For massless fermion, the free propagator in quantum field theory
is
\begin{eqnarray*}
& & \langle0|T\psi(x)\bar{\psi}(y)|0\rangle=\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}.
\end{eqnarray*}
In Peskin & Schroeder's book, An introduction to quantum field theory
(edition 1995, page 660, formula 19.40), they obtained the analytical
result for this propagator,
\begin{eqnarray*}
& & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}=-\frac{i}{2\pi^{2}}\frac{\gamma\cdot(x-y)}{(x-y)^{4}} .\tag{19.40}
\end{eqnarray*}
Question: Is this analytical result right? Actually I don't know
how to obtain it.
Best Answer
Method One:
\begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int dk_{0}e^{-ik_{0}x_{0}}\frac{1}{[k_{0}+(|\mathbf{k}|-i\epsilon)][k_{0}-(|\mathbf{k}|-i\epsilon)]}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\bigg(\theta(x_{0})(-2\pi i)\frac{1}{2|\mathbf{k}|}e^{-i|\mathbf{k}|x_{0}}+\theta(-x_{0})(2\pi i)\frac{1}{-2|\mathbf{k}|}e^{i|\mathbf{k}|x_{0}}\bigg)\\ & = & \frac{\pi}{(2\pi)^{4}}\int d^{3}k\frac{1}{|\mathbf{k}|}e^{i\mathbf{k}\cdot\mathbf{x}}e^{-i|\mathbf{k}||x_{0}|}\\ & = & \frac{\pi}{(2\pi)^{4}}2\pi\int_{0}^{\infty}d|\mathbf{k}||\mathbf{k}|^{2}\frac{1}{|\mathbf{k}|}e^{-i|\mathbf{k}||x_{0}|}\int_{-1}^{1}dye^{i|\mathbf{k}||\mathbf{x}|y}\\ & = & \frac{\pi}{(2\pi)^{4}}2\pi\int_{0}^{\infty}d|\mathbf{k}||\mathbf{k}|e^{-i|\mathbf{k}||x_{0}|}\frac{1}{i|\mathbf{k}||\mathbf{x}|}(e^{i|\mathbf{k}||\mathbf{x}|}-e^{-i|\mathbf{k}||\mathbf{x}|})\\ & = & \frac{\pi}{(2\pi)^{4}}2\pi\frac{1}{i|\mathbf{x}|}\int_{0}^{\infty}d|\mathbf{k}|(e^{-i|\mathbf{k}|(|x_{0}|-|\mathbf{x}|)}-e^{-i|\mathbf{k}|(|x_{0}|+|\mathbf{x}|)})\\ & = & \frac{\pi}{(2\pi)^{4}}2\pi\frac{1}{i|\mathbf{x}|}\bigg[\bigg(\pi\delta(|x_{0}|-|\mathbf{x}|)-\mathscr{P}\frac{i}{|x_{0}|-|\mathbf{x}|}\bigg)-\bigg(\pi\delta(|x_{0}|+|\mathbf{x}|)-\mathscr{P}\frac{i}{|x_{0}|+|\mathbf{x}|}]\bigg)\bigg]\\ & = & \frac{\pi}{(2\pi)^{4}}2\pi\frac{1}{i|\mathbf{x}|}\bigg[\pi\delta(|x_{0}|-|\mathbf{x}|)+\mathscr{P}\bigg(\frac{i}{|x_{0}|+|\mathbf{x}|}-\frac{i}{|x_{0}|-|\mathbf{x}|}\bigg)\bigg]\\ & = & \frac{\pi}{(2\pi)^{4}}2\pi\frac{1}{i|\mathbf{x}|}\bigg(2\pi|\mathbf{x}|\delta(x^{2})-2i|\mathbf{x}|\mathscr{P}\frac{1}{x^{2}}\bigg)\\ & = & \frac{\pi}{(2\pi)^{4}}2\pi\frac{1}{i|\mathbf{x}|}(-2i|\mathbf{x}|)\bigg(\mathscr{P}\frac{1}{x^{2}}+i\pi\delta(x^{2})\bigg)\\ & = & -\frac{1}{4\pi^{2}}\frac{1}{x^{2}-i\epsilon} \end{eqnarray*}
