed of light, relativistic length contraction will make it lookÂ like a pointlike particle.

What about the electron? Can we use similarÂ assumptions? For the photon, we can use the decay time to calculate the effective numberÂ of oscillations. What can we use for an electron? We will need to make some assumption about the phase velocity or, what amounts to the same, the group velocity of the particle. What formulas can we use? TheÂ p = mÂˇv is the relativistically correct formula for the momentum of an object if m = mv, so that’s the same m we use in the E =Â mc2Â formula. Of course,Â vÂ here is, obviously, the group velocity (vg), so that’s the classical velocity of our particle. Hence, we can write:

p = mÂˇvgÂ = (E/c2)ÂˇvgÂ âÂ vgÂ = p/m =Â Â pÂˇc2/E

This is just another way of writing thatÂ vgÂ =Â c2/vpÂ or vpÂ =Â c2/vgÂ so it doesn’t help, does it? Maybe. Maybe not. Let us substitute in our formula for the wavelength:

Îť =Â vp/fÂ =Â vpÂˇTÂ =Â vpâ(h/E) = (c2/vg)Âˇ(h/E) = h/(mÂˇvg) = h/pÂ

This gives us the otherÂ de BroglieÂ relation:Â Îť =Â h/p. This doesn’t help us much, although it is interesting to think about it. TheÂ fÂ = E/h relation is somewhat intuitive: higher energy, higher frequency. In contrast, what the Îť =Â h/p relation tells us that we get an infinite wavelength if the momentum becomes really small. What does this tell us? I am not sure. Frankly, I’ve look at the secondÂ de BroglieÂ relation like a zillion times now, and I think it’s rubbish. It’s meant to be used for the groupÂ velocity, I feel. I am saying that because we get a non-sensical energy formula out of it. Look at this:

1. E = hÂˇf and p = h/Îť. Therefore, f = E/h and Îť = p/h.
2. vÂ =Â fÂˇÎť = (E/h)â(p/h) = E/p
3. p = mÂˇv. Therefore, E = vÂˇp = mÂˇv2

E = mÂˇv2? This formula is only correct ifÂ vÂ =Â c, in which case it becomes theÂ E = mc2 equation. So it then describes a photon, or a massless matter-particle which… Well… That’s a contradictio in terminis. đ In all other cases, we get nonsense.

Let’s try something differently.Â  If our particle is at rest, then p = 0 and theÂ pÂˇx/Ä§ term in our wavefunction vanishes, so it’s just:

Ď =Â aÂˇeâiÂˇEÂˇt/Ä§ =Â aÂˇcos(Eât/Ä§) â iÂˇaÂˇsin(Eât/Ä§)

Hence, our wave doesn’t travel. It has the same amplitude at every point in space at any point in time. Both the phase and group velocity become meaningless concepts. TheÂ amplitude variesÂ – because of the sine and cosine – but the probability remains the same:Â |Ď|2Â  = a2. Hmm… So we need to find another way to define the size of our box. One of the formulas I jotted down in my paper in which I analyze the wavefunction as a gravitational waveÂ was this one:

It was a physicalÂ normalization condition: the energy contributions of the waves that make up a wave packet need to add up to the total energy of our wave. Of course, for our elementary wavefunction here, the subscripts vanish and so the formula reduces to E = (E/c2)Âˇa2Âˇ(E2/Ä§2), out of which we get our formula for the scattering radius: aÂ =Â Ä§/mc. Now how do we pack that energy in our cylinder?Â Assuming that energy is distributed uniformly, we’re tempted to write something like E =Â a2Âˇl or, looking at the geometry of the situation:

E = ĎÂˇa2Âˇl âÂ lÂ = E/(ĎÂˇa2)

It’s just the formula for the volume of a cylinder.Â Using the value we got for the Compton scattering radius (aÂ =Â 3.8616Ă10â13 m), we find anÂ l that’sÂ equal to (8.19Ă10â14)/(ĎÂˇ14.9Ă10â26) =â 0.175Ă1012Meter?Â Yes. We get the following formula:

0.175Ă1012Â m is 175 millionÂ kilometer. That’s – literally – astronomic. It corresponds to 583 light-seconds, or 9.7 light-minutes.Â So that’s about 1.17 times the (average) distance between the Sun and the Earth. You can see that we do need to build a wave packet: that space is a bit too large to look for an electron, right? đ

Could we possibly get some less astronomic proportions? What if weÂ imposeÂ thatÂ lÂ should equalÂ a? We get the following condition:We find that m would have to be equal to m â 1.11Ă10â36Â kg. That’s tiny. In fact, it’s equivalent to an energy of aboutÂ  equivalent to 0.623 eV (which you’ll see written as 623 milli-eV. This corresponds to light with a wavelength of about 2 micro-meter (Îźm), so that’s in the infrared spectrum. It’s a funny formula: we find, basically, that theÂ l/aÂ ratio is proportional to m4. Hmm… What should we think of this? If you have any ideas, let me know !

Post scriptum (3 October 2017):Â The paper is going well. Getting lots of downloads, and the views on my blog are picking up too. But I have been vicious. Substituting BÂ for (1/c)âiâEÂ or for â(1/c)âiâEÂ implies a very specific choice of reference frame. The imaginary unit is a two-dimensional concept: it only makes sense when giving it a planeÂ view. Literally. Indeed, myÂ formulas assume the iÂ (or âi) plane is perpendicular to the direction of propagation of the elementary quantum-mechanical wavefunction. So… Yes. The need for rotation matrices is obvious. But my physicalÂ interpretation of the wavefunction stands. đ

Wavefunctions as gravitational waves

This is the paper I always wanted to write. It is there now, and I think it is good – and that‘s an understatement. đ It is probably best to download it as a pdf-file from the viXra.org site because this was a rather fast ‘copy and paste’ job from the Word version of the paper, so there may be issues with boldface notation (vector notation), italics and, most importantly, with formulas – which I, sadly, have to ‘snip’ into this WordPress blog, as they don’t have an easy copy function for mathematical formulas.

It’s great stuff. If you have been following my blog – and many of you have – you will want to digest this. đ

Abstract : This paper explores the implications of associating the components of the wavefunction with a physical dimension: force per unit mass â which is, of course, the dimension of acceleration (m/s2) and gravitational fields. The classical electromagnetic field equations for energy densities, the Poynting vector and spin angular momentum are then re-derived by substituting the electromagnetic N/C unit of field strength (mass per unit charge) by the new N/kg = m/s2 dimension.

The results are elegant and insightful. For example, the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities, which establishes a physical normalization condition. Also, SchrĂśdingerâs wave equation may then, effectively, be interpreted as a diffusion equation for energy, and the wavefunction itself can be interpreted as a propagating gravitational wave. Finally, as an added bonus, concepts such as the Compton scattering radius for a particle, spin angular momentum, and the boson-fermion dichotomy, can also be explained more intuitively.

While the approach offers a physical interpretation of the wavefunction, the author argues that the core of the Copenhagen interpretations revolves around the complementarity principle, which remains unchallenged because the interpretation of amplitude waves as traveling fields does not explain the particle nature of matter.

Introduction

This is not another introduction to quantum mechanics. We assume the reader is already familiar with the key principles and, importantly, with the basic math. We offer an interpretation of wave mechanics. As such, we do not challenge the complementarity principle: the physical interpretation of the wavefunction that is offered here explains the wave nature of matter only. It explains diffraction and interference of amplitudes but it does not explain why a particle will hit the detector not as a wave but as a particle. Hence, the Copenhagen interpretation of the wavefunction remains relevant: we just push its boundaries.

The basic ideas in this paper stem from a simple observation: the geometric similarity between the quantum-mechanical wavefunctions and electromagnetic waves is remarkably similar. The components of both waves are orthogonal to the direction of propagation and to each other. Only the relative phase differs : the electric and magnetic field vectors (E and B) have the same phase. In contrast, the phase of the real and imaginary part of the (elementary) wavefunction (Ď = aÂˇeâiâÎ¸ = aâcosÎ¸ – aâsinÎ¸) differ by 90 degrees (Ď/2).[1] Pursuing the analogy, we explore the following question: if the oscillating electric and magnetic field vectors of an electromagnetic wave carry the energy that one associates with the wave, can we analyze the real and imaginary part of the wavefunction in a similar way?

We show the answer is positive and remarkably straightforward.  If the physical dimension of the electromagnetic field is expressed in newton per coulomb (force per unit charge), then the physical dimension of the components of the wavefunction may be associated with force per unit mass (newton per kg).[2] Of course, force over some distance is energy. The question then becomes: what is the energy concept here? Kinetic? Potential? Both?

The similarity between the energy of a (one-dimensional) linear oscillator (E = mÂˇa2ÂˇĎ2/2) and Einsteinâs relativistic energy equation E = mâc2 inspires us to interpret the energy as a two-dimensional oscillation of mass. To assist the reader, we construct a two-piston engine metaphor.[3] We then adapt the formula for the electromagnetic energy density to calculate the energy densities for the wave function. The results are elegant and intuitive: the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities. SchrĂśdingerâs wave equation may then, effectively, be interpreted as a diffusion equation for energy itself.

As an added bonus, concepts such as the Compton scattering radius for a particle and spin angular, as well as the boson-fermion dichotomy can be explained in a fully intuitive way.[4]

Of course, such interpretation is also an interpretation of the wavefunction itself, and the immediate reaction of the reader is predictable: the electric and magnetic field vectors are, somehow, to be looked at as real vectors. In contrast, the real and imaginary components of the wavefunction are not. However, this objection needs to be phrased more carefully. First, it may be noted that, in a classical analysis, the magnetic force is a pseudovector itself.[5] Second, a suitable choice of coordinates may make quantum-mechanical rotation matrices irrelevant.[6]

Therefore, the author is of the opinion that this little paper may provide some fresh perspective on the question, thereby further exploring Einsteinâs basic sentiment in regard to quantum mechanics, which may be summarized as follows: there must be some physical explanation for the calculated probabilities.[7]

We will, therefore, start with Einsteinâs relativistic energy equation (E = mc2) and wonder what it could possibly tell us.

