# Photon nonzero Spin (Helicity?)

I had a simple question - how many different values can a photon present for spin angular momentum, given that it's spin is 1?

And although I immediately thought "3, since j = 2s + 1" (j being the quantum number for anuglar momentum) - I saw the question had a hint:

"remember that photons have zero mass."

So I googled a bit and found that photons can only have spins +1 or -1, and also that helicity was their "useful property" opposed to spin.

Then, after looking into "similar questions" I found this:

Spin 1 just means that the spin in any direction can assume values out of {-1,0,1}. The 0 is only possible for massive particles, so the photon can have spin -1 or +1. (...) By StackUser: tonydo

So apparently having no mass affects the behavior of a particle's angular momentum. I assume it has to do with Helicity? How does that make photons not have null values for angular momentum?

Because from the classical view, I'd expect zero angular momentum, much like one would expect zero linear momentum on a first analysis of massless bodies. But I am aware of p = h/lambda. Is helicity something like that?

First off you should really specify spin angular momentum. Light can also have orbital angular momentum, given by the shape of the wavefront.

Not sure whether helicity can be used to prove it per se. But helicity is only defined for massless particles, which is much more to the core of the reason as to why photons only have 2 degrees of freedom.

Electromagnetism is a spin-1 gauge field theory, so you are working with a four-vector $$A^{\mu}$$. Having 4 entries, you start with 4 degrees of freedom, i.e. you have to make up 4 numbers in order to fully determine $$A^{\mu}$$.

You can reduce the number of degrees of freedom by finding equations that relate them among themselves.

The first is the gauge condition. Then, you can get rid of one more degree of freedom because the field is massless. Some maths can should you that the product $$k_{\mu} A^{\mu}$$ should be proportional to the mass $$m$$. Having $$m=0$$ entails you have yet another equation that relates the entries of $$A$$ among themselves.

So for a spin-1 massless gauge field theory, you only have 2 degrees of freedom.
Which means you only have two spin projections. Which also makes sense in the context of light in free space, since we know it can be described as L and R polarisations, or H and V.

• Thanks for the pointer - I editted the original question. On a sidenote - what else would I be looking for besides mass that could work as a constraint for the degrees of freedom? – 5Daydreams May 24 at 16:04
• The gauge condition. – SuperCiocia May 24 at 16:06