Feyerabend was a rather famous philosopher. He was of the opinion that ‘anything goes’. We disagree. Let me know your views on my latest paper. 🙂 Let me know your views on my latest paper. 🙂 Also check out this one: https://www.academia.edu/40226046/Neutrinos_as_the_photons_of_the_strong_force.

# Tag Archives: Copenhagen interpretation

# Philosophy and Science: Dirac’s Principles

# Quantum math: garbage in, garbage out?

This post is basically a continuation of my previous one but – as you can see from its title – it is much more aggressive in its language, as I was inspired by a very thoughtful comment on my previous post. Another advantage is that it avoids all of the math. 🙂 It’s… Well… I admit it: it’s just a rant. 🙂 [Those who wouldn’t appreciate the casual style of what follows, can download my paper on it – but that’s much longer and also has a lot more math in it – so it’s a much harder read than this ‘rant’.]

My previous post was actually triggered by an attempt to re-read Feynman’s Lectures on Quantum Mechanics, but in reverse order this time: from the last chapter to the first. [In case you doubt, I did follow the correct logical order when working my way through them for the first time because… Well… There is no other way to get through them otherwise. 🙂 ] But then I was looking at Chapter 20. It’s a Lecture on quantum-mechanical operators – so that’s a topic which, in other textbooks, is usually tackled earlier on. When re-reading it, I realize why people quickly turn away from the topic of physics: it’s a lot of mathematical formulas which are supposed to reflect reality but, in practice, few – if any – of the mathematical concepts are actually being explained. Not in the first chapters of a textbook, not in its middle ones, and… Well… Nowhere, really. Why? Well… To be blunt: I think most physicists themselves don’t really understand what they’re talking about. In fact, as I have pointed out a couple of times already, Feynman himself admits so much:

“Atomic behavior appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. *Even the experts do not understand it the way they would like to*.”

So… Well… If you’d be in need of a rather spectacular acknowledgement of the shortcomings of physics as a science, here you have it: if you don’t understand what physicists are trying to tell you, don’t worry about it, because they don’t really understand it themselves. 🙂

Take the example of a *physical state*, which is represented by a *state vector*, which we can combine and re-combine using the properties of an abstract *Hilbert space*. Frankly, I think the word is very misleading, because it actually doesn’t describe an *actual* physical state. Why? Well… If we look at this so-called physical state from another angle, then we need to *transform *it using a complicated set of transformation matrices. You’ll say: that’s what we need to do when going from one reference frame to another in classical mechanics as well, isn’t it?

Well… No. In classical mechanics, we’ll describe the physics using geometric vectors in three dimensions and, therefore, the *base *of our reference frame doesn’t matter: because we’re using *real *vectors (such as the electric of magnetic field vectors **E** and **B**), our orientation *vis-á-vis* the object – the *line of sight*, so to speak – doesn’t matter.

In contrast, in quantum mechanics, it does: Schrödinger’s equation – and the wavefunction – has only two degrees of freedom, so to speak: its so-called real and its imaginary dimension. Worse, physicists refuse to give those two dimensions any *geometric *interpretation. Why? I don’t know. As I show in my previous posts, it would be easy enough, right? We know both dimensions must be perpendicular to each other, so we just need to decide if *both *of them are going to be perpendicular to our line of sight. That’s it. We’ve only got two possibilities here which – in my humble view – explain why the matter-wave is different from an electromagnetic wave.

I actually can’t quite believe the craziness when it comes to interpreting the wavefunction: we get everything we’d want to know about our particle through these operators (momentum, energy, position, and whatever else you’d need to know), but mainstream physicists still tell us that the wavefunction is, somehow, not representing anything real. It might be because of that weird 720° symmetry – which, as far as I am concerned, confirms that those state vectors are not the right approach: you can’t represent a complex, asymmetrical shape by a ‘flat’ mathematical object!

* Huh? *Yes. The wavefunction is a ‘flat’ concept: it has two dimensions only, unlike the

*real*vectors physicists use to describe electromagnetic waves (which we may interpret as the wavefunction of the photon). Those have three dimensions, just like the mathematical space we project on events. Because the wavefunction is flat (think of a rotating disk), we have those cumbersome transformation matrices: each time we shift position

*vis-á-vis*the object we’re looking at (

*das Ding an sich*, as Kant would call it), we need to change our description of it. And our description of it – the wavefunction – is all we have, so that’s

*our*reality. However, because that reality changes as per our line of sight, physicists keep saying the wavefunction (or

*das Ding an sich*itself) is, somehow, not real.

Frankly, I do think physicists should take a basic philosophy course: you can’t describe what goes on in three-dimensional space if you’re going to use flat (two-dimensional) concepts, because the objects we’re trying to describe (e.g. non-symmetrical electron orbitals) aren’t flat. Let me quote one of Feynman’s famous lines on philosophers: “These philosophers are always with us, struggling in the periphery to try to tell us something, but they never really understand the subtleties and depth of the problem.” (Feynman’s Lectures, Vol. I, Chapter 16)

Now, I *love *Feynman’s Lectures but… Well… I’ve gone through them a couple of times now, so I do think I have an appreciation of the subtleties and depth of the problem now. And I tend to agree with some of the smarter philosophers: if you’re going to use ‘flat’ mathematical objects to describe three- or four-dimensional reality, then such approach will only get you where we are right now, and that’s a lot of mathematical* mumbo-jumbo* for the poor uninitiated. *Consistent* mumbo-jumbo, for sure, but mumbo-jumbo nevertheless. 🙂 So, yes, I do think we need to re-invent quantum math. 🙂 The description may look more complicated, but it would make more sense.

I mean… If physicists themselves have had continued discussions on the reality of the wavefunction for almost a hundred years now (Schrödinger published his equation in 1926), then… Well… Then the physicists have a problem. Not the philosophers. 🙂 As to how that new description might look like, see my papers on viXra.org. I firmly believe it can be done. This is just a hobby of mine, but… Well… That’s where my attention will go over the coming years. 🙂 Perhaps quaternions are the answer but… Well… I don’t think so either – for reasons I’ll explain later. 🙂

**Post scriptum**: There are many nice videos on Dirac’s belt trick or, more generally, on 720° symmetries, but this links to one I particularly like. It clearly shows that the 720° symmetry requires, in effect, a special relation between the observer and the object that is being observed. It is, effectively, like there is a leather belt between them or, in this case, we have an arm between the glass and the person who is holding the glass. So it’s not like we are walking around the object (think of the glass of water) and making a full turn around it, so as to get back to where we were. No. *We are turning it around by 360°! *That’s a very different thing than just looking at it, walking around it, and then looking at it again. That explains the 720° symmetry: we need to turn it around twice to get it back to its original state. So… Well… The description is more about us and what we do with the object than about the object itself. That’s why I think the quantum-mechanical description is defective.

# Wavefunctions, perspectives, reference frames, representations and symmetries

Ouff ! This title is quite a mouthful, isn’t it? 🙂 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 (or *physical*) interpretations of the wavefunction is the fact that a wavefunction does not seem to obey the classical 360° 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 (or representation, *tout court*) 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 frame—and so that explains the weird stuff (like that 720° 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 paper first.

### The reality of directions

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

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

*C*

_{+}and

*C*

_{−}and

*C’*

_{+}and

*C’*

_{−}in the

*S*and

*T*representation respectively)

**must be the same, but the amplitudes**, noting—in his typical style, mixing academic and colloquial language—that “there must be some way for a particle to tell that it has turned a corner in (b).”

*must*have different phasesThe various interpretations of what actually *happens* here may shed some light on the heated discussions on the *reality *of the wavefunction—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 wavefunction—which captures a continuum of possible states, so to speak—is introduced only later. However, we may look at the amplitude for a particle to be in the *up*– or *down*-state as a wavefunction and, therefore (but do note that’s my humble opinion once more), the analysis is actually *not *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 (90°), we *know* that the amplitudes are given by *C’*_{+} = *e ^{i}*

^{∙φ/2}∙

*C*

_{+}=

*e*

^{ i}^{∙π/4}∙

*C*

_{+}and

*C’*

_{−}=

*e*

^{−i∙φ/2}∙

*C*

_{+}=

*e*

^{− i∙π/4}∙

*C*

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

**—**

*we**not*the particle, Mr. Feynman!—

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

**know**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 “(a) and (b) are different”, “the probabilities are the same”. 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 turn—so it is just going in some other direction. That’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 might—or *must*—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 ‘might’ and ‘must’ 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… *Wait a minute…* Nothing is obvious in quantum mechanics, right? How can we *experimentally confirm *that they are different?

* Huh? *I must be joking, right? You can

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

*measurement*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 thing—same

*probabilities*, remember?—why are they different? Think of this: if we look at the two beam splitters as one single tube (an

*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 same

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

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

**(**, as shown below,

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

*relative*orientation to*T*is the same in both (a) and (b)Now, the axis of *U *is not the *z*-axis: it is the *x*-axis in (a), and the *y*-axis in (b). So what? Well… I will quote Feynman here—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 of *S *which produces a pure +*x* state. Such particles would be split into +*z* and −*z* into beams in *S*, but the two beams would be recombined to give a +*x* state again at P_{1}—the exit of *S*. The same thing happens again in *T*. If we follow *T *by a third apparatus *U*, whose axis is in the +*x* direction and, as shown in (a), all the particles would go into the + beam of *U*. Now imagine what happens if *T *and *U *are swung around *together* by 90° to the positions shown in (b). Again, the *T *apparatus puts out just what it takes in, so the particles that enter *U *are in a +*x *state ** with respect to S**, which is different. By symmetry, we would now expect only one-half of the particles to get through.”

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

Now, Feynman would be very angry with me—because, as mentioned, he hates philosophers—but I’d say: this experiment proves that a direction is something real. Of course, the next philosophical question then is: what *is *a direction? I could answer this by pointing to the experiment above: a direction is something that alters the probabilities between the *S**T**U* tube as set up in (a) versus the *S**T**U* tube in (b). In fact—but, I admit, that would be pretty ridiculous—we could use the varying probabilities as we wrench this tube over varying angles to *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. 🙂 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 720° 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 frame—or *representation*, as it’s referred to in quantum mechanics—to another?

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

Before talking representations, let us first think about what we really *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 *our *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 the *shape*, if that sounds easier to you)** of the measurement apparatus** (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 us—230 years ago—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. The apparatus gives us two *directions*:

(1) The *up *direction, which *we associate* with the positive direction of the *z*-axis, and

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

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

Of course, now you’ll say: if you think that ‘relative’ and ‘absolute’ are ambiguous terms and that we, therefore, may want to avoid them a bit more, then ‘real’ and its opposite (unreal?) are ambiguous terms too, right? Well… Maybe. What language would *you *suggest? 🙂 Just stick to the story for a while. I am not done yet. So… Yes… What *is *their *reality*? Let’s think about that in the next section.

