$\text{U}(1)$ 规范

考虑手性超场 ${\mathrm{\Phi }}_l$ 的 $\text{U}(1)$ 规范变换，其中 $t_l$ 为对应 $\text{U}(1)$ 荷，$\lambda$ 为旋转角，二者均为实常数，由于常数为手性超场，故 ${\mathrm{\Phi }}_l^{\prime }$ 也为手性超场

$${\mathrm{\Phi }}_l^{\prime }=e^{-i{t}_l\lambda }{\mathrm{\Phi }}_l$$

对于常数 $\lambda$，容易构造以下规范不变Lagrangian，其中 ${\mathcal{L}}_{P.E.}$ 通常称为超势；对于非零 $m_{ij}$ 或 $g_{ijk}$，$\text{U}(1)$ 规范不变性要求 $t_i+t_j=0$ 或 $t_i+t_j+t_k=0$

$$\mathcal{L}={\mathcal{L}}_{K.E.}+{\mathcal{L}}_{P.E.}$$

$${\mathcal{L}}_{K.E.}={\mathrm{\Phi }}_l^{†}{\mathrm{\Phi }}_l{|}_{\theta \theta \overline{\theta }\overline{\theta }}$$

$${\mathcal{L}}_{P.E.}={[\frac{1}{2}{m}_{ij}{\mathrm{\Phi }}_i{\mathrm{\Phi }}_j+\frac{1}{3}{g}_{ijk}{\mathrm{\Phi }}_i{\mathrm{\Phi }}_j{\mathrm{\Phi }}_k]|}_{\theta \theta }+\text{h.c.}$$

若 $\lambda$ 依赖于 $x$，则其应成为手性多重态 $\mathrm{\Lambda }$ 才能保持 ${\mathrm{\Phi }}_l^{\prime }$ 仍为手性超场

$${\mathrm{\Phi }}_l^{\prime }=e^{-i{t}_l\mathrm{\Lambda }}{\mathrm{\Phi }}_l{\overline{D}}_{\dot{\alpha }}\mathrm{\Lambda }=0$$

$${\mathrm{\Phi }}_l^{\prime †}=e^{i{t}_l{\mathrm{\Lambda }}^{†}}{\mathrm{\Phi }}_l^{†}{D}_{\alpha }{\mathrm{\Lambda }}^{†}=0$$

此时上式Lagrangian动能项不再满足规范不变

$${\mathrm{\Phi }}_l^{\prime †}{\mathrm{\Phi }}_l^{\prime }={\mathrm{\Phi }}_l^{†}{\mathrm{\Phi }}_l{e}^{i{t}_l({\mathrm{\Lambda }}^{†}-\mathrm{\Lambda })}$$

引进满足如下变换规律的矢量超场，可写出加上 $W_{\alpha },{\overline{W}}_{\dot{\alpha }}$ 项的完整 $\text{U}(1)$ 规范不变Lagrangian

$$V^{\prime }=V+i(\mathrm{\Lambda }-{\mathrm{\Lambda }}^{†})$$

$$\begin{array}{rl}\mathcal{L}& =\frac{1}{4}(W^{\alpha }{W}_{\alpha }{|}_{\theta \theta }+{\overline{W}}_{\dot{\alpha }}{\overline{W}}^{\dot{\alpha }}{|}_{\overline{\theta }\overline{\theta }})+{\mathrm{\Phi }}_l^{†}{e}^{t_{l}V}{\mathrm{\Phi }}_l{|}_{\theta \theta \overline{\theta }\overline{\theta }}\\ & +[{(\frac{1}{2}{m}_{ij}{\mathrm{\Phi }}_i{\mathrm{\Phi }}_j+\frac{1}{3}{g}_{ijk}{\mathrm{\Phi }}_i{\mathrm{\Phi }}_j{\mathrm{\Phi }}_k)|}_{\theta \theta }+\text{h.c.}]\end{array}$$

