两点关联函数里的其中一个时空点为啥是零 Källén–Lehmann 谱表示与色散关系

• 原文: https://zhuanlan.zhihu.com/p/396673735
• 发布日期: 2021-08-05
• 分类: 粒子物理 / QFT / 场论计算

从 LSZ 约化公式[1]可以看到相互作用理论下的关联函数是很重要的.

但两点关联函数究竟是怎样的一个概念呢? 见下文:

https://www.zhihu.com/question/266435083/answer/2447183807现在我们知道两点关联函数是长成这个样子的: \left\langle \Omega \right|\!\mathsf T[\mathcal{O}(x)\bar{\mathcal{O}(y)]\!\left| \Omega \right\rangle.

其中 $|\Omega ⟩$ 是相互作用理论下的真空态, 而 $\mathsf{T}$ 与 $\mathcal{O}(x)$ 则分别是时序算符与任意场算符.

但很多文献开局就给扔你一个 $\Pi ({p}^{2})=\text{i}\int{\text{d}^{4}x{\text{e}^{\text{i}p\cdot x}\langle \Omega |\mathsf T[\mathcal{O}(x)\overline{\mathcal{O}(0)]!\left| \Omega \right\rangle }.$

这, 这究竟是··· 没错! 这就是本文的主题··· 之一[2].

另一个主题就是色散关系 \Pi ({p^2}) = \frac{1}{\pi }\int_0^\infty {\rm{d}{m^2}\frac{\operatorname{Im} \Pi ({m^2})}{m^2} - {p^2} - {\rm{i}\varepsilon } 的来历.

本文省略了分离张量结构的过程, 因为这不是重点, 其实比如说核子本身就没有张量结构. 物理人口中的张量指的仅仅是 Lorentz 张量 btw.

其实这就是著名的 Källén–Lehmann 谱表示, 即:

$\langle \Omega |\mathsf T[\mathcal O (x)\bar {\mathcal O} (y)]\!\left| \Omega \right\rangle = - {\rm{i}\int {\frac{\rm{d}^4}p}{(2\pi )}^4}{\rm{e}^{\rm{i}p \cdot \left( {x - y} \right)} \Pi ({p^2}), $ 其中 \Pi ({p^2}) \equiv \int_0^\infty {\rm{d}{m^2}\frac{\rho ({m^2})}{m^2} - {p^2} - {\rm{i}\varepsilon }, 而 $\rho (m^2)$ 被称为物理质量谱密度.

有了上面那个等式你就知道文献里面的东西是怎么来的了, 其实就是个 Fourier 变换.

利用关系 \mathcal O (x) = {\rm{e}^{\rm{i}P \cdot x}\mathcal O (0){\rm{e}^{ - {\rm{i}P \cdot x} 我们可以得到:

$\langle \Omega |\mathsf T[\mathcal{O}(x)\overline{\mathcal{O}(y)]\!\left| \Omega \right\rangle =\langle \Omega |\mathsf T[\mathcal{O}(x){\text{e}^{\text{i}P\cdot y}\overline{\mathcal{O}(0){\text{e}^{-\text{i}P\cdot y}]\!\left| \Omega \right\rangle$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\langle \Omega |\mathsf T[\mathcal{O}(x){\text{e}^{\text{i}P\cdot y}\overline{\mathcal{O}(0)]\!\left| \Omega \right\rangle$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\langle \Omega |\mathsf T[{\text{e}^{-\text{i}P\cdot y}\mathcal{O}(x){\text{e}^{\text{i}P\cdot y}\overline{\mathcal{O}(0)]\!\left| \Omega \right\rangle$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\langle \Omega |\mathsf T[\mathcal{O}(x-y)\overline{\mathcal{O}(0)]\!\left| \Omega \right\rangle .$ 这个过程就是利用了空间的平移不变性, 或者说 Poincaré 对称性吧. 证明过程只用到了 {\text{e}^{-\text{i}P\cdot x}\left| \Omega \right\rangle ={\text{e}^{-\text{i}\left( H,\vec{P} \right)\cdot \left( t,\vec{x} \right)}\left| \Omega \right\rangle ={\rm e}^{-\text{i}0\cdot x}\left| \Omega \right\rangle =\left| \Omega \right\rangle . 至于说为何 {\text{e}^{-\text{i}P\cdot x} 能随意进出时序乘积, 你要想不通就写成 $\theta$ 函数的形式吧.

