# Transformation of the spinor indices of Hermitian $2\times 2$ matrices under the Lorentz group

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 $p_{\dot{\alpha}\alpha}$?

1. 4-vectors $x^{\mu}$ are realized as Hermitian $2\times 2$ matrices ${\bf x}$ with components$^1$ $$x^{\alpha\dot{\alpha}}~=~ x^{\mu} (\bar{\sigma}_{\mu})^{\alpha\dot{\alpha}}, \qquad x^{\mu}~=~\frac{1}{2}{\rm tr}(\sigma^{\mu}{\bf x}) ~=~\frac{1}{2}(\sigma^{\mu})_{\dot{\alpha}\alpha} x^{\alpha\dot{\alpha}}, \qquad \bar{\sigma}_{\mu} ~=~(1,\vec{\sigma})~=~\sigma^{\mu}, \tag{1}$$ cf. this Phys.SE post.

2. The quadratic form of the Minkowski metric $(+,-,-,-)$ becomes the determinant $$||x||^2~=~x^{\mu}\eta_{\mu\nu}x^{\nu} ~=~x^{\alpha\dot{\alpha}}\eta_{\dot{\alpha}\alpha,\dot{\beta}\beta}x^{\beta\dot{\beta}} ~=~\det({\bf x}), \qquad \eta_{\dot{\alpha}\alpha,\dot{\beta}\beta}~:=~\frac{1}{2} \epsilon_{\dot{\alpha}\dot{\beta}}\epsilon_{\alpha\beta} .\tag{2}$$

3. The Minkowski inner product is determined from the quadratic form (2) via polarization $$\langle x, y \rangle~=~x^{\mu}\eta_{\mu\nu}y^{\nu} ~=~x^{\alpha\dot{\alpha}}\eta_{\dot{\alpha}\alpha,\dot{\beta}\beta}y^{\beta\dot{\beta}} ~=~ -\frac{1}{2}{\rm tr}({\bf x}\epsilon{\bf y}^t\epsilon) ~=~\frac{1}{4}\sum_{\pm}|| x \pm y ||^2. \tag{3}$$

4. Lie group elements $g\in SL(2,\mathbb{C})$ act on Hermitian $2\times 2$ matrices as $${\bf x}^{\prime}~=~g{\bf x}g^{\dagger},\qquad x^{\prime \alpha\dot{\alpha}}~=~g^{\alpha}{}_{\beta} ~x^{\beta\dot{\beta}}(g^{\dagger})_{\dot{\beta}}{}^{\dot{\alpha}}. \tag{4}$$

5. Since the determinant is clearly preserved in eq. (4), $$||x^{\prime}||^2~\stackrel{(2)}{=}~\det({\bf x}^{\prime})~\stackrel{(4)}{=}~\det(g{\bf x}g^{\dagger})~=~\det({\bf x})~\stackrel{(2)}{=}~||x||^2, \tag{5}$$ the transformation (4) corresponds to a Lorentz transformation $$x^{\prime \mu}~=~\Lambda^{\mu}{}_{\nu} ~x^{\nu} , \qquad \Lambda^{\mu}{}_{\nu}~=~\frac{1}{2}{\rm tr}(\sigma^{\mu}g\bar{\sigma}_{\nu}g^{\dagger}), \qquad \Lambda~\in~O(1,3).\tag{6}$$ In fact, one may show that the Lorentz transformation $\Lambda\in SO^+(1,3)$ belongs to the restricted Lorentz group.

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$^1$ Be aware that different authors use different conventions. Here unbarred (barred) sigma matrices have dotted (undotted) row index and undotted (dotted) column index, respectively, which is usually the other way around, cf. e.g. A. Zee, QFT in a Nutshell, and M. D. Schwartz, QFT & SM.