### Perspectives, reference frames and symmetries

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

So… Well… **I want you to distinguish—just for a while—between the notion of a reference frame (think of the x–y–z reference frame that comes with the apparatus) and your 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 side—which, if you think about it, you can only

*define*in terms of the various positive and negative directions of the various axes. 🙂 If you think this is getting ridiculous… Well… Don’t. Feynman himself 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 the

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

*very*first illustration of his

*Lecture*on this:

He uses it to explain the apparatus (which we don’t do here because you’re supposed to already know how these (modified or improved) Stern-Gerlach apparatuses work). So let’s continue this story. Suppose that we are looking in the *positive* *y*-direction—so that’s the direction in which our particle is moving—then we might imagine how it would look like when *we *would make a 180° 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 coming *towards *us, out of the apparatus.

What happens—but that’s not scientific language, of course—is that left becomes right, and right becomes left. Top still is top, and bottom is bottom. We are looking now in the *negative y*-direction, and the positive direction of the *x*-axis—which pointed right when we were looking in the positive *y*-direction—now points left. I see you nodding your head now—because you’ve heard about parity inversions, mirror symmetries and what have you—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 the 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 to *this *story, which is about transformations of amplitudes (or wavefunctions). [If you really want to know—but I know this sounds counterintuitive—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 *mentally* 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 only *apparent*. It’s a common misconception, and… Well… I’ll let you figure this out yourself. I need to move on.] Just note the following:

- The
*xyz*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. Our*perspective*on it has changed. - However, if we think of the real and imaginary part of the wavefunction describing the electrons that are going through our apparatus as perpendicular oscillations (as shown below)—a cosine and sine function respectively—then our change in perspective
*might*, effectively, mess up our convention for measuring angles.

I am not saying it *does*. Not now, at least. I am just saying it *might*. It depends on the plane of the oscillation, as I’ll explain in a few moments. Think of this: we measure angles *counter*clockwise, right? As shown below… But… Well… If the thing below would be some funny clock going backwards—you’ve surely seen them in a bar or so, right?—then… Well… If they’d be transparent, and you’d go around them, you’d see them as going… Yes… Clockwise. 🙂 [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, *if *we would assume this clock represents something real—and, of course, **I am thinking of the elementary wavefunction e^{i}^{θ} = cosθ + i·sinθ now**—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…

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

[…]

We’d see it as *e*^{−i}^{θ} = *cos*(−θ) + *i*·*sin*(−θ) = *cos*θ − *i*·*sin*θ, right? The hand of our clock now goes clockwise, so that’s the *opposite *direction of our convention for measuring angles. Hence, instead of *e*^{i}^{θ}, we write *e*^{−i}^{θ}, right? So that’s the complex conjugate. So we’ve got a different *image *of the same thing here. *Not* good. *Not good at all.*

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

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

*if they are to represent something real*, our

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

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

*not…*Well… I am tempted to say…

*Not*relative at all but then… Well… We wanted to avoid that term, right? 🙂

[…]

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 symmetry—or *a*symmetry, I should say.

### The flywheel model of an electron

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

- The dimension of the matter-wave field vector is force per unit
*mass*(N/kg), as opposed to the force per unit*charge*(N/C) dimension of the electric field vector. This dimension is an acceleration (m/s^{2}), which is the dimension of the gravitational field. - We assume this gravitational disturbance causes our electron (or a charged
*mass*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, “when you do find the electron some place, the entire charge is there.” Of course, we cannot prove anything here, but our elegant yet simple derivation of the Compton radius of an electron is… Well… Just nice. 🙂 - Finally, and most importantly
*in the context of this discussion*, we noted that, in light of the direction of the magnetic moment of an electron in an inhomogeneous magnetic field,**the plane which circumscribes the circulatory motion of the electron should also**Hence, unlike an electromagnetic wave, the*comprise*the direction of its linear motion.*plane*of the two-dimensional oscillation (so that’s the polarization plane, really) can*not*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. The direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is traveling—can*not* be parallel to the direction of motion. On the contrary, it must be *perpendicular* 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. the *plane *of the polarization) has to *comprise *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 explanation—combined with the quantization hypothesis—goes a long way in explaining this: an object with an angular momentum ** J** and a magnetic moment

**that is**

*μ**not exactly*parallel to some magnetic field

**B**, will

*not*line up: it will

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

*spin*or

*angular momentum*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 the *up*-direction *only*, then it will be spinning *clockwise *if its angular momentum is down (so its *magnetic moment *is *up*). Conversely, it will be spinning *counter*clockwise if its angular momentum is *up*. Let us take the *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. 🙂 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 as—I am not ashamed to admit this—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 some *positive *angle. Walking around it—in whatever direction, clockwise or counterclockwise—doesn’t change the counterclockwise direction of our… Well… That weird object that might—just *might—*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 wrenched *e ^{i}*

^{θ}=

*cos*θ +

*i*·

*sin*θ function, right? The

*x-*and

*y*-axes

*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 180° angle so we’re looking in the *negative *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 the *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 the *y*-axis to point away, rather than towards us. In short, we have a transformation of the reference frame here: *z’* = *z*, *y’* = − *y*, and *x’* = − *x*. Mind you, this is still a regular right-handed reference frame. [That’s the difference with a *mirror *image: a *mirrored *right-hand reference frame is no longer right-handed.] So, in our new reference frame, that we choose to coincide with our *perspective*, we will now describe the same thing as some −*cos*θ − *i*·*sin*θ = −*e ^{i}*

^{θ}function. Of course, −

*cos*θ =

*cos*(θ + π) and −

*sin*θ =

*sin*(θ + π) so we can write this as:

−*cos*θ − *i*·*sin*θ = *cos*(θ + π) + *i*·*sin*θ = *e ^{i}*

^{·(}

^{θ+π)}=

*e*

^{i}^{π}·

*e*

^{i}^{θ}= −

*e*

^{i}^{θ}.

Sweet ! But… Well… First note this is *not *the complex conjugate: *e*^{−i}^{θ} = *cos*θ − *i*·*sin*θ ≠ −*cos*θ − *i*·*sin*θ = −*e ^{i}*

^{θ}. 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. 🙂 And, yes, let me lighten up the discussion with that painting here. 🙂 We need to have

*some*fun while torturing our brain, right?

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

^{θ}function instead of a

*e*

^{−i}

^{θ}function.

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

^{·(}

^{θ+π)}function. It’s the same as −

*e*

^{i}^{θ}but… Well… We walked around the

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

*phase shift*here.

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

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

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

*z*-axis by α and… Well… Of course, our

*e*

^{i}^{·}

^{θ}now becomes an our

*e*

^{i}^{·(}

^{θ+α)}function, right? We’ve just derived the transformation coefficient for a rotation about the

*z*-axis, didn’t we? It’s equal to

*e*

^{i}^{·}

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

*e*

^{i}^{·}

^{α}, right? It’s equal to

*e*

^{i}^{·}

^{α}·

*e*

^{i}^{·}

^{θ}=

*e*

^{i}^{·(}

^{θ+α)}, right?

Well…

[…]

No. The answer is: no. The transformation coefficient is not *e ^{i}*

^{·}

^{α}but

*e*

^{i}^{·}

^{α/2}. So we get an additional 1/2 factor in the

*phase shift*.

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

*half*of the rotation angle only.

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

*down*electron, the transformation coefficient is

*e*

^{−i·}

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

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

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:

**Our e^{i}^{·}^{α} coefficient describes a rotation of the reference frame. In contrast, the e^{i}^{·}^{α/2} and e^{−i·}^{α/2} coefficients describe what happens when we rotate the T apparatus! Now that is a very different proposition. **

Right! You got it! *Representations* and reference frames are different things. *Quite *different, I’d say: representations are *real*, reference frames aren’t—but then you don’t like philosophical language, do you? 🙂 But think of it. When we just go about the *z*-axis, a full 180°, but we don’t touch that *T*-apparatus, we don’t change *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 the *z*-axis, a full 180°, 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 frame—from *xyz* to *x’y’z’* to be precise: we do *not *change the representation.

In contrast, **when we rotate the T apparatus over a full 180°, 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 through

*S*, and now it goes in the opposite direction—

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

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

### Quantum-mechanical weirdness?

The transformation matrix for the amplitude of a system to be in an *up *or *down *state (and, hence, presumably, for a wavefunction) for a rotation about the *z*-axis is the following one:

Feynman derives this matrix in a rather remarkable intellectual *tour de force *in the 6th of his *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 how *I *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? 🙂 So… Well… Because he *knows*—from experiment—that the coefficient is *e ^{i}*

^{·}

^{α/2}instead of

*e*

^{i}^{·}

^{α}, he just says the phase shift—which he denotes by λ—must be some

*proportional*to the angle of rotation—which he denotes by φ rather than α (so as to avoid confusion with the

*Euler*angle α). So he writes:

λ = m·φ

Initially, he also tries the obvious thing: m should be one, right? So λ = φ, 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:

“Suppose *T* is rotated by 360°; then, clearly, it is right back at zero degrees, and we should have *C’*_{+} = *C*_{+} and *C’*_{−} = *C*_{−} or, what is the same thing, *e ^{i}*

^{·m·2π}= 1. We get m = 1. [But no!]

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

*T*is rotated by 180°. If m were equal to 1, we would have

*C’*

_{+}=

*e*

^{i}^{·π}

*C*

_{+}= −

*C*

_{+}and

*C’*

_{−}=

*e*

^{−}

^{i}^{·π}

*C*

_{−}= −

*C*

_{−}. [Feynman works with

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

*original*state all over again.

**amplitudes are just multiplied by −1 which gives back the original physical system. (It is again a case of a**

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

*common**T*and

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

*U*apparatus. At 180°, though, the (+) state of the

*U*apparatus is the (−

*x*) state of the original

*S*apparatus. So a (+

*x*) state would become a (−

*x*) state. But we have done nothing to

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

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

The result, of course, is this weird 720° symmetry. While we get the same *physics* after a 360° rotation of the *T* apparatus, we do *not *get the same amplitudes. We get the opposite (complex) number: *C’*_{+} = *e ^{i}*

^{·2π/2}

*C*

_{+}= −

*C*

_{+}and

*C’*

_{−}=

*e*

^{−}

^{i}^{·2π/2}

*C*

_{−}= −

*C*

_{−}. That’s OK, because… Well… It’s a

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

*reality.*But… Well…

*C’*

_{+}≠ −

*C*

_{+}and

*C’*

_{−}≠ −

*C*

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

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

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

^{θ}function we got when looking at the

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

*e*

^{i}^{θ}and −

*e*

^{i}^{θ}functions did

*not*describe a different reality. We just changed our perspective. The

*reference frame*. As such, the reference frame isn’t

*real*. The experimental set-up is. And—I know I will anger mainstream physicists with this—the

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

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

In contrast, a different perspective—or a different reference frame—does not.

