# What is the difference between the properties of Electron spin and Photon polarization/helicity?

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.

• 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. – Prastt 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: $\square A^\mu=0$ – Prastt 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. – Prastt Apr 5 '13 at 19:10

The two-state spin of an electron is controlled by the fundamental (weight-½) irrep. It means that 360° spatial rotation gives the same quantum state but with the opposite sign (180° phase shift). The projectivization of the ${\mathbb C}^2$ of state vectors gives the sphere $\mathrm{S}^2$; it is a usual sphere in the 3-dimensional real space.
The massive spin-1 particles a.k.a. vector bosons rely on the adjoint (weight-1) irrep of SU(2) that maps spatial rotations to rotations in ${\mathbb C}^3$ by the same angle, but the photon is not exactly a vector boson since it misses one spin state of the three. Moreover, the photon is massless; since it does not have any reference frame where it is at rest, one can’t understand anything about it in terms of SU(2) or SO(3).
A photon’s spin is similar to an electron’s spin in number of states – there are two. It means they carry the same amount of quantum information, the qubit, and are congruent informationally. But they are utterly dissimilar in terms of representations. You ask: what is the spin of a photon? In short: it is also an $\mathrm{S}^2$, but there are two distinct poles on it (left and right polarizations) and the equator between them (linear polarizations). You ask: why is it such a thing? Try to understand something in its (1, 0) ⊕ (0,1) representation. I do not understand this thing completely.