# A simple explanation of quantum-mechanical operators

I added an Annex to a paper that talks about all of the fancy stuff quantum physicists like to talk about, like scattering matrices and high-energy particle events. The Annex, however, is probably my simplest and shortest summary of the ordinariness of wavefunction math, including a quick overview of what quantum-mechanical operators actually are. It does not make use of state vector algebra or the usual high-brow talk about Gilbert spaces and what have you: you only need to know what a derivative is, and combine it with our realist interpretation of what the wavefunction actually represents.

I think I should do a paper on the language of physics. To show how (i) rotations (i, j, k), (ii) scalars (constants or just numerical values) and (iii) vectors (real vectors (e.g. position vectors) and pseudovectors (e.g. angular frequency or momentum)), and (iv) operators (derivatives of the wavefunction with respect to time and spatial directions) form ‘words’ (e.g. energy and momentum operators), and how these ‘words’ then combine into meaningful statements (e.g. Schroedinger’s equation).

All of physics can then be summed up in a half-page or so. All the rest is thermodynamics ðŸ™‚ JL

PS: You only get collapsing wavefunctions when adding uncertainty to the models (i.e. our own uncertainty about the energy and momentum). The ‘collapse’ of the wavefunction (let us be precise, the collapse of the (dissipating) wavepacket) thus corresponds to the ‘measurement’ operation. ðŸ™‚

PS2: Incidentally, the analysis also gives an even more intuitive explanation of Einstein’s mass-energy equivalence relation, which I summarize in a reply to one of the many ‘numerologist’ physicists on ResearchGate (copied below).

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# An intuitive interpretation of Einsteinâ€™s mass-energy equivalence relation

My dear readers – I haven’t published much lately, because I try to summarize my ideas now in short articles that might be suitable for publication in a journal. I think the latest one (on Einstein’s mass-energy relation) should be of interest. Let me just insert the summary here:

The radial velocity formula and the Planck-Einstein relation give us the Zitterbewegung (zbw)Â frequency (E = Ä§Ï‰ = E/Ä§) and zbw radius (a = c/Ï‰ = cÄ§/mc2 = Ä§/mc) ofÂ  the electron. We interpret this by noting that the c = aÏ‰ identity gives us the E = mc2 = ma2Ï‰2 equation, which suggests we should combine the total energy (kinetic and potential) of two harmonic oscillators to explain the electron mass. We do so by interpreting the elementary wavefunction as a two-dimensional (harmonic) electromagnetic oscillation in real space which drives the pointlike charge along the zbw current ring. This implies a dual view of the reality of the real and imaginary part of the wavefunction:

1. The x = acos(Ï‰t) and y = aÂ·sin(Ï‰t) equations describe the motion of the pointlike charge.
2. As an electromagnetic oscillation, we write it as E0 = E0cos(Ï‰t+Ï€/2) + iÂ·E0Â·sin(Ï‰t+Ï€/2).

The magnitudes of the oscillation a and E0 are expressed in distance (m) and force per unit charge (N/C) respectively and are related because the energy of both oscillations is one and the same. The model â€“ which implies the energy of the oscillation and, therefore, the effective mass of the electron is spread over the zbw disk â€“ offers an equally intuitive explanation for the angular momentum, magnetic moment and the g-factor of charged spin-1/2 particles. Most importantly, the model also offers us an intuitive interpretation of Einsteinâ€™s enigmatic mass-energy equivalence relation. Going from the stationary to the moving reference frame, we argue that the plane of the zbw oscillation should be parallel to the direction of motion so as to be consistent with the results of the Stern-Gerlach experiment.