然后对这个式子做 Fourier 变换:

$\,\ \ \ \ \int{\frac{\text{d}^{4}p}{\left( 2\pi \right)}^{4}{\text{e}^{\text{i}p\cdot \left( x-y \right)}\langle \Omega |\mathsf T[\mathcal{O}(x)\overline{\mathcal{O}(y)]\!\left| \Omega \right\rangle }$ $=\int{\frac{\text{d}^{4}p}{\left( 2\pi \right)}^{4}{\text{e}^{\text{i}p\cdot \left( x-y \right)}\langle \Omega |\mathsf T[\mathcal{O}(x-y)\overline{\mathcal{O}(0)]\!\left| \Omega \right\rangle }=\Pi (p).$

所以 $\Pi (p)$ 也被称作动量空间的关联函数, 你也可以轻易地推广到更多时空点的形式.

没了, 就这么回事, 但你会发现这基本上就是在说废话, 所以还是介绍一下 $\Pi (p^2)$ 是个啥吧.

我们先展开时序算符, 这很简单:

$\left\langle \Omega \right|\mathsf T[{\mathcal O} (x)\bar {\mathcal O} (y)]\left| \Omega \right\rangle = \left\langle \Omega \right|{\mathcal O} (x)\bar {\mathcal O} (y)\left| \Omega \right\rangle \theta ({x^0} - {y^0}) + \left\langle \Omega \right|\bar {\mathcal O} (y){\mathcal O} (x)\left| \Omega \right\rangle \theta ({y^0} - {x^0}).$

所以现在的目标自然就是要得到 $⟨\Omega |\mathcal{O}(x)\bar{\mathcal{O}}(y)|\Omega ⟩$ 的表达式.

利用关系 \mathcal O (x) = {\rm{e}^{\rm{i}P \cdot x}\mathcal O (0){\rm{e}^{ - {\rm{i}P \cdot x} 我们可以得到:

\langle \Omega |\mathcal{O}(x)\overline{\mathcal{O}(y)\left| \Omega \right\rangle =\langle \Omega |{\text{e}^{\text{i}P\cdot x}\mathcal{O}(0){\text{e}^{-\text{i}P\cdot x}{\text{e}^{\text{i}P\cdot y}\overline{\mathcal{O}(0){\text{e}^{-\text{i}P\cdot y}\left| \Omega \right\rangle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\langle \Omega |\mathcal{O}(0){\text{e}^{-\text{i}P\cdot (x-y)}\overline{\mathcal{O}(0)\left| \Omega \right\rangle.

接着要用到完备性关系式:

I = \sum\limits_X {\int {\rm{d}{\Pi _X}\left| X \right\rangle \langle X|} } \equiv \sum\limits_X {\int {\prod\limits_{i \in X} {\frac{\rm{d}^3}{p_i}{(2\pi )}^3}\frac{1}{2{E_i} \left| X \right\rangle \langle X|}. } 其中单粒子态 $|X⟩$ 是 Poincaré 群平移生成元 $P$ 的本征态, $\int \text{d}{\Pi }_X$ 是对相空间积分. 涉及的本征方程为: $P^{\mu }|X⟩=p^{\mu }|X⟩,p^0>0,p^2⩾0.$

插入关系式后得:

$⟨\Omega |\mathcal{O}(x)\bar{\mathcal{O}}(y)|\Omega ⟩$= \left\langle \Omega \right|{\mathcal O} (0){\rm{e}^{ - {\rm{i}P \cdot (x - y)}\bar {\mathcal O} (0)\left| \Omega \right\rangle= \sum\limits_X {\int {\rm{d}{\Pi _X}\left\langle \Omega \right|{\mathcal O} (0){\rm{e}^{ - {\rm{i}P \cdot (x - y)}\left| X \right\rangle \left\langle X \right|\bar {\mathcal O} (0)\left| \Omega \right\rangle } }= \sum\limits_X {\int {\rm{d}{\Pi _X}{\rm{e}^{ - {\rm{i}{p_X} \cdot (x - y)}\left\langle \Omega \right|{\mathcal O} (0)\left| X \right\rangle \left\langle X \right|\bar {\mathcal O} (0)\left| \Omega \right\rangle } }$= \sum\limits_X {\int {\rm{d}{\Pi _X}\left[ {\int {\frac{\rm{d}^4}p}{(2\pi )}^3}{(2\pi )}^3}{\delta ^{(4)}(p - {p_X})} } \right]{\rm{e}^{ - {\rm{i}{p_X} \cdot (x - y)}\left\langle \Omega \right|{\mathcal O} (0)\left| X \right\rangle \left\langle X \right|\bar {\mathcal O} (0)\left| \Omega \right\rangle } }$= \int {\frac{\rm{d}^4}p}{(2\pi )}^3}{\rm{e}^{ - {\rm{i}p \cdot (x - y)}\sum\limits_X {\int {\rm{d}{\Pi _X}{(2\pi )}^3}{\delta ^{(4)}(p - {p_X})\left\langle \Omega \right|{\mathcal O} (0)\left| X \right\rangle \left\langle X \right|\bar {\mathcal O} (0)\left| \Omega \right\rangle } } }= \int {\frac{\rm{d}^4}p}{(2\pi )}^3}{\rm{e}^{ - {\rm{i}p \cdot (x - y)}\theta ({p^0})\rho ({p^2})}.

