Exponential decay of propagator outside lightcone

Question

In Tong's lecture notes (http://www.damtp.cam.ac.uk/user/tong/qft.html) page 38, he calculates the following propagator:

$$D(x-y) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_\vec{p}} e^{-ip \cdot (x-y)}.\tag{2.90}$$

Then, he states that for spacelike separations,i.e. $(x-y)^2 < 0$, one can show that $$D(x-y) \sim e^{-m|\vec{x} - \vec{y} |}.\tag{2.91}$$

But how does he do this? There has been another post that addresses this (Exponential decay of Feynman propagator outside the lightcone) but the answers never go into how exactly Tong showed this.

This is my attempt:

Note that the propagator is Lorentz invariant, as well as $(x-y)^2$. So in order to have $(x-y)^2 < 0$, I pick the following events at equal times: $x = (t, \vec{x})$, $y = (t, \vec{y})$, $p = (p_0, \vec{p})$ such that $(x-y)^2 < 0$.

We then get $(x-y) = (0, \vec{x} - \vec{y})$ such that $p \cdot (x-y) = -\vec{p} \cdot (\vec{x} - \vec{y})$

and finally

$$D(x-y) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_\vec{p}} e^{i\vec{p} \cdot (\vec{x}-\vec{y})}.\tag{2.90}$$

Note that this is analogous to eq. 2.89 on page 37 of Tong's lecture notes. But then I do not get the exponential decay. How do I continue?

My expression should be correct since this can be used to show eq. 2.92.

Answer 1

Such asymptotic behaviour are typically calculated using the Laplace method (which is generalised to the saddle point method). It's worth looking into in depth, you'll use it again and again in QFT. You want to estimate (setting $m=1$ to fix the energy scale): $$ D(\vec x) = \int \frac{d^n\vec p}{(2\pi)^n2(\vec p^2+1)^s}e^{-i\vec p\cdot \vec x} $$ with $n=3,s=1/2$ in your case. Before applying Laplace, you'll first need to convert it to a 1D integral and make isotropy more evident by making it only depend on $x=|\vec x|$. In 3D, you can just do the angular integral which is accidentally easy, but instead I'll show you a method applicable in arbitrary dimensions and arbitrary powers. A general trick is to notice that (Mellin transform): $$ \frac1{x^s} = \frac1{\Gamma(s)}\int_0^{+\infty} e^{-xt}t^{s-1}dt $$ so this reduces the $n$-dimensional integral to a gaussian integral: $$ \begin{align} D(\vec x) &= \int_0^{+\infty} dt\frac{t^{s-1}e^{-t}}{\Gamma(s)}\int \frac{d^n\vec p}{(2\pi)^n2}e^{-i\vec p\cdot \vec x-t\vec p^2} \\ &= \frac{1}{2(4\pi)^{n/2}\Gamma(s)}\int_0^{+\infty} t^{s-1-n/2}e^{-t-x^2/4t}dt \\ &= \frac{1}{2(4\pi)^{n/2}\Gamma(s)}\left(\frac x2\right)^{s-n/2}\int_0^{+\infty} t^{s-1-n/2}e^{-\frac x2(t+1/t)}dt \end{align} $$ At this stage, you could recognise the integral representation of a Bessel function and look up its asymptotic behaviour and call it a day. The rest of the calculations will therefore be to compute the asymptotic behaviour of the Bessel function. The stage is set for applying Laplace's method, by sending $x\to\infty$. The argument of the exponential is where $t+1/t$ is maximised, i.e. $t=1$ and $t+1/t=2$. Therefore: $$ D(\vec x) \asymp e^{-x} $$ as claimed. You can even do better by Taylor expanding and doing the gaussian integral, which will give you an asymptotic equivalent: $$ D(\vec x) \sim \frac{1}{2(4\pi)^{n/2}\Gamma(s)}\left(\frac x2\right)^{s-n/2}e^{-x}\sqrt{\frac{2\pi}x} $$

Hope this helps.