此Lagrangian中项 ${\mathrm{\Phi }}^{†}{e}^{tV}\mathrm{\Phi }{|}_{\theta \theta }$ 看起来不可重整化，但在WZ规范下 $V^3=0$，计算可知此规范下其不含质量维数高于 $4$ 的项

$$\begin{array}{rl}{\mathrm{\Phi }}_l^{†}{e}^{t_{l}V}{\mathrm{\Phi }}_l{|}_{\theta \theta \overline{\theta }\overline{\theta }}& =F{F}^{\ast }+A◻A^{\ast }+i{\partial }_n\overline{\psi }{\overline{\sigma }}^n\psi \\ & +t{v}^n(\frac{1}{2}\overline{\psi }{\overline{\sigma }}^n\psi +\frac{i}{2}{A}^{\ast }{\partial }_{n}A-\frac{i}{2}{\partial }_n{A}^{\ast }A)\\ & -\frac{i}{\sqrt{2}}t(A\overline{\lambda }\overline{\psi }-A^{\ast }\lambda \psi )+\frac{1}{2}(tD-\frac{1}{2}{t}^2{v}_n{v}^n)A^{\ast }A\end{array}$$

电动力学的超对称拓展可以通过以下两个手性超场构造

$${\mathrm{\Phi }}_{+}^{\prime }=e^{-ie\mathrm{\Lambda }}{\mathrm{\Phi }}_{+}{\mathrm{\Phi }}_{-}^{\prime }=e^{ie\mathrm{\Lambda }}{\mathrm{\Phi }}_{-}$$

$$\begin{array}{rl}{\mathcal{L}}_{QED}& =\frac{1}{4}(WW{|}_{\theta \theta }+\overline{W}\overline{W}{|}_{\overline{\theta }\overline{\theta }})+{\mathrm{\Phi }}_{+}^{†}{e}^{eV}{\mathrm{\Phi }}_{+}{|}_{\theta \theta \overline{\theta }\overline{\theta }}+{\mathrm{\Phi }}_{-}^{†}{e}^{-eV}{\mathrm{\Phi }}_{-}{|}_{\theta \theta \overline{\theta }\overline{\theta }}\\ & +m({\mathrm{\Phi }}_{+}{\mathrm{\Phi }}_{-}{|}_{\theta \theta }+{\mathrm{\Phi }}_{+}^{†}{\mathrm{\Phi }}_{-}^{†}{|}_{\overline{\theta }\overline{\theta }})\end{array}$$

由 ${\mathcal{L}}_{QED}$ 分量形式可知两个Weyl旋量 ${\psi }_{+},{\psi }_{-}$ 合成一个有质量Dirac旋量，即电子

$$\begin{array}{rl}{\mathcal{L}}_{QED}& =\frac{1}{2}{D}^2-\frac{1}{4}{v}_{mn}{v}^{mn}-i\lambda {\sigma }^n{\partial }_n\overline{\lambda }\\ & +F_{+}{F}_{+}^{\ast }+F_{-}{F}_{-}^{\ast }+A_{+}^{\ast }◻A_{+}+A_{-}^{\ast }◻A_{-}\\ & +i({\partial }_n{\overline{\psi }}_{+}{\overline{\sigma }}^n{\psi }_{+}+{\partial }_n{\overline{\psi }}_{-}{\overline{\sigma }}^n{\psi }_{-})+e{v}^n[\frac{1}{2}{\overline{\psi }}_{+}{\overline{\sigma }}^n{\psi }_{+}-\frac{1}{2}{\overline{\psi }}_{-}{\overline{\sigma }}^n{\psi }_{-}\\ & +\frac{i}{2}{A}_{+}^{\ast }{\partial }_n{A}_{+}-\frac{i}{2}{\partial }_n{A}_{+}^{\ast }{A}_{+}-\frac{i}{2}{A}_{-}^{\ast }{\partial }_n{A}_{-}+\frac{i}{2}{\partial }_n{A}_{-}^{\ast }{A}_{-}]\\ & -\frac{ie}{\sqrt{2}}(A_{+}{\overline{\psi }}_{+}\overline{\lambda }-A_{+}^{\ast }{\psi }_{+}{\lambda }_{+}-A_{-}{\overline{\psi }}_{-}\overline{\lambda }+A_{-}^{\ast }{\psi }_{-}\lambda )\\ & \frac{e}{2}D(A_{+}^{\ast }{A}_{+}-A_{-}^{\ast }{A}_{-})-\frac{1}{4}{e}^2{v}_n{v}^n(A_{+}^{\ast }{A}_{+}+A_{-}^{\ast }{A}_{-})\\ & +m[A_{+}{F}_{-}+A_{-}{F}_{+}-{\psi }_{+}{\psi }_{-}-{\overline{\psi }}_{+}{\overline{\psi }}_{-}+A_{+}^{\ast }{F}_{-}^{\ast }+A_{-}^{\ast }{F}_{+}^{\ast }]\end{array}$$