### Conventions

While you might have had a lot of trouble going through all of the weird stuff above, the point is: it is *not *all that weird. We *can *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 formula—*e*^{i}^{θ} = *cos*θ + *i*·*sin*θ—would, one day, be used to represent *something real*: an electron, or any elementary particle, really. If he *would *have known, I am sure he would have noted what I am noting here: *Nature *can’t be bothered by our conventions. Hence, if *e*^{i}^{θ} represents something real, then *e*^{−i}^{θ} must also represent something real. [Coz I admire this genius so much, I can’t resist the temptation. Here’s his portrait. He looks kinda funny here, doesn’t he? :-)]

Frankly, he would probably have understood quantum-mechanical theory as easily and instinctively as Dirac, I think, and I am pretty sure he would have *noted*—and, if he would have known about circularly polarized waves, probably *agreed* to—that *alternative *convention for measuring angles: we could, effectively, measure angles clockwise *or* counterclockwise depending on the direction of our particle—as opposed to Euler’s ‘one-size-fits-all’ counterclockwise convention. But so we did *not *adopt that alternative convention because… Well… We want to keep honoring Euler, I guess. 🙂

So… Well… If we’re going to keep honoring Euler by sticking to that ‘one-size-fits-all’ counterclockwise convention, then **I do believe that e^{i}^{θ} and e^{−i}^{θ} represent two different 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 *see *something different when they go through a Stern-Gerlach apparatus. So it’s *not* just some convention to *count *things like 0, 1, 2, etcetera versus 0, −1, −2 etcetera. It’s the same story again: different but related *mathematical *notions are (often) related to different but related *physical *possibilities. So… Well… I think that’s what we’ve got here. Think of it. Mainstream quantum math treats all wavefunctions as right-handed but… Well… A particle with *up *spin is a different particle than one with *down *spin, right? And, again, *Nature* surely can*not* be bothered about our convention of measuring phase angles clockwise or counterclockwise, right? So… Well… Kinda obvious, right? 🙂

Let me spell out my conclusions here:

**1.** The angular momentum can be positive or, alternatively, negative: *J* = +ħ/2 or −ħ/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* particle—think of an electron, or whatever other particle you’d think of—comes in two *variants*: right-handed and left-handed. They will, therefore, *either* consist of (elementary) right-handed waves or, *else*, (elementary) left-handed waves. An elementary right-handed wave would be written as: ψ(θ* _{i}*)

*=*

*e*^{i}^{θi}

*= a*·(

_{i}*cos*θ

*+*

_{i}*i·sin*θ

*). In contrast, an elementary left-handed wave would be written as: ψ(θ*

_{i}*)*

_{i}*=*

*e*^{−i}^{θi}

*·(*

*=*a_{i}*cos*θ

*−*

_{i}*i·sin*θ

*). So that’s the complex conjugate.*

_{i}So… Well… Yes, I think complex conjugates are not just some *mathematical *notion: I believe they represent something real. It’s the usual thing: *Nature *has shown us that (most) mathematical possibilities correspond to *real *physical situations so… Well… Here you go. It is really 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 differences—different polarization plane and dimensions and what have you—I’ve already summed those up, so I won’t repeat myself here.] The point is: if we have two different *physical *situations, we’ll want to have two different functions to describe it. Think of it like this: why would we have *two*—yes, I admit, two *related—*amplitudes to describe the *up *or *down *state of the same system, but only one wavefunction for it? 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 (or *physical*) interpretation of the wavefunction might be, it won’t be compatible with the *isotropy *of space. You cannot *imagine *an object with a 720° symmetry. That’s *geometrically *impossible.”

Well… Almost three years ago, I wrote the following on this blog: “As strange as it sounds, a spin-1/2 particle needs *two *full rotations (2×360°=720°) until it is again in the same state. Now, in regard to that particularity, you’ll often read something like: “*There is **nothing** in our macroscopic world which has a symmetry like that.*” Or, worse, “*Common sense tells us that something like that cannot exist, that it simply is impossible.*” [I won’t quote the site from which I took this quotes, because it is, in fact, the site of a very respectable research center!]* Bollocks!* The Wikipedia article on spin has this wonderful animation: look at how the spirals flip between clockwise and counterclockwise orientations, and note that it’s only after spinning a full 720 degrees that this ‘point’ 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, we can 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 good *image*, I should say, because… Well… A representation is something real, remember? 🙂

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

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

The second question is more important. If we just rotate the reference frame over 360°, we see the same thing: some rotating object which we, vaguely, describe by some *e*^{+i}^{·θ} function—to be precise, I should say: by some *Fourier* sum of such functions—or, if the rotation is in the other direction, by some *e*^{−i}^{·θ} function (again, you should read: a *Fourier *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 same 360°, we get a *different *object: our *e*^{i}^{·θ} and *e*^{−i}^{·θ} function (again: think of a *Fourier *sum, so that’s a wave *packet*, really) becomes a −*e*^{±i}^{·θ} thing. We get a *minus *sign in front of it. 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 full 360°, then you’ll end up where you were when you started and, importantly, you’ll see the same thing. *Exactly *the same thing: if I was an *e*^{+i}^{·θ} wave packet, I am still an an *e*^{+i}^{·θ} wave packet now. Or if I was an *e*^{−i}^{·θ} wave packet, then I am still an an *e*^{−i}^{·θ} wave packet now. Easy. Logical. *Obvious*, right?

But so now we try something different: *I *turn around, over a full 360° turn, and *you *stay where you are. When I am back where I was—looking at you again, so to speak—then… Well… I am not quite the same any more. Or… Well… Perhaps I am but you *see *me differently. If I was *e*^{+i}^{·θ} wave packet, then I’ve become a −*e*^{+i}^{·θ} wave packet now. Not *hugely* different but… Well… That *minus *sign matters, right? Or If I was wave packet built up from elementary *a*·*e*^{−i}^{·θ} waves, then I’ve become a −*e*^{−i}^{·θ} wave packet now. What happened?

It makes me think of the twin paradox in special relativity. We know it’s a *paradox*—so that’s an *apparent *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 some *force*.

Can we relate this to the twin paradox? Maybe. Note that a *minus *sign in front of the *e*^{−±i}^{·θ} 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 180°: −*cos*θ = *cos*(θ ± π) and −*sin*θ = *sin*(θ ± π). Now, adding or subtracting a *common *phase factor to/from the argument of the wavefunction amounts to *changing *the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360° and 720° symmetries are, effectively, related. 🙂

# 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. 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 motion—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 is *very *intriguing, but let’s think about that later.

Let’s assume we’re looking at it from *some *specific direction. Then 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 respectively—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:

ψ = = *a·e*^{−i∙θ} = *a·e*^{−i∙E·t/ħ} = *a*·cos(−E∙t/ħ)* + i*·a·sin(−E∙t/ħ)* = a*·cos(E∙t/ħ) *−** i*·a·sin(E∙t/ħ)* *

So that’s the wavefunction in the reference frame of the particle itself. When we think of it as moving in some direction (so relativity kicks in), we need to add the * p*·

*term to the argument (θ = E·t −*

**x****p**∙

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

*= γ·E*

_{v }_{0}and so… Well… I’ll refer you to my post on this, in which I show the argument of the wavefunction is invariant under a Lorentz transformation: the way E

*and p*

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

_{v}*x*and

*t*relativistically

*transform*ensures the invariance.

In fact, I’ve always wanted to read *de Broglie*‘s 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 does * not *imply there is no need for a relativistic wave

*equation*: the wavefunction is a

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

*not*that Schrödinger’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 wave

*equation*for what it is and go back to our wave

*function*.

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

*J*= −ħ/2. [If you’ve read a thing or two about neutrinos, then… Well… They’re kinda special in this regard: they have no charge and neutrinos and antineutrinos are actually

_{z}*defined*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 to *calculate *the amplitude *a*. We got a result that tentatively confirms we’re on the right track with our interpretation: we found that *a *= ħ/m_{e}·*c*, so that’s the *Compton scattering radius* of our electron. All good ! But we were still a bit stuck—or *ambiguous*, I should say—on what the components of our wavefunction actually *are*. Are we really imagining the tip of that rotating arrow is a pointlike electric charge spinning around the center? [Pointlike or… Well… Perhaps we should think of the *Thomson *radius of the electron here, i.e. the so-called *classical *electron radius, which is equal to the Compton radius times the fine-structure constant:* r _{Thomson} = α·r_{Compton}* ≈ 3.86×10

^{−13}/137.]

So that would be the flywheel model.

In contrast, we may also think the whole arrow is some rotating *field vector*—something like the electric field vector, with the same or some other *physical *dimension, like newton per charge unit, or newton per mass unit? So that’s the *field *model. Now, these interpretations may or may not be compatible—or *complementary*, I should say. I sure *hope *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 the *interaction *between a photon and an electron, as I demonstrated in my previous post: the electromagnetic energy of the photon will *drive *the circulatory motion of our electron… So… Well… That’s a nice *physical *explanation for the transfer of energy. 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 flywheel *metaphor* to its logical limits. Let me remind you of what triggered it all: it was the *mathematical *equivalence of the energy equation for an oscillator (E = m·*a*^{2}·ω^{2}) and Einstein’s formula (E = m·*c*^{2}), which tells us energy and mass are *equivalent *but… Well… They’re not the same. So what *are *they then? What *is *energy, and what *is *mass—in the context of these matter-waves that we’re looking at. To be precise, the E = m·*a*^{2}·ω^{2} formula gives us the energy of *two *oscillators, so we need a *two*-spring model which—because I love motorbikes—I referred to as my V-twin engine model, but it’s not an *engine*, really: it’s two frictionless pistons (or springs) whose direction of motion is perpendicular to each other, so they are in a 90° degree angle and, therefore, their motion is, effectively, independent. In other words: they will not interfere *with each other*. It’s probably worth showing the illustration just one more time. And… Well… Yes. I’ll also briefly review the math one more time.

If the magnitude of the oscillation is equal to *a*, then the motion of these piston (or the mass on a spring) will be described by *x* = *a*·cos(ω·t + Δ). Needless to say, Δ is just a phase factor which defines our *t* = 0 point, and ω is the *natural angular *frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by *x* = *a*·cos(ω·t), while the motion of the other is given by *x* = *a*·cos(ω·t–π/2) = *a*·sin(ω·t). The kinetic and potential energy of *one *oscillator – think of one piston or one spring only – can then be calculated as:

- K.E. = T = m·
*v*^{2}/2 = (1/2)·m·ω^{2}·*a*^{2}·sin^{2}(ω·t + Δ) - P.E. = U = k·x
^{2}/2 = (1/2)·k·*a*^{2}·cos^{2}(ω·t + Δ)

The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω^{2}. Hence, the total energy—for *one *piston, or one spring—is equal to:

E = T + U = (1/2)· m·ω^{2}·*a*^{2}·[sin^{2}(ω·t + Δ) + cos^{2}(ω·t + Δ)] = m·*a*^{2}·ω^{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 = m·*a*^{2}·ω^{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, but 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 the E = m·*a*^{2}·ω^{2} and E = m·*c*^{2} formulas, we need to explain why we could, potentially, write *c *as *c *= *a*·ω = *a*·√(k/m). We’ve done that already—to some extent at least. The *tangential *velocity of a pointlike particle spinning around some axis is given by *v* = *r*·ω. Now, the radius *r *is given by *a *= ħ/(m·*c*), and ω = E/ħ = m·*c*^{2}/ħ, so *v *is equal to to *v *= [ħ/(m·*c*)]·[m·*c*^{2}/ħ] = *c*. Another beautiful result, but what does it *mean*? We need to think about the *meaning *of the ω = √(k/m) formula here. In the mentioned article, we boldly wrote that the speed of light is to be interpreted as the *resonant *frequency of spacetime, but so… Well… What do we really *mean *by that? Think of the following.