I. Energy as a two-dimensional oscillation of mass

The structural similarity between the relativistic energy formula, the formula for the total energy of an oscillator, and the kinetic energy of a moving body, is striking:

1. E = mc2
2. E = mĎ2/2
3. E = mv2/2

In these formulas, Ď, v and c all describe some velocity.[8] Of course, there is the 1/2 factor in the E = mĎ2/2 formula[9], but that is exactly the point we are going to explore here: can we think of an oscillation in two dimensions, so it stores an amount of energy that is equal to E = 2ÂˇmÂˇĎ2/2 = mÂˇĎ2?

That is easy enough. Think, for example, of a V-2 engine with the pistons at a 90-degree angle, as illustrated below. The 90Â° angle makes it possible to perfectly balance the counterweight and the pistons, thereby ensuring smooth travel at all times. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down and provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring. Hence, we can describe it by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs.

Figure 1: Oscillations in two dimensions

If we assume there is no friction, we have a perpetuum mobile here. The compressed air and the rotating counterweight (which, combined with the crankshaft, acts as a flywheel[10]) store the potential energy. The moving masses of the pistons store the kinetic energy of the system.[11]

At this point, it is probably good to quickly review the relevant math. If the magnitude of the oscillation is equal to a, then the motion of the piston (or the mass on a spring) will be described by x = aÂˇcos(ĎÂˇt + Î).[12] Needless to say, Î is just a phase factor which defines our t = 0 point, and Ď is the natural angular frequency of our oscillator. Because of the 90Â° angle between the two cylinders, Î would be 0 for one oscillator, and âĎ/2 for the other. Hence, the motion of one piston is given by x = aÂˇcos(ĎÂˇt), while the motion of the other is given by x = aÂˇcos(ĎÂˇtâĎ/2) = aÂˇsin(ĎÂˇt).

The kinetic and potential energy of one oscillator (think of one piston or one spring only) can then be calculated as:

1. K.E. = T = mÂˇv2/2 = (1/2)ÂˇmÂˇĎ2Âˇa2Âˇsin2(ĎÂˇt + Î)
2. P.E. = U = kÂˇx2/2 = (1/2)ÂˇkÂˇa2Âˇcos2(ĎÂˇt + Î)

The coefficient k in the potential energy formula characterizes the restoring force: F = âkÂˇx. From the dynamics involved, it is obvious that k must be equal to mÂˇĎ2. Hence, the total energy is equal to:

E = T + U = (1/2)Âˇ mÂˇĎ2Âˇa2Âˇ[sin2(ĎÂˇt + Î) + cos2(ĎÂˇt + Î)] = mÂˇa2ÂˇĎ2/2

To facilitate the calculations, we will briefly assume k = mÂˇĎ2 and a are equal to 1. The motion of our first oscillator is given by the cos(ĎÂˇt) = cosÎ¸ function (Î¸ = ĎÂˇt), and its kinetic energy will be equal to sin2Î¸. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:

d(sin2Î¸)/dÎ¸ = 2âsinÎ¸âd(sinÎ¸)/dÎ¸ = 2âsinÎ¸âcosÎ¸

Let us look at the second oscillator now. Just think of the second piston going up and down in the V-2 engine. Its motion is given by the sinÎ¸ function, which is equal to cos(Î¸âĎ /2). Hence, its kinetic energy is equal to sin2(Î¸âĎ /2), and how it changes â as a function of Î¸ â will be equal to:

2âsin(Î¸âĎ /2)âcos(Î¸âĎ /2) = = â2âcosÎ¸âsinÎ¸ = â2âsinÎ¸âcosÎ¸

We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the crankshaft will rotate with a constant angular velocity: linear motion becomes circular motion, and vice versa, and the total energy that is stored in the system is T + U = ma2Ď2.

We have a great metaphor here. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. We know the wavefunction consist of a sine and a cosine: the cosine is the real component, and the sine is the imaginary component. Could they be equally real? Could each represent half of the total energy of our particle? Should we think of the c in our E = mc2 formula as an angular velocity?

These are sensible questions. Let us explore them.

II. The wavefunction as a two-dimensional oscillation

The elementary wavefunction is written as:

Ď = aÂˇeâi[EÂˇt â pâx]/Ä§aÂˇeâi[EÂˇt â pâx]/Ä§ = aÂˇcos(pâx/Ä§ Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ Eât/Ä§)

When considering a particle at rest (p = 0) this reduces to:

Ď = aÂˇeâiâEÂˇt/Ä§ = aÂˇcos(Eât/Ä§) + iÂˇaÂˇsin(Eât/Ä§) = aÂˇcos(Eât/Ä§) iÂˇaÂˇsin(Eât/Ä§)

Let us remind ourselves of the geometry involved, which is illustrated below. Note that the argument of the wavefunction rotates clockwise with time, while the mathematical convention for measuring the phase angle (Ď) is counter-clockwise.

Figure 2: Eulerâs formula

If we assume the momentum p is all in the x-direction, then the p and x vectors will have the same direction, and pâx/Ä§ reduces to pâx/Ä§. Most illustrations â such as the one below â will either freeze x or, else, t. Alternatively, one can google web animations varying both. The point is: we also have a two-dimensional oscillation here. These two dimensions are perpendicular to the direction of propagation of the wavefunction. For example, if the wavefunction propagates in the x-direction, then the oscillations are along the y– and z-axis, which we may refer to as the real and imaginary axis. Note how the phase difference between the cosine and the sine  â the real and imaginary part of our wavefunction â appear to give some spin to the whole. I will come back to this.

Figure 3: Geometric representation of the wavefunction

Hence, if we would say these oscillations carry half of the total energy of the particle, then we may refer to the real and imaginary energy of the particle respectively, and the interplay between the real and the imaginary part of the wavefunction may then describe how energy propagates through space over time.

Let us consider, once again, a particle at rest. Hence, p = 0 and the (elementary) wavefunction reduces to Ď = aÂˇeâiâEÂˇt/Ä§. Hence, the angular velocity of both oscillations, at some point x, is given by Ď = -E/Ä§. Now, the energy of our particle includes all of the energy â kinetic, potential and rest energy â and is, therefore, equal to E = mc2.

Can we, somehow, relate this to the mÂˇa2ÂˇĎ2 energy formula for our V-2 perpetuum mobile? Our wavefunction has an amplitude too. Now, if the oscillations of the real and imaginary wavefunction store the energy of our particle, then their amplitude will surely matter. In fact, the energy of an oscillation is, in general, proportional to the square of the amplitude: E Âľ a2. We may, therefore, think that the a2 factor in the E = mÂˇa2ÂˇĎ2 energy will surely be relevant as well.

However, here is a complication: an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ak, and their own Ďi = -Ei/Ä§. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. To calculate the contribution of each wave to the total, both ai as well as Ei will matter.

What is Ei? Ei varies around some average E, which we can associate with some average mass m: m = E/c2. The Uncertainty Principle kicks in here. The analysis becomes more complicated, but a formula such as the one below might make sense:We can re-write this as:What is the meaning of this equation? We may look at it as some sort of physical normalization condition when building up the Fourier sum. Of course, we should relate this to the mathematical normalization condition for the wavefunction. Our intuition tells us that the probabilities must be related to the energy densities, but how exactly? We will come back to this question in a moment. Let us first think some more about the enigma: what is mass?

Before we do so, let us quickly calculate the value of c2Ä§2: it is about 1Â´1051 N2âm4. Let us also do a dimensional analysis: the physical dimensions of the E = mÂˇa2ÂˇĎ2 equation make sense if we express m in kg, a in m, and Ď in rad/s. We then get: [E] = kgâm2/s2 = (Nâs2/m)âm2/s2 = Nâm = J. The dimensions of the left- and right-hand side of the physical normalization condition is N3âm5.

III. What is mass?

We came up, playfully, with a meaningful interpretation for energy: it is a two-dimensional oscillation of mass. But what is mass? A new aether theory is, of course, not an option, but then what is it that is oscillating? To understand the physics behind equations, it is always good to do an analysis of the physical dimensions in the equation. Let us start with Einsteinâs energy equation once again. If we want to look at mass, we should re-write it as m = E/c2:

[m] = [E/c2] = J/(m/s)2 = NÂˇmâs2/m2 = NÂˇs2/m = kg

This is not very helpful. It only reminds us of Newtonâs definition of a mass: mass is that what gets accelerated by a force. At this point, we may want to think of the physical significance of the absolute nature of the speed of light. Einsteinâs E = mc2 equation implies we can write the ratio between the energy and the mass of any particle is always the same, so we can write, for example:This reminds us of the Ď2= C1/L or Ď2 = k/m of harmonic oscillators once again.[13] The key difference is that the Ď2= C1/L and Ď2 = k/m formulas introduce two or more degrees of freedom.[14] In contrast, c2= E/m for any particle, always. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in one physical space only: our spacetime. Hence, the speed of light c emerges here as the defining property of spacetime â the resonant frequency, so to speak. We have no further degrees of freedom here.

The Planck-Einstein relation (for photons) and the de Broglie equation (for matter-particles) have an interesting feature: both imply that the energy of the oscillation is proportional to the frequency, with Planckâs constant as the constant of proportionality. Now, for one-dimensional oscillations â think of a guitar string, for example â we know the energy will be proportional to the square of the frequency. It is a remarkable observation: the two-dimensional matter-wave, or the electromagnetic wave, gives us two waves for the price of one, so to speak, each carrying half of the total energy of the oscillation but, as a result, we get a proportionality between E and f instead of between E and f2.

However, such reflections do not answer the fundamental question we started out with: what is mass? At this point, it is hard to go beyond the circular definition that is implied by Einsteinâs formula: energy is a two-dimensional oscillation of mass, and mass packs energy, and c emerges us as the property of spacetime that defines how exactly.

When everything is said and done, this does not go beyond stating that mass is some scalar field. Now, a scalar field is, quite simply, some real number that we associate with a position in spacetime. The Higgs field is a scalar field but, of course, the theory behind it goes much beyond stating that we should think of mass as some scalar field. The fundamental question is: why and how does energy, or matter, condense into elementary particles? That is what the Higgs mechanism is about but, as this paper is exploratory only, we cannot even start explaining the basics of it.

