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?

  • $\begingroup$ The polarisation vector is a function of the momenta and of helicity: e = e(p, h), so its degrees of freedom are actually those of p plus the choice of helicity sign. $\endgroup$ – giulio bullsaver Jun 10 '15 at 6:44
  • $\begingroup$ For the second point, write the polarisation vectors as in 2.50 of arxiv.org/abs/1308.1697 (a nice review on helicity formalism btw) and convince yourself that the gauge transformation can be absorbed in a redefinition of |q] (or |q>). Hint: write p as p = -|p]<p| - |p>[p|... $\endgroup$ – giulio bullsaver Jun 10 '15 at 7:19

First Question

You are correct that we have "lost" two degrees of freedom in defining $\epsilon^+$ and $\epsilon^-$ as above. This because they are just choices of basis vectors.

In QFT we usually just work with some simple basis of polarization vectors, Indeed they'll be summed/averaged over anyway when calculating the cross-section.

The two "lost" degrees of freedom can be easily recovered by considering a general polarization vector $\epsilon=\alpha \epsilon^+ + \beta\epsilon^-$.

Second Question

Consider the two polarization vectors

$$\epsilon^+ = \frac{|p\rangle[q|}{[p \ q]} \quad\textrm{ and } \quad\tilde\epsilon^+ = \frac{|p\rangle[r|}{[p \ r]}$$

where $|r]$ and $|q]$ are linearly independent. Assume there is some constant $C$ such that $\epsilon^+ = \tilde\epsilon^+ +C|p\rangle[p|$. In other words

$$\frac{|p\rangle[q|}{[p \ q]}-\frac{|p\rangle[r|}{[p \ r]}=C|p\rangle[p| $$

Contracting both sides with $|q]$ shows that

$$C = \frac{[q \ r ]}{[p \ r][p \ q]}$$

We must check that this is a valid choice when contracting both sides with $|r]$ i.e.

$$\frac{[q \ r]}{[p \ q]}=\frac{[q \ r ]}{[p \ r][p \ q]}[p \ r]$$

which is trivially true! Hence our assumption of the existence of $C$ is justified a posteriori.


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