# Exponential decay of Feynman propagator outside the lightcone

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.

• Is your question about why we expect $e^{-mx}$ a priori or why we expect to get that from that particular integral? – Aaron Feb 25 '17 at 21:30
• 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 – Bob Bee Feb 25 '17 at 22:35
• @Aaron Why would we expect that from that particular integral. – alonso s Feb 25 '17 at 23:47
• @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 $D(x-y)\sim e^{-m|x-y|}$" – alonso s Feb 25 '17 at 23:47
• 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 – Bob Bee Feb 26 '17 at 1:15

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.
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...