上面的最后一步只是个定义, 其将一个 Lorentz 不变 (仅依赖于 $p^2$ ) 的量记作 $\theta (p^0)\rho (p^2).$

其中的 $\rho (p^2)$ 被称为物理质量谱密度, 在 $p^2>0$ 时有 $\rho (p^2)⩾0,$而在 ${p^2} \leqslant 0$ 时有 $\rho ({p^2}) = 0$, 总之就是非负 (后面会需要用到这个性质)[[3]](#ref_3). 最后需要注意的一点就是, 由于 $|X⟩$ 全都是实粒子的动量本征态, 所以对应的 $p_X$ 都是在壳的, 或者说有 $p_X^2⩾0,p_X^0>0,$ 然后因为存在 ${\delta }^{(4)}(p-p_X)$ 的限制, 所以 $p$ 中就只有满足$p^2⩾0,p^0>0$ 的部分才可能有贡献, 于是可以放一个 $\theta (p^0).$

然后再插入个 delta 函数:

$\,\ \ \ \int {\frac{\rm{d}^4}p}{(2\pi )}^3}{\rm{e}^{ - {\rm{i}p \cdot (x - y)}\theta ({p^0})\left[ {\int_0^\infty {\rm{d}{m^2}\delta ({p^2} - {m^2})} } \right]\rho ({p^2})}$ $= \int {\frac{\rm{d}^3}p}{(2\pi )}^3}\int {\rm{d}{p^0}{\rm{e}^{ - {\rm{i}p \cdot (x - y)}\theta ({p^0})\left[ {\int_0^\infty {\rm{d}{m^2}\delta ({p^2} - {m^2})} } \right]\rho ({p^2})} } $$= \int {\frac{\rm{d}^3}p}{(2\pi )}^3}\int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\rm{d}{p^0}{\rm{e}^{ - {\rm{i}p \cdot (x - y)}\theta ({p^0})\delta ({p^2} - {m^2})} } } $ [[4]](#ref_4)$= \int {\frac{\rm{d}^3}p}{(2\pi )}^3}\int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\rm{d}{p^0}{\rm{e}^{ - {\rm{i}p \cdot (x - y)}\theta ({p^0})\delta \left[ {({p^0})}^2} - ({\vec p}^2} + {m^2})} \right]} } } $

*****这里需要用到 delta 函数的性质: $ \delta [f(x)] = \sum\limits_i {\frac{\delta (x - {x_i})}{\left| {\left. {\partial _x}f(x)} \right|}_{x = {x_i} \right|} $[[5]](#ref_5).***** $= \int {\frac{\rm{d}^3}p}{(2\pi )}^3}\int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\rm{d}{p^0}{\rm{e}^{ - {\rm{i}p \cdot (x - y)}\theta ({p^0})\left[ {\frac{\delta ({p^0} + \sqrt {\vec p + {m^2} )}{2\sqrt {\vec p + {m^2} } + \frac{\delta ({p^0} - \sqrt {\vec p + {m^2} )}{2\sqrt {\vec p + {m^2} } \right]} } }$= \int {\frac{\rm{d}^3}p}{(2\pi )}^3}\int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\rm{d}{p^0}{\rm{e}^{ - {\rm{i}p \cdot (x - y)}\theta ({p^0})\frac{\delta ({p^0} - \sqrt {\vec p + {m^2} )}{2\sqrt {\vec p + {m^2} } } }= \int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\frac{\rm{d}^3}p}{(2\pi )}^3}\frac{\rm{e}^{ - {\rm{i}\sqrt {\vec p + {m^2} ({x^0} - {y^0})}{\rm{e}^{\rm{i}\vec p \cdot (\vec x - \vec y)}{2\sqrt {\vec p + {m^2} } }= \int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\frac{\rm{d}^3}p}{(2\pi )}^3}\frac{\rm{e}^{ - {\rm{i}{E_{\vec p}({x^0} - {y^0})}{\rm{e}^{\rm{i}\vec p \cdot (\vec x - \vec y)}{2{E_{\vec p} } .

