旋量

$\eta_{mn}\sim (-1,1,1,1)$，与度规相同，反称张量在$\text{SL}(2,\mathbb{C})$下不变

$$\varepsilon_{\alpha\beta}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\qquad\varepsilon^{\alpha\beta}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

$$\varepsilon_{\alpha\beta}=M_\alpha^{\;\gamma}M_\beta^{\;\delta}\varepsilon_{\gamma\delta}\qquad \varepsilon^{\alpha\beta}=\varepsilon^{\gamma\delta}M_\gamma^{\;\alpha}M_\delta^{\;\beta}$$

分别服从Lorentz群旋量表示$(1/2,0),(0,1/2)$下变换关系的 $\psi,\bar\psi$ 称为Weyl旋量

$$\psi_\alpha=\varepsilon_{\alpha\beta}\psi^\beta\qquad\psi^\alpha=\varepsilon^{\alpha\beta}\psi_\beta$$

$$\psi’_\alpha=M_\alpha^{\;\beta}\psi_\beta\qquad\bar\psi’_{\dot{\alpha}}=(M^*)_\dot{\alpha}^{\;\dot{\beta}}\bar\psi_{\dot{\beta}}$$

$$\psi’^\alpha={M^{-1}}_\beta^{\;\alpha}\psi^\beta\qquad\bar\psi’_{\dot{\alpha}}={(M^*)^{-1}}_\dot{\beta}^{\;\dot{\alpha}}\bar\psi^{\dot{\beta}}$$

Weyl旋量直和可得Dirac旋量和Majorana旋量，服从Lorentz群四维表示下变换关系

$$\Psi_D=\begin{pmatrix}\chi_\alpha\\\bar\psi^\dot{\alpha}\end{pmatrix}\qquad\Psi_M=\begin{pmatrix}\chi_\alpha\\\bar\chi^\dot{\alpha}\end{pmatrix}$$

约定以下记号，默认旋量分量乘积服从反交换律

$$\psi\chi=\psi^\alpha\chi_\alpha=-\psi_\alpha\chi^\alpha=\chi^\alpha\psi_\alpha=\chi\psi$$

$$\bar\psi\bar\chi=\bar\psi_{\dot\alpha}\bar\chi^{\dot\alpha}=-\bar\psi^{\dot\alpha}\bar\chi_{\dot\alpha}=\bar\chi_{\dot\alpha}\bar\psi^{\dot\alpha}=\bar\chi\bar\psi$$

$$(\chi\psi)^\dagger=(\chi^\alpha\psi_\alpha)^\dagger=\bar\psi_{\dot\alpha}\bar\chi^{\dot\alpha}=\bar\psi\bar\chi=\bar\chi\bar\psi$$

Sigma矩阵

$$\sigma^0=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\qquad
\sigma^1=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$

$$\sigma^2=\begin{pmatrix}0&-i\\i&0\end{pmatrix}\qquad
\sigma^3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

$$\gamma^m=\begin{pmatrix}0&\sigma^m\\\bar\sigma^m&0\end{pmatrix}\qquad\gamma^5=\gamma^0\gamma^1\gamma^2\gamma^3=\begin{pmatrix}-i&0\\0&i\end{pmatrix}$$

$2\times2$ Hermitian矩阵可分解为$\sigma^m$的实系数线性组合，洛伦兹变换保持其行列式不变

$$P=P_m\sigma^m$$

$$P’=MP_m\sigma^mM^\dagger$$

由此可得$\sigma$指标结构与升降规律

$$\bar\sigma^{m\dot{\alpha}\alpha}=\varepsilon^{\dot{\alpha}\dot{\beta}}\varepsilon^{\alpha\beta}\sigma_{\;\;\beta\dot{\beta}}^{m}$$

双旋量和矢量可以通过$\sigma$相互转化

$$v_{\alpha\dot{\alpha}}=\sigma_{\alpha\dot{\alpha}}^{\;\;\;m}v_m\qquad v^m=-\frac{1}{2}\bar\sigma^{m\dot{\alpha}\alpha}v_{\alpha\dot{\alpha}}$$

Lorentz群生成元在旋量表示下可写为

$$\sigma^{mn\;\;\beta}_{\;\;\;\;\alpha}=\frac{1}{4}(\sigma_{\alpha\dot{\alpha}}^{\;\;\;m}\bar\sigma^{n\dot{\alpha}\beta}-\sigma_{\alpha\dot{\alpha}}^{\;\;\;n}\bar\sigma^{m\dot{\alpha}\beta})$$

$$\bar\sigma^{mn\dot\alpha}_{\;\;\;\;\;\;\dot\beta}=\frac{1}{4}(\bar\sigma^{m\dot\alpha\alpha}\sigma_{\alpha\dot{\beta}}^{\;\;\;n}-\bar\sigma^{n\dot\alpha\alpha}\sigma_{\alpha\dot{\beta}}^{\;\;\;m})$$

