QED单圈重整化

这篇将会写比较长时间吧。一来因为计算量确实不小，二来……算了，也不要二来了。

高维球的体积和表面积公式直接给出吧。

$d$	0	1	2	3	4	$\cdots$	$d$
$V^d$	$1$	$2$	$\pi$	$\frac{4\pi}3$	$\frac{\pi^2}4$		$\pi^{d/2}/\Gamma\left(\frac d2+1\right)$
$S^d$	$0$	$2$	$2\pi$	$4\pi$	$2\pi^2$		$2\pi^{d/2}/\Gamma\left(\frac d2\right)$

以及所谓的”积木“

$$S(k)=\frac{m-ip\!\!\!\backslash}{p^2+m^2-i\epsilon},\Delta^{\mu\nu}(k)=\frac1{k^2-i\epsilon}\left[g^{\mu\nu}-(1-\xi)\frac{k^\mu k^\nu}{k^2}\right]$$

光子自能

光子自能图为

$$\Pi^{\mu\nu}(k)=(-1)\int\frac{-i\mathrm d^4l}{(2\pi)^4}\ \mathrm{Tr}\left[S(l+k)ieC^\mu S(l)ieC^\nu\right]=e^2\int\frac{\mathrm d^4l}{(2\pi)^4}\mathrm{Tr}\left[\frac{m-i(l\!\!\backslash+k\!\!\!\backslash)}{(l+k)^2+m^2}C^\mu\frac{m-il\!\!\backslash}{l^2+m^2}C^\nu\right]$$

求迹这种东西嘛，主要就是用$\{C^\mu,C^\nu\}=2g^{\mu\nu}\Rightarrow\mathrm{Tr}[C^\mu C^\nu]=4g^{\mu\nu}$

$$\begin{aligned} &\mathrm{Tr}[C^\rho C^\mu C^\sigma C^\nu] =\mathrm{Tr}[C^\rho C^\mu(2g^{\sigma\nu}-C^\nu C^\sigma)]\\ =&8g^{\rho\mu}g^{\sigma\nu}-\mathrm{Tr}[C^\rho C^\mu C^\nu C^\sigma]=8g^{\rho\mu}g^{\sigma\nu}-\mathrm{Tr}[C^\rho(2g^{\mu\nu}-C^\nu C^\mu)C^\sigma]\\ =&8g^{\rho\mu}g^{\sigma\nu}-8g^{\mu\nu}g^{\rho\sigma}+\mathrm{Tr}[C^\rho C^\nu C^\mu C^\sigma]\\ =&8(g^{\rho\mu}g^{\sigma\nu}-g^{\mu\nu}g^{\rho\sigma}+g^{\nu\rho}g^{\mu\sigma})-\mathrm{Tr}[C^\nu C^\rho C^\mu C^\sigma] \end{aligned}$$

所以$\mathrm{Tr}[C^\rho C^\mu C^\sigma C^\nu]=4(g^{\rho\mu}g^{\sigma\nu}-g^{\mu\nu}g^{\rho\sigma}+g^{\nu\rho}g^{\mu\sigma})$，因此

$$\Pi^{\mu\nu}(k)=4e^2\int\frac{-i\mathrm d^4l}{(2\pi)^4}\frac{[m^2+(l+k)_\rho l_\sigma g^{\rho\sigma}]g^{\mu\nu}-[(l+k)^\mu l^\nu+(l+k)^\nu l^\mu]g^{\mu\mu}g^{\nu\nu}}{[(l+k)^2+m^2][l^2+m^2]}$$

做完Wick转动$l^0=il^4,k^0=ik^4$

$$\Pi^{ij}(k_E)=4e^2\int\frac{\mathrm d^4l_E}{(2\pi)^4}\frac{[m^2+(l+k)\cdot l]\delta^{ij}-[(l+k)^\mu l^\nu+(l+k)^j l^i]}{[(l+k)^2+m^2][l^2+m^2]}$$

做Feynman参数化

$$\Pi^{ij}(k_E)=4e^2\int_0^1\mathrm dx\int\frac{\mathrm d^4l_E}{(2\pi)^4}\frac{[m^2+(l+k)\cdot l]\delta^{ij}-[(l+k)^il^j+(l+k)^jl^i]}{[(l+xk)^2+x(1-x)k^2+m^2]^2}$$

