场论算出 Delta 函数的导数我就你吗很警惕

• 原文: https://zhuanlan.zhihu.com/p/31173265290
• 发布日期: 2025-03-31
• 分类: 粒子物理 / QFT / 场论计算

就这个比计算每次碰到我都很不放心, 然后我就得迷思好一阵儿, 然后这结果还很不好验证, 然后我又忘了上次怎么搞的了. 然后我又碰到这样的计算, 然后我又迷思了好一阵儿, 然后就觉得这很不好验证呀, 然后我发现你吗的以前好像碰到过这样的计算吧? 然后现在又碰到了这样的计算, 这次我刚开始迷思就发现这你吗都记不清碰到过多少次了, 真得搞搞了, 故著此文，，，

这回可没撑死胆大的, 真有坑.

常用公式在下文亦有记载:

https://zhuanlan.zhihu.com/p/513594929就场论的计算中偶尔会碰到要将坐标或者动量改成微分的情形, 比如说 Fourier 变换的过程:

{\rm{i}\int {\rm{d}^d}x} \;{\rm{e}^{\rm{i}p \cdot x}\frac{x^\mu }{\left( {x^2} \right)}^a} = {\rm{i}{\left( { - 1} \right)^{a + 1}{2^{d - 2a + 1}{\pi ^{d/2}{\left( { - {p^2} \right)^{a - d/2 - 1}{p^\mu }\frac{\Gamma (d/2 - a + 1)}{\Gamma (a)}.

就上式来说 $a\le 0$ 时直接就归零了, 但这其实只是公式没能延拓过去. 我们可以先看看 $a=0$ 时该怎么处理, 就以下边儿这个计算为例:

\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right),\ \Pi \left( {k}_{1},{k}_{2} \right)\equiv \frac{k_{1}^{\mu }_{1}\cdots k_{1}^{\mu }_{s}k_{2}^{\nu }_{1}\cdots k_{2}^{\nu }_{t}{\left( k_{1}^{2}-{m}^{2} \right)\left( k_{2}^{2}-{m}^{2} \right)}.\

这是一个残缺的两圈图, 如果分母上有 $x^2$ 的话倒是能通过 Fourier 变换得到 ${\frac{1}{\left(p-k_1-k_2\right)}}^2$ 这样一条传播子, 但在没有的时候其实也不该归零呀, 应该退化成一圈图嘛:

• \ \ \ \ \int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)

\ \ \ \ \ \frac{\partial }{\partial {p}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\frac{\partial {\left( p-{k}_{1}-{k}_{2} \right)}^{\mu }{\partial {p}^{\mu }=\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Rightarrow {x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=-\text{i}\frac{\partial }{\partial {p}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}.\

• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ \left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right){\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right)
• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\left[ {\left( 2\pi \right)}^{d}\delta \left( p-{k}_{1}-{k}_{2} \right) \right]\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\text{i}\frac{\partial }{\partial {p}^{\mu }\Pi \left( {k}_{1},{k}_{2} \right)
• $=0.$

呃呃还是零···

这是因为这个算法就是错的, 你可以先自己看看错在哪, 要看不出来你可危险了, 赶紧退学吧. 这里实际上是因为我们在倒数第二步前用了分部积分, 但这里根本就没有关于 $p$ 的积分所以这个做法是毫无道理可言的. 从量子力学开始就一直都很常用的那个分部积分操作其实是 Leibniz 律加局域性要求的结果:

• \ \ \ \ \ \frac{\text{d}{\text{d}x}\left( AB \right)=\frac{\text{d}A}{\text{d}x}B+A\frac{\text{d}B}{\text{d}x}
• \Rightarrow \int_{-\infty }^{\infty }{\text{d}x}\frac{\text{d}A}{\text{d}x}B=-\int_{-\infty }^{\infty }{\text{d}x}A\frac{\text{d}B}{\text{d}x}+\int_{-\infty }^{\infty }{\text{d}x}\frac{\text{d}{\text{d}x}\left( AB \right)=-\int_{-\infty }^{\infty }{\text{d}x}A\frac{\text{d}B}{\text{d}x}.\

那为啥要拷打个退学级的小坑呢? 因为你是实际上是可以分部积分的, 且还有额外的好处, 但再说吧.

