分立对称性

分立对称性通常指的是空间反演、时间反演和电荷共轭，即$\mathcal{P,T,C}$。对于具体的拉氏量可能会有其他对称性。

此外有个重要定理CPT定理，主旨是作用量在三者的联合作用下不变。

空间反演

标量场

$U(\mathcal P)^{-1}\varphi(x)U(\mathcal P)=L(\mathcal P)\varphi(\mathcal P^{-1}x)$

一维的线性变换只能是常数，又根据两次变换回到自身的性质可知$L(\mathcal P)^2=1\Rightarrow L(\mathcal P)=\pm1$。所以

$U(\mathcal P)^{-1}\varphi(x)U(\mathcal P)=\pm\varphi(\mathcal Px)$

其中正号为普通标量场，负号为赝(pseudo)标量场

旋量场

旋量场的运动方程为

$(C^\mu\partial_\mu+m)\varphi(x)=0$

做空间反演变换$x\rightarrow y=\mathcal Px,\varphi(x)\rightarrow L(\mathcal P)\varphi(y)$，则

$0=\left(C^\mu\frac{\partial}{\partial \mathcal P^\mu_\nu y^\nu}+m\right)L(\mathcal P)\varphi(y)=L(\mathcal P)\left(L(\mathcal P)^{-1}C^\mu (P^{-1})^\nu_\mu L(\mathcal P)\frac{\partial}{\partial y^\nu}+m\right)\varphi(y)$

这说明

$L(\mathcal P)^{-1}C^0L(\mathcal P)=C^0,L(\mathcal P)^{-1}C^iL(\mathcal P)=-C^i$

解得$L(\mathcal P)=\sigma_PC^0$. 作用量为

$\begin{aligned} S=&-\frac i2\int\mathrm d^4x\ \varphi^T(x)C^0|(C^\mu\partial_\mu+m)\varphi(x)\\ =&-\frac i2\int\mathrm d^4y|\det(\mathcal P)|\ \varphi^T(y)L(\mathcal P)^TC^0 |L(\mathcal P)(C^\mu\partial_\mu+m)\varphi(y)\\ =&-\frac i2\int\mathrm d^4y\ \varphi^T(y)\sigma^2_P(C^0)^TC^0C^0(C^\mu\partial_\mu+m)\varphi(y)\\ =&-\frac i2\sigma_P^2\int\mathrm d^4y\ \varphi^T(y)C^0(C^\mu\partial_\mu+m)\varphi(y) =\sigma_P^2 S \end{aligned}\Rightarrow\sigma_P^2=1$

注意一个有趣的事，两次变换后$\varphi(x)\rightarrow\sigma_PC^0\varphi(y)\rightarrow\sigma_P^2(C^0)^2\varphi(x)=-\varphi(x)$并不回到自身

旋量场的算符都是双线性的，即$\bar\varphi X\varphi,X=1,C^\mu,\cdots$ 形式，计算第一个的变换关系

$\begin{aligned} U(\mathcal P)^{-1}\bar\varphi(x)\varphi(x)U(\mathcal P) =&U(\mathcal P)^{-1}\varphi^\dagger(x)U(\mathcal P)iC^0U(\mathcal P)^{-1}\varphi(x)U(\mathcal P)\\ =&[\varphi^\dagger(\mathcal Px)\sigma_P(C^0)^T]iC^0[\sigma_PC^0\varphi(\mathcal Px)]\\ =&\varphi^\dagger(\mathcal Px)iC^0\varphi(\mathcal Px)=\bar\varphi(y)\varphi(y) \end{aligned}$

同样我们可以得到$X\rightarrow(iC^0)^{-1}(C^0)^TiC^0XC^0=-C^0XC^0$，即

$\bar\varphi(x)X\varphi(x)$	$1$	$C_5=C^0C^1C^2C^3$	$iC^\mu$	$C^\mu C^5$	$S^{\mu\nu}=-\frac i4\left[C^\mu,C^\nu\right]$
$\bar\varphi(y)Y\varphi(y)$	$1$	$-C^5$	$\mathcal P^\mu_\nu iC^\nu$	$-\mathcal P^\mu_\nu C^\nu C^5$	$\mathcal P^\mu_\rho\mathcal P^\nu_\sigma S^{\rho\sigma}$

注意$g^\mu_\nu=\{-1,1,1,1\}=-\mathcal P^\mu_\nu$ ，以及最后一项的系数为$2(g^{\mu0}+g^{\nu0})+1$，当其中一者为零时为负.

