Calculating Cross Sections w/ Photon Spin States

[Bobhawke]

When calculating unpolarised cross sections with external fermions we average over initial spin states and sum over final spin states. The photon has 3 spin states; 1, 0 and -1. My question is why don't we average over initial boson spin states and sum over final boson spin states when calculating cross sections?

 

[malawi_glenn]

I thought the photon only had spin states -1 and +1.

 

[Meir Achuz]

I do avg over boson spin states as well as over fermions.
The photon is a spin one particle, but does not have m=0,
because it is massless.

 

[Bobhawke]

Why is it that the photon can only have +1 or -1 spin? Its total angular momentum number is l=1, I thought spin quantum number ranged from m= l, l-1, ... -l+1, -l, so m=0 should be a possible spin state too?

 

[malawi_glenn]

semiclassical: There is no frame where which the photon is at rest, it's spin axis is always along its direction of motion.

 

[Bobhawke]

Do all of the 4 polarisation states of the photon correspond to spin states, two of which are unphysical and can be eliminated using gauge invariance? This Then leaves us with two spin states corresponding to right and left circular polarisations of the photon.

 

[Bobhawke]

one last bump, someone answer please!