i.e. \begin{eqnarray*} \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x} & = & -\frac{1}{(2\pi)^{2}}\frac{1}{x^{2}-i\epsilon} \end{eqnarray*}
where we have used \begin{eqnarray*} \frac{1}{x+i\epsilon} & = & \mathscr{P}\frac{1}{x}-i\pi\delta(x) \end{eqnarray*}
\begin{eqnarray*} \int_{0}^{\infty}e^{ikx}dk & = & \lim_{\epsilon\to0^{+}}\int_{0}^{\infty}e^{ik(x+i\epsilon)}dk=\lim_{\epsilon\to0^{+}}\frac{i}{x+i\epsilon}=\pi\delta(x)+\mathscr{P}\frac{i}{x} \end{eqnarray*} \begin{eqnarray*} \delta(x^{2}) & = & \delta(x_{0}^{2}-\mathbf{x}^{2})=\frac{1}{2|\mathbf{x}|}[\delta(x_{0}-|\mathbf{x}|)+\delta(x_{0}+|\mathbf{x}|)]\\ & = & \frac{1}{2|\mathbf{x}|}[\theta(x_{0})\delta(x_{0}-|\mathbf{x}|)+\theta(-x_{0})\delta(x_{0}+|\mathbf{x}|)]\\ & = & \frac{1}{2|\mathbf{x}|}\delta(|x_{0}|-|\mathbf{x}|) \end{eqnarray*}
Method Two:
We can also see that \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+m^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int dk_{0}e^{-ik_{0}x_{0}}\frac{1}{[k_{0}+(E_{k}-i\epsilon)][k_{0}-(E_{k}-i\epsilon)]}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\bigg(\theta(x_{0})(-2\pi i)\frac{1}{2E_{k}}e^{-iE_{k}x_{0}}+\theta(-x_{0})(2\pi i)\frac{1}{-2E_{k}}e^{iE_{k}x_{0}}\bigg)\\ & = & \frac{i(-2\pi i)}{2(2\pi)^{4}}\int d^{3}k\frac{1}{E_{k}}e^{i\mathbf{k}\cdot\mathbf{x}}e^{-iE_{k}|x_{0}|}\\ & = & \frac{i(-2\pi i)(2\pi)}{2(2\pi)^{4}}\int_{0}^{\infty}dkk^{2}\frac{1}{E_{k}}e^{-iE_{k}|x_{0}|}\int_{-1}^{1}dye^{ik|\mathbf{x}|y}\\ & = & \frac{i(-2\pi i)(2\pi)}{2(2\pi)^{4}}\int_{0}^{\infty}dkk^{2}\frac{1}{E_{k}}e^{-iE_{k}|x_{0}|}\frac{1}{ik|\mathbf{x}|}(e^{ik|\mathbf{x}|}-e^{-ik|\mathbf{x}|})\\ & = & \frac{i(-2\pi i)(2\pi)2}{2(2\pi)^{4}}\int_{0}^{\infty}dkk\frac{1}{E_{k}}e^{-iE_{k}|x_{0}|}\frac{\mathrm{sin}(k|\mathbf{x}|)}{|\mathbf{x}|}\\ & = & \frac{1}{(2\pi)^{2}}\int_{0}^{\infty}dkk\frac{1}{E_{k}}e^{-iE_{k}|x_{0}|}\frac{\mathrm{sin}(k|\mathbf{x}|)}{|\mathbf{x}|}\\ & \equiv & \frac{1}{(2\pi)^{2}}\bigg(\theta(x^{2})\times\mathrm{I}+\theta(-x^{2})\times\mathrm{II}\bigg) \end{eqnarray*}