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

This effectively reminds us of the ω^{2} = *C*^{–}^{1}/*L* or ω^{2} = k/m formula for harmonic oscillators. The key difference is that the ω^{2}= *C*^{–}^{1}/*L* and ω^{2} = k/m formulas introduce *two *(or more) degrees of freedom. In contrast, *c*^{2}= E/m for *any *particle, *always*. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in *one *physical space only: *our *spacetime. Hence, the speed of light (*c*) emerges here as *the* defining property of 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 ω^{2}= 1/*LC* formula here. It’s basically the same concept: the ω^{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 ω^{2}= *C*^{−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 ω^{2}= *C*^{–}^{1}/*L* and ω^{2} = k/m sort of describe the same thing: harmonic oscillation. It’s just… Well… Unlike the ω^{2}= *C*^{–}^{1}/*L*, the ω^{2} = k/m is *directly *compatible with our V-twin engine metaphor, because it also involves *physical distances*, as I’ll show you here.] The *k *in the ω^{2} = k/m is, effectively, the stiffness of the spring. It is *defined *by Hooke’s Law, which states that the force that is needed to extend or compress a spring by some distance *x * is linearly proportional to that distance, so we write: F = k·*x*.

Now *that *is interesting, isn’t it? We’re talking *exactly *the same thing here: spacetime is, presumably, *isotropic*, so it should oscillate the same in any direction—I am talking those sine and cosine oscillations now, but in *physical *space—so there is nothing imaginary here: all is *real *or… Well… As real as we can imagine it to be. 🙂

We can elaborate the point as follows. The F = k·*x* equation implies k is a force *per unit distance*: k = F/*x*. Hence, its physical dimension is *newton per meter* (N/m). Now, the *x *in this equation may be equated to the *maximum *extension of our spring, or the *amplitude *of the oscillation, so that’s the radius *r *= *a *in the metaphor we’re analyzing here. Now look at how we can re-write the *c *= *a*·ω = *a*·√(k/m) equation:

In case you wonder about the E = F·*a* substitution: just remember that *energy is force times distance*. [Just do a dimensional analysis: you’ll see it works out.] So we have a spectacular result here, for several reasons. The first, and perhaps most obvious reason, is that we can actually *derive *Einstein’s E = m·*c*^{2} formula from our flywheel model. Now, that *is *truly glorious, I think. However, even more importantly, this equation suggests we do *not necessarily *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 a force acting over some distance**

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

*actually*acting*Now,*

**on a particle.***that*energy will have an

*equivalent*mass which is—or

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

* Huh? *Yes. In case you wonder what I am trying to get at, I am trying to convey the idea that the two interpretations—the field versus the flywheel model—are actually fully

*equivalent*, or

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

**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 the *Rydberg *energy, even Feynman casually writes that “we need not trust our answer [to questions like this] within factors like 2, π, etcetera.” Frankly, that’s disappointing. Factors like 2, 1/2, π or 2π 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 does *not *come into play: think of the Cooper pairs we use to explain superconductivity (so that’s the BCS theory). The 1/2 factor in Schrödinger’s equation also has weird consequences (when you plug in the elementary wavefunction and do the derivatives, you get a weird energy concept: E = m·*v*^{2}, to be precise). This problem may also be solved when assuming we’re actually calculating orbitals for a *pair *of electrons, rather than orbitals for just one electron only. [We’d get *twice *the mass (and, presumably, the charge, so… Well… It might work—but I haven’t done it yet. It’s on my agenda—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 *physical *interpretation of the wavefunction), you may or may not remember (if not, check my post on it) ‘ll remember I had to use the I = m·*r*^{2}/2 formula for the angular momentum, as opposed to the I = m·*r*^{2} formula. I = m·*r*^{2}/2 (*with* the 1/2 factor) gives us the angular momentum of a *disk *with radius *r*, as opposed to a *point *mass going around some circle with radius *r*. I noted that “the addition of this 1/2 factor may seem arbitrary”—and it totally *is*, of course—but so it gave us the result we wanted: the *exact *(Compton scattering) radius of our electron.

Now, the arbitrary 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 to *x *= *a*, we need to calculate it as the following integral:

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

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

OK. That’s it but… Well… 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 some *specific *direction. It could be *any *direction but… Well… It’s *some *direction. We have no *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 be *any *direction, our analysis is valid for any direction. Hence, *if* our interpretation would happen to be some *true*—and that’s a big *if*, of course—then our particle has to be *spherical*, right? Why? Well… Because we see this circular thing from any direction, so it *has *to be a sphere, right?

Well… Yes. But then… Well… While that logic seems to be *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. 😦

**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), the *a *= ħ/(m·*c*) formula gives a *much *smaller radius: 1835 times *smaller*, to be precise, so that’s around 2.1×10^{−16} m, which is about 1/4 of the so-called *charge *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 is *not *an elementary particle, so its mass incorporates other factors than what we’re considering here (two-dimensional oscillations).

# 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. 🙂 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*. 🙂

**Abstract : **This paper explores the implications of associating the components of the wavefunction with a physical dimension: force per unit *mass* – which is, of course, the dimension of acceleration (m/s^{2}) 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/s^{2} 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, Schrödinger’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 (ψ = *a·e*^{−i∙θ} = *a*∙cosθ – *a*∙sinθ) differ by 90 degrees (π/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 = m·*a*^{2}·ω^{2}/2) and Einstein’s relativistic energy equation E = m∙*c*^{2} 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. Schrödinger’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 Einstein’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 Einstein’s relativistic energy equation (E = m*c*^{2}) 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:

- E = m
*c*^{2} - E = mω
^{2}/2 - E = m
*v*^{2}/2

In these formulas, ω, *v *and *c *all describe some velocity.[8] Of course, there is the 1/2 factor in the E = mω^{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 = 2·m·ω^{2}/2 = m·ω^{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 90° 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* = *a*·cos(ω·t + Δ).[12] Needless to say, Δ is just a phase factor which defines our *t* = 0 point, and ω is the *natural angular *frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by *x* = *a*·cos(ω·t), while the motion of the other is given by *x* = *a*·cos(ω·t–π/2) = *a*·sin(ω·t).

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

- K.E. = T = m·
*v*^{2}/2 = (1/2)·m·ω^{2}·*a*^{2}·sin^{2}(ω·t + Δ) - P.E. = U = k·x
^{2}/2 = (1/2)·k·*a*^{2}·cos^{2}(ω·t + Δ)

The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω^{2}. Hence, the total energy is equal to:

E = T + U = (1/2)· m·ω^{2}·*a*^{2}·[sin^{2}(ω·t + Δ) + cos^{2}(ω·t + Δ)] = m·*a*^{2}·ω^{2}/2

To facilitate the calculations, we will briefly assume k = m·ω^{2} and *a* are equal to 1. The *motion *of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) *change *in kinetic energy at any point in time will be equal to:

d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ

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 sinθ function, which is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin^{2}(θ−π /2), and how it *changes *– as a function of θ – will be equal to:

2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ

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 = m*a*^{2}ω^{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 = m*c*^{2} 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:

ψ = *a·e*^{−i[E·t − p∙x]/ħ} = *a·e*^{−i[E·t − p∙x]/ħ} = *a·cos( p∙x/ħ *

*–*

*E∙t/ħ) + i·a·sin(*

**p**∙**x**/ħ*–*

*E∙t/ħ)*

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

ψ = *a·e*^{−i∙E·t/ħ} = *a·cos(**–**E∙t/ħ) + i·a·sin(**–**E∙t/ħ) = a·cos(E∙t/ħ) **–** i·a·sin(E∙t/ħ) *

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 (ϕ) is *counter*-clockwise.

**Figure 2**: Euler’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 **p***∙ x/ħ reduces to p∙x/ħ. Most illustrations – such as the one below – 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 – the real and imaginary part of our wavefunction – 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 ψ = *a·e*^{−i∙E·t/ħ}. Hence, the angular velocity of both oscillations, at some point **x**, is given by ω = -E/ħ. Now, the energy of our particle includes all of the energy – kinetic, potential and rest energy – and is, therefore, equal to E = m*c*^{2}.

Can we, somehow, relate this to the m·*a*^{2}·ω^{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 µ *a*^{2}. We may, therefore, think that the *a*^{2} factor in the E = m·*a*^{2}·ω^{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 *a*_{k}, and their own ω* _{i}* = -E

*/ħ. Each of these wavefunctions will*

_{i}*contribute*some energy to the total energy of the wave packet. To calculate the contribution of each wave to the total, both

*a*as well as E

_{i}*will matter.*

_{i}What is E* _{i}*? E

*varies around some average E, which we can associate with some*

_{i}*average mass*m: m = E/

*c*

^{2}. 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 *c*^{2}ħ^{2}: it is about 1´10^{–}^{51} N^{2}∙m^{4}. Let us also do a dimensional analysis: the physical dimensions of the E = m·*a*^{2}·ω^{2} equation make sense if we express m in kg, *a *in m, and ω in *rad*/s. We then get: [E] = kg∙m^{2}/s^{2} = (N∙s^{2}/m)∙m^{2}/s^{2} = N∙m = J. The dimensions of the left- and right-hand side of the physical normalization condition is N^{3}∙m^{5}.** **

# 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 Einstein’s energy equation once again. If we want to look at mass, we should re-write it as *m* = E/c^{2}:

[*m*] = [E/*c*^{2}] = J/(m/s)^{2} = N·m∙s^{2}/m^{2} = N·s^{2}/m = kg

This is not very helpful. It only reminds us of Newton’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. Einstein’s E = m*c*^{2} 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 ω^{2}= *C*^{–}^{1}/*L* or ω^{2} = *k*/*m* of harmonic oscillators once again.[13] The key difference is that the ω^{2}= *C*^{–}^{1}/*L* and ω^{2} = *k*/*m* formulas introduce *two *or more degrees of freedom.[14] In contrast, *c*^{2}= E/m for *any *particle, *always*. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in *one *physical space only: *our *spacetime. Hence, the speed of light *c* emerges here as *the* defining property of spacetime – 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 Planck’s constant as the constant of proportionality. Now, for *one-dimensional *oscillations – think of a guitar string, for example – 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 *f*^{2}.