设$q=l+xk,D=x(1-x)k^2+m^2$。略去一次项

$$\Pi^{ij}(k_E)=4e^2\int_0^1\mathrm dx\int\frac{\mathrm d^4q}{(2\pi)^4}\frac{[m^2+q^2-x(1-x)k^2]\delta^{ij}-2q^iq^j+2x(1-x)k^ik^j}{[q^2+x(1-x)k^2+m^2]^2}$$

利用$\int\mathrm d^4q\ q^iq^jf(q^2)=\frac14\delta^{ij}\int\mathrm d^4q\ q^2f(q^2)$得

$$\begin{aligned} \Pi^{ij}(k_E) =&4e^2\int_0^1\mathrm dx\int_0^\infty\frac{2\pi^2 q^3\mathrm dq}{(2\pi)^4}\frac{[m^2+q^2-x(1-x)k^2]\delta^{ij}-\frac12\delta^{ij}q^2+2x(1-x)k^ik^j}{[q^2+D]^2}\\ =&\frac{e^2}{4\pi^2}\int_0^1\mathrm dx\int_0^\infty q^2\mathrm dq^2\ \frac{[m^2+q^2-x(1-x)k^2]\delta^{ij}-\frac12\delta^{ij}q^2+2x(1-x)k^ik^j}{[q^2+D]^2}\\ =&\frac{e^2}{4\pi^2}\int_0^1\mathrm dx\int_0^\infty \mathrm dt\ \frac{[m^2\delta^{ij}-x(1-x)(k^2\delta^{ij}-2k^ik^j)]t+\frac12\delta^{ij}t^2}{[t+D]^2}\\ \end{aligned}$$

分别计算两项

$$\int_0^\infty\mathrm dt\frac t{(t+D)^2}=\int_0^Q\frac{t}{(t+D)^2}\mathrm dt+\int_Q^\infty\frac{\mathrm dt}t=-\frac{Q}{D+Q}+\ln\frac{D+Q}D-\ln\frac{\infty}{Q}=-\ln D+c_1$$ $$\int_0^\infty\frac{t^2\mathrm dt}{(t+D)^2}=\left[\frac{Q(2D+Q)}{D+Q}+2D\ln\frac{D}{D+Q}\right]+\infty-Q-2D\ln\frac{\infty}{Q}=2D\ln D+c_2+c_3D$$

所以

$$\Pi^{ij}(k_E)=\frac{e^2}{4\pi^2}\int_0^1\mathrm dx\ \left\{[m^2\delta^{ij}-x(1-x)(k^2\delta^{ij}-2k^ik^j)](-\ln D+c_1)+\delta^{ij}(D\ln D+\frac12c_2+\frac12c_3D)\right\}$$

整理得

$$\Pi^{ij}(k_E)=\frac{e^2}{4\pi^2}\int_0^1\mathrm dx\ \left\{2x(1-x)\ln D[k^2\delta^{ij}-k^ik^j]+2b_1k^ik^j+2(b_2+b_3k^2)\delta^{ij}\right\}$$

上式用到了$D=x(1-x)k^2+m^2$。变换回闵氏空间

$$\Pi^{\mu\nu}(k)=\frac{e^2}{2\pi^2}\int_0^1\mathrm dx[x(1-x)\ln D (k^2g^{\mu\nu}-k^\mu k^\nu)+b_1k^\mu k^\nu+(b_2+b_3k^2)g^{\mu\nu}]$$

On-shell 条件要求$k^2=0$时$\Pi^{\mu\nu}(k)=0,\partial\Pi^{\mu\nu}(k)=0$

$$\Pi^{\mu\nu}(k)=\frac{e^2}{2\pi^2}(k^2g^{\mu\nu}-k^\mu k^\nu)\int_0^1x(1-x)\ln \frac{x(1-x)k^2+m^2}{m^2}\mathrm dx$$

当$k^2\gg m^2$时

$$\Pi^{\mu\nu}(k)=\frac{e^2}{2\pi^2}\left[-\frac5{18}+\frac16\ln\frac{k^2}{m^2}\right]$$

