Quantum Mechanics and rotations

Let me warn you up front that I cannot yet reconcile Feynman’s answers with McMahon’s recipe. Well, this is supposed to be about the doing of math, not just about math itself.

For Feynman, i’m in “Lectures on Physics”, volume III, Quantum Mechanics. For McMahon, i’m using “Quantum Mechanics Demystified.” References to Schiff are to his “Quantum Mechanics”.

We have looked at the following here: if I know that a spin-1 particle is in the Jz state |+>, i.e.

$\left(\begin{array}{l} 1 \\ 0 \\ 0\end{array}\right)$

and I wanted to know: what are possible values of Jx (the x-component of angular momentum)? and with what probabilities would they occur? What we looked at was the relationship between Jz and Jx in a given coordinate system

I want to consider a different question; I want to rotate the coordinate system.

Physically, imagine that we have a “Stern-Gerlach” apparatus. The y-axis is the direction of motion of (undeflected) particles; the z-axis is up; and the x-axis forms a right-handed coordinate system. Particles are moved up or down, or not at all vertically, depending on whether they are in the Jz states |+>, |->, or |0>.

Whereas the previous problem related Jz and Jx in a one coordinate system, we are about to investigate just Jz in two coordinate systems.

Now we take a second apparatus, also on the y-axis, but we rotate it about the y-axis CCW thru an angle $\alpha$. In Volume III of the Feynman Lectures on Physics, Feynman does exactly this. His drawing is figure 5-6 on p. 5-5, and he says (p. 5-15) that the coordinate transformation is

$\begin{array}{c} x'=x \cos (\alpha )-z \sin (\alpha ) \\ y'=y \\ z'=z \cos (\alpha )+x \sin (\alpha )\end{array}$

where the primed coordinates are fixed in the rotated apparatus. All of that is consistent: first, his drawing shows a CCW rotation about the y-axis thru a positive angle (so my customary y-axis rotation

$Ry(\alpha) = \left(\begin{array}{lll} \cos (\alpha ) & 0 & -\sin (\alpha ) \\ 0 & 1 & 0 \\ \sin (\alpha ) & 0 & \cos (\alpha )\end{array}\right)$

should be used as the attitude matrix). second, his equations say that my $Ry(\alpha)$ is the inverse transition matrix.

But, for a rotation, that’s the attitude matrix, the same thing.

Feynman then hands us “the transition amplitudes”. he does not lay them out in an array, but he does label them. I believe his matrix is

$\left(\begin{array}{lll} \frac{1}{2} (\cos (\alpha )+1) & \frac{\sin (\alpha )}{\sqrt{2}} & \frac{1}{2} (1-\cos (\alpha )) \\ -\frac{\sin (\alpha )}{\sqrt{2}} & \cos (\alpha ) & \frac{\sin (\alpha )}{\sqrt{2}} \\ \frac{1}{2} (1-\cos (\alpha )) & -\frac{\sin (\alpha )}{\sqrt{2}} & \frac{1}{2} (\cos (\alpha )+1)\end{array}\right)$

and I believe that it is an inverse transition matrix: applied to old components, it gives new ones. (My only uncertainty was whether his matrix should have been laid out as the transpose of that; I may be wrong, but i’m no longer uncertain. Later.)

McMahon has a recipe (p. 293). He stated it for a rotation about the z-axis, but for a rotation about the y-axis, it implies that we should take the matrix exponential of $Jy\ \alpha$ (multiplied by some constants) namely,

$\text{MatrixExp}\left[-\frac{i\ \text{Jy}\ \alpha }{\hbar }\right]$

Recall that Jy is

$\left(\begin{array}{lll} 0 & -\frac{i \hbar }{\sqrt{2}} & 0 \\ \frac{i \hbar }{\sqrt{2}} & 0 & -\frac{i \hbar }{\sqrt{2}} \\ 0 & \frac{i \hbar }{\sqrt{2}} & 0\end{array}\right)$