What we can do, however, is look at the wave equation again (SchrĂśdingerâs equation), as we can now analyze it as an energy diffusion equation.

IV. SchrĂśdingerâs equation as an energy diffusion equation

The interpretation of SchrĂśdingerâs equation as a diffusion equation is straightforward. Feynman (Lectures, III-16-1) briefly summarizes it as follows:

âWe can think of SchrĂśdingerâs equation as describing the diffusion of the probability amplitude from one point to the next. [âŚ] But the imaginary coefficient in front of the derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of SchrĂśdingerâs equation are complex waves.â[17]

Let us review the basic math. For a particle moving in free space â with no external force fields acting on it â there is no potential (U = 0) and, therefore, the UĎ term disappears. Therefore, SchrĂśdingerâs equation reduces to:

âĎ(x, t)/ât = iÂˇ(1/2)Âˇ(Ä§/meff)Âˇâ2Ď(x, t)

The ubiquitous diffusion equation in physics is:

âĎ(x, t)/ât = DÂˇâ2Ď(x, t)

The structural similarity is obvious. The key difference between both equations is that the wave equation gives us two equations for the price of one. Indeed, because Ď is a complex-valued function, with a real and an imaginary part, we get the following equations[18]:

1. Re(âĎ/ât) = â(1/2)Âˇ(Ä§/meff)ÂˇIm(â2Ď)
2. Im(âĎ/ât) = (1/2)Âˇ(Ä§/meff)ÂˇRe(â2Ď)

These equations make us think of the equations for an electromagnetic wave in free space (no stationary charges or currents):

1. âB/ât = ââĂE
2. âE/ât = c2âĂB

The above equations effectively describe a propagation mechanism in spacetime, as illustrated below.

Figure 4: Propagation mechanisms

The Laplacian operator (â2), when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m2). In this case, it is operating on Ď(x, t), so what is the dimension of our wavefunction Ď(x, t)? To answer that question, we should analyze the diffusion constant in SchrĂśdingerâs equation, i.e. the (1/2)Âˇ(Ä§/meff) factor:

1. As a mathematical constant of proportionality, it will quantify the relationship between both derivatives (i.e. the time derivative and the Laplacian);
2. As a physical constant, it will ensure the physical dimensions on both sides of the equation are compatible.

Now, the Ä§/meff factor is expressed in (NÂˇmÂˇs)/(NÂˇ s2/m) = m2/s. Hence, it does ensure the dimensions on both sides of the equation are, effectively, the same: âĎ/ât is a time derivative and, therefore, its dimension is s1 while, as mentioned above, the dimension of â2Ď is m2. However, this does not solve our basic question: what is the dimension of the real and imaginary part of our wavefunction?

At this point, mainstream physicists will say: it does not have a physical dimension, and there is no geometric interpretation of SchrĂśdingerâs equation. One may argue, effectively, that its argument, (pâx – Eât)/Ä§, is just a number and, therefore, that the real and imaginary part of Ď is also just some number.

To this, we may object that Ä§ may be looked as a mathematical scaling constant only. If we do that, then the argument of Ď will, effectively, be expressed in action units, i.e. in NÂˇmÂˇs. It then does make sense to also associate a physical dimension with the real and imaginary part of Ď. What could it be?

We may have a closer look at Maxwellâs equations for inspiration here. The electric field vector is expressed in newton (the unit of force) per unit of charge (coulomb). Now, there is something interesting here. The physical dimension of the magnetic field is N/C divided by m/s.[19] We may write B as the following vector cross-product: B = (1/c)âexĂE, with ex the unit vector pointing in the x-direction (i.e. the direction of propagation of the wave). Hence, we may associate the (1/c)âexĂ operator, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by i also amounts to a rotation by 90Â° degrees. Hence, we may boldly write: B = (1/c)âexĂE = (1/c)âiâE. This allows us to also geometrically interpret SchrĂśdingerâs equation in the way we interpreted it above (see Figure 3).[20]

Still, we have not answered the question as to what the physical dimension of the real and imaginary part of our wavefunction should be. At this point, we may be inspired by the structural similarity between Newtonâs and Coulombâs force laws:Hence, if the electric field vector E is expressed in force per unit charge (N/C), then we may want to think of associating the real part of our wavefunction with a force per unit mass (N/kg). We can, of course, do a substitution here, because the mass unit (1 kg) is equivalent to 1 NÂˇs2/m. Hence, our N/kg dimension becomes:

N/kg = N/(NÂˇs2/m)= m/s2

What is this: m/s2? Is that the dimension of the aÂˇcosÎ¸ term in the aÂˇeâiÎ¸ aÂˇcosÎ¸ â iÂˇaÂˇsinÎ¸ wavefunction?

My answer is: why not? Think of it: m/s2 is the physical dimension of acceleration: the increase or decrease in velocity (m/s) per second. It ensures the wavefunction for any particle â matter-particles or particles with zero rest mass (photons) â and the associated wave equation (which has to be the same for all, as the spacetime we live in is one) are mutually consistent.

In this regard, we should think of how we would model a gravitational wave. The physical dimension would surely be the same: force per mass unit. It all makes sense: wavefunctions may, perhaps, be interpreted as traveling distortions of spacetime, i.e. as tiny gravitational waves.

V. Energy densities and flows

Pursuing the geometric equivalence between the equations for an electromagnetic wave and SchrĂśdingerâs equation, we can now, perhaps, see if there is an equivalent for the energy density. For an electromagnetic wave, we know that the energy density is given by the following formula:E and B are the electric and magnetic field vector respectively. The Poynting vector will give us the directional energy flux, i.e. the energy flow per unit area per unit time. We write:Needless to say, the ââ operator is the divergence and, therefore, gives us the magnitude of a (vector) fieldâs source or sink at a given point. To be precise, the divergence gives us the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. In this case, it gives us the volume density of the flux of S.

We can analyze the dimensions of the equation for the energy density as follows:

1. E is measured in newton per coulomb, so [EâE] = [E2] = N2/C2.
2. B is measured in (N/C)/(m/s), so we get [BâB] = [B2] = (N2/C2)Âˇ(s2/m2). However, the dimension of our c2 factor is (m2/s2) and so weâre also left with N2/C2.
3. The Ďľ0 is the electric constant, aka as the vacuum permittivity. As a physical constant, it should ensure the dimensions on both sides of the equation work out, and they do: [Îľ0] = C2/(NÂˇm2) and, therefore, if we multiply that with N2/C2, we find that is expressed in J/m3.[21]

Replacing the newton per coulomb unit (N/C) by the newton per kg unit (N/kg) in the formulas above should give us the equivalent of the energy density for the wavefunction. We just need to substitute Ďľ0 for an equivalent constant. We may to give it a try. If the energy densities can be calculated â which are also mass densities, obviously â then the probabilities should be proportional to them.

Let us first see what we get for a photon, assuming the electromagnetic wave represents its wavefunction. Substituting B for (1/c)âiâE or for â(1/c)âiâE gives us the following result:Zero!? An unexpected result! Or not? We have no stationary charges and no currents: only an electromagnetic wave in free space. Hence, the local energy conservation principle needs to be respected at all points in space and in time. The geometry makes sense of the result: for an electromagnetic wave, the magnitudes of E and B reach their maximum, minimum and zero point simultaneously, as shown below.[22] This is because their phase is the same.

Figure 5: Electromagnetic wave: E and B

Should we expect a similar result for the energy densities that we would associate with the real and imaginary part of the matter-wave? For the matter-wave, we have a phase difference between aÂˇcosÎ¸ and aÂˇsinÎ¸, which gives a different picture of the propagation of the wave (see Figure 3).[23] In fact, the geometry of the suggestion suggests some inherent spin, which is interesting. I will come back to this. Let us first guess those densities. Making abstraction of any scaling constants, we may write:We get what we hoped to get: the absolute square of our amplitude is, effectively, an energy density !

|Ď|2  = |aÂˇeâiâEÂˇt/Ä§|2 = a2 = u

This is very deep. A photon has no rest mass, so it borrows and returns energy from empty space as it travels through it. In contrast, a matter-wave carries energy and, therefore, has some (rest) mass. It is therefore associated with an energy density, and this energy density gives us the probabilities. Of course, we need to fine-tune the analysis to account for the fact that we have a wave packet rather than a single wave, but that should be feasible.

As mentioned, the phase difference between the real and imaginary part of our wavefunction (a cosine and a sine function) appear to give some spin to our particle. We do not have this particularity for a photon. Of course, photons are bosons, i.e. spin-zero particles, while elementary matter-particles are fermions with spin-1/2. Hence, our geometric interpretation of the wavefunction suggests that, after all, there may be some more intuitive explanation of the fundamental dichotomy between bosons and fermions, which puzzled even Feynman:

âWhy is it that particles with half-integral spin are Fermi particles, whereas particles with integral spin are Bose particles? We apologize for the fact that we cannot give you an elementary explanation. An explanation has been worked out by Pauli from complicated arguments of quantum field theory and relativity. He has shown that the two must necessarily go together, but we have not been able to find a way of reproducing his arguments on an elementary level. It appears to be one of the few places in physics where there is a rule which can be stated very simply, but for which no one has found a simple and easy explanation. The explanation is deep down in relativistic quantum mechanics. This probably means that we do not have a complete understanding of the fundamental principle involved.â (Feynman, Lectures, III-4-1)

The physical interpretation of the wavefunction, as presented here, may provide some better understanding of âthe fundamental principle involvedâ: the physical dimension of the oscillation is just very different. That is all: it is force per unit charge for photons, and force per unit mass for matter-particles. We will examine the question of spin somewhat more carefully in section VII. Let us first examine the matter-wave some more.

VI. Group and phase velocity of the matter-wave

The geometric representation of the matter-wave (see Figure 3) suggests a traveling wave and, yes, of course: the matter-wave effectively travels through space and time. But what is traveling, exactly? It is the pulse â or the signal â only: the phase velocity of the wave is just a mathematical concept and, even in our physical interpretation of the wavefunction, the same is true for the group velocity of our wave packet. The oscillation is two-dimensional, but perpendicular to the direction of travel of the wave. Hence, nothing actually moves with our particle.