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 Einstein’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 (Schrödinger’s equation), as we can now analyze it as an energy diffusion equation.** **

# IV. Schrödinger’s equation as an energy diffusion equation

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

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

Let us review the basic math. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Uψ term disappears. Therefore, Schrödinger’s equation reduces to:

∂ψ(**x**, t)/∂t = *i*·(1/2)·(ħ/m_{eff})·∇^{2}ψ(**x**, t)

The ubiquitous diffusion equation in physics is:

∂φ(**x**, t)/∂t = D·∇^{2}φ(**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 ψ is a complex-valued function, with a *real *and an *imaginary *part, we get the following equations[18]:

*Re*(∂ψ/∂t) = −(1/2)·(ħ/m_{eff})·*Im*(∇^{2}ψ)*Im*(∂ψ/∂t) = (1/2)·(ħ/m_{eff})·*Re*(∇^{2}ψ)

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

- ∂
**B**/∂t = –∇×**E** - ∂
**E**/∂t =*c*^{2}∇×**B**

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

**Figure 4**: Propagation mechanisms

The Laplacian operator (∇^{2}), when operating on a *scalar *quantity, gives us a flux density, i.e. something expressed per square meter (1/m^{2}). In this case, it is operating on ψ(**x**, t), so what is the dimension of our wavefunction ψ(**x**, t)? To answer that question, we should analyze the diffusion constant in Schrödinger’s equation, i.e. the (1/2)·(ħ/m_{eff}) factor:

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

Now, the ħ/m_{eff} factor is expressed in (N·m·s)/(N· s^{2}/m) = m^{2}/s. Hence, it does ensure the dimensions on both sides of the equation are, effectively, the same: ∂ψ/∂t is a time derivative and, therefore, its dimension is s^{–}^{1} while, as mentioned above, the dimension of ∇^{2}ψ is m^{–}^{2}. 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 Schrödinger’s equation. One may argue, effectively, that its argument, (**p**∙**x** – E∙t)/ħ, is just a number and, therefore, that the real and imaginary part of ψ is also just some number.

To this, we may object that ħ may be looked as a mathematical scaling constant only.** If **we do that,

**the argument of ψ will, effectively, be expressed in**

*then**action*units, i.e. in N·m·s. It then does make sense to also associate a physical dimension with the real and imaginary part of ψ. What could it be?

We may have a closer look at Maxwell’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*)∙**e****_{x}**×

**E**, with

**e****the unit vector pointing in the**

_{x}*x*-direction (i.e. the direction of propagation of the wave). Hence, we may associate the (1/

*c*)∙

**e****×**

_{x}*operator*, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by

*i*also amounts to a rotation by 90° degrees. Hence, we may boldly write:

**B**= (1/

*c*)∙

**e****×**

_{x}**E**= (1/

*c*)∙

*i*∙

**E**. This allows us to also geometrically interpret Schrödinger’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 Newton’s and Coulomb’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 N·s^{2}/m. Hence, our N/kg dimension becomes:

N/kg = N/(N·s^{2}/m)= m/s^{2}

What is this: m/s^{2}? Is *that *the dimension of the *a*·*cos*θ term in the *a*·*e*^{−iθ }= *a*·*cos*θ − *i*·*a*·*sin*θ wavefunction?

My answer is: **why not?** Think of it: m/s^{2} is the physical dimension of *acceleration*: the increase or decrease in velocity (m/s) per second. It ensures the wavefunction for *any *particle – matter-particles or particles with zero rest mass (photons) – 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 Schrödinger’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 **∇**∙ operator is the divergence and, therefore, gives us the magnitude of a (vector) field’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:

**E**is measured in*newton per coulomb*, so [**E**∙**E**] = [E^{2}] = N^{2}/C^{2}.**B**is measured in (N/C)/(m/s), so we get [**B**∙**B**] = [B^{2}] = (N^{2}/C^{2})·(s^{2}/m^{2}). However, the dimension of our*c*^{2}factor is (m^{2}/s^{2}) and so we’re also left with N^{2}/C^{2}.- The
*ϵ*_{0}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: [ε_{0}] = C^{2}/(N·m^{2}) and, therefore, if we multiply that with N^{2}/C^{2}, we find that*u*is expressed in J/m^{3}.[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 *ϵ*_{0} for an equivalent constant. We may to give it a try. If the energy densities can be calculated – which are also mass densities, obviously – 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*)∙*i*∙**E** or for −(1/*c*)∙*i*∙**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 *a*·*cos*θ and *a*·*sin*θ, 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 !

|ψ|^{2 } = |*a·e*^{−i∙E·t/ħ}|^{2 }= *a*^{2 }= *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:

“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.” (Feynman, *Lectures*, III-4-1)

The *physical* interpretation of the wavefunction, as presented here, may provide some better understanding of ‘the fundamental principle involved’:* 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 – or the *signal *– 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 – *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 Schrödinger’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:

ψ = *a·e*^{−i[E·t − p∙x]/ħ} = *a·e*^{−i[E·t − p∙x]/ħ} = *a·cos*(**p**∙**x**/ħ − E∙t/ħ)* + i·a·sin*(**p**∙**x**/ħ − E∙t/ħ)

The minus sign in the argument of our sine and cosine function defines the direction of travel: an F(x−*v∙*t) wavefunction will always describe some wave that is traveling in the *positive *x-direction (with *c *the wave velocity), while an F(x+*v∙*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 *counter*clockwise. 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:

ψ = *a·e*^{−i∙E·t/ħ} = *a·cos*(−E∙t/ħ)* + i·a·sin*(−E_{0}∙t/ħ)* = a·cos*(E_{0}∙t/ħ) −* i·a·sin*(E_{0}∙t/ħ)

E_{0} is, of course, the *rest *mass of our particle and, now that we are here, we should probably wonder *whose *time *t *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 t_{0}. It does not matter. You can see what we expect to see: E_{0}/ħ pops up as the *natural *frequency of our matter-particle: (E_{0}/ħ)∙t = ω∙t. Remembering the ω = 2π·*f* = 2π/T and T = 1/*f *formulas, we can associate a period and a frequency with this wave, using the ω = 2π·*f* = 2π/T. Noting that ħ = h/2π, we find the following:

T = 2π·(ħ/E_{0}) = h/E_{0} ⇔ *f *= E_{0}/h = m_{0}*c*^{2}/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 (*v*_{g}) should be zero here, because we assume our particle does not move. In contrast, the phase velocity is given by *v*_{p} = λ·*f *= (2π/k)·(ω/2π) = ω/k. In fact, we’ve got something funny here: the wavenumber k* = *p/ħ 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:

*v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = E/(m·*v*_{g}) = (m·*c*^{2})/(m·*v*_{g}) = *c*^{2}/*v*_{g}

This is interesting: it establishes a reciprocal relation between the phase and the group velocity, with *c *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:

*v*_{p}/*c *= β_{p} = *c*/*v*_{p} = 1/β_{g} = 1/(*c*/*v*_{p})

**Figure 6**: Reciprocal relation between phase and group velocity

We can also write the mentioned relationship as *v*_{p}·*v*_{g} = *c*^{2}, which reminds us of the relationship between the electric and magnetic constant (1/ε_{0})·(1/μ_{0}) = *c*^{2}. This is interesting in light of the fact we can re-write this as (*c*·ε_{0})·(*c*·μ_{0}) = 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/ħ. 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 *c*^{2}! Such interpretation is also consistent with the Uncertainty Principle: if Δx·Δp ≥ ħ, then *neither* Δx *nor* Δp 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 = m·*c* = m·*c*^{2}/*c *= E/*c*. Using the relationship above, we get:

*v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = *c* ⇒ *v*_{g} = *c*^{2}/*v*_{p} = *c*^{2}/*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 ψ = *a·e*^{−i[E·t − p∙x]/ħ} or, for a particle at rest, the ψ = *a·e*^{−i∙E·t/ħ} function. We must build a wave *packet* for that: a sum of wavefunctions, each with their own amplitude *a _{i}*, and their own ω

*= −E*

_{i}*/ħ. Indeed, in section II, we showed that each of these wavefunctions will*

_{i}*contribute*some energy to the total energy of the wave packet and that, to calculate the contribution of each wave to the total, both

*a*as well as E

_{i}*matter. This may or may not resolve the apparent paradox. Let us look at the group velocity.*

_{i}To calculate a meaningful group velocity, we must assume the *v*_{g} = ∂ω* _{i}*/∂k

*= ∂(E*

_{i}*/ħ)/∂(p*

_{i}*/ħ) = ∂(E*

_{i}*)/∂(p*

_{i}*) exists. So we must have some*

_{i}*dispersion relation*. How do we calculate it? We need to calculate ω

*as a function of k*

_{i}

_{i}*here, or E*

_{ }*as a function of p*

_{i}*. 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 Schrödinger’s equation as we did, i.e. by distinguishing the real and imaginary parts of the ∂ψ/∂t =*

_{i}*i*·[ħ/(2m)]·∇

^{2}ψ wave equation and, hence, re-write it as the following

*pair*of two equations:

*Re*(∂ψ/∂t) = −[ħ/(2m_{eff})]·*Im*(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·cos(kx − ωt)*Im*(∂ψ/∂t) = [ħ/(2m_{eff})]·*Re*(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·sin(kx − ωt)

Both equations imply the following dispersion relation:

ω = ħ·k^{2}/(2m_{eff})

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

*, but… What about m*

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

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

_{i}*equivalent*mass of E

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

_{i}*= E*

_{i}*/*

_{i}*c*

^{2}. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too. Here, I should refer back to Section II: E

*varies around some*

_{i}*average*energy E and, therefore, the Uncertainty Principle kicks in.

