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  Counting degrees of freedom in spinor-helicity formalism

+ 1 like - 0 dislike

Just a couple of quick questions about the spinor-helicity formalism.

We start with $p^\mu$ and $\epsilon^\mu$, so we have eight degrees of freedom. Then we have that $p^\mu p_\mu = 0$ and that $\epsilon^\mu p_\mu = 0$, which reduce this to six. Also there is a freedom to choose the gauge, which means that $\epsilon^\mu \rightarrow \epsilon^\mu + c p^\mu$ for $c \in \mathbb{R}$ can't affect the physics, which I think reduces the system by one degree of freedom more. So this gives us five remaining degrees of freedom.

Then we go to the spinor-helicity notation to reduce this redundancy, and instead we use $P_{a\dot{a}} = p_\mu \sigma^\mu_{a\dot{a}} = \lambda_a \tilde{\lambda}_\dot{a}$. Because the lambdas are constructed from $p^\mu$, I conclude that they encode the three degrees of freedom of our null momentum vector.
We now construct two different helicity polarisation vectors, using the lambdas
$\epsilon^{-}_{a\dot{a}} = -\sqrt{2}\frac{\lambda_a \tilde{\mu}_\dot{a}}{[\tilde{\lambda}\tilde{\mu}]}$
$\epsilon^{+}_{a\dot{a}} = -\sqrt{2}\frac{\mu_a \tilde{\lambda}_\dot{a}}{\langle\lambda\mu\rangle}$

Where $\mu$ and $\tilde{\mu}$ are aribitary reference spinors, representing the choice of gauge. Two questions:
1) What happened to the two degrees of freedom of the original polisation vector? We appear to have constructed our new polarisation vectors completely from the three degrees of freedom of our momentum
2) I'm happy to accept that the choice of $\mu$ corresponds to the choice of gauge as written for the polarisation vector, but I can't see it explicitly. Can someone help to show me why that is?

This post imported from StackExchange Physics at 2015-06-03 19:31 (UTC), posted by SE-user Joe
asked Jun 2, 2015 in Theoretical Physics by Joe (40 points) [ no revision ]

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