Quantum Mechanics and rotations

May 21, 2008 — rip

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.

rotate about the y axis

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.

rotate about the z axis

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.

His answer is

\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!)

Posted in math QM. Tags: , , , . 2 Comments »

Happenings – 24 March (3) Quantum Mechanics

March 29, 2008 — rip
this is the third post of its kind. as i said in the first happenings post, i had three breakthroughs last weekend: in PCA, controls, and quantum mechanics. the controls was in post (2), the PCA is coming; this is the quantum mechanics.
at the beginning of what is now last weekend, i had a breakthru in quantum mechanics. i found a simple example of one of the things i was trying to calculate. this came from McMahon’s “Quantum Mechanics Demystified”. he is also the author of “Relativity Demystified”. both books have some really excellent examples in them, things not usually found in introductory books. the bad news is, his quantum mechanics (henceforth QM) text is marred by an awful lot of typos. the good news is, it’s cheap.
i can summarize my overall reaction as follows. i just checked for a 2nd edition which might have fixed the typos. there isn’t one. but he does have “quantum field theory demystified” and i’m going to order it, just because he’s the author. i’ll take my chances with typos. i’m expecting to see some informative examples anyway.
ok, i’ve ordered it, and a few other books on my shopping list. oh, and there’s a “string theory demystified” by mcmahon notyet published. for under $15, how can i go wrong?
for more on this kind of stuff, you want volume III of “The Feynman Lectures on Physics”. possibly Dirac’s “Quantum Mechanics”. for this, you want bras and kets, rather than wave functions and the schrodinger equation.
i’m not going to try to teach you quantum mechanics, but maybe i’ll whet your appetite. i’m also not going to prove everything in what follows.
consider the following matrix:
Jz = \left(\begin{array}{ccc} \hbar &O&O\\ O&O&O\\ O&O&-\hbar \end{array}\right)
it is diagonal. the diagonal entries must be the eigenvalues, and the identity matrix is an orthogonal eigenvector matrix for Jz. (go ahead, work that out if you need to.)
we interpret Jz as a QM operator which can produce one of 3 measurements of “spin”, namely 0 or \pm \hbar (which we usually think of as \pm 1). and after a measurement the system is in a “state” described by the eigenvector associated with the eigenvalue we obtained from our measurement.
that is, if i say that we measured Jz and got + \hbar , then the system is in the state
\left(\begin{array}{c} 1\\ O\\ O\end{array}\right)
which is the eigenvector associated to +\hbar .
now, we measure Jx, the component of spin along the x-axis. for now, i accept the author’s claim that the operator matrix for Jx is given by
Jx = \left(\begin{array}{ccc} O&\frac{\hbar }{\sqrt{2}}&O\\ \frac{\hbar }{\sqrt{2}}&O&\frac{\hbar }{\sqrt{2}}\\ O&\frac{\hbar }{\sqrt{2}}&O\end{array}\right)
wrt the Jz eigenvector basis. i know how to work it out in principle, but i need to go back to Lie algebras. what i want to shout about is the following calculations.
we believe that a measurement of Jx must be one of the eigenvalues of Jx, and that after such a measurement, the system will be in a state described by one of the Jx eigenvectors.
we ask two questions: what are the possible values of Jx “the spin along the x-axis”? and what is the probability of obtaining each one of those values?
well, we need to get the eigendecomposition of Jx. mathematica returns an eigenvector matrix: that is not orthonormal. “make it so.” we get the following orthonormal eigenvector matrix, which is also a transition matrix to the eigenvector basis of Jx:
\left(\begin{array}{ccc} -\frac{1}{\sqrt{2}}&\frac{1}{2}&\frac{1}{2}\\ O&-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{2}&\frac{1}{2}\end{array}\right)
yo, look at the eigenvalues:
{0,\ -\hbar ,\ \hbar}
i have them in non-customary order. the 1st eigenvector is for the 0 state, the 3rd is for the +1 state. it’s very good that the eigenvalues are 0 and \pm \hbar , the same possibilities as for Jz. to be specific, for example, the first eigenvalue is 0 and the first eigenvector is
\left(\begin{array}{c} -\frac{1}{\sqrt{2}}\\ O\\ \frac{1}{\sqrt{2}}\end{array}\right)
so that is the eigenvector (wrt the Jz basis) associated to a measurement of 0 of Jx.
what we have in hand, after measuring Jz = +\hbar , is a vector u
u = \left(\begin{array}{c} 1\\ O\\ O\end{array}\right)
in the Jz (original) basis. what are its components v wrt the new basis? from
u = U\ v,
v = U^{-1}\ u = U^T\ u
(U is orthogonal, i.e. the inverse is the transpose).
so we compute
v = U^T\ u = \left(\begin{array}{ccc} -\frac{1}{\sqrt{2}}&O&\frac{1}{\sqrt{2}}\\ \frac{1}{2}&-\frac{1}{\sqrt{2}}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{\sqrt{2}}&\frac{1}{2}\end{array}\right) x \left(\begin{array}{c} 1\\ O\\ O\end{array}\right)
= \left(\begin{array}{c} -\frac{1}{\sqrt{2}}\\ \frac{1}{2}\\ \frac{1}{2}\end{array}\right)
so our system, in the state (1,0,0) after a measurement of Jz, is in the state
(-\frac{1}{\sqrt{2}},\frac{1}{2},\frac{1}{2})
with respect to the Jx-eigenvector basis (not wrt to the Jz basis). the result of measuring Jx, however, must be one of the eigenvalues of Jx, and the state of the system after the measurement must be one of the eigenvectors of Jx.
(incidentally, you may not recognize my computation using the transition matrix in McMahon’s solution. he got my answer, but in a more round-about way.)
how do we get from a non-eigenvector to an eigenvector? QM tells us that the probabilitiy of finding the system in a particluar eigenvector state is given by the squared magnitude of the component wrt that eigenvector.
we take the state
(-\frac{1}{\sqrt{2}},\frac{1}{2},\frac{1}{2})
and square each component, getting
(\frac{1}{2},\frac{1}{4},\frac{1}{4}).
those are the probabilities of finding the system in the associated eigenvector state. given my ordering for the eigenvalues,
{0,\ -\hbar ,\ \hbar}
the probabilities must be read off as 1/2 for 0, and 1/4 for each of \hbar and - \hbar .
yes! that’s his answer.
(because all the elements were real, i simply squared the components; in general, we could have complex entries and we would need to compute the squared magnitude of complex numbers.)
Posted in diary, math QM. Tags: , , . Leave a Comment »