# A Transformation of the spinor indices of the Weyl operator under the Lorentz group

1. Nov 18, 2016

### spaghetti3451

The left-handed Weyl operator is defined by the $2\times 2$ matrix

$$p_{\mu}\bar{\sigma}_{\dot{\beta}\alpha}^{\mu} = \begin{pmatrix} p^0 +p^3 & p^1 - i p^2\\ p^1 + ip^2 & p^0 - p^3 \end{pmatrix},$$

where $\bar{\sigma}^{\mu}=(1,-\vec{\sigma})$ are sigma matrices.

One can use the sigma matrices to go back and forth between four-vectors and $2\times 2$ matrices:

$$p_{\mu} \iff p_{\dot{\beta}\alpha}\equiv p_{\mu}\bar{\sigma}^{\mu}_{\dot{\beta}\alpha}.$$

Given two four-vectors $p$ and $q$ written as $2\times 2$ matrices,

$$\epsilon^{\dot{\alpha}\dot{\beta}}\epsilon^{\alpha\beta}p_{\dot{\alpha}\alpha}q_{\dot{\beta}\beta} = 2p^{\mu}q_{\mu}.$$

Given a complex $2\times 2$ matrix $\Lambda_{L}$ with unit determinant, it can be shown that the transformation $p_{\dot{\beta}\alpha} \rightarrow (\Lambda_{L}^{-1\dagger}p\Lambda_{L}^{-1})_{\dot{\beta}\alpha}$ preserves the product $\epsilon^{\dot{\alpha}\dot{\beta}}\epsilon^{\alpha\beta}p_{\dot{\alpha}\alpha}q_{\dot{\beta}\beta}$.

How does it then follow that $\Lambda_{L}$ is a Lorentz transformation? Do we have to use the fact that $\epsilon^{\dot{\alpha}\dot{\beta}}\epsilon^{\alpha\beta}p_{\dot{\alpha}\alpha}q_{\dot{\beta}\beta} \sim p^{\mu}q_{\mu}$? What is the Lorentz transformation for $p^{\mu}$ due to the transformation $\Lambda_{L}$ for $p_{\dot{\alpha}\alpha}$?

2. Nov 18, 2016

### vanhees71

You find a very clear treatment about anything connected with the Poincare group in

R. U. Sexl and H. K. Urbandtke, Relativity, Groups, Particles, Springer, Wien, 2001.

3. Nov 18, 2016

### spaghetti3451

I think the second author's name is Urbantke, not Urbandtke.

4. Nov 18, 2016

### spaghetti3451

Is this the ultimate guide for anything related to the Poincare group?

5. Nov 18, 2016

### vanhees71

True, it's Urbantke. There's no "ultimate guide" to anything, but it's a very good book to get the group-theoretical foundations needed to study relativistic QFT more easily than without this basis.