电磁场

矢量变换规律为

$U(\mathcal P)^{-1}A^\mu(x)U(\mathcal P)=\mathcal P^\mu_\nu A^\nu(\mathcal P^{-1}x)$

所以电磁张量变换规律为

$\begin{aligned} U(\mathcal P)^{-1}F^{\mu\nu}(x)U(\mathcal P)=&U(\mathcal P)^{-1}\left(\frac{\partial}{\partial x^\mu}A^\nu(x)-\frac{\partial}{\partial x^\nu}A^\mu(x)\right)U(\mathcal P)\\ =&\frac{\partial}{\partial x^\mu}\mathcal P^\nu_\sigma A^\sigma(\mathcal P^{-1}x)-\frac{\partial}{\partial x^\nu}\mathcal P^\mu_\rho A^\rho(\mathcal P^{-1}x)\\ =&\mathcal P^\nu_\sigma(\mathcal P^{-1})^\mu_\rho\partial^\rho A^\sigma(y)-\mathcal P^\mu_\rho(\mathcal P^{-1})^\nu_\sigma\partial^\sigma A^\rho(y)\\ =&\mathcal P^\mu_\rho\mathcal P^\nu_\sigma F^{\rho\sigma}(y) \end{aligned}$

时间反演

标量场

$U(\mathcal T)^{-1}\varphi(x)U(\mathcal T)=\pm\varphi^*(\mathcal Tx)$

旋量场

和空间反演类似，我们可以得到

$L(\mathcal T)^{-1}C^0L(\mathcal T)=-C^0,L(\mathcal T)^{-1}C^iL(\mathcal T)=C^i\quad\Rightarrow L(\mathcal T)=\sigma_TC^0C_5$

再根据作用量不变得到$\sigma_T^2=1$.

$\begin{aligned} U(\mathcal T)^{-1}\bar\varphi(x)X\varphi(x)U(\mathcal T) =&\varphi^\dagger(y)L(\mathcal T)^T(-iC^0)X^*L(\mathcal T)\varphi(y)\\ =&\varphi^\dagger(y)(C^0C_5)^T(-iC^0)X^*(C^0C_5)\varphi(y)\\ =&\bar\varphi(y)iC^0C_5^T(C^0)^T(-iC^0)X^*(C^0C_5)\varphi(y)\\ =&\bar\varphi(y)C_5C^0X^*C^0C_5\varphi(y) \end{aligned}$

因此

$\bar\varphi(x)X\varphi(x)$	$1$	$C_5$	$iC^\mu$	$C^\mu C^5$	$S^{\mu\nu}$
$\bar\varphi(y)Y\varphi(y)$	$1$	$-C^5$	$-\mathcal T^\mu_\nu iC^\nu$	$-\mathcal T^\mu_\nu C^\nu C^5$	$-\mathcal T^\mu_\rho\mathcal T^\nu_\sigma S^{\rho\sigma}$

作为例子，守恒荷$Q=-\int\mathrm d^3\vec x\ \bar\varphi(x)iC^0\varphi(x)$，变换为$-\int\mathrm d^3\vec x\ \bar\varphi(y)iC^0\varphi(y)=Q$

流$J=\int\mathrm d^3\vec x\ \bar\varphi(x)iC^i\varphi(x)$，变换为$-J$。

电磁场

为了使$J_\mu A^\mu$为Lorentz标量，电磁势的变换为

$U(\mathcal T)^{-1}A^\mu(x)U(\mathcal T)=-\mathcal T^\mu_\nu A^\nu(\mathcal x)$

电荷共轭

复标量场

对于复标量场，根据$U(1)$对称性，取$\varphi(x)\rightarrow e^{+i\alpha(x)}\varphi(x),\alpha\in\mathbb R$

$\begin{aligned} \mathcal L =&-\partial_\mu(e^{-i\alpha(x)}\varphi^\dagger(x))\partial^\mu(e^{+i\alpha(x)}\varphi(x))-m^2\varphi^\dagger(x)\varphi(x)\\ =&-(\partial_\mu-i\partial_\mu\alpha(x))\varphi^\dagger(x)(\partial^\mu+i\partial^\mu\alpha(x))\varphi(x)-m^2\varphi^\dagger(x)\varphi(x) \end{aligned}$