电子自能

$$\Sigma(p)=\int\frac{-i\mathrm d^4l}{(2\pi)^4}ieC^\nu S(l)ieC^\mu\Delta_{\mu\nu}(p-l) =-e^2\int\frac{-i\mathrm d^4l}{(2\pi)^4}\frac{C^\nu(m-iC^\rho l_\rho)C^\mu}{l^2+m^2-i\epsilon}\frac{g_{\mu\nu}}{(p-l)^2-i\epsilon}$$

做Wick转动和变量代换

$$\begin{aligned} \Sigma(p)=&-e^2\int\frac{\mathrm d^4l_E}{(2\pi)^4}\frac{C^i(m-iC^al_a)C^j\delta_{ij}}{(l^2+m^2)(p-l)^2}\\ =&-e^2\int\frac{\mathrm d^4l_E}{(2\pi)^4}\frac{4m-i\sum_jC^jC^aC^jl_a}{(l^2+m^2)(p-l)^2}\\ =&-e^2\int\frac{\mathrm d^4l_E}{(2\pi)^4}\frac{4m+2iC^al_a}{(l^2+m^2)(p-l)^2}\\ \end{aligned}$$

做Feynman参数化

$$\begin{aligned} \Sigma(p)=&-e^2\int_0^1\mathrm dx\int\frac{\mathrm d^4l_E}{(2\pi)^4}\frac{4m+2iC^al_a} {[(l-xp)^2+(1-x)m^2+x(1-x)p^2]^2}\\ =&-e^2\int_0^1\mathrm dx\int\frac{\mathrm d^4q}{(2\pi)^4}\frac{4m+2xiC^ap_a} {[q^2+D]^2}\\ =&-e^2\int_0^1\mathrm dx\int_0^\infty\frac{2\pi^2q^3\mathrm dq}{(2\pi)^4}\frac{4m+2xiC^ap_a} {[q^2+D]^2}\\ =&-\frac{e^2}{8\pi^2}\int_0^1\mathrm dx\int_0^\infty t\mathrm dt\frac{4m+2xiC^ap_a} {[t+D]^2}\\ =&-\frac{e^2}{8\pi^2}\int_0^1\mathrm dx\left[-\frac{Q}{D+Q}+\ln\frac{D+Q}D+\ln\frac{\infty}{Q}\right](4m+2xiC^ap_a)\\ =&-\frac{e^2}{8\pi^2}\int_0^1\mathrm dx\left[-\ln D+c\right](4m+2xiC^ap_a)\\ \end{aligned}$$

当$p^2+m^2=0$时$\Sigma(p)=0$

$$\Sigma(p)=\frac{e^2}{16\pi^2}\int_0^1(4m+2xip\!\!\!\backslash)\left[\ln\frac{(1-x)m^2+x(1-x)p^2}{m^2}+2\right]\mathrm dx$$

同样，$p^2\gg m^2$时

$$\Sigma(p)=\frac{e^2}{16\pi^2}\int(m+2xip\!\!\!\backslash)\left[\ln x(1-x)+2+\ln\frac{p^2}{m^2}\right]\mathrm dx =\frac{e^2}{16\pi^2}(m+ip\!\!\!\backslash)\ln\frac{p^2}{m^2}$$

顶点

$$\mathbf V(p,p',k)=\int\frac{-i\mathrm d^4l}{(2\pi)^4} ieC^\rho S(p'-l)ieC^\nu S(p-l)ieC^\mu \Delta_{\mu\rho}(l)$$

Wick转动和变量代换

$$\mathbf V(p,p',(k,\mu))=-i\frac{e^3}{(2\pi)^4}\int\mathrm d^4l_E\ \frac{C^\rho(m-i(p\!\!\!\backslash'-l\!\!\backslash))C^\mu(m-i(p\!\!\!\backslash-l\!\!\backslash))C^\nu\delta_{\nu\rho}}{[(p'-l)^2+m^2][(p-l)^2+m^2]l^2}$$ $$\mathbf V(p,p',(k,\mu))=-i\frac{e^3}{(2\pi)^4}\int\mathrm d^4l_E\ \frac{-2C^\mu m^2-i8m(p^\mu+p'^\mu-2l^\mu)+(p\!\!\!\backslash-l\!\!\backslash)C^\mu(p\!\!\!\backslash'-l\!\!\backslash)}{[(p'-l)^2+m^2][(p-l)^2+m^2]l^2}$$