\Rightarrow \left\langle \Omega \right|{\mathcal O} (x)\bar {\mathcal O} (y)\left| \Omega \right\rangle = \int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\frac{\rm{d}^3}p}{(2\pi )}^3}\frac{\rm{e}^{ - {\rm{i}{E_{\vec p}({x^0} - {y^0})}{\rm{e}^{\rm{i}\vec p \cdot (\vec x - \vec y)}{2{E_{\vec p} } . 其中 $E_{\vec{p}}\equiv {\sqrt{\vec{p}}}^2+m^2$ , 仅是个记号.

将上述结论代入两点关联函数 $⟨\Omega |\mathsf{T}[\mathcal{O}(x)\bar{\mathcal{O}}(y)]|\Omega ⟩$ 得:

$\ \ \ \,\left\langle \Omega \right|\mathsf T[{\mathcal O} (x)\bar {\mathcal O} (y)]\left| \Omega \right\rangle $$=⟨\Omega |\mathcal{O}(x)\bar{\mathcal{O}}(y)|\Omega ⟩\theta (x^0-y^0)+⟨\Omega |\bar{\mathcal{O}}(y)\mathcal{O}(x)|\Omega ⟩\theta (y^0-x^0)$=\int_{0}^{\infty }{\text{d}{m}^{2}\rho ({m}^{2})\int{\frac{\text{d}^{3}p}{(2\pi )}^{3}\frac{\text{e}^{-\text{i}{E}_{\vec{p}({x}^{0}-{y}^{0})}{\text{e}^{\text{i}\vec{p}\cdot (\vec{x}-\vec{y})}{2{E}_{\vec{p}\theta ({x}^{0}-{y}^{0})\ \ +\int_{0}^{\infty }{\text{d}{m}^{2}\rho ({m}^{2})\int{\frac{\text{d}^{3}p}{(2\pi )}^{3}\frac{\text{e}^{\text{i}{E}_{\vec{p}({x}^{0}-{y}^{0})}{\text{e}^{-\text{i}\vec{p}\cdot (\vec{x}-\vec{y})}{2{E}_{\vec{p}\theta ({y}^{0}-{x}^{0}) $=\int_{0}^{\infty }{\text{d}{m}^{2}\rho ({m}^{2})\int{\frac{\text{d}^{3}p}{(2\pi )}^{3}\left[ \begin{align} & \ \ \ \ \, \frac{\text{e}^{-\text{i}{E}_{\vec{p}({x}^{0}-{y}^{0})}{\text{e}^{\text{i}\vec{p}\cdot (\vec{x}-\vec{y})}{2{E}_{\vec{p}\theta ({x}^{0}-{y}^{0}) \\ & +\frac{\text{e}^{\text{i}{E}_{\vec{p}({x}^{0}-{y}^{0})}{\text{e}^{-\text{i}\vec{p}\cdot (\vec{x}-\vec{y})}{2{E}_{\vec{p}\theta ({y}^{0}-{x}^{0}) \\ \end{align} \right]}$ $=\int_{0}^{\infty }{\text{d}{m}^{2}\rho ({m}^{2})\int{\frac{\text{d}^{3}p}{(2\pi )}^{3}\left[ \begin{align} & \ \ \ \ \ \frac{\text{e}^{-\text{i}{E}_{\vec{p}({x}^{0}-{y}^{0})}{\text{e}^{-\text{i}\vec{p}\cdot (\vec{x}-\vec{y})}{2{E}_{\vec{p}\theta ({x}^{0}-{y}^{0}) \\ & +\frac{\text{e}^{\text{i}{E}_{\vec{p}({x}^{0}-{y}^{0})}{\text{e}^{-\text{i}\vec{p}\cdot (\vec{x}-\vec{y})}{2{E}_{\vec{p}\theta ({y}^{0}-{x}^{0}) \\ \end{align} \right]}$ $=\int_{0}^{\infty }{\text{d}{m}^{2}\rho ({m}^{2})\int{\frac{\text{d}^{3}p}{(2\pi )}^{3}{\text{e}^{-\text{i}\vec{p}\cdot (\vec{x}-\vec{y})}\frac{1}{2{E}_{\vec{p}\left[ \begin{align} & \ \ \ \, \ {\text{e}^{-\text{i}{E}_{\vec{p}({x}^{0}-{y}^{0})}\theta ({x}^{0}-{y}^{0}) \\ & +{\text{e}^{\text{i}{E}_{\vec{p}({x}^{0}-{y}^{0})}\theta ({y}^{0}-{x}^{0}) \\ \end{align} \right]}$

