If photons are spin-1 bosons, then doesn't quantum mechanics imply that the allowed values for the z-component of spin (in units of $\hbar$) are -1, 0, and 1?
Why then in practice do we only use the $\pm 1$ states?
I have been told that this is directly related to the two polarizations of the photon. This seems to be more of a classical argument however, arising from the fact that Maxwell's equations do not permit longitudinal EM waves in a vacuum.
I have also heard that it is related to the fact that photons have no rest mass, although I understand far less of this reasoning.
What I'm looking for are elaborations on these two arguments (if I have them correct), and perhaps an argument as to how these two are equivalent (if one exists).
Answer
I can't improve on KDN's answer, but given Todd's comments this is an attempt to rephrase KDN's answer in layman's terms.
A system is only in an eigenstate of spin around an axis if a rotation about the axis doesn't change the system. Take $z$ to be the direction of travel, then for a spin 1 system the $S_z$ = 0 state would be symmetric to a rotation about an axis normal to the direction of travel. But this can only be the case if the momentum is zero i.e. in the rest frame. If the system has a non-zero momentum any rotation will change the direction of the momentum so it won't leave the system unchanged.
For a massive particle we can always find a rest frame, but for a massless particle there is no rest frame and therefore it is impossible to find a spin eigenfunction about any axis other than along the direction of travel. This applies to all massless particles e.g. gravitons also have only two spin states.
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