If $x^{2}>0$, we can choose a frame with $x^{\mu}=(x_{0},\mathbf{0})$ and $x^{2}=x_{0}^{2}$. We can see \begin{eqnarray*} \mathrm{I} & = & \int_{0}^{\infty}dkk\frac{1}{E_{k}}e^{-iE_{k}|x_{0}|}\frac{\mathrm{sin}(k|\mathbf{x}|)}{|\mathbf{x}|}=\int_{0}^{\infty}dkk^{2}\frac{1}{E_{k}}e^{-iE_{k}|x_{0}|}\\ & = & \int_{m}^{\infty}dE_{k}\sqrt{E_{k}^{2}-m^{2}}e^{-iE_{k}|x_{0}|}\\ & = & m^{2}\int_{1}^{\infty}dt\sqrt{t^{2}-1}e^{-im|x_{0}|t},\ \bigg[\text{note: }a\equiv m|x_{0}|=m\sqrt{x^{2}}\bigg]\\ & = & m^{2}\int_{1}^{\infty}dt(t^{2}-1)\frac{e^{-iat}}{\sqrt{t^{2}-1}}\\ & = & -m^{2}(\frac{\partial^{2}}{\partial a^{2}}+1)\int_{1}^{\infty}dt\frac{e^{-iat}}{\sqrt{t^{2}-1}}\\ & = & \frac{i\pi m^{2}}{2}[H_{0}^{(2)\prime\prime}(a)+H_{0}^{(2)}(a)] \end{eqnarray*}
where we have used \begin{eqnarray*} \int_{1}^{\infty}dt\frac{e^{-iat}}{\sqrt{t^{2}-1}} & = & -\frac{i\pi}{2}J_{0}(a)-\frac{\pi}{2}N_{0}(a)=-\frac{i\pi}{2}H_{0}^{(2)}(a),\ (a>0) \end{eqnarray*}
From
\begin{eqnarray*} Z_{\nu}^{\prime} & = & Z_{\nu-1}-\frac{\nu}{x}Z_{\nu}\\ Z_{\nu}^{\prime} & = & -Z_{\nu+1}+\frac{\nu}{x}Z_{\nu} \end{eqnarray*}
we can see \begin{eqnarray*} Z_{0}^{\prime} & = & -Z_{1}\\ Z_{0}^{\prime\prime} & = & -Z_{1}^{\prime}=-(Z_{0}-\frac{1}{x}Z_{1}) \end{eqnarray*}
i.e. \begin{eqnarray*} Z_{0}^{\prime\prime}+Z_{0} & = & \frac{1}{x}Z_{1} \end{eqnarray*}
So we have \begin{eqnarray*} \mathrm{I} & = & \frac{i\pi m^{2}}{2}[H_{0}^{(2)\prime\prime}(a)+H_{0}^{(2)}(a)]=\frac{i\pi m^{2}}{2a}H_{1}^{(2)}(a)=\frac{i\pi m}{2\sqrt{x^{2}}}H_{1}^{(2)}(m\sqrt{x^{2}}) \end{eqnarray*}
If $x^{2}<0$, we can choose a frame with $x^{\mu}=(0,\mathbf{x})$ and $x^{2}=-|\mathbf{x}|^{2}$. We can see \begin{eqnarray*} \mathrm{II} & = & \int_{0}^{\infty}dkk\frac{1}{E_{k}}e^{-iE_{k}|x_{0}|}\frac{\mathrm{sin}(k|\mathbf{x}|)}{|\mathbf{x}|}=\frac{1}{|\mathbf{x}|}\int_{0}^{\infty}dkk\frac{\mathrm{sin}(k|\mathbf{x}|)}{\sqrt{m^{2}+k^{2}}}\\ & = & \frac{m}{|\mathbf{x}|}\int_{0}^{\infty}dt\frac{t\mathrm{sin}(m|\mathbf{x}|t)}{\sqrt{1+t^{2}}},\ \bigg[\text{note: }b\equiv m|\mathbf{x}|=m\sqrt{-x^{2}}\bigg]\\ & = & \frac{m^{2}}{b}\int_{0}^{\infty}dt\frac{t\mathrm{sin}(bt)}{\sqrt{1+t^{2}}}\\ & = & -\frac{m^{2}}{b}\frac{\partial}{\partial b}\int_{0}^{\infty}dt\frac{\mathrm{cos}(bt)}{\sqrt{1+t^{2}}}\\ & = & -\frac{m^{2}}{b}K_{0}^{\prime}(b) \end{eqnarray*}
