# Fourier Transform of Feynman Propagator [closed]

I have been trying to Fourier Transform Feynman Propagator for scalar fields. Below is the expression of Feynman Propagator in momentum space:

$$\bigtriangleup_{F}(k)=\frac{1}{k^{2}-m^{2}+i\epsilon}$$ Can someone show all steps of the Fourier Transform of this propagator to the position space?

Let me show my understanding of the problem. The following relation gives Fourier Transform:

$$\bigtriangleup_{F}(x)=\int \frac{d^{4}p}{{2\pi}^4}\bigtriangleup_{F}(k)e^{ikx}$$

Do I only have to insert the value of the $$\bigtriangleup_{F}(k)$$ in the second expression or is there a more elegant way to do it?

• Could someone explain why this question got downvoted? It appropriately had the homework-and-exercises tag, was phrased concisely, and wrote down the equation in momentum space. I'm not the original poster, but I am curious about the downvotes and would like to know how this question can be improved. Sep 21, 2021 at 17:15
• The improvement would be an attempted solution which points out where the OP got stuck. Sep 21, 2021 at 17:22
• I am stuck from the start. I know how to move from position to momentum space but completely stuck here. I know some of relations of delta functions and momentum eigen states in position representation, but completely failed to put the pieces together. Sep 21, 2021 at 17:44
• @LittlePhysicist I guess you have already obtained the Fourier transform. Sep 21, 2021 at 18:24
• Thanks for the explanation @Connor Behan! Sep 21, 2021 at 18:49

## Definitions

We will perform this in general dimension $$d$$, setting d=4 only when required; this makes some steps clearer. Let us begin with a definition of what it we want to calculate; note that we use the mostly negative metric signature ($$+-\ldots-$$ here). \begin{align} \Delta_F(x) &= \int \frac{d^d k}{(2\pi)^d} \frac{1}{k^2-m^2+i\epsilon}e^{-i k x} \end{align} Our goal is to make the $$k^0$$ integral equal to a contour integral, which we evaluate via the residue theorem, rather than manually evaluating it. We then do the easy (d-2) spatial integrals, leaving the final one in a form that we recognise as the integral representation of a special function, which for $$d=4$$ is $$K_1(z)$$.

We will also assume that $$x^2 \ge 0$$; the $$x^2 <0$$ case is similar (see Huang's QFT, section 2.9).

## Contour integration

We will close the real $$k^0$$ integration from $$(-\infty, \infty)$$ via an infinitely large semicircle. Ideally we want the contribution of these semicircles to be zero, such that $$\int_{-\infty}^\infty dk^0 \, f(k) = \int_{-\infty}^\infty dk^0 \, f(k) + \int_{\text{semicircle}} dk^0 \, f(k) = \int_{\gamma} dk^0 \, f(k) = \text{ residues at poles}.$$

In our case, this will only happen if we have $$\mathrm{Re}[-ik_0 x_0] < 0$$ for the $$k_0$$ points on this semicircle, so that the exponent does not blow up; this means that we demand that $$\mathrm{Im}[k_0]x_0 <0$$. This means that:

• If $$x^0$$ > 0, we need to close where $$\mathrm{Im}[k_0] <0$$, i.e. the lower half-plane.
• If $$x^0$$ < 0, we need to close where $$\mathrm{Im}[k_0] > 0$$, i.e. the upper half-plane.

In both cases, the semicircular arc contributes zero to the integral, by construction.

We need to choose different contours depending on the sign of $$x^0$$; we will implement this by the step function $$\theta$$. The two contours completed by semicircles above and below the line are $$\gamma^+$$ and $$\gamma^-$$, and they enclose the poles at $$k=-\omega$$ and $$\omega$$ respectively, where $$\omega = \sqrt{|\vec{k}|^2+m^2-i\epsilon}$$.

\begin{align} \Delta_F (x)&= \theta(x^0) \int_{\mathbb{R}^{d-1}}\frac{d^{d-1} k}{(2\pi)^{d}} \int_{\gamma^-} d k^0 \frac{e^{-ikx}}{k^2-m^2+i\epsilon}+\theta(-x^0) \int_{\mathbb{R}^{d-1}}\frac{d^{d-1} k}{(2\pi)^{d}} \int_{\gamma^+} d k^0\frac{e^{-ikx}}{k^2-m^2+i\epsilon}\\ &= \Delta^{+}_F(x) + \Delta^{-}_F(x)\\ \Rightarrow \Delta^{\pm}_F (x)&=\theta(\pm x^0) \int_{\mathbb{R}^{d-1}}\frac{d^{d-1} k}{(2\pi)^d} \int_{\gamma^\mp } d k^0 \frac{1}{2\omega}\left[\frac{1}{k^0-\omega}-\frac{1}{k^0+\omega} \right]e^{-ikx} \end{align} This is a trivial exercise in contour integration for $$k^0$$. \begin{align} \Delta^{\pm}_F (x)&=-2\pi i \theta(\pm x^0) \int_{\mathbb{R}^{d-1}} \frac{d^{d-1}k}{(2\pi)^{d}}\frac{1}{2\omega}e^{\mp i \omega x^0}e^{i\vec{k} \cdot \vec{x}}\\ &=-i \theta(\pm x^0) \int_{\mathbb{R}^{d-1}} \frac{d^{d-1} k}{(2\pi)^{d-1}}\frac{1}{2\omega}e^{- i \omega |x^0|}e^{i\vec{k} \cdot \vec{x}}\\ \end{align} Adding the two back together, since the expressions are identical the $$\theta$$s disappear: \begin{align} \Delta_F (x)&=-\frac{i}{2}\int_{\mathbb{R}^{d-1}} \frac{d^{d-1} k}{(2\pi)^{d-1}}\frac{1}{\omega}e^{- i \omega |x^0|}e^{i\vec{k} \cdot \vec{x}} \end{align}