Here, we should also reiterate that we did not answer the question as to what is oscillating up and down and/or sideways: we only associated a physical dimension with the components of the wavefunction â newton per kg (force per unit mass), to be precise. We were inspired to do so because of the physical dimension of the electric and magnetic field vectors (newton per coulomb, i.e. force per unit charge) we associate with electromagnetic waves which, for all practical purposes, we currently treat as the wavefunction for a photon. This made it possible to calculate the associated energy densities and a Poynting vector for energy dissipation. In addition, we showed that SchrĂśdinger’s equation itself then becomes a diffusion equation for energy. However, let us now focus some more on the asymmetry which is introduced by the phase difference between the real and the imaginary part of the wavefunction. Look at the mathematical shape of the elementary wavefunction once again:

Ď = aÂˇeâi[EÂˇt â pâx]/Ä§aÂˇeâi[EÂˇt â pâx]/Ä§ = aÂˇcos(pâx/Ä§ â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ â Eât/Ä§)

The minus sign in the argument of our sine and cosine function defines the direction of travel: an F(xâvât) wavefunction will always describe some wave that is traveling in the positive x-direction (with the wave velocity), while an F(x+vât) wavefunction will travel in the negative x-direction. For a geometric interpretation of the wavefunction in three dimensions, we need to agree on how to define i or, what amounts to the same, a convention on how to define clockwise and counterclockwise directions: if we look at a clock from the back, then its hand will be moving counterclockwise. So we need to establish the equivalent of the right-hand rule. However, let us not worry about that now. Let us focus on the interpretation. To ease the analysis, we’ll assume we’re looking at a particle at rest. Hence, p = 0, and the wavefunction reduces to:

Ď = aÂˇeâiâEÂˇt/Ä§ = aÂˇcos(âEât/Ä§) + iÂˇaÂˇsin(âE0ât/Ä§) = aÂˇcos(E0ât/Ä§) â iÂˇaÂˇsin(E0ât/Ä§)

E0 is, of course, the rest mass of our particle and, now that we are here, we should probably wonder whose time we are talking about: is it our time, or is the proper time of our particle? Well… In this situation, we are both at rest so it does not matter: t is, effectively, the proper time so perhaps we should write it as t0. It does not matter. You can see what we expect to see: E0/Ä§ pops up as the natural frequency of our matter-particle: (E0/Ä§)ât = Ďât. Remembering the Ď = 2ĎÂˇf = 2Ď/T and T = 1/formulas, we can associate a period and a frequency with this wave, using the Ď = 2ĎÂˇf = 2Ď/T. Noting that Ä§ = h/2Ď, we find the following:

T = 2ĎÂˇ(Ä§/E0) = h/E0 â = E0/h = m0c2/h

This is interesting, because we can look at the period as a natural unit of time for our particle. What about the wavelength? That is tricky because we need to distinguish between group and phase velocity here. The group velocity (vg) should be zero here, because we assume our particle does not move. In contrast, the phase velocity is given by vp = ÎťÂˇ= (2Ď/k)Âˇ(Ď/2Ď) = Ď/k. In fact, we’ve got something funny here: the wavenumber k = p/Ä§ is zero, because we assume the particle is at rest, so p = 0. So we have a division by zero here, which is rather strange. What do we get assuming the particle is not at rest? We write:

vp = Ď/k = (E/Ä§)/(p/Ä§) = E/p = E/(mÂˇvg) = (mÂˇc2)/(mÂˇvg) = c2/vg

This is interesting: it establishes a reciprocal relation between the phase and the group velocity, with as a simple scaling constant. Indeed, the graph below shows the shape of the function does not change with the value of c, and we may also re-write the relation above as:

vp/= Î˛p = c/vp = 1/Î˛g = 1/(c/vp)

Figure 6: Reciprocal relation between phase and group velocity

We can also write the mentioned relationship as vpÂˇvg = c2, which reminds us of the relationship between the electric and magnetic constant (1/Îľ0)Âˇ(1/Îź0) = c2. This is interesting in light of the fact we can re-write this as (cÂˇÎľ0)Âˇ(cÂˇÎź0) = 1, which shows electricity and magnetism are just two sides of the same coin, so to speak.[24]

Interesting, but how do we interpret the math? What about the implications of the zero value for wavenumber k = p/Ä§. We would probably like to think it implies the elementary wavefunction should always be associated with some momentum, because the concept of zero momentum clearly leads to weird math: something times zero cannot be equal to c2! Such interpretation is also consistent with the Uncertainty Principle: if ÎxÂˇÎp âĽ Ä§, then neither Îx nor Îp can be zero. In other words, the Uncertainty Principle tells us that the idea of a pointlike particle actually being at some specific point in time and in space does not make sense: it has to move. It tells us that our concept of dimensionless points in time and space are mathematical notions only. Actual particles – including photons – are always a bit spread out, so to speak, and – importantly – they have to move.

For a photon, this is self-evident. It has no rest mass, no rest energy, and, therefore, it is going to move at the speed of light itself. We write: p = mÂˇc = mÂˇc2/= E/c. Using the relationship above, we get:

vp = Ď/k = (E/Ä§)/(p/Ä§) = E/p = c â vg = c2/vp = c2/c = c

This is good: we started out with some reflections on the matter-wave, but here we get an interpretation of the electromagnetic wave as a wavefunction for the photon. But let us get back to our matter-wave. In regard to our interpretation of a particle having to move, we should remind ourselves, once again, of the fact that an actual particle is always localized in space and that it can, therefore, not be represented by the elementary wavefunction Ď = aÂˇeâi[EÂˇt â pâx]/Ä§ or, for a particle at rest, the Ď = aÂˇeâiâEÂˇt/Ä§ function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ai, and their own Ďi = âEi/Ä§. Indeed, in section II, we showed that each of these wavefunctions will contribute some energy to the total energy of the wave packet and that, to calculate the contribution of each wave to the total, both ai as well as Ei matter. This may or may not resolve the apparent paradox. Let us look at the group velocity.

To calculate a meaningful group velocity, we must assume the vg = âĎi/âki = â(Ei/Ä§)/â(pi/Ä§) = â(Ei)/â(pi) exists. So we must have some dispersion relation. How do we calculate it? We need to calculate Ďi as a function of ki here, or Ei as a function of pi. How do we do that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write SchrĂśdinger’s equation as we did, i.e. by distinguishing the real and imaginary parts of the âĎ/ât =iÂˇ[Ä§/(2m)]Âˇâ2Ď wave equation and, hence, re-write it as the following pair of two equations:

1. Re(âĎ/ât) = â[Ä§/(2meff)]ÂˇIm(â2Ď) â ĎÂˇcos(kx â Ďt) = k2Âˇ[Ä§/(2meff)]Âˇcos(kx â Ďt)
2. Im(âĎ/ât) = [Ä§/(2meff)]ÂˇRe(â2Ď) â ĎÂˇsin(kx â Ďt) = k2Âˇ[Ä§/(2meff)]Âˇsin(kx â Ďt)

Both equations imply the following dispersion relation:

Ď = Ä§Âˇk2/(2meff)

Of course, we need to think about the subscripts now: we have Ďi, ki, but… What about meff or, dropping the subscript, m? Do we write it as mi? If so, what is it? Well… It is the equivalent mass of Ei obviously, and so we get it from the mass-energy equivalence relation: mi = Ei/c2. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too. Here, I should refer back to Section II: Ei varies around some average energy E and, therefore, the Uncertainty Principle kicks in.

VII. Explaining spin

The elementary wavefunction vector â i.e. the vector sum of the real and imaginary component â rotates around the x-axis, which gives us the direction of propagation of the wave (see Figure 3). Its magnitude remains constant. In contrast, the magnitude of the electromagnetic vector â defined as the vector sum of the electric and magnetic field vectors â oscillates between zero and some maximum (see Figure 5).

We already mentioned that the rotation of the wavefunction vector appears to give some spin to the particle. Of course, a circularly polarized wave would also appear to have spin (think of the E and B vectors rotating around the direction of propagation – as opposed to oscillating up and down or sideways only). In fact, a circularly polarized light does carry angular momentum, as the equivalent mass of its energy may be thought of as rotating as well. But so here we are looking at a matter-wave.

The basic idea is the following: if we look at Ď = aÂˇeâiâEÂˇt/Ä§ as some real vector â as a two-dimensional oscillation of mass, to be precise â then we may associate its rotation around the direction of propagation with some torque. The illustration below reminds of the math here.

Figure 7: Torque and angular momentum vectors

A torque on some mass about a fixed axis gives it angular momentum, which we can write as the vector cross-product L = rĂp or, perhaps easier for our purposes here as the product of an angular velocity (Ď) and rotational inertia (I), aka as the moment of inertia or the angular mass. We write:

L = IÂˇĎ

Note we can write L and Ď in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = IÂˇĎ (no boldface). We can now do some calculations. Let us start with the angular velocity. In our previous posts, we showed that the period of the matter-wave is equal to T = 2ĎÂˇ(Ä§/E0). Hence, the angular velocity must be equal to:

Ď = 2Ď/[2ĎÂˇ(Ä§/E0)] = E0/Ä§

We also know the distance r, so that is the magnitude of r in the LrĂp vector cross-product: it is just a, so that is the magnitude of Ď = aÂˇeâiâEÂˇt/Ä§. Now, the momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = mÂˇv. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = rÂˇĎ. So now we only need to think about what we should use for m or, if we want to work with the angular velocity (Ď), the angular mass (I). Here we need to make some assumption about the mass (or energy) distribution. Now, it may or may not sense to assume the energy in the oscillation â and, therefore, the mass â is distributed uniformly. In that case, we may use the formula for the angular mass of a solid cylinder: I = mÂˇr2/2. If we keep the analysis non-relativistic, then m = m0. Of course, the energy-mass equivalence tells us that m0 = E0/c2. Hence, this is what we get:

L = IÂˇĎ = (m0Âˇr2/2)Âˇ(E0/Ä§) = (1/2)Âˇa2Âˇ(E0/c2)Âˇ(E0/Ä§) = a2ÂˇE02/(2ÂˇÄ§Âˇc2)

Does it make sense? Maybe. Maybe not. Let us do a dimensional analysis: that wonât check our logic, but it makes sure we made no mistakes when mapping mathematical and physical spaces. We have m2ÂˇJ2 = m2ÂˇN2Âˇm2 in the numerator and NÂˇmÂˇsÂˇm2/s2 in the denominator. Hence, the dimensions work out: we get NÂˇmÂˇs as the dimension for L, which is, effectively, the physical dimension of angular momentum. It is also the action dimension, of course, and that cannot be a coincidence. Also note that the E = mc2 equation allows us to re-write it as:

L = a2ÂˇE02/(2ÂˇÄ§Âˇc2)

Of course, in quantum mechanics, we associate spin with the magnetic moment of a charged particle, not with its mass as such. Is there way to link the formula above to the one we have for the quantum-mechanical angular momentum, which is also measured in NÂˇmÂˇs units, and which can only take on one of two possible values: J = +Ä§/2 and âÄ§/2? It looks like a long shot, right? How do we go from (1/2)Âˇa2Âˇm02/Ä§ to Âą (1/2)âÄ§? Let us do a numerical example. The energy of an electron is typically 0.510 MeV Âť 8.1871Ă10â14 Nâm, and aâŚ What value should we take for a?