与$\gamma$相似，$\sigma$具有以下性质，式中 $\varepsilon^{0123}=1$

$$\bar\sigma^0=\sigma^0\qquad\bar\sigma^{1,2,3}=-\sigma^{1,2,3}$$

$$\tr{\sigma^m\bar\sigma^n}=-2\eta^{mn}$$

$$\sigma_{\alpha\dot{\alpha}}^{\;\;\;m}\bar\sigma_m^{\;\;\dot{\beta}\beta}=-2\delta_\alpha^{\;\beta}\delta_{\dot\alpha}^{\;\dot\beta}$$

$$(\sigma^m\bar\sigma^n+\sigma^n\bar\sigma^m)_\alpha^{\;\beta}=-2\eta^{mn}\delta_\alpha^{\;\beta}$$

$$(\bar\sigma^m\sigma^n+\bar\sigma^n\sigma^m)_{\;\dot\beta}^\dot{\alpha}=-2\eta^{mn}\delta_{\;\dot\beta}^\dot{\alpha}$$

$$\sigma^{mn\;\;\alpha}_{\;\;\;\;\alpha}=0$$

$$\sigma^{mn\;\;\beta}_{\;\;\;\;\alpha}\varepsilon_{\beta\gamma}=\sigma^{mn\;\;\beta}_{\;\;\;\;\gamma}\varepsilon_{\beta\alpha}$$

$$\varepsilon^{abcd}\sigma_{cd}=-2i\sigma^{ab}$$

$$\varepsilon^{abcd}\bar\sigma_{cd}=2i\bar\sigma^{ab}$$

$$\sigma_{\alpha\dot{\alpha}}^{\;\;\;m}\sigma_{\beta\dot{\beta}}^{\;\;\;n}-\sigma_{\alpha\dot{\alpha}}^{\;\;\;n}\sigma_{\beta\dot{\beta}}^{\;\;\;m}=2[(\sigma^{mn}\varepsilon)_{\alpha\beta}\varepsilon_{\dot\alpha\dot\beta}+(\varepsilon\bar\sigma^{mn})_{\dot\alpha\dot\beta}\varepsilon_{\alpha\beta}]$$

$$\sigma_{\alpha\dot{\alpha}}^{\;\;\;m}\sigma_{\beta\dot{\beta}}^{\;\;\;n}+\sigma_{\alpha\dot{\alpha}}^{\;\;\;n}\sigma_{\beta\dot{\beta}}^{\;\;\;m}=-\eta^{mn}\varepsilon_{\alpha\beta}\varepsilon_{\dot\alpha\dot\beta}+4(\sigma^{lm}\varepsilon)_{\alpha\beta}(\varepsilon\bar\sigma^{ln})_{\dot\alpha\dot\beta}$$

$$\tr{\sigma^{mn}\sigma^{kl}}=-\frac{1}{2}(\eta^{mk}\eta^{nl}-\eta^{ml}\eta^{nk})-\frac{i}{2}\varepsilon^{mnkl}$$

$$\sigma^a\bar\sigma^b\sigma^c+\sigma^c\bar\sigma^b\sigma^a=2(\eta^{ac}\sigma^b-\eta^{bc}\sigma^a-\eta^{ab}\sigma^c)$$

$$\bar\sigma^a\sigma^b\bar\sigma^c+\bar\sigma^c\sigma^b\bar\sigma^a=2(\eta^{ac}\bar\sigma^b-\eta^{bc}\bar\sigma^a-\eta^{ab}\bar\sigma^c)$$

$$\sigma^a\bar\sigma^b\sigma^c-\sigma^c\bar\sigma^b\sigma^a=2i\varepsilon^{abcd}\sigma_d$$

$$\bar\sigma^a\sigma^b\bar\sigma^c-\bar\sigma^c\sigma^b\bar\sigma^a=-2i\varepsilon^{abcd}\bar\sigma_d$$

旋量代数

$$\theta^\alpha\theta^\beta=-\frac{1}{2}\varepsilon^{\alpha\beta}\theta\theta\qquad \theta_\alpha\theta_\beta=\frac{1}{2}\varepsilon_{\alpha\beta}\theta\theta$$

$$\bar\theta^{\dot\alpha}\bar\theta^{\dot\beta}=\frac{1}{2}\varepsilon^{\dot\alpha\dot\beta}\bar\theta\bar\theta\qquad \bar\theta_{\dot\alpha}\bar\theta_{\dot\beta}=-\frac{1}{2}\varepsilon_{\dot\alpha\dot\beta}\bar\theta\bar\theta$$

$$\theta\sigma^m\bar\theta\theta\sigma^n\bar\theta=-\frac{1}{2}\theta\theta\bar\theta\bar\theta\eta^{mn}$$

$$(\theta\phi)(\theta\psi)=-\frac{1}{2}(\phi\psi)(\theta\theta)\qquad (\bar\theta\bar\phi)(\bar\theta\bar\psi)=-\frac{1}{2}(\bar\phi\bar\psi)(\bar\theta\bar\theta)$$

$$\varepsilon^{\alpha\beta}\frac{\partial}{\partial\theta^\beta}=-\frac{\partial}{\partial\theta_\alpha}$$

$$\varepsilon^{\alpha\beta}\frac{\partial}{\partial\theta^\alpha}\frac{\partial}{\partial\theta^\beta}\theta\theta=4\qquad \varepsilon_{\dot\alpha\dot\beta}\frac{\partial}{\partial\bar\theta_{\dot\alpha}}\frac{\partial}{\partial\bar\theta_{\dot\beta}}\bar\theta\bar\theta=4$$

$$\chi\sigma^m\bar\psi=-\bar\psi\bar\sigma^m\chi\qquad (\chi\sigma^m\bar\psi)^\dagger=\psi\sigma^m\bar\chi$$

$$\chi\sigma^m\bar\sigma^n\psi=\psi\sigma^n\bar\sigma^m\chi\qquad (\chi\sigma^m\bar\sigma^n\psi)^\dagger=\bar\psi\bar\sigma^n\sigma^m\bar\chi$$

$$(\phi\psi)\bar\chi_{\dot\alpha}=-\frac{1}{2}(\phi\sigma^m\bar\chi)(\psi\sigma_m)_{\dot\alpha}$$