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

Just to recap the essentials, I part ways with mainstream physicists in regard to the *interpretation *of the wavefunction. For mainstream physicists, the wavefunction is just some mathematical construct. Nothing *real*. Of course, I acknowledge mainstream physicists have very good reasons for that, but… Well… I believe that, if there is interference, or diffraction, then *something *must be interfering, or something must be diffracting. I won’t dwell on this because… Well… I have done that too many times already. My *hypothesis *is that the wavefunction is, in effect, a *rotating **field vector*, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).

Of course, it must be different, and it is. First, the (physical) dimension of the field vector of the matter-wave must be different. So what is it? Well… I am tempted to associate the real and imaginary component of the wavefunction with a force *per unit mass *(as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s^{2}), so that’s the dimension of a gravitational field.

Second, I also am tempted to think that this gravitational disturbance causes an electron (or any matter-particle) to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves do *not *involve any mass: they’re just an oscillating *field*. Nothing more. Nothing less. Why would I believe there must still be some pointlike particle involved? Well… As Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s *Lectures*, III-21-4) So… Well… That’s why.

The third difference is one that I thought of only recently: the *plane *of the oscillation can*not *be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus. I am more explicit on that in the mentioned post, so you may want to check there. 🙂

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

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

So I think the magnetic moment—or the angular momentum, really—comes from an oscillatory motion in the *x*– and *y*-directions. One is the *real *component (the cosine function) and the other is the imaginary component (the sine function), as illustrated below.

So the crucial difference with the animations above (which illustrate left- and a right-handed polarization respectively) is that we, somehow, need to imagine the circular motion is *not *in the *xz*-plane, but in the *yz*-plane. Now what happens if we change the reference frame?

Well… That depends on what you mean by changing the reference frame. Suppose we’re looking in the positive *y*-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. Now, I did a post on that earlier this year, which you may want to re-read. When we’re looking at the same thing from the other side (from the back side, so to speak), we will want to use our familiar reference frame. So we will want to keep the *z*-axis as it is (pointing upwards), and we will also want to define the *x*– and *y-*axis using the familiar right-hand rule for defining a coordinate frame. So our new *x*-axis and our new *y-*axis will the same as the old *x-* and *y-*axes but with the sign reversed. In short, we’ll have the following mini-transformation: (1) *z*‘ = *z*, (2) *x’* = −*x*, and (3) *y’* = −*y*.

So… Well… If we’re effectively looking at something *real *that was moving along the *y*-axis, then it will now still be moving along the *y’*-axis, but in the *negative *direction. Hence, our elementary wavefunction *e ^{i}*

^{θ}=

*cos*θ +

*i*·

*sin*θ will

*transform*into −

*cos*θ −

*i*·

*sin*θ = −

*cos*θ −

*i*·

*sin*θ =

*cos*θ −

*i*·

*sin*θ. It’s the same wavefunction. We just… Well… We just changed our reference frame. We didn’t change reality.

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

Now, if φ is equal to 180° (so that’s π in radians), then these *e*^{i}^{φ/2} and *e*^{−i}^{φ/2}/√2 factors are equal to *e*^{i}^{π/2} = *+i* and *e*^{−i}^{π/2} = −*i* respectively. Hence, our *e ^{i}*

^{θ}=

*cos*θ +

*i*·

*sin*θ becomes…

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

*very*different things here, right? The

*e*

^{i}^{θ}=

*cos*θ +

*i*·

*sin*θ is an

*elementary*wavefunction which, we presume, describes some real-life particle—we talked about an electron with its spin in the

*up*-direction—while these transformation matrices are to be applied to amplitudes describing… Well… Either an

*up*– or a

*down*-state, right?

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

^{θ}=

*cos*θ +

*i*·

*sin*θ wavefunction describes an

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

*e*

^{i}^{φ/2}=

*e*

^{i}^{π/2}=

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

*e*

^{i}^{φ/2}·

*e*

^{i}^{θ}=

*e*

^{i}^{π/2}·

*e*

^{i}^{θ}=

*+i*·

*e*

^{i}^{θ}=

*i*·

*cos*θ +

*i*

^{2}·

*sin*θ =

*sin*θ −

*i*·

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

*cos*θ −

*i*·

*sin*θ function we thought we’d find?

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

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

Hmm… Interesting. Let’s think about the difference between the *sin*θ − *i*·*cos*θ and *cos*θ − *i*·*sin*θ functions. First, note that they will give us the same probabilities: the square of the absolute value of both complex numbers is the same. [It’s equal to 1 because we didn’t bother to put a coefficient in front.] Secondly, we should note that the sine and cosine functions are essentially the same. They just differ by a phase factor: *cos*θ = *sin*(θ + π/2) and −*sin*θ = *cos*(θ + π/2). Let’s see what we can do with that. We can write the following, for example:

*sin*θ − *i*·*cos*θ = −*cos*(θ + π/2) − *i*·*sin*(θ + π/2) = −[*cos*(θ + π/2) + *i*·*sin*(θ + π/2)] = −*e ^{i}*

^{·(θ + π/2)}

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

^{·θ}and −

*e*

^{i}^{·(θ + π/2)}functions differ by a phase shift

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

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