So… Well… Have fun with it ! I think I am going to sign off. ðŸ™‚ Yours – JL

# 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:

Ïˆ = =Â aÂ·eâˆ’iâˆ™Î¸Â =Â aÂ·eâˆ’iâˆ™EÂ·t/Ä§ = aÂ·cos(âˆ’Eâˆ™t/Ä§) + iÂ·aÂ·sin(âˆ’Eâˆ™t/Ä§) = aÂ·cos(Eâˆ™t/Ä§) âˆ’ iÂ·aÂ·sin(Eâˆ™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 pÂ·x term to the argument (Î¸ = EÂ·t âˆ’Â pâˆ™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 Î³Â·E0Â 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 SchrÃ¶dinger equation has some obvious issues, which I briefly touch upon in one of my other postsâ€”and which is why SchrÃ¶dinger himself and other contemporaries came up with a relativistic wave equation (Oskar Klein and Walter Gordon got the credit but others (including Louis de Broglie) also suggested a relativistic wave equation when SchrÃ¶dinger published his). In my humble opinion, the key issue is notÂ that SchrÃ¶dinger’s equation is non-relativistic. It’s that 1/2 factor again but… Well… I won’t dwell on that here. We need to move on. So let’s leave the 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 âˆ’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§/meÂ·c, 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 Î±Â·rComptonÂ â‰ˆ 3.86Ã—10âˆ’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 mÂ·a2Â·Ï‰2) and Einstein’s formula (E =Â mÂ·c2), 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 mÂ·a2Â·Ï‰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 = aÂ·cos(Ï‰Â·t + Î”).Â Needless to say, Î” is just a phase factor which defines our t = 0 point, and Ï‰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€“Ï€/2 for the other. Hence, the motion of one piston is given by x = aÂ·cos(Ï‰Â·t), while the motion of the other is given by x = aÂ·cos(Ï‰Â·tâ€“Ï€/2) = aÂ·sin(Ï‰Â·t). 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 = mÂ·v2/2 =Â (1/2)Â·mÂ·Ï‰2Â·a2Â·sin2(Ï‰Â·t + Î”)
2. P.E. = U = kÂ·x2/2 = (1/2)Â·kÂ·a2Â·cos2(Ï‰Â·t + Î”)

The coefficient k in the potential energy formula characterizes the restoring force: F = âˆ’kÂ·x. From the dynamics involved, it is obvious that k must be equal to mÂ·Ï‰2. Hence, the total energyâ€”forÂ oneÂ piston, or one springâ€”is equal to:

E = T + U = (1/2)Â· mÂ·Ï‰2Â·a2Â·[sin2(Ï‰Â·t + Î”) + cos2(Ï‰Â·t + Î”)] = mÂ·a2Â·Ï‰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 = mÂ·a2Â·Ï‰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 mÂ·a2Â·Ï‰2Â and E =Â mÂ·c2Â formulas, we need to explain why we could, potentially, writeÂ cÂ asÂ cÂ =Â aÂ·Ï‰Â =Â aÂ·âˆš(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 rÂ·Ï‰. Now, the radiusÂ rÂ is given byÂ aÂ =Â Ä§/(mÂ·c), andÂ Ï‰ = E/Ä§ =Â mÂ·c2/Ä§, soÂ vÂ is equal to toÂ v = [Ä§/(mÂ·c)]Â·[mÂ·c2/Ä§] =Â c. Another beautiful result, but what does itÂ mean? We need to think about theÂ meaning of theÂ Ï‰ =Â âˆš(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 Ï‰2 = C1/L or Ï‰2 = k/mÂ formula for harmonic oscillators.Â The key difference is that the Ï‰2= C1/L and Ï‰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 Ï‰2= 1/LC formula here. It’s basically the same concept:Â the Ï‰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 Ï‰2= Câˆ’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 Ï‰2= C1/L and Ï‰2 = k/m sort of describe the same thing: harmonic oscillation. It’s just… Well… Unlike theÂ Ï‰2= C1/L, theÂ Ï‰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 Ï‰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 = kÂ·x.

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. ðŸ™‚

We can elaborate the point as follows. TheÂ F = kÂ·xÂ 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 aÂ·Ï‰Â =Â aÂ·âˆš(k/m) equation:

In case you wonder about the E =Â FÂ·a 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 = mÂ·c2Â 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” ðŸ™‚ 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. ðŸ™‚ 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. ðŸ™‚ So it’s just like Einstein’s equation. ðŸ™‚

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 Ï€, etcetera.” Frankly, that’s disappointing. Factors like 2, 1/2, Ï€ or 2Ï€ 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 SchrÃ¶dinger’s equation also has weird consequences (when you plug in the elementary wavefunction and do the derivatives, you get a weird energy concept: E = mÂ·v2, 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 = mÂ·r2/2 formula for the angular momentum, as opposed to the I = mÂ·r2Â formula.Â I = mÂ·r2/2 (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. ðŸ™‚ 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… ðŸ™‚

Talking food for thought, let me make one final point here. TheÂ c2Â = a2Â·k/m relation implies a value for k which is equal to k = mÂ·c2/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/Ï‰ ratioÂ asÂ c/Ï‰ =Â aÂ·Ï‰/Ï‰ =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 = aÂ·k =Â aÂ·F/xÂ (just re-writing…) implies that the force is proportional to the energyâ€” F = (x/a)Â·E 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. ðŸ™‚ When everything is said and done, one only truly understands stuff when one is able to explain it to someone else, right? ðŸ™‚ Please do think of more innovative or creative ways if you can! ðŸ™‚

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. ðŸ˜¦

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§/(mÂ·c) formula gives a muchÂ smaller radius: 1835 timesÂ smaller, to be precise, so that’s around 2.1Ã—10âˆ’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).