*****这里需要运用恒等式: $- \frac{1}{2E}\left[ {\theta (t){\rm{e}^{ - {\rm{i}Et} + \theta ( - t){\rm{e}^{\rm{i}Et} \right] = \frac{1}{2\pi {\rm{i}\int {\rm{d}\omega \frac{\rm{e}^{\rm{i}\omega t}{\omega ^2} - {E^2} + {\rm{i}\varepsilon }.$来源参考 [[此文](https://zhuanlan.zhihu.com/p/409397025)].*****

= \int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\frac{\rm{d}^3}p}{(2\pi )}^3}{\rm{e}^{ - {\rm{i}\vec p \cdot (\vec x - \vec y)}\frac{\rm{i}{2\pi }\int {\rm{d}\omega \frac{\rm{e}^{\rm{i}\omega ({x^0} - {y^0})}{\omega ^2} - E_{\vec p}^2 + {\rm{i}\varepsilon } } }= \int_0^\infty {\rm{d}{m^2}\rho ({m^2})\int {\frac{\rm{d}^3}p}{(2\pi )}^3}{\rm{e}^{ - {\rm{i}\vec p \cdot (\vec x - \vec y)}\frac{\rm{i}{2\pi }\int {\rm{d}{p^0}\frac{\rm{e}^{\rm{i}{p^0}({x^0} - {y^0})}{({p^0})}^2} - E_{\vec p}^2 + {\rm{i}\varepsilon } } }=\int{\frac{\text{d}^{4}p}{(2\pi )}^{4}{\text{e}^{\text{i}p\cdot (x-y)}\text{i}\int_{0}^{\infty }{\text{d}{m}^{2}\frac{\rho ({m}^{2})}{p}^{2}-{m}^{2}+\text{i}\varepsilon }=\int_{0}^{\infty }{\text{d}{m}^{2}\rho ({m}^{2}){D}_{\text{F}\left( x-y \right)}上面的第二个等号只是想说明下这里面藏了一个 Feynman 传播子, 就像前面说的那样.=-\text{i}\int{\frac{\text{d}^{4}p}{(2\pi )}^{4}{\text{e}^{\text{i}p\cdot (x-y)}\Pi ({p}^{2})}, 证毕.

得到 Källén–Lehmann 谱表示 \langle \Omega |\mathsf T[{\mathcal O} (x)\bar {\mathcal O} (y)]\!\left| \Omega \right\rangle = - {\rm{i}\int {\frac{\rm{d}^4}p}{(2\pi )}^4}{\rm{e}^{\rm{i}p \cdot \left( {x - y} \right)} \Pi ({p^2}), 其中 \Pi ({p^2}) \equiv \int_0^\infty {\rm{d}{m^2}\frac{\rho ({m^2})}{m^2} - {p^2} - {\rm{i}\varepsilon }.

有些地方的定义会与此处相差一个负号, but you know, same shit.

若令谱表示中的 $y=0$ 则可得关系 $⟨\Omega |\mathsf{T}[\mathcal{O}(x)\bar{\mathcal{O}}(0)]|\Omega ⟩=-\mathrm{i}\int \frac{{\mathrm{d}}^4}{\mathrm{p}}{(2\pi )}^4{\mathrm{e}}^{\mathrm{i}\mathrm{p}\cdot \mathrm{x}}\mathrm{\Pi }({\mathrm{p}}^2).$

故 $\Pi (p^2)$ 是 {\rm{i}\langle \Omega |\mathsf T[{\mathcal O} (x)\bar {\mathcal O} (0)]\!\left| \Omega \right\rangle 的 Fourier 变换.

