The analytical result for free massless fermion propagator

Question

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.

Answer 1

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 $\frac{1}{x^{2}}$.

Answer 2

Yes it is correct. The derivation in P&S is straightforward but I will expand on it a bit. The key observation is that \begin{equation} \int\frac{d^4k}{(2\pi)^4}e^{-ik\cdot(y-z)}\frac{i\gamma^{\mu}k_{\mu}}{k^2+i\epsilon} =-\gamma^{\mu}\partial_{\mu}\int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2+i\epsilon}e^{-ik\cdot(y-z)}, \end{equation} where the integral on the right hand side is the Feynman propagator for a massless scalar. Performing the $k$ integrals to get to position space yields \begin{equation} \int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2+i\epsilon}e^{-ik\cdot(y-z)}=\frac{i}{4\pi^2}\frac{1}{(y-z)^2-i\epsilon}. \end{equation} If you aren't sure about this last step, it is easier to consider the massive case and then take the limit as $m\rightarrow 0$ at the end. Schwinger parameters are also helpful for proving this identity.

Once you have transformed to position space you simply act with $-\gamma^{\mu}\partial_{\mu}$ to arrive at the final expression.

Answer 3

As alluded to in the other answer here, the integral can basically be evaluated in several ways. One of them is the one that OP has himself followed. (PS - OP, congratulations on completing that feat!)

Let me present here a way of computing this using Schwinger parameterization. We will use $$ \frac{1}{a} = \int_0^\infty d\tau e^{- \tau a} ~, a > 0~. $$ We want to compute $$ I = \int \frac{d^4k}{(2\pi)^4} \frac{i e^{- i k \cdot x}}{k^2+ i \epsilon} = \int \frac{dk^0}{2\pi} \frac{d^3k}{(2\pi)^3} \frac{i e^{- i k^0 t + i \vec{k} \cdot \vec{x} }}{(k^0)^2 - \vec{k}^2+ i \epsilon} $$ Do a Wick rotation $k^0 \to i k^0_E$, $t \to - i t_E$. Then, $$ I = \int \frac{dk_E^0}{2\pi} \frac{d^3k}{(2\pi)^3} \frac{ e^{ k_E^0 t + i \vec{k} \cdot \vec{x} }}{(k_E^0)^2 + \vec{k}^2+ i \epsilon} = \int \frac{d^4k}{(2\pi)^3} \frac{ e^{ - i k_E^0 t_E + i \vec{k} \cdot \vec{x} }}{ k^2 } $$ where in the last equation, we now have a Euclidean $k^2$ that is always positive over the range of integration. Now, we may use the Schwinger parameterization so that $$ I = \int_0^\infty d\tau \int \frac{d^4k}{(2\pi)^4} e^{- k^2 \tau - i k_E^0 t_E + i \vec{k} \cdot \vec{x}} $$ Now, we can do the integral of $k$ quite easily since $k^2 = \sum_i k_i^2$. This gives $$ I = \int_0^\infty d\tau \frac{1}{(4\pi)^2} \frac{e^{ - \frac{1}{4\tau} ( t_E^2 + \vec{x}^2 ) } }{\tau^2} $$ Now to perform the integral over $\tau$, define new integration variable $y = \frac{1}{4\tau} $. Then $$ I = \int_0^\infty dy \frac{1}{(2\pi)^2} e^{ - y ( t_E^2 + \vec{x}^2 ) } $$ This last integral is again the Schwinger parameter one. It converges and is nice so we compute it and find $$ I = \frac{1}{4\pi^2 ( t_E^2 + \vec{x}^2 ) } = - \frac{1}{4\pi^2 ( t^2 - \vec{x}^2 ) } = - \frac{1}{4\pi^2 x^2 } ~. $$ where in the last step, we have performed the inverse Wick rotation to go back to Lorentzian time.

Answer 4

In the following, I will carefully deal with Wick rotation. In the end, I have found that I was confused.

The integration is \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{1}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-(E_{k}-i\epsilon)^{2}}e^{-ik_{0}t}\\ & \equiv & \frac{1}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\times\mathrm{I} \end{eqnarray*}

with \begin{eqnarray*} \mathrm{I} & = & \int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-a^{2}}e^{-ik_{0}t}\\ a & = & E_{k}-i\epsilon=\sqrt{m^{2}+\mathbf{k}^{2}}-i\epsilon \end{eqnarray*}

Now we will use Wick rotation to calculate $\mathrm{I}$. Note that $\pm a$ are two singularities of the integrand. Consider following contour. The radii of coutours $l_{5},l_{6}$ are both $R$ and $R\rightarrow\infty$.

