Recent content by joneall

I'm arriving late for this discussion, but it is just in time as I am now reading about the subject in Sakurai and Napolitano. Just to get things straighter, let me quote more from them, p. 128. "In general, the canonical momentum p is not a gauge-invariant quantity; its numerical value depends...

I have not given up on this subject. I have purchased and am now reading Ballentine. I will resist the temptation to continue the dialog until I have finished at least chapter 7. (Chapter 9 looks quite interesting too.) Thanks to all who have contributed to raising my understanding, which...

Is there a way to preview a post, or do I have to post it and then do an edit?

See I bumbled several things there. Let's start over. A (z-up) spinor rotated about the x-axis undergoes a transformation $$ R_1(\theta) = e^{i\theta \frac{\sigma_1}{2}} = \begin{pmatrix} cos(\frac{\theta}{2}) & i sin(\frac{\theta}{2}) i sin(\frac{\theta}{2}) & cos(\frac{\theta}{2})...

Sorry, that should have been ##\sigma_1## not 3. $$ R_1(\theta) = e^{i\theta \frac{\sigma_1}{2}} = = \begin{pmatrix} cos(\frac{\theta}{2}) & i sin(\frac{\theta}{2}) \\ i sin(\frac{\theta}{2}) & cos(\frac{\theta}{2}) \end{pmatrix} $$

Rotation by ##\pi## about the x-axis multiplies by ##i##. Isn't that a phase, ##e^{i\pi/2}##?

One more point. I quoted Sean Carroll, "A particle of spin s is invariant under a rotation by ##2\pi/s## radians." Would it not be more precise to say "particle wave function" instead of particle? The observable particle is the modulus square of the wave function and that is invariant under a...

Yes! Using $$ R_x(\theta) = e^{i\theta \frac{\sigma_3}{2}} = \begin{pmatrix} cos(\frac{\theta}{2}) & i sin(\frac{\theta}{2}) \\ i sin(\frac{\theta}{2}) & cos(\frac{\theta}{2}) \end{pmatrix} $$ multiplies by -1 for ##\theta=2\pi##. This also says that rotation thru any angle has the result...

Reading now. Thanks.

This may bore you. Short answer: I've studied and read QM a lot, including in graduate school in the 60s. You can skip the rest except the last paragraph, which is important. I once (before 1972) was a particle physicist, post-docs at BNL and College de France. But I haven't worked in physics...

I agree, but we don't seem to be getting anywhere.

Sorry, I just do not get it. It seems to me if wave function ##\ket{z+}## with z-spin up, with which you agree, then the rotation to get wave function ##- \ket{z+}## very definitely represents spin down. It seems to me that ##- \ket{z+}## actually IS ##\ket{z-}##. I mean, we've turned something...

So what, finally, was my 'significant misunderstanding'? Confusing the rotation (operator) and the wave being rotated?

Aha, now we are getting somewhere. This was what my original question (remember that?) was about. Are you then saying that when I rotate a spinor by a given angle, the change in the spinor is what you would expect given half that angle? If so, then all this about the spinor only being rotated...

I don't believe I said they were wave functions, since I know perfectly well they represent rotations of spinors or vectors. If I expressed that badly, well, my bad. Why "heuristically"? Is it more complicated than that?