非Abelian规范

将超场规范变换推广到非Abelian情况

$${\mathrm{\Phi }}^{\prime }=e^{i\mathrm{\Lambda }}\mathrm{\Phi }{\mathrm{\Phi }}^{\prime †}={\mathrm{\Phi }}^{†}{e}^{i{\mathrm{\Lambda }}^{†}}$$

$$e^{V^{\prime }}=e^{-i{\mathrm{\Lambda }}^{†}}{e}^V{e}^{i\mathrm{\Lambda }}$$

式中 $\mathrm{\Lambda },V$ 均为矩阵，$T^a$ 为规范群的hermitian生成元

$${\mathrm{\Lambda }}_{ij}=T_{ij}^a{\mathrm{\Lambda }}_a{V}_{ij}=T_{ij}^a{V}_a$$

将生成元作如下归一化，注意结构常数为反对称张量

$$\mathrm{Tr}{T}^a{T}^b=k{\delta }^{ab}$$

$$[T^a,T^b]=i{t}^{abc}{T}^c$$

由Hausdorff公式可计算 $V^{\prime }=V+i(\mathrm{\Lambda }-{\mathrm{\Lambda }}^{†})+\cdots$，记Lie导数 $L_{A}B=[A,B]$

$$e^A{e}^B=e^{A+L_{A/2}\cdot [B+\coth \left(L_{A/2}\right)\cdot B]+\cdots }$$

$$(L_A{)}^n\cdot B=[A,[A,[\cdots ,[A,B]\cdots ]]]$$

$$\delta V=V^{\prime }-V=i{L}_{V/2}\cdot [(\mathrm{\Lambda }+{\mathrm{\Lambda }}^{†})+\coth \left(L_{V/2}\right)\cdot (\mathrm{\Lambda }-{\mathrm{\Lambda }}^{†})]$$

同样对超对称场强 $W^{\alpha }$ 作非Abelian推广，可得其变换规律

$$W_{\alpha }=-\frac{1}{4}\overline{D}\overline{D}{e}^{-V}{D}_{\alpha }{e}^V$$

$$W_{\alpha }^{\prime }=e^{-i\mathrm{\Lambda }}{W}_{\alpha }{e}^{i\mathrm{\Lambda }}$$

现在包含可重整化标量、旋量和矢量场相互作用的最普遍形式的Lagrangian可写为

$$\begin{array}{rl}\mathcal{L}& =\frac{1}{16k{g}^2}\mathrm{Tr}(W^{\alpha }{W}_{\alpha }{|}_{\theta \theta }+{\overline{W}}_{\dot{\alpha }}{\overline{W}}^{\dot{\alpha }}{|}_{\overline{\theta }\overline{\theta }})+{\mathrm{\Phi }}^{†}{e}^V\mathrm{\Phi }{|}_{\theta \theta \overline{\theta }\overline{\theta }}\\ & +[{(\frac{1}{2}{m}_{ij}{\mathrm{\Phi }}_i{\mathrm{\Phi }}_j+\frac{1}{3}{g}_{ijk}{\mathrm{\Phi }}_i{\mathrm{\Phi }}_j{\mathrm{\Phi }}_k)|}_{\theta \theta }+\text{h.c.}]\end{array}$$

还原耦合常数 $V\to 2gV$ 后，规范场动能项的归一化应与分量场的正则归一化相一致