设$eA^\mu=i\partial^\mu\alpha(x)$，则

$\mathcal L=-(\partial_\mu-eA_\mu)\varphi^\dagger(x)(\partial^\mu+eA^\mu)\varphi(x)-m^2\varphi^\dagger(x)\varphi(x)$

当$e\rightarrow-e$时可知

$U(\mathcal C)^{-1}\varphi(x)U(\mathcal C)=\pm\varphi^\dagger(x),\quad U(\mathcal C^{-1})\varphi^\dagger U(\mathcal C)=\pm\varphi(x)$

旋量场

即便我们不知道具体形式，我们希望仍然是取共轭

$U(\mathcal C)^{-1}\varphi_a(x)U(\mathcal C)=\varphi_a^*(x)$

则对于$\bar\varphi(x)X\varphi(x)$变换为

$\begin{aligned} \rightarrow &U(\mathcal C)^{-1}\varphi^\dagger(x)iC^0X\varphi(x)U(\mathcal C) =\varphi^T(x)iC^0X\varphi^*(x)\\ =&\varphi_a iC^0_{ab}X_{bc}\varphi^*_c =iC^0_{ab}X_{bc}(-\varphi^*_c\varphi_a+\{\varphi_a,\varphi_c^*\})\\ =&-\varphi^\dagger X^T(iC^0)^T\varphi+\sum_{a,b,c}iC^0_{ab}X_{bc}\{\varphi_a,\varphi^*_c\}\\ =&-\bar\varphi(x)C^0X^TC^0\varphi(x)+\mathrm{Tr}[iC^0X]\delta^3(0) \end{aligned}$

因此

$\bar\varphi(x)X\varphi(x)$	$1$	$C_5$	$iC^\mu$	$C^\mu C^5$	$S^{\mu\nu}$
$\bar\varphi(x)Y\varphi(x)$	$1$	$C^5$	$-iC^\mu$	$C^\mu C^5$	$-S^{\mu\nu}$

特殊的在于$X=iC^0$时，变换为$-\bar\varphi(x)iC^0\varphi(x)+4\delta^3(0)$

电磁场

为了使$J_\mu A^\mu$ 为标量

$U(\mathcal C)^{-1}A^\mu(x)U(\mathcal C)=-A^\mu(x)\quad\Rightarrow\quad U(\mathcal C)^{-1}F^{\mu\nu}(x)U(\mathcal C)=-F^{\mu\nu}(x)$

总结

	标量场$\varphi(x)$	旋量场$\varphi(x)$	$\bar\varphi(x)X\varphi(x)$	电磁势$A^\mu(x)$	电磁张量$F^{\mu\nu}(x)$
$\mathcal P$	$\varphi(\mathcal Px)$	$C^0\varphi(\mathcal Px)$	$-\bar\varphi(y)C^0XC^0\varphi(y)$	$\mathcal P^\mu_\rho A^\rho(y)$	$\mathcal P^\mu_\rho\mathcal P^\nu_\sigma F^{\rho\sigma}(y)$
$\mathcal T$	$\varphi(\mathcal Tx)$	$C^0C_5\varphi(\mathcal Tx)$	$\bar\varphi(y)C_5C^0X^*C^0C_5\varphi(y)$	$-\mathcal T^\mu_\rho A^\rho(y)$	$-\mathcal T^\mu_\rho\mathcal T^\nu_\sigma F^{\rho\sigma}(y)$
$\mathcal C$	$\varphi^*(x)$	$\varphi^*(x)$	$-\bar\varphi(x)C^0X^TC^0\varphi(x)+\mathrm{Tr}[iC^0X]\delta^3(0)$	$-A^\mu(x)$	$-F^{\mu\nu}(x)$

Dirac流为$J^\mu(x)=\bar\varphi(x)iC^\mu\varphi(x)$
对于标量场变换后均有个$\pm$，对于空间反演区分了标量场和赝标量场
旋量场变换后符号也未定，但因为物理量都是双线性的，所以就不重要了

CPT定理

既然$\mathcal{C,P,T}$作用下都能使作用量保持不变，那么其联合作用自然也能保持不变。

更一般的，如今已知的所有过程都满足联合作用不变，但对于单一过程可能有变化。