# What is the spin of photon [duplicate]

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I am reading a book. It says that spin $s$ describes the symmetry property of a particle. Basically, the number of different wavefunctions which are transformed into linear combination of one another is $2s+1$, under rotation.

Since photon wavefunction is a vector, therefore it has spin $s=1$. So it has $2s+1=3$ components under transformation.

But photon is massless. So it doesn't have rest frame. It always moves with light speed in any frame. As a result, it has a preferred direction which is direction of the momentum of the photon.

What I don't understand is since photon can never be at rest, it doesn't have the rotation symmetry of spin = 1 (seems to me). Indeed, when writing the polarization in density matrix, it is always 2 components. And I thought this corresponds to spin = 1/2 (i.e. 2s+1=2).

So, it seems photon behaves as spin=1/2. What's my misconception here? Is it possible to have longitudinal photon? Thanks a lot!

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## 1 Answer

Massless particles are a representation of a different little group than massive particles. That's the fancy group theory way of saying that photons have no rest frame. More formally, the little group is the subgroup of Lorentz transformations that leaves the 4-momentum invariant.

It is true that the little group of massive particles, $SO(3)$, has $2s+1$ states associated with a spin $s$ particle.

The little group of massless participles is $ISO(2)$, which is different. Massless spin-0 particles have 1 degree of freedom. All higher spin-$s$ particles have 2 degrees of freedom. (There's a tricky topological argument for this given for example in Weinberg Vol 1, chapter 2). A massless spin $s>0$ particle has helicity $h=\pm s$ states. Helicity is defined by transformation under rotation about the 3-momentum. Under rotation by an angle $\theta$ about the 3-momentum, a helicity $h$ state transforms as $|h\rangle\rightarrow e^{i h \theta} |h\rangle$.

(The above answer works in 4 spacetime dimensions only)