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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Prastt
    Apr 4 '13 at 22:53
  • $\begingroup$ 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/… $\endgroup$ Apr 4 '13 at 23:49
  • $\begingroup$ do these quantities both occupy an equivalent 2D Hilbert space? $\endgroup$
    – Ben Steen
    Apr 5 '13 at 2:25
  • $\begingroup$ @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: $\square A^\mu=0$ $\endgroup$
    – Prastt
    Apr 5 '13 at 19:07
  • $\begingroup$ ....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. $\endgroup$
    – Prastt
    Apr 5 '13 at 19:10

First of all, you need to understand from which root cause the spin appears in general. This root cause is a symmetry of the physical space-time.

Particles with different spins (I mean, spin-0 particles, spin-½ particles, spin-1 particles, and so on) use different representations of a symmetry group to map geometry of the space-time to their quantum spin states. Also, there is a difference between spin of a massive particle and of a massless one. The relevant part of the symmetry can be thought of as a Spin group, but in relativistic description of a massless particle it practically means SL(2, ℂ), whereas its subgroup SU(2) is appropriate for a frame of reference where the particle is at rest. In the latter case one can understand something about spin thinking about SU(2) only, that doubly covers the rotation group SO(3).

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.

If you also are going to ask me “does a relativistic theory of spin of massive particles make sense?”, then I can answer: it does, but it is rather complex thing. You can read also about the bispinor representation and Dirac equation. One can describe the spin of a massive particle relativistically, but unlikely one can understand about it anything more than from a SU(2) theory.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.