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. 🙂

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