In Chapter three (I.3) of A. Zee's Quantum Field Theory in a Nutshell, the author derives the Feynman propagator for a scalar field: $$ \begin{aligned} D(x)&=\int \frac{\operatorname{d}^4 \mathbf{k}}{(2\pi)^4} \frac{e^{ikx}}{k^2-m^2+i\epsilon} \\ &=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\omega_k} \left[e^{-i(\omega_kt-\mathbf{k}\cdot \mathbf{x})}\theta(t)+ e^{i(\omega_kt-\mathbf{k}\cdot \mathbf{x})}\theta(-t)\right] \end{aligned} $$ where $\omega_k=\sqrt{\mathbf{k}^2+m^2}$.

Without working through the $\mathbf{k}$ integral, the behavior of the propagator for events inside and outside the light-cone can be roughtly analyzed (or so the text states): for time-like events in the future cone, e.g., $x=(t,\mathbf{x}=0)$, with $t>0$, the propagator is a sum of plane waves $$D(t,0)=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\omega_k} e^{-i\omega_kt}$$ Likewise, for time-like events in the past cone ($t<0$) the propagator is a sum of plane waves with the opposite phase.

Now, for space-like events, e.g., $x=(0,\mathbf{x})$, after interpreting $\theta(0)=\frac{1}{2}$ and observing the propagator allows for the exchange $\mathbf{k}\rightarrow -\mathbf{k}$, we obtain $$D(0,\mathbf{x})=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\sqrt{\mathbf{k}^2+m^2}} e^{-i\mathbf{k}\cdot \mathbf{x}}$$

The author then states that "...the square root cut starting at $\pm im$ leads to an exponential decay $\sim e^{-m|\mathbf{x}|}$, as we would expect." It is left to the reader to verify this as a later problem.

The question is: how can I see that the above is true, without going through the $\mathbf{k}$ integral?

Secondarily, what does "the square root cut starting at $\pm im$" mean? I know that one must supply the complex square root with a branch cut, but said branch cut must be a whole ray of the plane, not just a segment.

I have tried going through the integral; by rotating the $\mathbf{k}$ so that $\mathbf{x}$ points along the $k^3$ direction and switching to spherical coordinates ($k=|\mathbf{k}|, x=|\mathbf{x}|$) the integral becomes:

$$ \begin{aligned} D(0,\mathbf{x})&=-i\int_0^\infty \operatorname{d}k \int_0^\pi \operatorname{d}\theta \int_0^{2\pi} \operatorname{d}\varphi \left( \frac{k^2 \sin{\theta}e^{-i kx\cos{\theta}}}{(2\pi)^3 2\sqrt{k^2+m^2}} \right)\\ &=-i\int_0^\infty \operatorname{d}k \int_0^\pi \operatorname{d}\theta \left( \frac{k^2 \sin{\theta}e^{-i kx\cos{\theta}}}{(2\pi)^2 2\sqrt{k^2+m^2}} \right)\\ &=\frac{-i}{(2\pi)^2}\int_0^\infty \operatorname{d}k \left( \frac{k}{2ix\sqrt{k^2+m^2}}\right)e^{ikx}-e^{-ikx}\\ &=\frac{-i}{(2\pi)^2}\int_0^\infty \operatorname{d}k \frac{k\sin{kx}}{x\sqrt{k^2+m^2}}\sim \frac{1}{|\mathbf{x}|} \end{aligned} $$ Which is not the desired result.

  • 2
    $\begingroup$ Is your question about why we expect $e^{-mx}$ a priori or why we expect to get that from that particular integral? $\endgroup$
    – Aaron
    Commented Feb 25, 2017 at 21:30
  • $\begingroup$ It is explained well in Tong's QFT lectures at damtp.cam.ac.uk/user/tong/qft.html. I would write the answer if I remembered, it was not hard to follow. I believe it is in Lecture 2, but cannot be sure. It's a little bit of a shock when one first sees it, you'd think it should be 0 so c is not exceeded, but the exponential decay is close enough. If you replace k= +/- im, you see the exp(-ikx) becomes exp(-m|x|), but (and too lazy to reason it exactly now) you need to integrate around infinity and around the cut of the singularities in a certain way (up/down or reverse). See Tong $\endgroup$
    – Bob Bee
    Commented Feb 25, 2017 at 22:35
  • $\begingroup$ @Aaron Why would we expect that from that particular integral. $\endgroup$
    – alonso s
    Commented Feb 25, 2017 at 23:47
  • $\begingroup$ @BobBee From D. Tong's notes: "The function $D(x-y)$ is called the propagator. For spacelike separation $(x-y)^2<0$, one can show that $D(x-y)$ decays like $D(x-y)\sim e^{-m|x-y|}$" $\endgroup$
    – alonso s
    Commented Feb 25, 2017 at 23:47
  • $\begingroup$ Maybe it was in the actual video lectures, I was convinced. Never saw his lecture notes, but if I remember right he showed how to derive it per the integral around the singularities and at infinity. Thanks $\endgroup$
    – Bob Bee
    Commented Feb 26, 2017 at 1:15

2 Answers 2


See the Wikipedia article on the Feynman propagator. It's real-space form is: $$G_F(x,y) = \left\{\begin{array}{cc} -\frac{1}{4\pi}\delta(\tau^2) + \frac{m}{8\pi \tau} H_1^{(2)}(m\tau) & \tau^2 \ge 0 \\ -\frac{im}{4\pi^2 \sqrt{|\tau|}} K_1(m|\tau|) & \tau^2 < 0, \end{array}\right.$$ where $\tau^2 \equiv (x^0 - y^0)^2 - (\vec{x} - \vec{y})^2$, $H_1^{(2)}$ is a Hankel function and $K_1$ is a modified Bessel function of the second kind. The desired result follows directly from the asymptotic properties of $K_1$ for large arguments.

If the desired result is to explore the properties of the integral, then you may find an answer digging around in the integral representations of the modified Bessel functions.

If you are fine with a proof that the modified Bessel function is part of a propagator, though not particularly the Feynman propagator, and aren't concerned about the light cone delta function, it follows from analytic continuation of the 4-d Euclidean Green's function.


What I will do isn't mathematicaly valid, but I think it works a little like physical intuition: $$\oint \frac{e^{-ikx} \frac{k^2}{\sqrt{k-im}}}{\sqrt{k+im}} \approx \frac{-m^2}{\sqrt{-2im}} \oint \frac{e^{-ikx} }{\sqrt{k+im}}$$ Usin fractional calculus generalization for cauchy's integral formula: $$\frac{-m^2}{\sqrt{-2im}} \frac{2\pi i}{\Gamma(1/2)} \frac{\partial^{\frac{1}{2}} e^{-ikx}}{\partial k^{\frac{1}{2}}} \approx \frac{-m^2}{\sqrt{-2im}} \frac{2\pi i}{\sqrt{-i\pi x}} e^{-mx}$$... well... I got a $1/\sqrt{x}$ but it has the exponential...


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