一道积分证明

在之前的一篇文章里提到一个积分

$$[\phi(x),\phi(y)]=c_+\int\frac{\mathrm d^4k}{(2\pi)^4}(2\pi)\left[e^{ik\cdot x}\theta(k^0)-e^{ik\cdot x}\theta(-k^0)\right]\delta(k^2+\omega_0^2)$$ $$[\phi(x),\phi(y)]=c_+\mathrm{sign}(x^0-y^0)\left[i\omega_0\theta(\tau^2)\frac{J_1(\omega_0\tau)}{4\pi\tau}-\frac i{2\pi}\delta(\tau^2)\right]$$

本文用于证明此积分。

先将$k^0$积掉，只计算第一项

$$\int_{-\infty}^\infty\mathrm dk^0\ e^{-ik^0x^0}\theta(k^0)\delta\left[-(k^0)^2+\vec k^2+\omega_0^2\right]=\frac{e^{-i\sqrt{\vec k^2+\omega_0^2}x^0}}{2\sqrt{\vec k^2+\omega_0^2}}$$

再将积分变换到球坐标，定义$|\vec x|=r,r^0=x^0-y^0$，那么 $\tau^2=(r^0)^2-r^2$

$$\begin{aligned} &\frac1{c_+}[\phi(x),\phi(y)]\\ =&\int\frac{k^2\sin\theta\mathrm{d}k\mathrm{d}\theta\mathrm{d}\phi}{(2\pi)^32\sqrt{k^2+\omega_0^2}}\left(e^{-i\sqrt{k^2+\omega_0^2}r^0}-e^{i\sqrt{k^2+\omega_0^2}r^0}\right)e^{ikr\cos\theta}\\ =&\int_0^\infty\frac{k^2\mathrm{d}k}{(2\pi)^22\sqrt{k^2+\omega_0^2}}\left(e^{-i\sqrt{k^2+\omega_0^2}r^0}-e^{i\sqrt{k^2+\omega_0^2}r^0}\right)\frac{2\sin kr}{kr}\\ =&\int_0^\infty\frac{\mathrm{d}k}{(2\pi)^2r}\left(e^{-i\sqrt{k^2+\omega_0^2}r^0}-e^{i\sqrt{k^2+\omega_0^2}r^0}\right)\frac{k\sin kr}{\sqrt{k^2+\omega_0^2}}\\ =&\int_0^\infty\frac{\mathrm{d}k}{(2\pi)^2r}\left[\sum_{n=0}^\infty\frac{(r^0)^{2n+1}}{(2n+1)!}(-1)^{n+1}2i\sqrt{k^2+\omega_0^2}^{2n+1}\right]\frac{k\sin kr}{\sqrt{k^2+\omega_0^2}}\\ =&\sum_{n=0}^\infty\frac{(r^0)^{2n+1}(-1)^{n+1}i}{(2n+1)!2\pi^2r}\int_0^\infty(k^2+\omega_0^2)^nk\sin kr\mathrm{d}k\\ =&\sum_{n=0}^\infty\sum_{m=0}^n\frac{(r^0)^{2n+1}(-1)^{n+1}i}{(2n+1)!2\pi^2r}\frac{n!}{m!(n-m)!}\int_0^\infty k^{2(n-m)+1}\omega_0^{2m}\sin kr\mathrm{d}k\\ =&\sum_{m=0}^\infty\sum_{n=m}^\infty\frac{(r^0)^{2n+1}(-1)^{n+1}i}{(2n+1)!2\pi^2r}\frac{n!}{m!(n-m)!}\int_0^\infty k^{2(n-m)+1}\omega_0^{2m}\sin kr\mathrm{d}k\\ \end{aligned}\tag{1}$$

求和变换 $(n,m)\rightarrow(l=n-m,m)$

$$\begin{aligned} &\frac1{c_+}[\phi(x),\phi(y)]\\ =&\sum_{m=0}^\infty\sum_{l=0}^\infty\frac{(r^0)^{2(l+m)+1}(-1)^{m+l+1}i}{(2m+2l+1)!2\pi^2r}\frac{(m+l)!}{m!l!}\int_0^\infty k^{2l+1}\omega_0^{2m}\sin kr\mathrm{d}k\\ =&i\sum_{m=0}^\infty\frac{(-1)^{m+1}(\omega_0r^0)^{2m}}{2\pi^2m!r}\int_0^\infty \sum_{l=0}^\infty\frac{(-1)^l(m+l)!}{(2m+2l+1)!l!} (kr^0)^{2l+1}\sin kr\mathrm{d}k\\ =&i\sum_{m=0}^\infty\frac{(-1)^{m+1}(\omega_0r^0)^{2m}}{2\pi^2m!r}\int_0^\infty 2^{-m}(kr^0)^{1-m}j_m(kr^0)\sin kr\mathrm{d}k\\ \end{aligned}\tag{2}$$