where we have used \begin{eqnarray*} \int_{0}^{\infty}dt\frac{\mathrm{cos}(bt)}{\sqrt{1+t^{2}}} & = & K_{0}(b),\ (x>0) \end{eqnarray*}
From
\begin{eqnarray*} Z_{\nu}^{\prime} & = & Z_{\nu-1}-\frac{\nu}{x}Z_{\nu}\\ Z_{\nu}^{\prime} & = & -Z_{\nu+1}+\frac{\nu}{x}Z_{\nu} \end{eqnarray*}
we can see \begin{eqnarray*} Z_{0}^{\prime} & = & -Z_{1} \end{eqnarray*}
So we get \begin{eqnarray*} \mathrm{II} & = & -\frac{m^{2}}{b}K_{0}^{\prime}(b)=\frac{m^{2}}{b}K_{1}(b)=\frac{m}{\sqrt{-x^{2}}}K_{1}(m\sqrt{-x^{2}}) \end{eqnarray*}
Finally we have \begin{eqnarray*} \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+m^{2}+i\epsilon}e^{-ik\cdot x} & = & \frac{1}{(2\pi)^{2}}\bigg(\theta(x^{2})\times\frac{i\pi m}{2\sqrt{x^{2}}}H_{1}^{(2)}(m\sqrt{x^{2}})+\theta(-x^{2})\times\frac{m}{\sqrt{-x^{2}}}K_{1}(m\sqrt{-x^{2}})\bigg)\\ & = & \frac{m}{(2\pi)^{2}\sqrt{|x^{2}|}}\bigg(\theta(x^{2})\frac{i\pi}{2}H_{1}^{(2)}(m\sqrt{x^{2}})+\theta(-x^{2})K_{1}(m\sqrt{-x^{2}})\bigg) \end{eqnarray*}
then we have \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \lim_{m\to0}\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+m^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \lim_{m\to0}\frac{m}{(2\pi)^{2}\sqrt{|x^{2}|}}\bigg(\theta(x^{2})\frac{i\pi}{2}\frac{2i}{\pi m\sqrt{x^{2}}}+\theta(-x^{2})\frac{1}{m\sqrt{-x^{2}}}\bigg)\\ & = & \frac{1}{(2\pi)^{2}}\bigg(-\theta(x^{2})\frac{1}{x^{2}}+\theta(-x^{2})\frac{1}{-x^{2}}\bigg)\\ & = & -\frac{1}{(2\pi)^{2}}\frac{1}{x^{2}} \end{eqnarray*}
where we have used \begin{eqnarray*} H_{1}^{(2)}(x) & = & J_{1}(x)-iN_{1}(x)\rightarrow\frac{2i}{\pi x}\ \ \text{as}\ \ x\rightarrow0\\ K_{1}(x) & \rightarrow & \frac{1}{x}\ \ \text{as}\ \ x\rightarrow0 \end{eqnarray*}
So we get \begin{eqnarray*} \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x} & = & -\frac{1}{(2\pi)^{2}}\frac{1}{x^{2}} \end{eqnarray*}
with $\frac{1}{x^{2}-i\epsilon}$ replaced by $\frac{1}{x^{2}}$.