We have an obvious trio of candidates here: the Bohr radius, the classical electron radius (aka the Thompon scattering length), and the Compton scattering radius.

Let us start with the Bohr radius, so that is about 0.Ă10â10 Nâm. We get L = a2ÂˇE02/(2ÂˇÄ§Âˇc2) = 9.9Ă10â31 Nâmâs. Now that is about 1.88Ă104 times Ä§/2. That is a huge factor. The Bohr radius cannot be right: we are not looking at an electron in an orbital here. To show it does not make sense, we may want to double-check the analysis by doing the calculation in another way. We said each oscillation will always pack 6.626070040(81)Ă10â34 joule in energy. So our electron should pack about 1.24Ă10â20 oscillations. The angular momentum (L) we get when using the Bohr radius for a and the value of 6.626Ă10â34 joule for E0 and the Bohr radius is equal to 6.49Ă10â59 Nâmâs. So that is the angular momentum per oscillation. When we multiply this with the number of oscillations (1.24Ă10â20), we get about 8.01Ă10â51 Nâmâs, so that is a totally different number.

The classical electron radius is about 2.818Ă10â15 m. We get an L that is equal to about 2.81Ă10â39 Nâmâs, so now it is a tiny fraction of Ä§/2! Hence, this leads us nowhere. Let us go for our last chance to get a meaningful result! Let us use the Compton scattering length, so that is about 2.42631Ă10â12 m.

This gives us an L of 2.08Ă10â33 Nâmâs, which is only 20 times Ä§. This is not so bad, but it is good enough? Let us calculate it the other way around: what value should we take for a so as to ensure L = a2ÂˇE02/(2ÂˇÄ§Âˇc2) = Ä§/2? Let us write it out:

In fact, this is the formula for the so-called reduced Compton wavelength. This is perfect. We found what we wanted to find. Substituting this value for a (you can calculate it: it is about 3.8616Ă10â33 m), we get what we should find:

This is a rather spectacular result, and one that would â a priori â support the interpretation of the wavefunction that is being suggested in this paper.

VIII. The boson-fermion dichotomy

Let us do some more thinking on the boson-fermion dichotomy. Again, we should remind ourselves that an actual particle is localized in space and that it can, therefore, not be represented by the elementary wavefunction Ď = aÂˇeâi[EÂˇt â pâx]/Ä§ or, for a particle at rest, the Ď = aÂˇeâiâEÂˇt/Ä§ function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ai, and their own Ďi = âEi/Ä§. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. Now, we can have another wild but logical theory about this.

Think of the apparent right-handedness of the elementary wavefunction: surely, Nature can’t be bothered about our convention of measuring phase angles clockwise or counterclockwise. Also, the angular momentum can be positive or negative: J = +Ä§/2 or âÄ§/2. Hence, we would probably like to think that an actual particle – think of an electron, or whatever other particle you’d think of – may consist of right-handed as well as left-handed elementary waves. To be precise, we may think they either consist of (elementary) right-handed waves or, else, of (elementary) left-handed waves. An elementary right-handed wave would be written as:

Ď(Î¸i= aiÂˇ(cosÎ¸i + iÂˇsinÎ¸i)

In contrast, an elementary left-handed wave would be written as:

Ď(Î¸i= aiÂˇ(cosÎ¸i â iÂˇsinÎ¸i)

How does that work out with the E0Âˇt argument of our wavefunction? Position is position, and direction is direction, but time? Time has only one direction, but Nature surely does not care how we count time: counting like 1, 2, 3, etcetera or like â1, â2, â3, etcetera is just the same. If we count like 1, 2, 3, etcetera, then we write our wavefunction like:

Ď = aÂˇcos(E0ât/Ä§) â iÂˇaÂˇsin(E0ât/Ä§)

If we count time like â1, â2, â3, etcetera then we write it as:

Ď = aÂˇcos(âE0ât/Ä§) â iÂˇaÂˇsin(âE0ât/Ä§)= aÂˇcos(E0ât/Ä§) + iÂˇaÂˇsin(E0ât/Ä§)

Hence, it is just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! This, then, should explain why we can have either positive or negative quantum-mechanical spin (+Ä§/2 or âÄ§/2). It is the usual thing: we have two mathematical possibilities here, and so we must have two physical situations that correspond to it.

It is only natural. If we have left- and right-handed photons – or, generalizing, left- and right-handed bosons – then we should also have left- and right-handed fermions (electrons, protons, etcetera). Back to the dichotomy. The textbook analysis of the dichotomy between bosons and fermions may be epitomized by Richard Feynmanâs Lecture on it (Feynman, III-4), which is confusing and â I would dare to say â even inconsistent: how are photons or electrons supposed to know that they need to interfere with a positive or a negative sign? They are not supposed to know anything: knowledge is part of our interpretation of whatever it is that is going on there.

Hence, it is probably best to keep it simple, and think of the dichotomy in terms of the different physical dimensions of the oscillation: newton per kg versus newton per coulomb. And then, of course, we should also note that matter-particles have a rest mass and, therefore, actually carry charge. Photons do not. But both are two-dimensional oscillations, and the point is: the so-called vacuum – and the rest mass of our particle (which is zero for the photon and non-zero for everything else) – give us the natural frequency for both oscillations, which is beautifully summed up in that remarkable equation for the group and phase velocity of the wavefunction, which applies to photons as well as matter-particles:

(vphaseÂˇc)Âˇ(vgroupÂˇc) = 1 â vpÂˇvg = c2

The final question then is: why are photons spin-zero particles? Well… We should first remind ourselves of the fact that they do have spin when circularly polarized.[25] Here we may think of the rotation of the equivalent mass of their energy. However, if they are linearly polarized, then there is no spin. Even for circularly polarized waves, the spin angular momentum of photons is a weird concept. If photons have no (rest) mass, then they cannot carry any charge. They should, therefore, not have any magnetic moment. Indeed, what I wrote above shows an explanation of quantum-mechanical spin requires both mass as well as charge.[26]

IX. Concluding remarks

There are, of course, other ways to look at the matter â literally. For example, we can imagine two-dimensional oscillations as circular rather than linear oscillations. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation â around any axis â will be some combination of a rotation around the two other axes. Hence, we may want to think of a two-dimensional oscillation as an oscillation of a polar and azimuthal angle.

Figure 8: Two-dimensional circular movement

The point of this paper is not to make any definite statements. That would be foolish. Its objective is just to challenge the simplistic mainstream viewpoint on the reality of the wavefunction. Stating that it is a mathematical construct only without physical significance amounts to saying it has no meaning at all. That is, clearly, a non-sustainable proposition.

The interpretation that is offered here looks at amplitude waves as traveling fields. Their physical dimension may be expressed in force per mass unit, as opposed to electromagnetic waves, whose amplitudes are expressed in force per (electric) charge unit. Also, the amplitudes of matter-waves incorporate a phase factor, but this may actually explain the rather enigmatic dichotomy between fermions and bosons and is, therefore, an added bonus.

The interpretation that is offered here has some advantages over other explanations, as it explains the how of diffraction and interference. However, while it offers a great explanation of the wave nature of matter, it does not explain its particle nature: while we think of the energy as being spread out, we will still observe electrons and photons as pointlike particles once they hit the detector. Why is it that a detector can sort of âhookâ the whole blob of energy, so to speak?

The interpretation of the wavefunction that is offered here does not explain this. Hence, the complementarity principle of the Copenhagen interpretation of the wavefunction surely remains relevant.

Appendix 1: The de Broglie relations and energy

The 1/2 factor in SchrĂśdingerâs equation is related to the concept of the effective mass (meff). It is easy to make the wrong calculations. For example, when playing with the famous de Broglie relations â aka as the matter-wave equations â one may be tempted to derive the following energy concept:

1. E = hÂˇf and p = h/Îť. Therefore, f = E/h and Îť = p/h.
2. v = fÂˇÎť = (E/h)â(p/h) = E/p
3. p = mÂˇv. Therefore, E = vÂˇp = mÂˇv2

E = mÂˇv2? This resembles the E = mc2 equation and, therefore, one may be enthused by the discovery, especially because the mÂˇv2 also pops up when working with the Least Action Principle in classical mechanics, which states that the path that is followed by a particle will minimize the following integral:Now, we can choose any reference point for the potential energy but, to reflect the energy conservation law, we can select a reference point that ensures the sum of the kinetic and the potential energy is zero throughout the time interval. If the force field is uniform, then the integrand will, effectively, be equal to KE â PE = mÂˇv2.[27]

However, that is classical mechanics and, therefore, not so relevant in the context of the de Broglie equations, and the apparent paradox should be solved by distinguishing between the group and the phase velocity of the matter wave.