其中 $j_m$ 是球贝塞尔函数。简单起见，定义

$$I[m;r,\epsilon]=\int_0^\infty2^{-m}(kr^0)^{1-m}j_m(kr^0)\sin kr e^{-k\epsilon}\mathrm{d}k\tag{3}$$

利用Mathematica，计算得

$$I[0;r,\epsilon]=\int_0^\infty\sin kr^0\sin kr e^{-k\epsilon}\mathrm{d}k=\frac{2\epsilon r^0r}{\epsilon^4+2\epsilon^2(r^2+(r^0)^2)+((r^0)^2-r^2)^2}$$

也就是

$$\begin{aligned} I[0;r,0^+]&=\pi rr^0\sqrt{\frac2{(r^0)^2+r^2}}\delta((r^0)^2-r^2)\\ &=\pi r\mathrm{sign}(r^0)\delta((r^0)^2-r^2)\\ &=\pi r\mathrm{sign}(r^0)\delta(\tau^2) \end{aligned}\tag{4}$$

再次利用Mathematica，有公式

$$\int_0^{\pm\infty}\mathrm{d}x\ x^{1-m}j_m(x)\sin xy=\pm2^{-m}\pi y\frac{(1-y^2)^{-1+m}}{\Gamma(m)}\theta(1-y^2)$$

令$x=kr^0,y=r/r^0$，得到

$$I[m;r,0]=2^{-2m}\pi r\frac{(\tau^2)^{-1+m}}{(m-1)!(r^0)^{2m}}\theta(\tau^2)\mathrm{sign}(r^0)\tag{5}$$

(4)(5)代入(2)

$$\begin{aligned} &\frac 1c_+[\phi(x),\phi(y)]=i\sum_{m=0}^\infty\frac{(-1)^{m+1}(\omega_0r^0)^{2m}}{2\pi^2m!r}I[m;r,0]\\ =&i\frac{-1}{2\pi^2r}\pi r\mathrm{sign}(r^0)\delta(\tau^2)\\ +&i\sum_{m=1}^\infty\frac{(-1)^{m+1}(\omega_0r^0)^{2m}}{2\pi^2m!r}2^{-2m}\pi r\frac{(\tau^2)^{-1+m}}{(m-1)!(r^0)^{2m}}\theta(\tau^2)\mathrm{sign}(r^0)\\ =&i\frac{-1}{2\pi}\mathrm{sign}(r^0)\delta(\tau^2)\\ +&i\sum_{m=1}^\infty\frac{(-1)^{m+1}}{2\pi m!(m-1)!\tau^2}\left(\frac{\omega_0\tau}2\right)^{2m}\theta(\tau^2)\mathrm{sign}(r^0)\\ =&i\frac{-1}{2\pi}\mathrm{sign}(r^0)\delta(\tau^2)\\ +&i\sum_{m=0}^\infty\frac{(-1)^m}{2\pi m!(m+1)!\tau^2}\left(\frac{\omega_0\tau}2\right)^{2m+2}\theta(\tau^2)\mathrm{sign}(r^0)\\ \end{aligned}\tag{6}$$

注意到

$$J_1(x)=\sum_{m=0}^\infty\frac{(-1)^m}{m!(m+1)!}\left(\frac x2\right)^{2m+1}$$

(6)式变为

$$\frac 1{c_+}[\phi(x),\phi(y)]=i\frac1{2\pi}\mathrm{sign}(r^0)\left[-\delta(\tau^2)+\frac{1}{\tau^2}\frac{\omega_0\tau}2J_1(\omega_0\tau)\theta(\tau^2)\right]$$

这正是我们要证明的

$$[\phi(x),\phi(y)]=ic_+\mathrm{sign}(r^0)\left[\omega_0\theta(\tau^2)\frac{J_1(\omega_0\tau)}{4\pi\tau}-\frac1{2\pi}\delta(\tau^2)\right]\tag{7}$$