Dirac旋量

由三维空间角动量表达式可得Lorentz群生成元，进而得到Lorentz群代数
满足变换规律 $\phi(x)\to \Lambda\phi(\Lambda^{-1}x)$ 的$\phi$称为矢量，此变换下Klein-Gordon方程不变

$$J=x\times p\qquad J^{\mu\nu}=i(x^\mu\partial^\nu-x^\nu\partial^\mu)$$

$$[J^{\mu\nu},J^{\rho\sigma}]=i(g^{\mu\sigma}J^{\nu\rho}+g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho})$$

$$\Lambda=\exp\left(-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}\right)\qquad \Lambda^{\dagger}\Lambda=I$$

$$(\Box+m^2)\phi(x)\to\Lambda(\Box+m^2)\phi(\Lambda^{-1}x)$$

若取以下矩阵$S^{\mu\nu}$，可验证其满足Lorentz代数，因此也为Lorentz群生成元
满足变换规律 $\psi(x)\to \Lambda_{1/2}\psi(\Lambda^{-1}x)$ 的$\psi$称为Dirac旋量，此变换下Dirac方程不变

$$S^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]\qquad\gamma^\mu=\begin{pmatrix}0&\sigma^\mu\\\bar\sigma^\mu&0\end{pmatrix}$$

$$S^{0i}=-\frac{i}{2}\begin{pmatrix}\sigma^i&0\\0&-\sigma^i\end{pmatrix}\qquad S^{ij}=\frac{1}{2}\epsilon^{ijk}\begin{pmatrix}\sigma^k&0\\0&\sigma^k\end{pmatrix}$$

$$\Lambda_{1/2}=\exp\left(-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}\right)\qquad\Lambda_{1/2}^{-1}\gamma^\mu\Lambda_{1/2}=\Lambda^\mu_{\;\nu}\gamma^\nu$$

$$(i\gamma^\mu\partial_\mu-m)\psi(x)\to\Lambda_{1/2}(i\gamma^\nu\partial_\nu-m)\psi(\Lambda^{-1}x)$$

由于 $\Lambda_{1/2}$ 非酉矩阵，故 $\psi^\dagger\psi$ 不是标量，可定义

$$\bar\psi=\psi^\dagger\gamma^0\qquad \bar\psi\to \bar\psi\Lambda_{1/2}^{-1}$$

$$\text{scalar}:\quad\bar\psi\psi\quad \bar\psi\gamma^\mu\partial_\mu\psi$$

$$\text{vector}:\quad\bar\psi\gamma^\mu\psi$$

$$\mathcal{L}_\text{Dirac}=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi$$

$$(i\gamma^\mu\partial_\mu-m)\psi=0\qquad -i\partial_\mu\bar\psi\gamma^\mu-m\bar\psi=0$$

$$(-i\gamma^\mu\partial_\mu-m)(i\gamma^\mu\partial_\mu-m)\psi=(\Box+m^2)\psi$$

Weyl旋量

由$\Lambda_{1/2}$为分块对角阵知Dirac表示为可约表示，可将Dirac旋量拆分为两个Weyl旋量

$$\psi=\begin{pmatrix}\psi_L\\\psi_R\end{pmatrix}\to\begin{pmatrix}\exp\left(\displaystyle -i\boldsymbol{\theta}\cdot\frac{\boldsymbol{\sigma}}{2}-\boldsymbol{\beta}\cdot\frac{\boldsymbol{\sigma}}{2}\right)&0\\0&\exp\left(\displaystyle -i\boldsymbol{\theta}\cdot\frac{\boldsymbol{\sigma}}{2}+\boldsymbol{\beta}\cdot\frac{\boldsymbol{\sigma}}{2}\right)\end{pmatrix}\begin{pmatrix}\psi_L\\\psi_R\end{pmatrix}$$

可写出相应的Dirac方程，$m=0$时称为Weyl方程

$$(i\gamma^\mu\partial_\mu-m)\psi=\begin{pmatrix}-m&i\sigma\cdot\partial\\i\bar\sigma\cdot\partial&-m\end{pmatrix}\begin{pmatrix}\psi_L\\\psi_R\end{pmatrix}=0$$

平面波解

由于$\psi$也满足Klein-Gordon方程，因此可写为平面波线性组合，取以下平面波解

$$\psi=u(p)e^{-ipx}\quad\text{or}\quad\psi=v(p)e^{ipx}$$

解Dirac方程可得平面波表达式，其中$s=1,2$表示Weyl旋量$\xi$的两个基，要求$\xi$归一化

$$u^s(p)=\begin{pmatrix}\sqrt{p\cdot\sigma}\;\xi^s\\\sqrt{p\cdot\bar\sigma}\;\xi^s\end{pmatrix}\qquad v^s(p)=\begin{pmatrix}\sqrt{p\cdot\sigma}\;\xi^s\\-\sqrt{p\cdot\bar\sigma}\;\xi^s\end{pmatrix}$$

$$(p\cdot\sigma)(p\cdot\bar\sigma)=m^2\qquad\xi^\dagger\xi=1$$

同样定义 $\bar u(p),\bar v(p)$

$$\bar u(p)=u^\dagger(p)\gamma^0\qquad\bar v(p)=v^\dagger(p)\gamma^0$$

$$u^{r\dagger}(p)u^s(p)=2E_\boldsymbol{p}\delta^{rs}\qquad\bar u^r(p)u^s(p)=2m\delta^{rs}$$

$$v^{r\dagger}(p)v^s(p)=2E_\boldsymbol{p}\delta^{rs}\qquad\bar v^r(p)v^s(p)=-2m\delta^{rs}$$

$$\bar u^r(p)v^s(p)=\bar v^r(p)u^s(p)=0$$

$$ u^{r\dagger}(\boldsymbol{p})v^s(-\boldsymbol{p})=v^{r\dagger}(\boldsymbol{p})u^s(-\boldsymbol{p})=0$$

$$\sum_s u^s(p)\bar u^s(p)=\gamma\cdot p+m\qquad \sum_s v^s(p)\bar v^s(p)=\gamma\cdot p-m$$

由于 $\gamma\cdot p$ 频繁出现，故将其简记为 $\cancel{p}=\gamma^\mu p_\mu$

Dirac场的量子化

$$\psi(x)=\int\frac{\text{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2E_\boldsymbol{p}}}\sum_s\left[a_\boldsymbol{p}^s u(p)e^{-ipx}+b_\boldsymbol{p}^{s\dagger} v(p)e^{ipx}\right]$$

$$\bar\psi(x)=\int\frac{\text{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2E_\boldsymbol{p}}}\sum_s\left[b_\boldsymbol{p}^s \bar v(p)e^{-ipx}+a_\boldsymbol{p}^{s\dagger} \bar u(p)e^{ipx}\right]$$

$$\{a_\boldsymbol{p}^r,a_\boldsymbol{p}^{s\dagger}\}=\{b_\boldsymbol{p}^r,b_\boldsymbol{p}^{s\dagger}\}=(2\pi)^2\delta^3(\boldsymbol{p}-\boldsymbol{q})\delta^{rs}$$