I just started studying Quantum Field Theory in a Nutshell By Zee to quench my thirst for physics. But I got stuck on one of the early exercises. In exercise 1.3.1 (and 1.3.2), the question asks to work out the propagator $D(x)$ in 1+1 dimensional spacetime with $t=0$ where general $D(x)$ is defined as:

$$D(x) = −i \int{ \frac{d^3k}{(2π)^32ω_k} [e^{−i(ω_kt−k.x)}θ(x^0) + e^{i(ω_kt−k.x)}θ(−x^0)] }$$
where $k$ and $x$ are vectors. and $x^0$ is time dimension.

Using some help from the text itself, I figured for 1+1 Dimensional spacetime at t=0, the expression for propagator would be:

$$D(x) = -i \int^{-\infty}_{\infty}\frac{dk}{2\pi\sqrt{k^2+m^2}}e^{-ikx}$$
with all scalars.
Now, how do I further solve it? I have following two questions to answer:

  1. To show that it decays exponentially
  2. To study its behaviour for large $x$

Can the above questions be solved without further solving the integral?
I tried using $k=m*tan(z)$ substitution but that didn't work too well.
Any hint on further solving this would be helpful.


The integral is non trivial, and I must admit that it is quite mean of Zee to give such a difficult exercise to the reader early on. But I'll give it a try to explain to you how to solve this integral

We begin by simply rewriting the integral that you have stated $$ D(x)=-i\int^\infty_{-\infty}\frac{dk}{2\pi}\frac{e^{-ikx}}{\sqrt{k^2+m^2}} =-\frac{i}{m}\int^\infty_{-\infty}\frac{dk}{2\pi}\frac{e^{-ikx}}{\sqrt{(\frac{k}{m})^2+1}} $$ Next we substitute variables $k = m t$, where $t$ is our new variable, so $$ D(x)=-i\int^\infty_{-\infty}\frac{dt}{2\pi}\frac{e^{-imtx}}{\sqrt{t^2+1}} $$ Now we use Eulers identity to expand the exponential into trigonometric functions $$ D(x)=-i\int^\infty_{-\infty}\frac{dt}{2\pi}\frac{\cos(mtx)}{\sqrt{t^2+1}} -\int^\infty_{-\infty}\frac{dt}{2\pi}\frac{\sin(mtx)}{\sqrt{t^2+1}} $$ Note that the second term vanish because it's an odd function over an even interval. Likewise, the first term is an even function over an even interval so we can modify the integration limits and write $$ D(x) =-2i\int^\infty_0\frac{dt}{2\pi}\frac{\cos(mtx)}{\sqrt{t^2+1}} $$ We identify this integral as a Modified Bessel Function of the Second Kind Reference eq. (6), hence $$ D(x)=-2i\int^\infty_0\frac{dt}{2\pi}\frac{\cos(mtx)}{\sqrt{t^2+1}} =-i\frac{K_0(mx)}{\pi} $$ It is the last part that I find tricky. I hope this exercise doesn't scare you away from this beautiful field. I promise the upcoming exercises in the book aren't usually this difficult to solve.

Finally, to study the behavior for large $x$, you have too look up the asymptotic behavior of the $K_n(x)$ Bessel function. Me myself use a mathematics handbook to look it up $$ x\rightarrow\infty:\\ K_n(x)=\sqrt{\frac{\pi}{2x}}e^{-x}\Big[1+\mathcal{O}\Big(\frac{1}{x}\Big)\Big] $$

  • $\begingroup$ Thank you so much for the detailed integral. I am definitely not going to stop at this point, I am too enthusiastic about physics to stop now. Would you recommend some online resource or any specific book to look up such difficult integrals? $\endgroup$ – lucifer Apr 9 '18 at 15:33
  • $\begingroup$ You're very welcome! I would maybe recommend Wolfram Alpha or Mathematica, as a first step to look up the solution to difficult integrals. Actually, as a first step I plugged in the integral that you stated into Mathematica (just to make sure that it had a solution). And when I knew what the solution was going to be I kinda knew which mathematical tricks to use to analytically obtain that solution. $\endgroup$ – Turbotanten Apr 9 '18 at 16:51
  • $\begingroup$ As for specific books to look up integrals, I don't really have one for recommendation. The one I use is called Mathematics Handbook for Science and Engineering, and it has all the necessities that I've ever needed. There is also this very useful Wiki page goo.gl/gMvWm7 with common integrals in quantum field theory. $\endgroup$ – Turbotanten Apr 9 '18 at 16:52
  • $\begingroup$ oh, I didn't know Mathematica (or wolfram alpha) could do such complex integrals! I will try using those. Thank you again for the information. $\endgroup$ – lucifer Apr 10 '18 at 2:39
  • $\begingroup$ @Turbotanten Thanks! Now I know that this integral can be written in a closed-form otherwise I used limiting behavior of integrand to do this question $\endgroup$ – aitfel Jan 16 '20 at 14:20

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