# VII. Explaining spin

The elementary wavefunction *vector* – i.e. the vector sum of the real and imaginary component – 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 – defined as the vector sum of the electric and magnetic field vectors – 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 ψ =

*a·e*

^{−i∙E·t/ħ}as some

*real*vector – as a two-dimensional oscillation of mass, to be precise –

*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** = ** r**×

**p**or, perhaps easier for our purposes here as the product of an

*angular*velocity (

**ω**) and rotational inertia (I), aka as the

*moment of inertia*or the

*angular mass*. We write:

**L** = I·**ω**

Note we can write **L** and **ω** in **boldface** here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (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 = 2π·(ħ/E_{0}). Hence, the angular velocity must be equal to:

ω = 2π/[2π·(ħ/E_{0})] = E_{0}/ħ

We also know the distance *r*, so that is the magnitude of *r** *in the **L** = * r*×

**p**vector cross-product: it is just

*a*, so that is the

*magnitude*of ψ =

*a·e*

^{−i∙E·t/ħ}. Now, the momentum (

**p**) is the product of a

*linear*velocity (

*) – in this case, the*

**v***tangential*velocity – and some mass (m):

**p**= m·

*. If we switch to*

**v***scalar*instead of vector quantities, then the (tangential) velocity is given by

*v*=

*r*·ω. So now we only need to think about what we should use for m or, if we want to work with the

*angular*velocity (ω), 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 – and, therefore, the mass – is distributed uniformly. In that case, we may use the formula for the angular mass of a solid cylinder: I = m·

*r*

^{2}/2. If we keep the analysis non-relativistic, then m = m

_{0}. Of course, the energy-mass equivalence tells us that m

_{0}= E

_{0}/

*c*

^{2}. Hence, this is what we get:

L = I·ω = (m_{0}·*r*^{2}/2)·(E_{0}/ħ) = (1/2)·*a*^{2}·(E_{0}/*c*^{2})·(E_{0}/ħ) = *a*^{2}·E_{0}^{2}/(2·ħ·*c*^{2})

Does it make sense? Maybe. Maybe not. Let us do a dimensional analysis: that won’t check our logic, but it makes sure we made no mistakes when mapping mathematical and physical spaces. We have m^{2}·J^{2} = m^{2}·N^{2}·m^{2} in the numerator and N·m·s·m^{2}/s^{2} in the denominator. Hence, the dimensions work out: we get N·m·s 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 = m*c*^{2} equation allows us to re-write it as:

L = *a*^{2}·E_{0}^{2}/(2·ħ·*c*^{2})

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 N·m·s units, and which can only take on one of two possible values: *J* = +ħ/2 and −ħ/2? It looks like a long shot, right? How do we go from (1/2)·*a*^{2}·m_{0}^{2}/ħ to ± (1/2)∙ħ? Let us do a numerical example. The energy of an electron is typically 0.510 MeV » 8.1871×10^{−14} N∙m, and *a*… 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.×10^{−10} N∙m. We get L = *a*^{2}·E_{0}^{2}/(2·ħ·*c*^{2}) = 9.9×10^{−31} N∙m∙s. Now that is about 1.88×10^{4} *times *ħ/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)×10^{−34} *joule *in energy. So our electron should pack about 1.24×10^{−20} oscillations. The angular momentum (L) we get when using the Bohr radius for *a* and the value of 6.626×10^{−34} *joule *for E_{0} and the Bohr radius is equal to 6.49×10^{−59} N∙m∙s. So that is the angular momentum per oscillation. When we multiply this with the number of oscillations (1.24×10^{−20}), we get about 8.01×10^{−51} N∙m∙s, so that is a totally different number.

The classical electron radius is about 2.818×10^{−15} m. We get an L that is equal to about 2.81×10^{−39} N∙m∙s, so now it is a tiny *fraction *of ħ/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 2.42631×10^{−12} m.

This gives us an L of 2.08×10^{−33} N∙m∙s, which is only 20 times ħ. 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 = *a*^{2}·E_{0}^{2}/(2·ħ·*c*^{2}) = ħ/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 3.8616×10^{−33} m), we get what we should find:

This is a rather spectacular result, and one that would – a priori – 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 ψ = *a·e*^{−i[E·t − p∙x]/ħ} or, for a particle at rest, the ψ = *a·e*^{−i∙E·t/ħ} function. We must build a wave *packet* for that: a sum of wavefunctions, each with their own amplitude *a _{i}*, and their own ω

*= −E*

_{i}*/ħ. Each of these wavefunctions will*

_{i}*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* = +ħ/2 or −ħ/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:

ψ(θ* _{i}*)

*= a*·(

_{i}*cos*θ

*+*

_{i}*i·sin*θ

*)*

_{i}In contrast, an elementary left-handed wave would be written as:

ψ(θ* _{i}*)

*= a*·(

_{i}*cos*θ

*−*

_{i}*i·sin*θ

*)*

_{i}How does that work out with the E_{0}·t 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 −1, −2, −3, etcetera is just the same. If we count like 1, 2, 3, etcetera, then we write our wavefunction like:

ψ = *a·cos*(E_{0}∙t/ħ)* − i·a·sin*(E_{0}∙t/ħ)

If we count time like −1, −2, −3, etcetera then we write it as:

ψ = *a·cos*(*−*E_{0}∙t/ħ)* − i·a·sin*(*−*E_{0}∙t/ħ)= *a·cos*(E_{0}∙t/ħ)* + i·a·sin*(E_{0}∙t/ħ)

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 (+ħ/2 or −ħ/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 Feynman’s *Lecture *on it (Feynman, III-4), which is confusing and – I would dare to say – 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:

(*v _{phase}*·

*c*)·(

*v*·

_{group}*c*) = 1 ⇔

*v*·

_{p}*v*=

_{g}*c*

^{2}

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 – 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 – around any axis – 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 ‘hook’ 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 Schrödinger’s equation is related to the concept of the *effective *mass (m_{eff}). It is easy to make the wrong calculations. For example, when playing with the famous *de Broglie *relations – aka as the matter-wave equations – one may be tempted to *derive* the following energy concept:

- E = h·
*f*and p = h/λ. Therefore,*f*= E/h and λ = p/h. *v*=*f·*λ = (E/h)∙(p/h) = E/p- p = m·
*v*. Therefore, E =*v*·p = m·*v*^{2}

E = m·*v*^{2}? This *resembles *the E = m*c*^{2} equation and, therefore, one may be enthused by the discovery, especially because the m·*v*^{2} 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 − PE *= m·v*^{2}.[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 – as used in Schrödinger’s equation – is a rather enigmatic concept. To make sure we are making the right analysis here, I should start by noting you will usually see Schrödinger’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:

∂ψ(**x**, t)/∂t = *i*·(1/2)·(ħ/m_{eff})·∇^{2}ψ(**x**, t)

We just moved the *i*·ħ coefficient to the other side, noting that 1/*i *= –*i*. Now, in one-dimensional space, and assuming ψ is just the elementary wavefunction (so we substitute *a·e*^{−i∙[E·t − p∙x]/ħ} for ψ), this implies the following:

−*a*·*i*·(E/ħ)·*e*^{−}*i∙*^{[E·t − p∙x]/ħ} = −*i*·(ħ/2m_{eff})·*a*·(p^{2}/ħ^{2})·* e*^{−i∙[E·t − p∙x]/ħ }

⇔ E = p^{2}/(2m_{eff}) ⇔ m_{eff} = m∙(*v/c*)^{2}/2 = m∙β^{2}/2

It is an ugly formula: it *resembles *the kinetic energy formula (K.E. = m∙*v*^{2}/2) but it is, in fact, something completely different. The β^{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 m_{eff} as *two *times the old m_{eff} (hence, m_{eff}^{NEW} = 2∙m_{eff}^{OLD}), as a result of which the formula will look somewhat better:

m_{eff} = m∙(*v/c*)^{2} = m∙β^{2}

We know β varies between 0 and 1 and, therefore, m_{eff} will vary between 0 and m. Feynman drops the subscript, and just writes m_{eff} 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 – which is the equivalent of a *source *term in a diffusion equation – and that may explain why the above-mentioned m_{eff} = m∙(*v/c*)^{2} = m∙β^{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: Feynman’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 ψ = *a·e*^{−i∙θ} = *a·e*^{−i[E·t − p∙x]/ħ} = *a·(cosθ **–** i·a·sinθ).* We must build a wave *packet* for that: a sum of wavefunctions, each with its own amplitude *a*_{k} and its own argument θ_{k} = (E_{k}∙t – **p**_{k}∙**x**)/ħ. 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/s^{2}), thereby facilitating a direct interpretation in terms of Newton’s force law.

[3] In physics, a two-*spring *metaphor is more common. Hence, the pistons in the author’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 Schrödinger’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: “God does not play dice.”The actual quote comes out of one of Einstein’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: ω is an *angular *velocity, while *v *is a *linear *velocity: ω 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 = m·*a*^{2}∙ω^{2}/2. The additional factor (*a*) is the (maximum) amplitude of the oscillator.

[9] We also have a 1/2 factor in the E = m*v*^{2}/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 (γ), we can write the relativistically correct formula for the kinetic energy as K.E. = E − E_{0} = m_{v}*c*^{2} − m_{0}*c*^{2} = m_{0}γ*c*^{2} − m_{0}*c*^{2} = m_{0}*c*^{2}(γ − 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 = m*v*^{2}. 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 ω. The *kinetic *energy of a rotating object is then given by K.E. = (1/2)·I·ω^{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 ω^{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 *ω^{2}= *C*^{–}^{1}/*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 γ*m* and as R = γ*L* 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 10^{8}, which means the wave train will last about 10^{–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 3.2×10^{–8 }seconds (this is the so-called decay time τ). Now, because the frequency of sodium light is some 500 THz (500×10^{12 }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 – riding along the wave as it is being emitted, so to speak – 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)·m·ω^{2}·*a*^{2}·sin^{2}(ω·t + Δ) and the P.E. = U = k·x^{2}/2 = (1/2)· m·ω^{2}·*a*^{2}·cos^{2}(ω·t + Δ) formulas for the linear oscillator.

[17] Feynman further formalizes this in his *Lecture on Superconductivity *(Feynman, III-21-2), in which he refers to Schrödinger’s equation as the “equation for continuity of probabilities”. The analysis is centered on the *local *conservation of energy, which confirms the interpretation of Schrödinger’s equation as an energy diffusion equation.

[18] The m_{eff} is the *effective* mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write m_{eff} = m. Appendix 2 provides some additional notes on the concept. As for the equations, they are easily derived from noting that two complex numbers a + *i*∙b and c + *i*∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = *i*∙(ħ/m_{eff})∙∇^{2}ψ equation amounts to writing something like this: a + *i*∙b = *i*∙(c + *i*∙d). Now, remembering that *i*^{2} = −1, you can easily figure out that *i*∙(c + *i*∙d) = *i*∙c + *i*^{2}∙d = − d + *i*∙c.

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

[20] Of course, multiplication with* i *amounts to a *counter*clockwise 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 Schrödinger’s equation in three-dimensional space: if we look at a clock from the back, then its hand will be moving *counter*clockwise. When writing **B** = (1/*c*)∙*i*∙**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*)∙*i*∙**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 C^{2}/(N·m^{2}) with N^{2}/C^{2}, we get N/m^{2}, 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: sinθ = cos(θ−π /2).

[24] I must thank a physics blogger for re-writing the 1/(ε_{0}·μ_{0}) = *c*^{2} 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 90° 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).

# Re-visiting the matter wave (II)

My previous post was, once again, littered with formulas – even if I had not intended it to be that way: I want to convey some kind of *understanding *of what an electron – or any particle at the atomic scale – actually *is *– with the minimum number of formulas necessary.

