For photons (and any massless particle) we consider only a spin projection into the direction of motion (helicity). Why it's meaningless to talk about projection of photon's spin into some arbitrary direction? Is this because we can't measure it (photon does not have a rest frame)?
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2 Answers
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The reason is indeed kind of related to the absence of the rest frame.
The angular momentum with respect to axis $x$ acting on a state (object) $|\psi\rangle$ is given by $$ J_x |\psi\rangle = \lim_{\Delta \phi\to 0} \frac{i\hbar}{\Delta\phi} \left(|\psi\rangle_{{\rm rotated\,\,by\,\,}\Delta\phi\,\,{\rm around\,\,}x} - |\psi\rangle \right) $$ So the state $|\psi\rangle$ may only be an eigenstate of $J_x$ – it has a well-defined value of $J_x$ – if its wave function remains essentially unchanged, up to a modified phase (which is not directly physically measurable) when we rotate the object around the $x$ axis.
If the particle is in the rest frame, its momentum is $\vec p=0$. In that case, the only changes of $|\psi\rangle$ induced by rotations are those that have something to do with the polarization vectors or spinors, i.e. with the intrinsic (spin) part of the angular momentum.
However, if $\vec p\neq 0$, and the momentum of a photon is inevitably nonzero because photons can't have a rest frame, then the rotation around $x$ also changes the value of $\vec p$, assuming that $\vec p$ is pointing in a different direction than $x$. This is equivalent to saying that there is also a nonzero orbital angular momentum, $\vec L=\vec r\times \vec p$.
So the rotation maps $|\psi\rangle$ into a completely different state, one with a different direction of $\vec p$. Consequently, the state isn't an eigenstate of rotations around $x$ and it is therefore not an eigenstate of $J_x$, either.
For massless particles, one may only find eigenvectors of $J_x$ for particle states whose momentum $\vec p$ goes along the same axis $x$, i.e. one may find eigenvalues of $\vec J\cdot \vec p / |\vec p|$, known as the helicity.
Only the total angular momentum is conserved. But even if you tried to artificially separate the spin of the photon (and similarly Weyl neutrinos) from its orbital angular momentum, you will fail to define the spin with respect to other directions than the direction of motion. It's because the polarization vectors $\vec \epsilon$ of photons are transverse to the direction of motion so there don't exist any physical states of the photons at all that would have $\vec \epsilon$ parallel to $\vec p$. Such longitudinal states would be needed to define all $SO(3)$ rotations of a given photon state, i.e. to study the transformation of the state under all components of $\vec J$.
answered Jan 7, 2012 at 12:46
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$\begingroup$ You say "if $\vec{p} \neq 0$ ... [the state] is therefore not an eigenstate of $J_x$". But massive spin-1 particles can be in an eigenstate of $J_x$ even if they have nonzero momentum. Doesn't your argument lead to the too-strong conclusion that even if there exists a rest frame, if you're not in it, then you can't have $S_z = 0$? $\endgroup$– tparkerCommented Jul 5, 2016 at 13:31
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$\begingroup$ No, my argument is correct so it doesn't lead to any wrong conclusions. The first problem of photons - the absence of the rest frame - is avoided for massive spin-one bosons because they do have a rest frame. The second problem with the photons, the fact that they only have 2 transverse polarizations (to the momentum), is also avoided for massive vector bosons because those have all 3 polarizations, not just 2. But if you ask whether states of vector-one bosons with a well-defined $\vec p\neq 0$ in a non-$z$ direction may be eigenstates of $J_z$, my argument holds and the answer is No. $\endgroup$ Commented Jul 5, 2016 at 16:47
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1$\begingroup$ Massive vector bosons' states which are eigenstates of $\vec p$ with some eigenvalue along a different axis than $z$ cannot be simultaneously eigenstates of $J_z$, the total spin, $z$ component. But for massive vector particles, the Hilbert space basically tensor factorizes to the spin degrees of freedom and the center-of-mass (or momentum) ones, so even for $\vec p\neq 0$, it may be an eigenstate of $S_z$, the internal part of the spin, because $\vec S$ and $\vec p$ commute with each other. But that separation to 2 parts isn't possible for massless photons. $\endgroup$ Commented Jul 5, 2016 at 16:51
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Sorry for reiteration of some points already made by Luboš, but in short, the spin operator of a massless particle isn’t a vector and hence, one can’t define its spatial projection.
To understand the difference it is important to point out why said operator for a massive particle is a vector. It is so for the reason explained by Luboš: the center-of-mass frame admits the rotation symmetry expressed with the group SU(2) (a.k.a. Spin(3), a covering group of SO(3)) and respective Lie algebra ${\mathfrak su}(2) = {\mathfrak so}(3)$, the algebra of 3-dimensional Euclidean vectors. The action of said Lie algebra on states of the particle determines what is known as the spin operator. You can also refer to Why, for a spin-½ particle, are the possible outcomes of measuring spin projection along any direction the same? thread for details about spin-½ particles.
A massless particle with given 4-momentum doesn’t admit an SO(3) symmetry (primarily because there are no CoM frames). Its “little” symmetry group is E(2), where a physical interpretation of generators (of the Lie group) is one rotation and two Lorentz boosts. It doesn’t lead to Euclidean vectors. See also the comment at Why does photon have only two possible eigenvalues of helicity? for group action details.