(Incidentally, we might recognize that as an “infinitesimal rotation”.) Then I claim that the resulting matrix exponential we have been told to compute is

$\left(\begin{array}{lll} \frac{1}{2} (\cos (\alpha )+1) & -\frac{\sin (\alpha )}{\sqrt{2}} & \frac{1}{2} (1-\cos (\alpha )) \\ \frac{\sin (\alpha )}{\sqrt{2}} & \cos (\alpha ) & -\frac{\sin (\alpha )}{\sqrt{2}} \\ \frac{1}{2} (1-\cos (\alpha )) & \frac{\sin (\alpha )}{\sqrt{2}} & \frac{1}{2} (\cos (\alpha )+1)\end{array}\right)$

Unfortunately, that is the transpose of Feynman’s answer; more importantly, I believe, it is Feynman’s answer with $\alpha$ replaced by $- \alpha$.

it sure looks like we have a sign convention problem.

You might have noticed that although I was careful to specify Feynman’s rotation convention, I did no such thing for McMahon. I don’t actually know that McMahon takes a positive angle as a CCW rotation of the coordinate system.

I do know that the recipe is standard, including the minus sign inside the matrix exponential. I do know that Jy matches Schiff. I also know that McMahon’s answer is an inverse transition matrix: applied to old components, it gives us new.

Feynman also gives us the answer for a rotation about the z-axis. (the second Stern-Gerlach apparatus changes the direction of motion, the y-axis.) He also explicitly gives us the transformation of coordinates

$\begin{array}{c} x'=x \cos (\theta )+y \sin (\theta ) \\ y'=y \cos (\theta )-x \sin (\theta ) \\ z'=z\end{array}$

That is, he has given us the inverse transition matrix, and it is my attitude matrix for $Rz(\theta)$, so Feynman is still using a positive angle for CCW of the axes.

$\left(\begin{array}{lll} e^{i \theta } & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^{-i \theta }\end{array}\right)$

That is interesting because it’s a complex matrix. It is unitary rather than orthogonal. (Its conjugate transpose is its inverse.)

It’s also informative: it is its own transpose.

Let’s apply McMahon’s recipe: we have Jz:

$\left(\begin{array}{lll} \hbar & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -\hbar \end{array}\right)$

We compute

$\text{MatrixExp}\left[-\frac{i\ \text{Jz}\ \theta }{\hbar }\right]$

and get

$\left(\begin{array}{lll} e^{-i \theta } & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^{i \theta }\end{array}\right)$

and, once again, we are off by the sign of the angle. And this time I can’t possibly fix it by transposing, which I could have done for the previous answer. In other words, i’m sure of the layout of Feynman’s first answer because then both of his answers differ from McMahon’s by the sign of the angle.

Incidentally, it is no accident that the matrix exponential of a diagonal matrix D is again diagonal, and that its entries are the exponentials of the diagonal elements of D.

I’ve delayed this post long enough. It was this use of the matrix exponential that got me going on it again, and the investigation of Feynman’s transformation that got me to look at rotations in detail. I think it is really spiffy that the matrix exponential gets us from Jz in one coordinate system to Jz in a rotated coordinate system.

What’s going on? my fervent hope is that McMahon’s recipe is for a positive angle being a CCW rotation of a vector, rather than of coordinate axes. I have found a derivation of the recipe in Schiff (pp. 196-197), but I haven’t sorted it out. I have to be careful: because I want to believe that Schiff uses the other sign convention, I have to be very careful not to err in favor of that.

Feynman has a derivation, too, but it’s drawn-out physical arguments rather than nice compact math. Still, he has a derivation.

Plan A is to discover that McMahon’s recipe uses the other sign convention.

Plan B is to hope that the sign of the angle is immaterial.

Plan C is to talk to someone.

(What, you think I should stop and ask for directions now? How unmanly! How boring!)