## Rewriting the spatial integrations

Now let's cheat, and use the fact that the answer must be Lorentz symmetric, and therefore only a function of $$x^2$$. Assuming that $$x^2 >0$$, we can therefore boost to a frame where $$\vec{x}=0$$, evaluate the integral, and boost back later. We will use spherical coordinates in $$(d-1)=3$$ dimensions, with $$l=|\vec{k}|$$, where we note that

$$\int d \Omega_{d-2} = \Omega_{d-2} = \frac{2 \pi^{\frac{d-1}{2}}}{\Gamma(\frac{d-1}{2})}\, \text{= surface area of a (d-2)-sphere}$$

This does not actually converge, so let's shift $$|x^0| \to |x^0| -i\delta$$, $$\delta \rightarrow 0$$; this to ensure convergence of the integral, by making the integrand vanish for large $$l$$. We are left with: \begin{align} \Delta_F(x)&=-\frac{i}{2} \frac{1}{(2\pi)^{d-1}}\left(\int d \Omega_{d-2}\right) \lim_{\delta \rightarrow 0} \int_{0}^\infty dl \frac{l^{d-2}}{\sqrt{l^2+m^2}}e^{-i \sqrt{l^2+m^2}(|x^0| - i \delta)}\\ &=-\frac{1}{2} \frac{\Omega_{d-2}}{(2\pi)^{d-1}} (d-3) \lim_{\delta \rightarrow 0} \frac{1}{|x^0|-i\delta}\int_{\mathbb{R}^+} dl \, l^{d-4} e^{- i \sqrt{l^2+m^2}(|x^0| - i \delta)} \end{align} Where to reach the second line we just integrated by parts.

## Setting $$d=1+3$$

We now set $$d=4$$, and use the usual $$\Omega_2 = 4\pi$$. Now making a change of variable $$\frac{|\vec{k}|}{m}=\sinh \alpha$$ we obtain: \begin{align} \Delta_F (x)= -\frac{1}{2} \frac{m}{2 \pi^2} \lim_{\delta \rightarrow 0}\frac{1}{|x^0|-i\delta}\int_{\mathbb{R}^+}d\alpha \cosh \alpha\,e^{- i m \cosh \alpha\,(|x^0| - i \delta)} \end{align} This is a Bessel function in disguise, since if the following integral exists, we have \begin{align} \int_{\mathbb{R}^+}d\alpha \cosh \alpha\,e^{- z \cosh \alpha} = K_1(z) \end{align} To ensure the integral converges, we need to close the contour of integration by using the trick of "go to the infinity, make a small angle upward and then go back to the origin".

We finally boost back to a frame where $$\vec{x} \neq \vec{0}$$; by Lorentz invariance of the final answer, all we need to do is modify $$|x^0| \rightarrow \sqrt{(x^0)^2-|\vec{x}|^2}\equiv \sqrt{x^2}$$. One finds: $$\begin{equation} \Delta_F(x)= -\frac{m}{4\pi^2} \lim_{\delta \rightarrow 0} \frac{1}{\sqrt{x^2}-i\delta} K_1(im (\sqrt{x^2}-i\delta)) \end{equation}$$

As described nicely here, this can then be rewritten without the $$\delta$$ limit using the principal-value prescription (See Weinberg QFT I, around equation 3.1.22); in doing we obtain the delta function on the light-cone mentioned in the comments.

## Coefficient cross-check, via the massless limit.

We might have lost some factors of $$i$$, $$-1$$, 2, or $$\pi$$ in the above process, so it is always worth a cross-check. As usual, we compare our result to a limit where we know (or know how to find) the answer.

Taking the limit as $$m\to 0^+$$, we obtain the following $$\begin{equation} \Delta^{\text{massless}}_F(x)= \frac{i}{4\pi^2} \frac{1}{(\sqrt{x^2}-i\delta)^2} = \frac{i}{4\pi^2} \frac{1}{x^2-i\delta} \end{equation}$$ which we can cross-check, say, with Peskin & Schroder equation 19.40, which calculates the position space free massless fermion propagator with our conventions, using in the process exactly the result we want to check in the process. \begin{align} \int \frac{d^4 k}{(2\pi)^4} \frac{i \not{k}}{k^2}e^{-i k (y-z)} &= -\not{\partial_y}\left( \int \frac{d^4 k}{(2\pi)^4} \frac{1}{k^2}e^{-i k (y-z)}\right)\\ & = -\not{\partial_y} \left(\frac{i}{4\pi} \frac{1}{(y-z)^2}\right)\\ & = -\frac{i}{2\pi^2} \frac{\gamma^\mu (y-z)_\mu}{(y-z)^4} \end{align}

• @SeanE.Lake ludoft edited the answer to address your comment. Note that this is not my answer anymore because this person has completely rewritten it. Jun 21 at 17:42
• @ludoft I accepted this complete edit because it added a lot to my initial answer. But you should know that instead of rewriting it completely, it would be better to write another answer, even if the OP is one year old. Jun 21 at 17:44