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What is the difference between a photon's polarization/helicity and an electrons spin half? I know that the photon is spin 1 but isn't its polarization analogous to spin half?

This question stems from wondering why there isn't a classical wave equation like maxwell's equation for the electron.

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Photons don't have spin, they have helicity. In mathematical terms, the photon being massless has $SO(2)$ as it's little group, so it has one generator, $J^3$. Thus the representation is labeled by its eigenvalue, the projection of angular momentum in the dir. of propagation. The photon has two polarizations (helicities) $h=\pm 1/2$. On the other hand, the electron is massive so it's little group is $SU(2)$. The eigenvalue of the Casimir operator happens to be $1/2$ for the electron so the possible projections are $s=\pm 1/2$. That the numbers $\pm 1/2$ are the same is a coincidence. –  Barefeg Apr 4 '13 at 22:53
    
The Stern-Gerlach experiment (en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment) won't work with photons because they have spin but no magnetic moment. You have a good point that polarization with its two states does seem to be analogous to spin-1/2 for electrons. That was discussed a bit here physics.stackexchange.com/questions/45877/… –  Brandon Enright Apr 4 '13 at 23:49
    
do these quantities both occupy an equivalent 2D Hilbert space? –  Ben Steen Apr 5 '13 at 2:25
    
@BenSteen, the states of the electron are labeled by $|M,s\rangle$ and the ones for the photon by $|h \rangle$. But the algebra of the operators is different. For example, for the electron you can construct ladder operators from the complexification of $SU(2)$ to change the projection of the spin. While for the photon, each helicity is basically a different particle unless you include parity transformations which are the ones that go from one helicity to the other. But I don't quite understand your question. The E-L eq. that comes from Maxwell's theory is the wave eq: $\square A^\mu=0$ –  Barefeg Apr 5 '13 at 19:07
    
....where $A^\mu$ is the EM vector potential. While the eq. that comes from the Dirac lagrangian for electrons is $(i\gamma^\mu \partial_\mu-M)\psi=0$ where $\psi$ is the four-component spinor field. All of this classically, I don't know if this is what you are looking for. –  Barefeg Apr 5 '13 at 19:10
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