Appendix 2: The concept of the effective mass

The effective mass â as used in SchrĂśdingerâs equation â is a rather enigmatic concept. To make sure we are making the right analysis here, I should start by noting you will usually see SchrĂśdingerâs equation written as:This formulation includes a term with the potential energy (U). In free space (no potential), this term disappears, and the equation can be re-written as:

âĎ(x, t)/ât = iÂˇ(1/2)Âˇ(Ä§/meff)Âˇâ2Ď(x, t)

We just moved the iÂˇÄ§ coefficient to the other side, noting that 1/i = –i. Now, in one-dimensional space, and assuming Ď is just the elementary wavefunction (so we substitute aÂˇeâiâ[EÂˇt â pâx]/Ä§ for Ď), this implies the following:

âaÂˇiÂˇ(E/Ä§)Âˇeâiâ[EÂˇt â pâx]/Ä§ = âiÂˇ(Ä§/2meff)ÂˇaÂˇ(p2/Ä§2)Âˇ eâiâ[EÂˇt â pâx]/Ä§

â E = p2/(2meff) â meff = mâ(v/c)2/2 = mâÎ˛2/2

It is an ugly formula: it resembles the kinetic energy formula (K.E. = mâv2/2) but it is, in fact, something completely different. The Î˛2/2 factor ensures the effective mass is always a fraction of the mass itself. To get rid of the ugly 1/2 factor, we may re-define meff as two times the old meff (hence, meffNEW = 2âmeffOLD), as a result of which the formula will look somewhat better:

meff = mâ(v/c)2 = mâÎ˛2

We know Î˛ varies between 0 and 1 and, therefore, meff will vary between 0 and m. Feynman drops the subscript, and just writes meff as m in his textbook (see Feynman, III-19). On the other hand, the electron mass as used is also the electron mass that is used to calculate the size of an atom (see Feynman, III-2-4). As such, the two mass concepts are, effectively, mutually compatible. It is confusing because the same mass is often defined as the mass of a stationary electron (see, for example, the article on it in the online Wikipedia encyclopedia[28]).

In the context of the derivation of the electron orbitals, we do have the potential energy term â which is the equivalent of a source term in a diffusion equation â and that may explain why the above-mentioned meff = mâ(v/c)2 = mâÎ˛2 formula does not apply.

References

This paper discusses general principles in physics only. Hence, references can be limited to references to physics textbooks only. For ease of reading, any reference to additional material has been limited to a more popular undergrad textbook that can be consulted online: Feynmanâs Lectures on Physics (http://www.feynmanlectures.caltech.edu). References are per volume, per chapter and per section. For example, Feynman III-19-3 refers to Volume III, Chapter 19, Section 3.

Notes

[1] Of course, an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction Ď = aÂˇeâiâÎ¸aÂˇeâi[EÂˇt â pâx]/Ä§ = aÂˇ(cosÎ¸ iÂˇaÂˇsinÎ¸). We must build a wave packet for that: a sum of wavefunctions, each with its own amplitude ak and its own argument Î¸k = (Ekât – pkâx)/Ä§. This is dealt with in this paper as part of the discussion on the mathematical and physical interpretation of the normalization condition.

[2] The N/kg dimension immediately, and naturally, reduces to the dimension of acceleration (m/s2), thereby facilitating a direct interpretation in terms of Newtonâs force law.

[3] In physics, a two-spring metaphor is more common. Hence, the pistons in the authorâs perpetuum mobile may be replaced by springs.

[4] The author re-derives the equation for the Compton scattering radius in section VII of the paper.

[5] The magnetic force can be analyzed as a relativistic effect (see Feynman II-13-6). The dichotomy between the electric force as a polar vector and the magnetic force as an axial vector disappears in the relativistic four-vector representation of electromagnetism.

[6] For example, when using SchrĂśdingerâs equation in a central field (think of the electron around a proton), the use of polar coordinates is recommended, as it ensures the symmetry of the Hamiltonian under all rotations (see Feynman III-19-3)

[7] This sentiment is usually summed up in the apocryphal quote: âGod does not play dice.âThe actual quote comes out of one of Einsteinâs private letters to Cornelius Lanczos, another scientist who had also emigrated to the US. The full quote is as follows: “You are the only person I know who has the same attitude towards physics as I have: belief in the comprehension of reality through something basically simple and unified… It seems hard to sneak a look at God’s cards. But that He plays dice and uses ‘telepathic’ methods… is something that I cannot believe for a single moment.” (Helen Dukas and Banesh Hoffman, Albert Einstein, the Human Side: New Glimpses from His Archives, 1979)

[8] Of course, both are different velocities: Ď is an angular velocity, while v is a linear velocity: Ď is measured in radians per second, while v is measured in meter per second. However, the definition of a radian implies radians are measured in distance units. Hence, the physical dimensions are, effectively, the same. As for the formula for the total energy of an oscillator, we should actually write: E = mÂˇa2âĎ2/2. The additional factor (a) is the (maximum) amplitude of the oscillator.

[9] We also have a 1/2 factor in the E = mv2/2 formula. Two remarks may be made here. First, it may be noted this is a non-relativistic formula and, more importantly, incorporates kinetic energy only. Using the Lorentz factor (Îł), we can write the relativistically correct formula for the kinetic energy as K.E. = E â E0 = mvc2 â m0c2 = m0Îłc2 â m0c2 = m0c2(Îł â 1). As for the exclusion of the potential energy, we may note that we may choose our reference point for the potential energy such that the kinetic and potential energy mirror each other. The energy concept that then emerges is the one that is used in the context of the Principle of Least Action: it equals E = mv2. Appendix 1 provides some notes on that.

[10] Instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft.

[11] It is interesting to note that we may look at the energy in the rotating flywheel as potential energy because it is energy that is associated with motion, albeit circular motion. In physics, one may associate a rotating object with kinetic energy using the rotational equivalent of mass and linear velocity, i.e. rotational inertia (I) and angular velocity Ď. The kinetic energy of a rotating object is then given by K.E. = (1/2)ÂˇIÂˇĎ2.

[12] Because of the sideways motion of the connecting rods, the sinusoidal function will describe the linear motion only approximately, but you can easily imagine the idealized limit situation.

[13] The Ď2= 1/LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor (R), an inductor (L), and a capacitor (C). Writing the formula as Ď2= C1/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring.

[14] The resistance in an electric circuit introduces a damping factor. When analyzing a mechanical spring, one may also want to introduce a drag coefficient. Both are usually defined as a fraction of the inertia, which is the mass for a spring and the inductance for an electric circuit. Hence, we would write the resistance for a spring as Îłm and as R = ÎłL respectively.

[15] Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. Feynman (Lectures, I-33-3) shows us how to calculate the Q of these atomic oscillators: it is of the order of 108, which means the wave train will last about 10â8 seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/e). For example, for sodium light, the radiation will last about 3.2Ă10â8 seconds (this is the so-called decay time Ď). Now, because the frequency of sodium light is some 500 THz (500Ă1012 oscillations per second), this makes for some 16 million oscillations. There is an interesting paradox here: the speed of light tells us that such wave train will have a length of about 9.6 m! How is that to be reconciled with the pointlike nature of a photon? The paradox can only be explained by relativistic length contraction: in an analysis like this, one need to distinguish the reference frame of the photon â riding along the wave as it is being emitted, so to speak â and our stationary reference frame, which is that of the emitting atom.

[16] This is a general result and is reflected in the K.E. = T = (1/2)ÂˇmÂˇĎ2Âˇa2Âˇsin2(ĎÂˇt + Î) and the P.E. = U = kÂˇx2/2 = (1/2)Âˇ mÂˇĎ2Âˇa2Âˇcos2(ĎÂˇt + Î) formulas for the linear oscillator.

[17] Feynman further formalizes this in his Lecture on Superconductivity (Feynman, III-21-2), in which he refers to SchrĂśdingerâs equation as the âequation for continuity of probabilitiesâ. The analysis is centered on the local conservation of energy, which confirms the interpretation of SchrĂśdingerâs equation as an energy diffusion equation.

[18] The meff is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write meff = m. Appendix 2 provides some additional notes on the concept. As for the equations, they are easily derived from noting that two complex numbers a + iâb and c + iâd are equal if, and only if, their real and imaginary parts are the same. Now, the âĎ/ât = iâ(Ä§/meff)ââ2Ď equation amounts to writing something like this: a + iâb = iâ(c + iâd). Now, remembering that i2 = â1, you can easily figure out that iâ(c + iâd) = iâc + i2âd = â d + iâc.

[19] The dimension of B is usually written as N/(mâA), using the SI unit for current, i.e. the ampere (A). However, 1 C = 1 Aâs and, hence, 1 N/(mâA) = 1 (N/C)/(m/s).

[20] Of course, multiplication with i amounts to a counterclockwise rotation. Hence, multiplication by –i also amounts to a rotation by 90 degrees, but clockwise. Now, to uniquely identify the clockwise and counterclockwise directions, we need to establish the equivalent of the right-hand rule for a proper geometric interpretation of SchrĂśdingerâs equation in three-dimensional space: if we look at a clock from the back, then its hand will be moving counterclockwise. When writing B = (1/c)âiâE, we assume we are looking in the negative x-direction. If we are looking in the positive x-direction, we should write: B = -(1/c)âiâE. Of course, Nature does not care about our conventions. Hence, both should give the same results in calculations. We will show in a moment they do.

[21] In fact, when multiplying C2/(NÂˇm2) with N2/C2, we get N/m2, but we can multiply this with 1 = m/m to get the desired result. It is significant that an energy density (joule per unit volume) can also be measured in newton (force per unit area.

[22] The illustration shows a linearly polarized wave, but the obtained result is general.

[23] The sine and cosine are essentially the same functions, except for the difference in the phase: sinÎ¸ = cos(Î¸âĎ /2).

[24] I must thank a physics blogger for re-writing the 1/(Îľ0ÂˇÎź0) = c2 equation like this. See: http://reciprocal.systems/phpBB3/viewtopic.php?t=236 (retrieved on 29 September 2017).

[25] A circularly polarized electromagnetic wave may be analyzed as consisting of two perpendicular electromagnetic plane waves of equal amplitude and 90Â° difference in phase.

[26] Of course, the reader will now wonder: what about neutrons? How to explain neutron spin? Neutrons are neutral. That is correct, but neutrons are not elementary: they consist of (charged) quarks. Hence, neutron spin can (or should) be explained by the spin of the underlying quarks.