We know particles display wave behavior: when an electron beam encounters an obstacle or a slit that is somewhat comparable in size to its wavelength, we’ll observe diffraction, or interference. [I have to insert a quick note on terminology here: the terms diffraction and interference are often used interchangeably, but there is a tendency to use interference when we have more than one wave source and diffraction when there is only one wave source. However, I’ll immediately add that distinction is somewhat artificial. Do we have one or two wave sources in a double-slit experiment? There is one beam but the two slits break it up in two and, hence, we would call it interference. If it’s only one slit, there is also an interference pattern, but the phenomenon will be referred to as diffraction.]

We also know that the wavelength we are talking about it here is *not *the wavelength of some electromagnetic wave, like light. It’s the wavelength of a *de Broglie *wave, i.e. a** matter wave**: such wave is represented by an (oscillating) complex number – so we need to keep track of a real and an imaginary part – representing a so-called probability *amplitude* Ψ(x, t) whose modulus squared (│Ψ(x, t)│^{2}) is the *probability* of actually detecting the electron at point x at time t. [The purists will say that complex numbers can’t oscillate – but I am sure you get the idea.]

You should read the phrase above twice: **we cannot know where the electron actually is. We can only calculate probabilities **(and, of course, compare them with the probabilities we get from experiments). Hence, when the wave function tells us the probability is greatest at point x at time t, then we may be lucky when we actually probe point x at time t and find it there, but it may also

*not*be there. In fact, the probability of finding it

*exactly*at some point x at some

*definite*time t is zero. That’s just a characteristic of such probability density functions: we need to probe some

*region*Δx in some

*time interval*Δt.

If you think that is not very satisfactory, there’s actually a very common-sense explanation that has nothing to do with quantum mechanics: our scientific instruments do not allow us to go beyond a certain scale anyway. Indeed, the resolution of the best electron microscopes, for example, is some 50 *pico*meter (1 pm = 1×10^{–12} m): that’s small (and resolutions get higher by the year), but so it implies that we are not looking at *points* – as defined in math that is: so that’s something with *zero *dimension – but at *pixels* of size Δx = 50×10^{–12} m.

The same goes for time. Time is measured by atomic clocks nowadays but even these clocks do ‘tick’, and these ‘ticks’ are discrete. Atomic clocks take advantage of the property of atoms to *resonate* at *extremely* consistent frequencies. I’ll say something more about resonance soon – because it’s very relevant for what I am writing about in this post – but, for the moment, just note that, for example, Caesium-133 (which was used to build the first atomic clock) oscillates at 9,192,631,770 *cycles per second*. In fact, the *International Bureau of Standards and Weights *re-defined the (time) second in 1967 to correspond to “the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the Caesium-133 atom at rest at a temperature of 0 K.”

Don’t worry about it: the point to note is that when it comes to measuring time, we also have an uncertainty. Now, when using this Caesium-133 atomic clock, this uncertainty would be in the range of ±9.2×10^{–9} seconds (so that’s *nano*seconds: 1 ns = 1×10^{–9} s), because that’s the rate at which this clock ‘ticks’. However, there are other (much more plausible) ways of measuring time: some of the *unstable* baryons* *have lifetimes in the range of a few *pico*seconds only (1 ps = 1×10^{–12} s) and the *really *unstable ones – known as baryon *resonances* – have lifetimes in the 1×10^{–22} to 1×10^{–24} s range. This we can only measure because they leave some trace after these particle collisions in particle accelerators and, because we have some idea about their speed, we can calculate their lifetime from the (limited) distance they travel before disintegrating. The thing to remember is that for time also, we have to make do with time pixels instead of time points, so there is a Δt as well. [In case you wonder what baryons are: they are particles consisting of three quarks, and the proton and the neutron are the most prominent (and most stable) representatives of this family of particles.]

So what’s the size of an electron? Well… It depends. We need to distinguish two very different things: (1) the size of the area where we are *likely to find *the electron, and (2) the size of the electron itself. Let’s start with the latter, because that’s the easiest question to answer: there is a so-called *classical* electron radius *r*_{e}, which is also known as *the Thompson scattering length*, which has been calculated as:

As for the constants in this formula, you know these by now: the speed of light *c*, the electron charge *e*, its mass *m*_{e}, and the permittivity of free space *ε*_{e}. For whatever it’s worth (*because you should note that, in quantum mechanics, electrons do not have a size: they are treated as point-like particles, so they have a point charge and zero dimension*), that’s small. It’s in the *femtometer* range (1 fm = 1×10^{–15} m). You may or may not remember that the size of a proton is in the femtometer range as well – 1.7 fm to be precise – and we had a femtometer size estimate for quarks as well: 0.7 m. So we have the rather remarkable result that the much heavier proton (its rest mass is 938 MeV/*c*^{2 }sas opposed to only 0.511 MeV MeV/*c*^{2}, so the proton is 1835 times *heavier*) is 1.65 times *smaller *than the electron. That’s something to be explored later: for the moment, we’ll just assume the electron wiggles around a bit more – exactly *because *it’s lighter*. *Here you just have to note that this ‘classical’ electron radius does measure something: it’s something ‘hard’ and ‘real’ because it scatters, absorbs or deflects photons (and/or other particles). In one of my previous posts, I explained how particle accelerators probe things at the femtometer scale, so I’ll refer you to that post (*End of the Road to Reality?*) and move on to the next question.

The question concerning *the area where we are likely to detect the electron* is more interesting in light of the topic of this post (the *nature* of these matter waves). It is given by that wave function and, from my previous post, you’ll remember that we’re talking the *nano*meter scale here (1 nm = 1×10^{–9} m), so that’s a *million *times** larger** than the femtometer scale. Indeed, we’ve calculated a *de Broglie *wavelength of 0.33 nanometer for relatively slow-moving electrons (electrons in orbit), and the slits used in single- or double-slit experiments with electrons are also *nanotechnology*. In fact, now that we are here, it’s probably good to look at those experiments *in detail*.

The illustration below relates the *actual experimental set-up* of a double-slit experiment performed in 2012 to Feynman’s 1965 *thought experiment. *Indeed, in 1965, the nanotechnology you need for this kind of experiment was not yet available, although the phenomenon of electron diffraction had been confirmed experimentally already in 1925 in the famous Davisson-Germer experiment. [It’s famous not only because electron diffraction was a weird thing to contemplate at the time but also because it confirmed the *de Broglie *hypothesis only two years after *Louis de Broglie *had advanced it!]. But so here is the experiment which Feynman thought would never be possible because of technology constraints:

The insert in the upper-left corner shows the two slits: they are each 50 nanometer wide (50×10^{–9} m) and 4 micrometer tall (4×10^{–6} m). [The thing in the middle of the slits is just a little support. Please do take a few seconds to contemplate the technology behind this feat: **50 nm is 50 millionths of a millimeter**. Try to imagine dividing one millimeter in ten, and then one of these tenths in ten again, and again, and once again, again, and again. You just can’t

*imagine*that, because our mind is used to addition/subtraction and – to some extent – with multiplication/division: our mind can’t deal with with exponentiation really – because it’s not a everyday phenomenon.] The second inset (in the upper-right corner) shows the mask that can be moved to close one or both slits partially or completely.

Now, 50 nanometer is 150 times larger than the 0.33 nanometer range we got for ‘our’ electron, but it’s small enough to show diffraction and/or interference. [In fact, in this experiment (done by *Bach*, *Pope*, *Liou* and *Batelaan* from the University of Nebraska-Lincoln less than two years ago indeed), the beam consisted of electrons with an (average) energy of 600 eV and a *de Broglie *wavelength of 50 *pico*meter. So that’s like the electrons used in electron microscopes. 50 pm is 6.6 times smaller than the 0.33 nm wavelength we calculated for our low-energy (70 eV) electron – but then the energy and the fact these electrons are guided in electromagnetic fields explain the difference. Let’s go to the results.

The illustration below shows the *predicted *pattern next to the *observed *pattern for the two scenarios:

- We
*first*close slit 2, let a lot of electrons go through it, and so we get a pattern described by the probability*density*function P_{1}= │Φ_{1}│^{2}. Here we see no interference but a typical*diffraction*pattern: the*intensity*follows a more or less normal (i.e. Gaussian) distribution. We*then*close slit 1 (and open slit 2 again), again let a lot of electrons through, and get a pattern described by the probability*density*function P_{2}= │Φ_{2}│^{2}._{1}and P_{2}. - We then open
*both*slits, let a whole electrons through, and get according to the pattern described by probability density function P_{12}= │Φ_{1}+Φ_{2}│^{2}, which we get*not*from adding the probabilities P_{1}and P_{2 }(hence, P_{12}≠ P_{1}+ P_{2}) – as one would expect if electrons would behave like particles – but from adding the probability*amplitudes*. We have*interference*, rather than diffraction.

But so *what* exactly is interfering? Well… The electrons. But that can’t be, can it?

The electrons are obviously *particles*, as evidenced from the impact they make – ** one by one** – as they hit the screen

**as shown below**. [If you want to know what screen, let me quote the researchers: “The resulting patterns were magnified by an electrostatic quadrupole lens and imaged on a two-dimensional microchannel plate and phosphorus screen, then recorded with a charge-coupled device camera. […] To study the build-up of the diffraction pattern, each electron was localized using a “blob” detection scheme: each detection was replaced by a blob, whose size represents the error in the localization of the detection scheme. The blobs were compiled together to form the electron diffraction patterns.” So there you go.]

Look carefully at how this interference pattern becomes ‘reality’ as the electrons hit the screen ** one by one**. And then say it:

*WAW !*Indeed, as predicted by Feynman (and any other physics professor at the time), ** even if the electrons go through the slits one by one, they will interfere** – with themselves so to speak. [In case you wonder if these electrons really went through one by one, let me quote the researchers once again: “The electron source’s intensity was reduced so that the electron detection rate in the pattern was about 1 Hz. At this rate and kinetic energy, the average distance between consecutive electrons was 2.3 × 106 meters. This ensures that only one electron is present in the 1 meter long system at any one time, thus eliminating electron-electron interactions.” You don’t need to be a scientist or engineer to understand that, isn’t it?]

While this is very spooky, I have not seen any better way to describe the *reality *of the *de Broglie *wave: **the particle is not some point-like thing but a matter wave, as evidenced from the fact that it does interfere with itself** when forced to move through two slits – or through one slit, as evidenced by the diffraction patterns built up in this experiment when closing one of the two slits: the electrons went through one by one as well!