所以正确的做法应当是改成对积分变量 ${\mathbf{k}}_1$ 或 ${\mathbf{k}}_2$ 的微分:

• \ \ \ \ \int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)

\ \ \ \ \ \frac{\partial }{\partial k_{1}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\frac{\partial {\left( p-{k}_{1}-{k}_{2} \right)}^{\mu }{\partial k_{1}^{\mu }=-\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Rightarrow {x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=\text{i}\frac{\partial }{\partial k_{1}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}.\

• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ \text{i}\frac{\partial }{\partial k_{1}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\frac{\partial }{\partial k_{1}^{\mu }\left[ {\left( 2\pi \right)}^{d}\delta \left( p-{k}_{1}-{k}_{2} \right) \right]\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right).
• 但要注意别提前用掉 $\delta$ 函数, 这显然是不对的
• \ne \int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\partial }{\partial k_{1}^{\mu }\Pi \left( {k}_{1},p-{k}_{1} \right)
• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\partial }{\partial k_{1}^{\mu }\left[ \Pi \left( {k}_{1},p-{k}_{1} \right)\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right) \right]
• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\partial }{\partial k_{1}^{\mu }\left[ \int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\Pi \left( {k}_{1},{k}_{2} \right) \right]
• \not =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\frac{\partial }{\partial k_{1}^{\mu }\Pi \left( {k}_{1},{k}_{2} \right).\

这个坑也很显然嘛, 就你微分都没有作用在 $\delta$ 上怎么就敢直接把 $\delta$ 用掉呢? 顺带一提还曾有教授人跟我稍微 argue 了下这个操作, 我的评价事降职称嗷.

值得注意的是这里无论是对 $k_1$ 还是 $k_2$ 微分都应该等价才对, 因为二者地位完全对称:

• \ \ \ \ \int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)

\ \ \ \ \ \frac{\partial }{\partial k_{2}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\frac{\partial {\left( p-{k}_{1}-{k}_{2} \right)}^{\mu }{\partial k_{2}^{\mu }=-\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Rightarrow {x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=\text{i}\frac{\partial }{\partial k_{2}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}.\

• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ \text{i}\frac{\partial }{\partial k_{2}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\frac{\partial }{\partial k_{2}^{\mu }\left[ {\left( 2\pi \right)}^{d}\delta \left( p-{k}_{1}-{k}_{2} \right) \right]\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{2}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right).

所以竟然有 \left\{ \begin{align} & \ \ \ \ \ \text{i}\int{\frac{\text{d}^{d}{k}_{1}{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right) \\ & =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{2}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right). \\ \end{align} \right.

显然吗? 我是觉得不太显然··· ? 其实也显然, 你再把微分分部回去, 到 $\delta$ 身上不就对称了吗? 但这··· 感觉就很微妙你知道吗? 你吗我好不放心啊. 但这东西能验证吗? 其实也很很难, 毕竟你要参数化的, 那可就五花八门了. 嘛, 再说吧.

要觉得搞分部积分可能不一定靠谱的话其实还有个看起来似乎更保险点儿的做法, 就是说首先这种全空间积分显然就都应该满足这种平移不变性:

• \ \ \ \ \ \int{\text{d}^{d}{k}_{1}\ f\left( x,p,{k}_{1}+\lambda X,{k}_{2} \right)=\int{\text{d}^{d}{k}_{1}\ f\left( x,p,{k}_{1},{k}_{2} \right)
• \Rightarrow \frac{\partial }{\partial \lambda }\int{\text{d}^{d}{k}_{1}\ f\left( x,p,{k}_{1}+\lambda X,{k}_{2} \right)=\frac{\partial }{\partial \lambda }\int{\text{d}^{d}{k}_{1}\ f\left( x,p,{k}_{1},{k}_{2} \right)=0
• =\int{\text{d}^{d}{k}_{1}\frac{\partial f\left( x,p,{k}_{1}+\lambda X,{k}_{2} \right)}{\partial {\left( {k}_{1}+\lambda X \right)}^{\mu }\frac{\partial {\left( {k}_{1}+\lambda X \right)}^{\mu }{\partial \lambda }
• =\int{\text{d}^{d}{k}_{1}\frac{\partial f\left( x,p,{k}_{1}+\lambda X,{k}_{2} \right)}{\partial {\left( {k}_{1}+\lambda X \right)}^{\mu }{X}^{\mu }\
• \Rightarrow \int{\text{d}^{d}{k}_{1}\frac{\partial f\left( x,p,{k}_{1}+\lambda X,{k}_{2} \right)}{\partial {\left( {k}_{1}+\lambda X \right)}^{\mu }{X}^{\mu }=0.\