[27] We detailed the mathematical framework and detailed calculations in the following online article: https://readingfeynman.org/2017/09/15/the-principle-of-least-action-re-visited.

[28] https://en.wikipedia.org/wiki/Electron_rest_mass (retrieved on 29 September 2017).

Math, physics and reality

This blog has been nice. It doesn’t get an awful lot of traffic (about a thousand visitors a week) but, from time to time, I do get a response or a question that fires me up, if only because it tells me someone is actuallyÂ reading what I write.

Looking at the site now, I feel like I need to reorganize it completely. It’s justÂ chaos, right? But then that’s what gets me the positive feedback: my readers are in the same boat. We’re trying to make sense of what physicists tell us is reality. TheÂ interference modelÂ I presented in my previous post is really nice. It has all the ingredients of quantum mechanics, which I would group under two broad categories: uncertainty and duality. Both are related, obviously. I will not talk about theÂ realityÂ of the wavefunction here, because I am biased: I firmly believe the wavefunction represents something real. Why? Because Einstein’s E = mÂˇc2Â formula tells us so: energy is a two-dimensional oscillation of mass. Two-dimensional, because it’s gotÂ twiceÂ the energy of the classroom oscillator (think of a mass on a spring). More importantly, the real and imaginary dimension of the oscillation are both real: they’re perpendicular to the direction of motion of the wave-particle. Photon or electron. It doesn’t matter. Of course, we have all of the transformation formulas, but… Well… These areÂ notÂ real: they are only there to accommodateÂ ourÂ perspective: the state of the observer.

The distinction between theÂ groupÂ andÂ phaseÂ velocity of a wave packet is probably the best example of the failure of ordinary words to describe reality: particles are not waves, and waves are not particles. They are both… Well… Both at the same time. To calculate theÂ actionÂ along someÂ path, we assume there is some path, and we assume there is some particle following some path. The path and the particle are just figments of our mind. Useful figments of the mind, but… Well… There is no such thing as an infinitesimally small particle, and the concept of some one-dimensional line in spacetime does not make sense either. Or… Well… They do. Because they helpÂ usÂ to make sense of the world. Of whatÂ is, whatever it is. đ

The mainstream views on the physical significance of the wavefunction are probably best summed up in the EncyclopĂŚdia Britannica, which says the wavefunction has no physical significance. Let me quote the relevant extract here:

“TheÂ wave function,Â in quantum mechanics, is a variable quantity that mathematically describes the wave characteristics of a particle. The value of the wave function of a particle at a given point of space and time is related to the likelihood of the particleâs being there at the time. By analogy with waves such as those of sound, a wave function, designated by the Greek letter psi, Î¨, may be thought of as an expression for the amplitude of the particle wave (or de Broglie wave), although for such waves amplitude has no physical significance. The square of the wave function, Î¨2, however, does have physical significance: the probability of finding the particle described by a specific wave function Î¨ at a given point and time is proportional to the value of Î¨2.”

Really? First, this is factuallyÂ wrong: the probability is given by the square of theÂ absoluteÂ value of the wave function. These are twoÂ veryÂ different things:

1. The square of a complex number is just another complex number:Â (aÂ + ib)2Â = a2Â + (ib)2Â + 2iab = a2Â +Â i2b2Â + 2iab = a2Â â b2Â + 2iab.
2. In contrast, the square of the absolute value always gives us a realÂ number, to which we assign the mentioned physical interpretation:|aÂ + ib|2Â = [â(a2Â + b2)]2Â =Â a2Â + b2.

But it’s not only position: using the right operators, we can also get probabilities on momentum, energy and other physical variables. Hence, the wavefunction is so much more than what theÂ EncyclopĂŚdia BritannicaÂ suggests.

More fundamentally, what is written there is philosophicallyÂ inconsistent.Â SquaringÂ something – the number itself or its norm –Â is a mathematical operation. How can a mathematical operation suddenly yield something that has physical significance, if none of the elements it operates on, has any. One cannot just go from the mathematical to the physical space. The mathematical space describesÂ the physical space. Always. In physics, at least. đ

So… Well… There is too much nonsense around. Disgusting. And theÂ EncyclopĂŚdia BritannicaÂ should not just present the mainstream view. The truth is: the jury is still out, and there are many guys like me. We think the majority view is plain wrong. In this case, at least. đ

Playing with amplitudes

Let’s play a bit with the stuff we found in our previous post. This is going to be unconventional, or experimental, if you want. The idea is to give you… Well… Some ideas. So you can play yourself. đ Let’s go.

Let’s first look at Feynman’s (simplified) formula for the amplitude of a photon to go from point aÂ to point b. If we identify point aÂ by the position vector r1Â and point bÂ by the position vectorÂ r2, and using Dirac’s fancyÂ bra-ketÂ notation, then it’s written as:

So we have a vector dot product here: pâr12Â = |p|â|r12|ÂˇÂ cosÎ¸ = pâr12ÂˇcosÎą. The angle here (Îą) is the angle between theÂ pÂ andÂ r12Â vector. All good. Well… No. We’ve got a problem. When it comes to calculating probabilities, the Îą angle doesn’t matter: |eiÂˇÎ¸/r|2Â = 1/r2. Hence, for the probability, we get: P = |Â âŠr2|r1âŞ |2Â =Â 1/r122. Always ! Now that’s strange. The Î¸ =Â pâr12/Ä§Â argument gives us a different phase depending on the angle (Îą) between p and r12. But… Well… Think of it:Â cosÎą goes from 1 to 0 when Îą goes from 0 to Âą90Â° and, of course, is negative when p and r12Â have opposite directions but… Well… According to this formula, the probabilitiesÂ doÂ not depend on the direction of the momentum. That’s just weird, I think. Did Feynman, in his iconicÂ Lectures, give us a meaningless formula?

Maybe. We may also note this function looks like the elementary wavefunction for any particle, which we wrote as:

Ď(x, t) = aÂˇeâiâÎ¸Â =Â aÂˇeâiâ(Eât âÂ pâx)/Ä§= aÂˇeâiâ(Eât)/Ä§Âˇeiâ(pâx)/Ä§

The only difference is that the âŠr2|r1âŞ sort of abstracts away from time, so… Well… Let’s get a feel for the quantities. Let’s think of a photon carryingÂ some typical amount of energy. Hence, let’s talk visible light and, therefore, photons of a few eV only – say 5.625 eV = 5.625Ă1.6Ă10â19Â J = 9Ă10â19Â J. Hence, their momentum is equal to p = E/c = (9Ă10â19Â NÂˇm)/(3Ă105Â m/s) = 3Ă10â24Â NÂˇs. That’s tiny but that’s only becauseÂ newtonsÂ andÂ secondsÂ are enormous units at the (sub-)atomic scale. As for the distance, we may want to use the thickness of a playing card as a starter, as that’s what Young used when establishing the experimentalÂ fact of light interfering with itself. Now, playing cards in Young’s time were obviously rougher than those today, but let’s take the smaller distance: modern cards are as thin as 0.3 mm. Still, that distance is associated with a value ofÂ Î¸ that is equal to 13.6 million. Hence, theÂ densityÂ of our wavefunction is enormous at this scale, and it’s a bit of a miracle that Young could see any interference at all ! As shown in the table below, we only get meaningful values (remember:Â Î¸ is a phase angle) when we go down to the nanometerÂ scale (10â9Â m) or, even better, theÂ angstroms scale ((10â9Â m).Â

So… Well… Again: what can we do with Feynman’s formula? Perhaps he didn’t give us a propagatorÂ function but something that is more general (read: more meaningful) at our (limited) level of knowledge. As I’ve been reading Feynman for quite a while now – like three or four years đ – I think… Well… Yes. That’s it. Feynman wants us to think about it. đ Are you joking again, Mr. Feynman?Â đÂ So let’s assume the reasonable thing: let’s assume it gives us the amplitude to go from point a toÂ point bÂ by the position vectorÂ along some path r.Â So, then, in line with what we wrote in our previous post, let’s say pÂˇrÂ (momentum over a distance) is the action (S) we’d associate with this particular path (r) and then see where we get. So let’s writeÂ the formula like this:

ĎÂ =Â aÂˇeiÂˇÎ¸Â = (1/r)ÂˇeiÂˇS/Ä§Â =Â eiÂˇpâr/Ä§/r

We’ll use an index to denote the various paths: r0Â is the straight-line path and riÂ is any (other) path.Â Now, quantum mechanics tells us we should calculate this amplitudeÂ for every possible path. The illustration below shows the straight-line path and two nearby paths. So each of these paths is associated with some amount of action, which we measure in PlanckÂ units:Â Î¸ =Â S/Ä§.Â

The time interval is given by tÂ = t0Â =Â r0/c, for all paths. Why is the time interval the same for all paths? Because we think of a photon going from some specificÂ point in space and in timeÂ to some otherÂ specificÂ point in space and in time. Indeed, when everything is said and done, we do think of light as traveling from pointÂ a to pointÂ bÂ at the speed of light (c). In fact, all of the weird stuff here is all about trying to explain howÂ it does that. đ

Now, if we would think of the photon actually traveling along this or that path, then this implies its velocityÂ along any of the nonlinear paths will be largerÂ thanÂ c, which is OK. That’s just the weirdness of quantum mechanics, and you should actuallyÂ notÂ think of the photon actually traveling along one of these paths anyway although we’ll often put it that way. Think of something fuzzier, whatever that may be. đ

So the action is energy times time, or momentum times distance. Hence, the difference in action between two paths iÂ andÂ jÂ is given by:

Î´SÂ = pÂˇrjÂ âÂ pÂˇriÂ = pÂˇ(rjÂ â ri) = pÂˇÎr

I’ll explain theÂ Î´S <Â 2ĎÄ§/3 thing in a moment. Let’s first pause and think about theÂ uncertainty and how we’re modeling it. We can effectively think of the variation in SÂ as some uncertaintyÂ in the action: Î´SÂ = ÎS = pÂˇÎr. However, if SÂ is also equal to energy times time (SÂ = EÂˇt), and we insist tÂ is the same for all paths, then we must have some uncertainty in the energy, right? Hence, we can write Î´SÂ as ÎS = ÎEÂˇt. But, of course, E =Â E =Â mÂˇc2Â = pÂˇc, so we will have an uncertainty in the momentum as well. Hence, the variation inÂ SÂ should be written as:

Î´SÂ = ÎSÂ = ÎpÂˇÎr

That’s just logical thinking: if we, somehow, entertain the idea of a photon going from someÂ specificÂ point in spacetime to some otherÂ specificÂ point in spacetime along various paths, then the variation, or uncertainty,Â in the action will effectively combine some uncertainty in the momentum and the distance. We can calculate Îp as ÎE/c, so we get the following:

Î´SÂ = ÎSÂ = ÎpÂˇÎr =Â ÎEÂˇÎr/c = ÎEÂˇÎt with Ît =Â Îr/c

So we have the two expressions for the Uncertainty Principle here: ÎSÂ = ÎpÂˇÎr =Â ÎEÂˇÎt. Just be careful with the interpretation of Ît: it’s just the equivalent of Îr. We just express the uncertainty in distance in secondsÂ using the (absolute) speed of light. We are notÂ changing our spacetime interval: we’re still looking at a photon going fromÂ aÂ toÂ bÂ inÂ tÂ seconds,Â exactly. Let’s now look at theÂ Î´S <Â 2ĎÄ§/3 thing. If we’re addingÂ twoÂ amplitudes (twoÂ arrowsÂ or vectors, so to speak) and we want the magnitude of the result to be larger than the magnitude of the two contributions, then the angle between them should be smaller than 120 degrees, so that’s 2Ď/3 rad. The illustration below shows how you can figure that out geometrically.Hence, if S0Â is the action for r0, then S1Â = S0Â + Ä§Â and S2Â = S0Â + 2ÂˇÄ§ are still good, but S3Â = S0Â + 3ÂˇÄ§Â isÂ notÂ good. Why? Because the difference in the phase angles is ÎÎ¸Â =Â S1/Ä§Â âÂ S0/Ä§Â = (S0Â + Ä§)/Ä§Â âÂ S0/Ä§ = 1 andÂ ÎÎ¸ =Â S2/Ä§Â âÂ S0/Ä§Â = (S0Â + 2ÂˇÄ§)/Ä§Â âÂ S0/Ä§ = 2 respectively, so that’s 57.3Â°Â and 114.6Â°Â respectively and that’s, effectively,Â lessÂ than 120Â°. In contrast,Â for the next path, we find that ÎÎ¸Â =Â S3/Ä§Â âÂ S0/Ä§Â = (S0Â + 3ÂˇÄ§)/Ä§Â âÂ S0/Ä§ = 3, so that’s 171.9Â°. So that amplitude gives us a negative contribution.

Let’s do some calculations using a spreadsheet. To simplify things, we will assume we measure everything (time, distance, force, mass, energy, action,…) in Planck units. Hence, we can simply write:Â SnÂ = S0Â + n. Of course, nÂ = 1, 2,… etcetera, right? Well… Maybe not. We areÂ measuringÂ action in units ofÂ Ä§, butÂ do we actually think actionÂ comesÂ in units ofÂ Ä§?Â I am not sure. It would make sense, intuitively, butâŚ WellâŚ Thereâs uncertainty on the energy (E) and the momentum (p) of our photon, right? And how accurately can we measure the distance? So thereâs some randomness everywhere. đŚ So let’s leave that question open as for now.

We will also assume that the phase angle forÂ S0Â is equal to 0 (or some multiple of 2Ď, if you want). That’s just a matter of choosing the origin of time. This makes it really easy: ÎSnÂ =Â SnÂ â S0Â = n, and the associated phase angle Î¸nÂ = ÎÎ¸nÂ is the same. In short, the amplitude for each path reduces to ĎnÂ = eiÂˇn/r0. So we need to add these firstÂ andÂ thenÂ calculate the magnitude, which we can then square to get a probability. Of course, there is also the issue of normalization (probabilities have to add up to one) but let’s tackle that later. For the calculations, we use Euler’s rÂˇeiÂˇÎ¸Â = rÂˇ(cosÎ¸ + iÂˇsinÎ¸) = rÂˇcosÎ¸ + iÂˇrÂˇsinÎ¸ formula. Needless to say, |rÂˇeiÂˇÎ¸|2Â = |r|2Âˇ|eiÂˇÎ¸|2Â = |r|2Âˇ(cos2Î¸ + sin2Î¸) = r. Finally, when adding complex numbers, we add the real and imaginary parts respectively, and we’ll denote the Ď0Â + Ď1Â +Ď2Â + … sum as Î¨.

Now, we also need to see how our ÎSÂ = ÎpÂˇÎrÂ works out. We may want to assume that the uncertainty in p and in r will both be proportional to the overall uncertainty in the action. For example, we could try writing the following:Â ÎSnÂ = ÎpnÂˇÎrnÂ =Â nÂˇÎp1ÂˇÎr1. It also makes sense that you may want ÎpnÂ and ÎrnÂ to be proportional to Îp1Â and Îr1Â respectively. Combining both, the assumption would be this:

ÎpnÂ =Â ânÂˇÎp1Â andÂ ÎrnÂ =Â ânÂˇÎr1

So now we just need to decide how we will distribute ÎS1Â =Â Ä§Â = 1 over Îp1Â and Îr1Â respectively. For example, if we’d assume Îp1Â = 1, then Îr1Â = Ä§/Îp1Â = 1/1 = 1. These are the calculations. I will let you analyze them. đWell… We get a weird result. It reminds me ofÂ Feynman’s explanation of the partial reflection of light, shown below, but… Well… That doesn’t make much sense, does it?

Hmm… Maybe it does. đ Look at the graph more carefully. The peaks sort of oscillate out so… Well… That might make sense… đ

Does it? Are we doingÂ something wrongÂ here? These amplitudes should reflect the ones that are reflected in those nice animations (like this one, for example, which is part of thatâs part of the Wikipedia article on Feynmanâs path integral formulation of quantum mechanics). So what’s wrong, if anything? Well… Our paths differ by some fixed amount of action, which doesn’t quite reflect the geometric approach that’s used in those animations. The graph below shows how the distanceÂ rÂ varies as a function ofÂ n.Â

If we’d use a model in which the distance wouldÂ increaseÂ linearly or, preferably, exponentially, then we’d get the result we want to get, right?

Well… Maybe. Let’s try it.Â Hmm… We need to think about the geometry here. Look at the triangle below.Â IfÂ bÂ is the straight-line path (r0), thenÂ acÂ could be one of the crooked paths (rn). To simplify, we’ll assume isosceles triangles, soÂ aÂ equalsÂ cÂ and, hence, rnÂ = 2Âˇa = 2Âˇc. We will also assume theÂ successive paths are separated by the same vertical distance (h =Â h1) right in the middle, so hbÂ =Â hnÂ = nÂˇh1.Â It is then easy to show the following:This gives the following graph for rnÂ = 10 and h1Â = 0.01.

Is this the right step increase? Not sure. We can vary the values in our spreadsheet. Let’s first build it. TheÂ photon will have to travel faster in order to cover the extra distance in the same time, so its momentum will be higher. Let’s think about the velocity. Let’s start with the first path (nÂ = 1). In order to cover the extraÂ distance Îr1, the velocity c1Â must be equal to (r0Â + Îr1)/tÂ = r0/tÂ + Îr1/t =Â cÂ + Îr1/tÂ = c0Â + Îr1/t. We can write c1Â as c1Â =Â c0Â + Îc1, so Îc1Â = Îr1/t.Â Now, theÂ ratioÂ of p1Â  and p0Â will be equal to theÂ ratioÂ of c1Â andÂ c0Â because p1/p0Â = (mc1)/mc0) = c1/c0. Hence, we have the following formula for p1:

p1Â = p0Âˇc1/c0Â = p0Âˇ(c0Â + Îc1)/c0Â = p0Âˇ[1 + Îr1/(c0Âˇt) = p0Âˇ(1 + Îr1/r0)

ForÂ pn, the logic is the same, so we write:

pnÂ = p0Âˇcn/c0Â = p0Âˇ(c0Â + Îcn)/c0Â = p0Âˇ[1 + Îrn/(c0Âˇt) = p0Âˇ(1 + Îrn/r0)

Let’s do the calculations, and let’s use meaningful values, so the nanometer scale and actual values for Planck’s constant and the photon momentum. The results are shown below.Â

Pretty interesting. In fact, this looksÂ reallyÂ good. TheÂ probabilityÂ first swings around wildly, because of these zones of constructive and destructive interference, but then stabilizes. [Of course, I would need to normalize the probabilities, but you get the idea, right?] So… Well… I think we get a veryÂ meaningful result with this model. Sweet ! đ I’m lovin’ it ! đ And, here you go, this is (part of) the calculation table, so you can see what I am doing. đ

The graphs below look even better: I just changed the h1/r0Â ratio from 1/100 to 1/10. The probability stabilizes almost immediately. đ So… Well… It’s not as fancy as the referenced animation, but I think the educational value of this thing here is at least as good ! đ

đ This is good stuff… đ

Post scriptum (19 September 2017): There is an obvious inconsistency in the model above, and in the calculations. We assume there is a path r1Â = ,Â r2, r2,etcetera, and then we calculate the action for it, and the amplitude, and then we add the amplitude to the sum. But, surely, we should count these paths twice, in two-dimensional space, that is. Think of the graph: we have positive and negative interference zones that are sort of layered around the straight-line path, as shown below.

In three-dimensional space, these lines become surfaces. Hence, rather than adding oneÂ arrow for everyÂ Î´Â Â having oneÂ contribution only, we may want to add… Well… In three-dimensional space, the formula for the surface around the straight-line path would probably look like ĎÂˇhnÂˇr1, right? Hmm… Interesting idea. I changed my spreadsheet to incorporate that idea, and I got the graph below. It’s a nonsensical result, because the probability does swing around, but it gradually spins out of control: it never stabilizes.That’s because we increase theÂ weightÂ of the paths that are further removed from the center. So… Well… We shouldn’t be doing that, I guess. đ I’ll you look for the right formula, OK? Let me know when you found it. đ