According to contour integral theorem, we can see \begin{eqnarray*} \mathrm{I} & = & \int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-a^{2}}e^{-ik_{0}t}\\ & = & \int_{l_{1}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}+\int_{l_{2}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}\\ & = & \bigg(\int_{l_{5}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}+\int_{l_{3}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}\bigg)+\bigg(\int_{l_{4}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}+\int_{l_{6}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}\bigg)\\ & & \bigg[\text{note: set }z=ik_{E}^{0}\text{ in }l_{3},l_{4}\text{ and combine }l_{5},l_{6}\bigg]\\ & = & (-i)\int_{-\infty}^{\infty}dk_{E}^{0}\frac{1}{(k_{E}^{0})^{2}+a^{2}}e^{tk_{E}^{0}}+\int_{l_{6}}dz\frac{1}{z^{2}-a^{2}}(e^{-izt}+e^{izt}),\ \bigg[\text{set }z=Re^{i\phi}\text{ in }l_{6}\bigg]\\ & = & (-i)\int_{-\infty}^{\infty}dk_{E}^{0}\frac{1}{(k_{E}^{0})^{2}+a^{2}}e^{tk_{E}^{0}}\\ & & -iR\int_{0}^{\frac{\pi}{2}}d\phi e^{i\phi}\frac{1}{R^{2}e^{2i\phi}-a^{2}}(e^{-itR\cos\phi+tR\sin\phi}+e^{itR\cos\phi-tR\sin\phi})\\ & \equiv & (-i)\int_{-\infty}^{\infty}dk_{E}^{0}\frac{1}{(k_{E}^{0})^{2}+a^{2}}e^{tk_{E}^{0}}+\mathrm{II} \end{eqnarray*}

with \begin{eqnarray*} \mathrm{II} & = & -iR\int_{0}^{\frac{\pi}{2}}d\phi e^{i\phi}\frac{1}{R^{2}e^{2i\phi}-a^{2}}(e^{-itR\cos\phi+tR\sin\phi}+e^{itR\cos\phi-tR\sin\phi}) \end{eqnarray*}

Actually, I do not know how to prove $\mathrm{II}=0$ as $R\to\infty$. But if $\mathrm{II}\neq0$ as $R\to\infty$, then we can not simply obtain \begin{eqnarray*} \int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-a^{2}}e^{-ik_{0}t} & = & (-i)\int_{-\infty}^{\infty}dk_{E}^{0}\frac{1}{(k_{E}^{0})^{2}+a^{2}}e^{tk_{E}^{0}} \end{eqnarray*}

I am just confused at this point.

Answer 5

The calculation of the propagator in four dimensions is as follows.

\begin{eqnarray*} \int\frac{d^4 k}{(2\pi)^4}e^{-ik\cdot (x-y)}\frac{1}{k^2} &=& i\int \frac{d^4 k_E}{(2\pi)^4}e^{ik_E\cdot (x_E-y_E)}\frac{1}{-k_E^2} \\ &=& \frac{-i}{(2\pi)^4} \left( \int_0^{2\pi}d\theta_3 \int_0^{\pi}d\theta_2 \sin \theta_2 \right) \int_0^{\infty} dk_E k_E^3 \frac{1}{k_E^2} \int_0^{\pi}d\theta_1 \sin^2 \theta_1 e^{ik_E | x_E-y_E | \cos \theta_1} \\ &=& \frac{-i4\pi}{(2\pi)^4} \int_0^{\infty} dk_E k_E \int_0^{\pi}d\theta_1 \frac{1-\cos 2\theta_1}{2} e^{ik_E | x_E-y_E | \cos \theta_1} \\ &=& \frac{-i}{4\pi^3} \frac{1}{| x_E-y_E |^2} \int_0^{\infty} ds s (\frac{\pi}{2} J_0(s)- \frac{\pi i^2}{2} J_2(s)) \end{eqnarray*} where $s\equiv k_E\| x_E-y_E \| $, and $J_n(s)$'s are bessel functions and I made use of Hansen-Bessel Formula.
\begin{eqnarray*} &=& \frac{-i}{4\pi^3} \frac{1}{| x_E-y_E |^2} \int_0^{\infty} ds s \frac{\pi}{2} \frac{2}{s} J_1(s) \\ &=& -\frac{i}{4\pi^2} \frac{1}{| x_E-y_E|^2} \int_0^{\infty} ds \, J_1(s) \\ &=& -\frac{i}{4\pi^2} \frac{1}{| x_E-y_E |^2} \\ &=& \frac{i}{4\pi^2} \frac{1}{(x-y)^2} \end{eqnarray*}