令 $\lambda =0$ 得到 {X}^{\mu }\int{\text{d}^{d}{k}_{1}\frac{\partial }{\partial k_{1}^{\mu }f\left( x,p,{k}_{1},{k}_{2} \right)=0, 其实这应该也是局域性的体现.

现在令 f\left( x,p,{k}_{1},{k}_{2} \right)\equiv {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right) 得到:

• \ \ \ \ \ {X}^{\mu }\int{\text{d}^{d}{k}_{1}\ \frac{\partial }{\partial k_{1}^{\mu }\left[ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right) \right]=0
• \Rightarrow {X}^{\mu }\int{\text{d}^{d}{k}_{1}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\left( -\text{i}{x}_{\mu }+\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right)=0
• *考虑到 $X^{\mu }$ 的任意性, 可以直接给它扔了. *
• \Rightarrow \int{\text{d}^{d}{k}_{1}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\left( -\text{i}{x}_{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right)=-\int{\text{d}^{d}{k}_{1}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\frac{\partial }{\partial k_{1}^{\mu }\Pi \left( {k}_{1},{k}_{2} \right)
• \Rightarrow \int{\text{d}^{d}{k}_{1}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)=\int{\text{d}^{d}{k}_{1}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right).\

这里也无论是对 $k_1$ 还是 $k_2$ 微分都应该等价才对, 毕竟我流程都同样的走呀.

那现在利用上面这条公式就可以绕开对 $\delta$ 函数的求导:

• \ \ \ \ \int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\text{d}^{d}x\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)

\int{\text{d}^{d}{k}_{1}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)=\int{\text{d}^{d}{k}_{1}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right).\

• =\text{i}\int{\text{d}^{d}x\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right).\

这个做法和前面所得到的结果是相同的, 这样就稍微放心了些吧? 至少俩不同的思路算出来的东西都能对上了, 还有就是也注意别提前把 $\delta$ 函数给用了.

然而开头的错其实也未必就那么错, 只要别扔了全微分项就行:

• \ \ \ \ \int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)

\ \ \ \ \ \frac{\partial }{\partial {p}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\frac{\partial {\left( p-{k}_{1}-{k}_{2} \right)}^{\mu }{\partial {p}^{\mu }=\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Rightarrow {x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=-\text{i}\frac{\partial }{\partial {p}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}.\

• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ \left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right){\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}x}\ \left( -\text{i} \right)\left\{ \frac{\partial }{\partial {p}^{\mu }\left[ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right) \right]-{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\frac{\partial }{\partial {p}^{\mu }\Pi \left( {k}_{1},{k}_{2} \right) \right\}
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}x}\ \left( -\text{i} \right)\frac{\partial }{\partial {p}^{\mu }\left[ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right) \right]
• =\left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right)
• =\left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}{\left( 2\pi \right)}^{d}\delta \left( p-{k}_{1}-{k}_{2} \right)\Pi \left( {k}_{1},{k}_{2} \right)
• =\left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\ \Pi \left( {k}_{1},p-{k}_{1} \right).

这样有个好处是可以全搞完了再求导, 然后还能提前用掉 $\delta$ 函数, 那写程序就轻松得多.

但为啥 \bm{\left[ \begin{align} & \ \ \ \ \left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\ \Pi \left( {k}_{1},p-{k}_{1} \right) \\ & =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right) \\ \end{align} \right]} 呢?

