†
Jazzdude said:
Sorry for the late response, I was quite busy.
I don't quite see what you mean with this. First of all, if you have a reflection, then it's not specific to one operator but acts like a reflection on all operators. And secondly, \sigma_1\sigma_2 is a rotation by the way it transforms. So I guess everything that follows is quite questionable.
Cheers,
Jazz
Jazz,
My mistake for using the wrong operator and limiting it to one dimension. So I try again. The reflection operator sending \sigma to \sigma' by \sigma'=-n\sigma n, is
n, where
n is the unit normal to the plane of reflection. If \sigma is a normized vector \sigma=s_1\sigma_1+s_2\sigma_2+s_3\sigma_3 (scalar coefficients satisfying s_1^2+s_2^2+s_3^2=1) , and the plane of reflection is the 1-2 plane, this becomes \sigma'=-\sigma_3(s_1\sigma_1+s_2\sigma_2+s_3\sigma_3)\sigma_3=s_1\sigma_1+s_2\sigma_2-s_3\sigma_3. As expected, only the component normal to the plane changes.
Now suppose |\uparrow_\sigma\rangle is an eigenvector of \sigma satisfying \sigma|\uparrow_\sigma\rangle=|\uparrow_\sigma\rangle. Let us reflect \sigma in the plane orthogonal to it to obtain \sigma"=(-\sigma)\sigma(\sigma), where now the reflection operator is just \sigma, and substitute it into the eigenvalue equation. My argument above runs \sigma"|\uparrow_\sigma\rangle=-\sigma\sigma\sigma|\uparrow_\sigma\rangle=-\sigma|\uparrow_\sigma\rangle=-|\uparrow_\sigma\rangle. Well, the reflection changed the sign of the spin operator (which was what was bothering me since spin has even parity), but nevertheless the eigenvalue equation yields the expected answer: if a measurement in the positive \sigma direction yields spin angular momentum in that direction, then a measurement in the negative \sigma direction still yields spin angular momentum whose direction is positive \sigma. Think of this as measuring electrons prepared in a spin-up date with a Stern Gerlach apparatus resulting in all the electrons being deflected upwards. Now if you rotate the apparatus 180 degrees about the beam axis, but don't change how the states are prepared, the electrons will be deflected downward. You still interpret the electrons as spin-up per the original convention.
This makes it seem to me like the axial character of spin is imparted, in quantum mechanical formalism, by the states, not the operators. In that formalism, spin 1/2 operators are \hbar/2 times the the Pauli matrices \sigma_1,\sigma_2,\sigma_3. I think what you're saying is that in quantum mechanical formalism, these \sigma s are not vectors, but Hodge duals of bivectors, and their axial character is hidden in the formalism. And furthermore, had we avoided the Hodge duals by staying explicitly with the bivectors (retaining the factor
i), the commutation relations are preserved but the axial character would become explicit in the operators. But I'm not sure how to do that since it would make the operators non-Hermitian while nevertheless they're interpreted as observables.
Another thing bothering me. The reflection operator is unitary (but not special unitary). I've always been able to stick with special unitary operators, and the transformation looks like U^†\sigma U. But here, the reflection operator,
n, is also Hermitian, so we'd have U^†=n^†=n=U, and we can't write this as -n\sigma n, which is how I got tripped up in my earlier post.
Well, I've probably really wrapped myself around the axle this time, but so be it.
Cheers,
Elemental