QED重整化

高能QED裸Lagrangian

$$\mathcal{L}=-\frac{1}{4}(F_{\mu \nu }{)}^2+\overline{\psi }(i\cancel{\partial }-m_0)\psi -e_0\overline{\psi }{\gamma }^{\mu }\psi A_{\mu }$$

单电子和单光子传播子费曼图

定义重整化场，消去Feymann规则中的 $Z_2,Z_3$

$$\psi =Z_2^{1/2}{\psi }_r{A}^{\mu }=Z_3^{1/2}{A}_r^{\mu }$$

$$\mathcal{L}=-\frac{Z_3}{4}(F_{\mu \nu }{)}_r^2+Z_2{\overline{\psi }}_r(i\cancel{\partial }-m_0){\psi }_r-Z_2{Z}_3^{1/2}{e}_0{\overline{\psi }}_r{\gamma }^{\mu }{\psi }_r{A}_{\mu }^r$$

利用低能下的观测电荷 $e$ 与质量 $m$，省略重整化角标 $r$，可将Lagrangian写为低能物理项与微扰项之和

$$Z_2{Z}_3^{1/2}{e}_0=e^{\prime }=Z_{1}e{Z}_2{m}_0=m+{\delta }_m$$

$$Z_i=1+{\delta }_i$$

$$\mathcal{L}=(-\frac{1}{4}{F}_{\mu \nu }^2+\overline{\psi }(i\cancel{\partial }-m)\psi -e\overline{\psi }{\gamma }^{\mu }\psi A_{\mu })+(-\frac{{\delta }_3}{4}{F}_{\mu \nu }^2+\overline{\psi }(i{\delta }_2\cancel{\partial }-{\delta }_m)\psi -e{\delta }_1\overline{\psi }{\gamma }^{\mu }\psi A_{\mu })$$

下文计算单圈修正下的 $Z_i$，可得出电荷与质量对能标的依赖关系

反常磁矩

计算以下Feymann图，图中均为重整化场，虚光子 $q^2<0$ ，令 ${\mathrm{\Gamma }}^{\mu }={\gamma }^{\mu }+{\delta }_{\alpha }{\mathrm{\Gamma }}^{\mu }$，在 $\alpha$ 一阶修正下

$$\begin{array}{rl}& =\overline{u}(p^{\prime }){\delta }_{\alpha }{\mathrm{\Gamma }}^{\mu }(p^{\prime },p)u(p)\\ \\ & =\int \frac{{\text{d}}^{4}k}{(2\pi {)}^4}\frac{-i{g}_{\nu \rho }}{(k-p{)}^2}\overline{u}(p^{\prime })(-ie{\gamma }^{\nu })\frac{i(\cancel{k^{\prime }}+m)}{k^{\prime 2}-m^2}{\gamma }^{\mu }\frac{i(\cancel{k}+m)}{k^2-m^2}(-ie{\gamma }^{\rho })u(p)\\ \\ & =2i{e}^2\int \frac{{\text{d}}^{4}k}{(2\pi {)}^4}\frac{\overline{u}(p^{\prime })[\cancel{k}{\gamma }^{\mu }\cancel{k^{\prime }}+m^2{\gamma }^{\mu }-2m(k+k^{\prime }{)}^{\mu }]u(p)}{(k-p{)}^2(k^{\prime 2}-m^2)(k^2-m^2)}\end{array}$$

计算中用到了以下等式

$$\{\cancel{a},\cancel{b}\}=2ab{\gamma }^{\nu }{\gamma }^{\mu }{\gamma }_{\nu }=-2{\gamma }^{\mu }\{{\gamma }^{\mu },\cancel{a}\}=2a^{\mu }$$

在计算积分之前，可将 ${\mathrm{\Gamma }}^{\mu }$ 进一步简化，将 ${\mathrm{\Gamma }}^{\mu }$ 写为以下形式

$${\mathrm{\Gamma }}^{\mu }=A{\gamma }^{\mu }+B(p^{\prime }+p{)}^{\mu }+C(p^{\prime }-p{)}^{\mu }$$

由Dirac方程，低能下可用 $m$ 替代 $\cancel{p^{\prime }},\cancel{p}$，因此 ${\mathrm{\Gamma }}^{\mu }$ 系数仅与 $q^2$ 有关；由Ward等式，${\mathcal{M}}^{\mu }{q}_{\mu }=0$，因此 $C=0$；再由Gordon等式，可将 ${\mathrm{\Gamma }}^{\mu }$ 写为以下形式

$$\overline{u}(p^{\prime }){\gamma }^{\mu }u(p)=\overline{u}(p^{\prime })[\frac{p^{\prime \mu }+p^{\mu }}{2m}+\frac{i{\sigma }^{\mu \nu }{q}_{\nu }}{2m}]u(p){\sigma }^{\mu \nu }=\frac{i}{2}[{\gamma }^{\mu },{\gamma }^{\nu }]$$

$${\mathrm{\Gamma }}^{\mu }={\gamma }^{\mu }{F}_1(q^2)+i\frac{{\sigma }^{\mu \nu }{q}_{\nu }}{2m}{F}_2(q^2)$$

由 $e^{\prime }=Z_{1}e$ 可得重整化条件

$$-i{Z}_{1}e{\mathrm{\Gamma }}^{\mu }(q=0)=-ie{\gamma }^{\mu }{\mathrm{\Gamma }}^{\mu }(q=0)=Z_1^{-1}{\gamma }^{\mu }$$

用 $F_1,F_2$ 表示重整化 Dirac Lagrangian

$$\begin{array}{rl}{\mathcal{L}}_{\text{Dirac}}& =\overline{\psi }(i{\gamma }^{\mu }({\partial }_{\mu }+ie{A}_{\mu })-m)\psi \\ & =\overline{\psi }(i\cancel{\partial }-m)\psi -e\overline{\psi }{\gamma }^{\mu }{F}_1(0)A_{\mu }\psi -e\overline{\psi }(\frac{{\sigma }^{\mu \nu }}{2m}{F}_2(0){\partial }_{\nu }{A}_{\mu })\psi \end{array}$$

Dirac方程平方成为

$$[(i{\partial }_{\mu }-e{F}_1(0)A_{\mu }{)}^2-m^2+\frac{e}{2}{\sigma }^{\mu \nu }{F}_{\mu \nu }(1+F_2(0))+\cdots ]\psi =0$$

Feymann参数

$$\frac{1}{A_1{A}_2\cdots A_n}={\int }_0^1\mathrm{d}{x}_1\mathrm{d}{x}_2\cdots \mathrm{d}{x}_n\delta (\sum _{i=1}^n{x}_i-1)\frac{(n-1)!}{[x_1{A}_1+x_2{A}_2+\cdots +x_n{A}_n{]}^n}$$

$$\frac{1}{(k-p{)}^2(k^{\prime 2}-m^2)(k^2-m^2)}={\int }_0^1\mathrm{d}x\mathrm{d}y\mathrm{d}z\delta (x+y+z-1)\frac{2}{D^3}$$

$$D=l^2-\mathrm{\Delta }$$

$$l=k+yq-zp\mathrm{\Delta }=-xy{q}^2+(1-z{)}^2{m}^2$$