即是说 $\int{\text{d}^{4}x{\text{e}^{-\text{i}q\cdot x}\langle \Omega |\mathsf T[\mathcal{O}(x)\overline{\mathcal{O}(0)]\!\left| \Omega \right\rangle$=-\text{i}\int{\frac{\text{d}^{4}p}{(2\pi )}^{4}\int{\text{d}^{4}x{\text{e}^{\text{i}(p-q)\cdot x}\Pi ({p}^{2})}$=-\text{i}\int {\frac{{\text{d}}^{4}p}{(2\pi )}}^4\Pi (p^2){(2\pi )}^4{\delta }^{(4)}(p-q)=-\text{i}\Pi (q^2).$

接着只需再令 $q=-p$ 即可得文献上常见的那个式子:

$\Pi ({p}^{2})\equiv \int_{0}^{\infty }{\text{d}{m}^{2}\frac{\rho ({m}^{2})}{m}^{2}-{p}^{2}-\text{i}\varepsilon }=\text{i}\int{\text{d}^{4}x{\text{e}^{\text{i}px}\langle \Omega |\mathsf T[\mathcal{O}(x)\overline{\mathcal{O}(0)]\!\left| \Omega \right\rangle }.\$

下面介绍一下所谓的色散关系 \Pi ({p^2}) = \frac{1}{\pi }\int_0^\infty {\rm{d}{m^2}\frac{\operatorname{Im} \Pi ({m^2})}{m^2} - {p^2} - {\rm{i}\varepsilon }.

其中 $2\mathrm{Im}\Pi (p^2)=\sum _X\int \mathrm{d}{\mathrm{\Pi }}_{\mathrm{X}}{(2\pi )}^4{\delta }^{(4)}(p-p_X)⟨\Omega |\mathcal{O}(0)|X⟩⟨X|\bar{\mathcal{O}}(0)|\Omega ⟩.$

其实只需恒等式 $\mathrm{Im}\frac{1}{m^2}-p^2-\mathrm{i}\epsilon =\pi \delta (m^2-p^2)$ , 来源参考 [此文].

对定义式 \Pi ({p^2}) \equiv \int_0^\infty {\rm{d}{m^2}\frac{\rho ({m^2})}{m^2} - {p^2} - {\rm{i}\varepsilon } 使用恒等式得:

\operatorname{Im} \Pi ({p^2}) \equiv -\int_0^\infty {\rm{d}{m^2}\operatorname{Im} \frac{\rho ({m^2})}{m^2} - {p^2} - {\rm{i}\varepsilon }$=-{\int }_0^{\infty }\mathrm{d}{\mathrm{m}}^2\pi \delta ({\mathrm{m}}^2-{\mathrm{p}}^2)\rho ({\mathrm{m}}^2)=\pi \rho (p^2).$$\Rightarrow \Pi (p^2)\equiv {\int }_0^{\infty }\text{d}{m}^2{\frac{\rho (m^2)}{m}}^2-p^2-\text{i}\epsilon =\frac{1}{\pi }{\int }_0^{\infty }\text{d}{m}^2{\frac{\mathrm{Im}\Pi (m^2)}{m}}^2-p^2-\text{i}\epsilon$, 证毕.

同时由谱密度的定义可知:

$2\mathrm{Im}\Pi (p^2)=2\pi \rho (p^2)\equiv \sum _X\int \mathrm{d}{\mathrm{\Pi }}_{\mathrm{X}}{(2\pi )}^4{\delta }^{(4)}(p-p_X)⟨\Omega |\mathcal{O}(0)|X⟩⟨X|\bar{\mathcal{O}}(0)|\Omega ⟩.$

色散关系其实还可以从两点关联函数的解析性得到:

https://zhuanlan.zhihu.com/p/422978333## 参考

• ^将来若有空的话会试着用最直白的语言介绍一下这个公式的用处.
• ^写完突然想顺便给出色散关系, 就也顺便写写吧.
• ^其实它还很显然是一个实函数, 它具有正定实函数的性质是显然的, 因为它的每个组成部分都是非负实数.
• ^写到这里插入 delta 函数的原因就体现出来了, 除去谱函数的部分, 剩下的其实就和自由理论下的半个 Feynman 传播子是一样的, 所以其实后面的推导就是重复了一遍当年 Feynman 传播子的推导.
• ^其中 x_i 为方程 f(x) = 0 的根