这里其实挺微妙的:

• \ \ \ \left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\ \Pi \left( {k}_{1},p-{k}_{1} \right)
• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\partial {\left( p-{k}_{1} \right)}^{\mu }{\partial {p}^{\mu }\frac{\partial }{\partial {\left( p-{k}_{1} \right)}^{\mu }\Pi \left( {k}_{1},p-{k}_{1} \right)
• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\partial }{\partial {\left( p-{k}_{1} \right)}^{\mu }\Pi \left( {k}_{1},p-{k}_{1} \right)
• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}{\left. \frac{\partial }{\partial k_{2}^{\mu }\Pi \left( {k}_{1},{k}_{2} \right) \right|}_{k}_{2}=p-{k}_{1}
• *一大关键点就在于这一步. *
• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\frac{\partial }{\partial k_{2}^{\mu }\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{2}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right)
• *到这里考虑到 $k_1,k_2$ 我两个都同样的操就已经证完了. *
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \text{i}\frac{\partial }{\partial k_{2}^{\mu }\delta \left( p-{k}_{1}-{k}_{2} \right)\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \text{i}\frac{\partial }{\partial k_{2}^{\mu }\int{\frac{\text{d}^{d}x}{\left( 2\pi \right)}^{d}{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\ \Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \text{i}\frac{\partial }{\partial k_{1}^{\mu }\int{\frac{\text{d}^{d}x}{\left( 2\pi \right)}^{d}{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\ \Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \text{i}\frac{\partial }{\partial k_{1}^{\mu }\delta \left( p-{k}_{1}-{k}_{2} \right)\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right).\

也就是说虽然你不能提前用掉 $\delta$ 函数, 但换个写法总是没问题的.

****所以结论就是:

• \ \ \ \ \int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{1}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right)
• =\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\int{\text{d}^{d}{k}_{2}\ \delta \left( p-{k}_{1}-{k}_{2} \right)\left( -\text{i}\frac{\partial }{\partial k_{2}^{\mu } \right)\Pi \left( {k}_{1},{k}_{2} \right)
• =\left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\ \Pi \left( {k}_{1},p-{k}_{1} \right).\

而我则一般用外动量的形式, 写程序啥的也简洁些:

• \ \ \ \ \int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}{x}_{\mu }\cdots {x}_{\nu }\Pi \left( {k}_{1},{k}_{2} \right)

\ \ \ \ \ \frac{\partial }{\partial {p}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\frac{\partial {\left( p-{k}_{1}-{k}_{2} \right)}^{\mu }{\partial {p}^{\mu }=\text{i}{x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Rightarrow {x}_{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}=-\text{i}\frac{\partial }{\partial {p}^{\mu }{\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}.\

• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ \left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\cdots \left( -\text{i}\frac{\partial }{\partial {p}^{\nu } \right){\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right)
• =\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}\int{\text{d}^{d}x}\ \left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\cdots \left( -\text{i}\frac{\partial }{\partial {p}^{\nu } \right)\left[ {\text{e}^{\text{i}\left( p-{k}_{1}-{k}_{2} \right)\cdot x}\Pi \left( {k}_{1},{k}_{2} \right) \right]
• =\left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\cdots \left( -\text{i}\frac{\partial }{\partial {p}^{\nu } \right)\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\frac{\text{d}^{d}{k}_{2}{\left( 2\pi \right)}^{d}\ \text{i}{\left( 2\pi \right)}^{d}\delta \left( p-{k}_{1}-{k}_{2} \right)\Pi \left( {k}_{1},{k}_{2} \right)
• =\left( -\text{i}\frac{\partial }{\partial {p}^{\mu } \right)\cdots \left( -\text{i}\frac{\partial }{\partial {p}^{\nu } \right)\text{i}\int{\frac{\text{d}^{d}{k}_{1}{\left( 2\pi \right)}^{d}\ \Pi \left( {k}_{1},p-{k}_{1} \right).\

注意这里的 $\Pi \left(k_1,k_2\right)$ 是不包含 $p^{\mu }$ 这样的东西的, 如果里面有外动量的话请务必拎出去, 可千万别让它参与偏导了.

你这真的假的呀, 我咋那么不信呢?

那就验证下? 好你吗麻烦啊我日··· 哎就先写个 Fourier 变换的自动程序吧, 写完了:

然后再写个 Feynman 积分的自动程序吧, 写完了:

好的开始验证，，，

果然事等价得, めでたしめでたし～はいパチパチパチパチ～

爷得评价事, 真得打 Maimai:

https://zhuanlan.zhihu.com/p/136542995