# A question about photon helicity

1. Oct 14, 2012

### Comanche

Dear everyone,

I have a simple question about the helicity of photon. The helicity operator is defined as
$$\hat{\mathbf{S}} \hat{\mathbf{p}}/|\mathbf{p}|$$. How to show the photon has +1/-1 helicity eigenvalue from this definition?

Thank you ~~

2. Oct 15, 2012

### tom.stoer

It's

$$\chi = \vec{\mathbf{S}} \hat{\mathbf{p}} = \vec{\mathbf{S}} \frac{\vec{\mathbf{p}}}{|\vec{\mathbf{p}}|}$$

The argument is rather simple. First we know that for the photon we have spin s = 1 and therefore sz = 0, ±1. But the longitudinal polarization is absent for massles vector particles, therefore only sz = ±1 remains. (*)

Now the next step is to fix the arbitary axis ez w.r.t. which we define the spin orientation; we set

$$\hat{{e}}_z = \hat{{p}}$$

Now the photon is moving into z-direction and we have

$$\vec{\mathbf{S}} \vec{\mathbf{p}} = \mathbf{S}_x\mathbf{p}_x + \mathbf{S}_y\mathbf{p}_y + \mathbf{S}_z\mathbf{p}_z \to s_z p_z = \pm p$$

where on the r.h.s sz and pz are the eigenvalues w.r.t. the new reference frame with px = py = 0

(*) But I guess this simple algebra is not really your question I and I gues you want to understand why for massless particles only these two spin orientations survive

EDIT: There are two different ways to find the two physical helicity states of the photon, but I don't see how they are related; therefore I started a new thread https://www.physicsforums.com/showthread.php?t=644149

Last edited: Oct 15, 2012