I have a problem with proof of causality in Peskin & Schroeder, *An Introduction to QFT,* page 28. To avoid confusion I use three vectors notation, rewriting the Eq. (2.53) for $y=0$ as follows:

$[\phi(x,t),\phi(0,0)]=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2\sqrt{p^2+m^2}}\left(e^{-i\mathrm{p}.\mathrm{x}-it\sqrt{p^2+m^2}}-e^{i\mathrm{p}.\mathrm{x}+it\sqrt{p^2+m^2}}\right)$

The book goes on about how the integrand being Lorentz invariant makes this integral zero for the x out of the light cone. But I (not being a special relativity expert) want to see it more rigorously:

after changing variables $p\to-p$ in the first term, the equation simplifies to:

$[\phi(x,t),\phi(0,0)]=\int \frac{d^3p}{(2\pi)^3}\frac{-2i}{2\sqrt{p^2+m^2}}e^{i\mathrm{p}.\mathrm{x}}\sin\left(t\sqrt{p^2+m^2}\right)$

using spherical coordinates:

$[\phi(x,t),\phi(0,0)]=\int \frac{dpd\phi d\theta p^2\sin\theta}{(2\pi)^3}\frac{-i}{\sqrt{p^2+m^2}}e^{ipx\cos\theta}\sin\left(t\sqrt{p^2+m^2}\right)\\
[\phi(x,t),\phi(0,0)]=\int_0^{\infty}\frac{dpp}{(2\pi)^2}\frac{-2i}{x\sqrt{p^2+m^2}}\sin (px)\sin\left(t\sqrt{p^2+m^2}\right)$

again after another change of variables $u=\sqrt{p^2+m^2}$,

$[\phi(x,t),\phi(0,0)]=\frac{-2i}{x}\int_m^{\infty}\frac{du}{(2\pi)^2}\sin (x\sqrt{u^2-m^2})\sin\left(tu\right)$

I cannot see how this integral should be zero for $x>t$ !!! Can somebody please explain this to me?

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user Blackie