But so how does it relate to the characteristics of that wave packet which I described in my previous post? Let me sum up the salient conclusions from that discussion:

- The wavelength λ of a wave
*packet*is calculated directly from the momentum by using*de Broglie*‘s second relation: λ = h/p. In this case, the wavelength of the electrons averaged 50*pico*meter. That’s relatively small as compared to the width of the slit (50 nm) –**a thousand times smaller actually!**– but, as evidenced by the experiment, it’s small enough to show the ‘reality’ of the*de Broglie*wave. - From a math point (but, of course, Nature does not care about our math), we can decompose the wave packet in a finite or infinite number of component waves. Such decomposition is referred to, in the first case (finite number of composite waves or
*discrete*calculus) as a*Fourier analysis*, or, in the second case, as a*Fourier transform*. A Fourier transform maps our (continuous) wave function, Ψ(x), to a (continuous) wave function in the momentum space, which we noted as φ(p). [In fact, we noted it as Φ(p) but I don’t want to create confusion with the Φ symbol used in the experiment, which is actually the wave function in space, so Ψ(x) is Φ(x) in the experiment – if you know what I mean.] The point to note is that uncertainty about momentum is related to uncertainty about position. In this case, we’ll have pretty standard electrons (so not much variation in momentum), and so the location of the wave packet in space should be fairly precise as well. - The
*group*velocity of the wave*packet*(*v*_{g}) – i.e. the*envelope*in which our Ψ wave oscillates – equals the speed of our electron (*v*), but the*phase*velocity (i.e. the speed of our Ψ wave itself) is superluminal: we showed it’s equal to (*v*_{p}) = E/p =*c*^{2}/*v = c*/β, with β =*v*/c, so that’s the ratio of the speed of our electron and the speed of light. Hence, the phase velocity will always be superluminal but will approach*c*as the speed of our particle approaches*c*. For slow-moving particles, we get astonishing values for the phase velocity, like more than a hundred times the speed of light for the electron we looked at in our previous post. That’s weird but it does not contradict relativity: if it helps, one can think of the wave packet as a modulation of an incredibly fast-moving ‘carrier wave’.

Is any of this relevant? Does it help you to *imagine *what the electron actually *is*? Or what that matter wave actually *is*? Probably not. You will still wonder: **How does it look like? What is it in reality?**

That’s hard to say. If the experiment above does not convey any ‘reality’ according to you, then perhaps the illustration below will help. It’s one I have used in another post too (*An Easy Piece: Introducing Quantum Mechanics and the Wave Function*). I took it from Wikipedia, and it represents “the (likely) space in which a single electron on the 5d atomic orbital of an atom would be found.” The solid body shows the places where the electron’s probability density (so that’s the squared modulus of the probability *amplitude*) is above a certain value – so it’s basically the area where the likelihood of finding the electron is higher than elsewhere. The hue on the colored surface shows the complex phase of the wave function.

So… Does this help?

You will wonder why the shape is so complicated (but it’s beautiful, isn’t it?) but that has to do with quantum-mechanical calculations involving quantum-mechanical quantities such as spin and other machinery which I don’t master (yet). I think there’s always a bit of a gap between ‘first principles’ in physics and the ‘model’ of a real-life situation (like a real-life electron in this case), but it’s surely the case in quantum mechanics! That being said, when looking at the illustration above, you should be aware of the fact that you are actually looking at **a 3D representation of the wave function of an electron in orbit. **

Indeed, wave functions of electrons in orbit are somewhat less random than – let’s say – the wave function of one of those baryon resonances I mentioned above. As mentioned in my *Not So Easy Piece*, in which I introduced the Schrödinger equation (i.e. one of my previous posts), they are solutions of a second-order partial differential equation – known as the **Schrödinger wave equation** indeed – which basically incorporates one key condition: these solutions – which are (atomic or molecular) ‘orbitals’ indeed – have to correspond to so-called stationary states or *standing waves*. Now what’s the ‘reality’ of that?

The illustration below comes from Wikipedia once again (Wikipedia is an incredible resource for autodidacts like me indeed) and so you can check the article (on stationary states) for more details if needed. Let me just summarize the basics:

- A stationary state is called
*stationary*because the system remains in the same ‘state’*independent of time*. That does*not*mean the wave function is stationary. On the contrary, the wave function changes as function of both time and space – Ψ = Ψ(x, t) remember? – but it represents a so-called.*standing wave* - Each of these
*possible*states corresponds to an energy state, which is given through the*de Broglie*relation: E = h*f*. So the energy of the state is proportional to the oscillation frequency of the (standing) wave, and Planck’s constant is the factor of proportionality. From a formal point of view, that’s actually the one and only condition we impose on the ‘system’, and so it*immediately*yields the so-called*time-independent Schrödinger equation*, which I briefly explained in the above-mentioned*Not So Easy Piece*(but I will not write it down here because it would only confuse you even more). Just look at these so-called*harmonic*oscillators below:

A and B represent a harmonic oscillator in *classical mechanics*: a ball with some *mass* m (mass is a measure for inertia, remember?) on a spring oscillating back and forth. In case you’d wonder what the difference is between the two: both the amplitude as well as the frequency of the movement are different. 🙂 A spring and a ball?

It represents a simple system. A harmonic oscillation is basically a *resonance *phenomenon: springs, electric circuits,… * anything that swings, moves or oscillates *(including large-scale things such as bridges and what have you – in his 1965

*Lectures*(Vol. I-23),

*Feynman even discusses resonance phenomena in the atmosphere in his*

*Lectures*)

**, also referred to as the resonance frequency, at which it oscillates**

*has some natural frequency*ω_{0}*naturally*indeed: that means it requires (relatively) little energy to keep it going. How much energy it takes

*exactly*to keep them going depends on the frictional forces involved: because the springs in A and B keep going, there’s obviously no friction involved at all. [In physics, we say there is no

*damping*.] However, both springs do have a different

*k*(that’s the key characteristic of a spring in Hooke’s Law, which describes how springs work), and the mass m of the ball might be different as well. Now, one can show that the

*period*of this ‘natural’ movement will be equal to t

_{0}= 2π/ω

_{0 }= 2π(m/k)

^{1/2 }or that ω

_{0 }= (m/k)

^{–1/2}. So we’ve got a A and a B situation which differ in k and m. Let’s go to the so-called quantum oscillator, illustrations C to H.

C to H in the illustration are six possible solutions to the Schrödinger Equation for this situation. The horizontal axis is position (and so time is the variable) – but we could switch the two independent variables easily: as I said a number of times already, time and space are interchangeable in the argument representing the phase (θ) of a wave provided we use the right units (e.g. light-seconds for distance and seconds for time): θ = ωt – kx. Apart from the nice animation, the other great thing about these illustrations – and the main difference with resonance frequencies in the classical world – is that they show both the real part (**blue**) as well as the imaginary part (**red**) of the wave function as a function of space (fixed in the x axis) and time (the animation).

Is this ‘real’ enough? If it isn’t, I know of no way to make it any more ‘real’. Indeed, that’s key to understanding the *nature *of matter waves: we have to come to terms with the idea that these strange fluctuating mathematical quantities actually represent something. What? Well… The spooky thing that leads to the above-mentioned experimental results: electron diffraction and interference.

Let’s explore this quantum oscillator some more. Another key difference between natural frequencies in atomic physics (so the atomic scale) and resonance phenomena in ‘the big world’ is that there is more than one possibility: each of the six *possible *states above corresponds to a solution and an energy state indeed, which is given through the *de Broglie *relation: E = h*f*. However, in order to be fully complete, I have to mention that, while G and H are also solutions to the wave equation, they are actually not *stationary *states. The illustration below – which I took from the same Wikipedia article on stationary states – shows why. For stationary states, all *observable properties* of the state (such as the probability that the particle is at location x) are *constant*. For non-stationary states, the probabilities *themselves* fluctuate as a function of time (and space of obviously), so the *observable properties* of the system are* not *constant. These solutions are solutions to the time-*dependent* Schrödinger equation and, hence, they are, obviously, *time-dependent *solutions.

We can find these time-dependent solutions by *superimposing* two stationary states, so we have a new wave function Ψ_{N} which is the sum of two others: Ψ_{N} = Ψ_{1} + Ψ_{2}. [If you include the normalization factor (as you should to make sure all probabilities add up to 1), it’s actually Ψ_{N} = (2^{–1/2})(Ψ_{1} + Ψ_{2}).] So G and H above still represent a state of a *quantum harmonic oscillator* (with a specific energy level proportional to h), but so they are *not* standing waves.

Let’s go back to our electron traveling in a more or less straight path. What’s the shape of the solution for that one? It could be anything. Well… Almost anything. As said, the only condition we can impose is that the *envelope* of the wave packet – its ‘general’ shape so to say – should not change. That because we should not have dispersion – as illustrated below. [Note that this illustration only represent the real or* *the imaginary part – not both – but you get the idea.]

That being said, if we *exclude* dispersion (because a *real-life* electron traveling in a straight line doesn’t just disappear – as do dispersive wave packets), then, *inside of that envelope*, the weirdest things are possible – in theory that is. Indeed, Nature does not care much about our Fourier transforms. So the example below, which shows a theoretical wave packet (again, the real or imaginary part only) based on some theoretical distribution of the wave numbers of the (infinite number) of component waves that make up the wave packet, may or may not represent our real-life electron. However, if our electron has any resemblance to real-life, then I would expect it to *not *be as well-behaved as the theoretical one that’s shown below.

The shape above is usually referred to as a Gaussian wave packet, because of the nice normal (Gaussian) probability density functions that are associated with it. But we can also imagine a ‘square’ wave packet: a somewhat weird shape but – in terms of the math involved – as consistent as the smooth Gaussian wave packet, in the sense that we can demonstrate that the wave packet is made up of an infinite number of waves with an angular frequency ω that is linearly related to their wave number *k*, so the dispersion relation is ω = a*k* + b. [Remember we need to impose that condition to ensure that our wave packet will not dissipate (or disperse or *disappear* – whatever term you prefer.] That’s shown below: a Fourier analysis of a square wave.

While we can construct many theoretical shapes of wave packets that respect the *‘no dispersion!’* condition, we cannot know which one will *actually *represent that electron we’re trying to visualize. Worse, if push comes to shove, we don’t know if these matter waves (so these wave packets) actually consist of component waves (or time-independent stationary states or whatever).

[…] OK. Let me finally admit it: while I am trying to explain you the ‘reality’ of these matter waves, we actually don’t *know* how *real *these matter waves actually are. We cannot ‘see’ or ‘touch’ them indeed. All that we know is that (i) assuming their existence, and (ii) assuming these matter waves are more or less well-behaved (e.g. that actual particles will be represented by a composite wave characterized by a *linear* dispersion relation between the angular frequencies and the wave numbers of its (theoretical) component waves) allows us to do all that arithmetic with these (complex-valued) probability* *amplitudes. *More importantly, all that arithmetic with these complex numbers actually yields (real-valued) probabilities that are consistent with the probabilities we obtain through repeated experiments.* So that’s what’s real and ‘not so real’ I’d say.

Indeed, the bottom-line is that **we do not know what goes on inside that envelope.** Worse, according to the commonly accepted Copenhagen interpretation of the Uncertainty Principle (and tons of experiments have been done to try to overthrow that interpretation – all to no avail),

*we never will*.