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?


2 Answers 2


I'll address your point about why the integral is Lorentz invariant, as from the comments to cduston's answer, I think this is your sticking point:

You can see the relation between a manifestly Lorentz invariant form like this $$\int d^4p \frac{e^{-ipx}}{(p^2-m^2)} \ \ \ (1) $$ and the not-so-obviously Lorentz invariant form $$ \int d^3p \frac{1}{E_{{\bf{p}}}}{e^{-ipx}} \ \ \ (2)$$ by using the identity $$\frac{1}{(p^2-m^2)}= \frac{1}{2E_{{\bf{p}}}}\{\frac{1}{(E_{{\bf{p}}}+p_0)}-\frac{1}{(E_{{\bf{p}}}-p_0)}\} $$ Here $E_{{\bf{p}}} = \sqrt{{\bf{p}}^2+m^2}$ is the on-shell time component of the momentum four vector, and $p_0$ is the "generic" time component - not necessarily on-shell.

If you substitute this in (1) and do the $p_0$ integral using the appropriate contour, you'll get (2).

What's actually going on is explained in the discussion near equation (2.40), you're doing a 4 momentum integral, but just restricting it to the mass shell using a delta function. Restriction to a mass shell is a Lorentz invariant operation, so you're maintaining Lorentz invariance throughout (even though with the three momentum integral it doesn't look like it!).


In the text he says the two terms vanish under $(x-y)\rightarrow -(x-y)$. In other words, there is a Lorentz transformation which takes $(x-y)\rightarrow -(x-y)$ in the second term when the separation is spacelike ($(x-y)^2<0$ using the wrong sign...). Do that, and the commutator vanishes.

  • $\begingroup$ I agree with your comment in the text says when the separation is space like one can do such a transformation and get zero. Provided that the term is Lorentz invariant. I cannot see how this term is Lorentz invariant. I can see how $\int d^3pf(p)/\sqrt{p^2+m^2}$ is Lorentz invariant but not a function like $\int d^3pf(p,x,t)/\sqrt{p^2+m^2}$. Could you please explain to me why $\int d^3p e^{-ip.x-it\sqrt{p^2+m^2}}/\sqrt{p^2+m^2}$ is Lorentz invariant? $\endgroup$
    – Lawless
    Mar 3, 2013 at 5:59
  • 3
    $\begingroup$ The thing in the exponent is just the Lorentz-invariant inner product between the 4-momentum $p_\mu = (\vec{p}, \sqrt{\vec{p}^2 + m^2})$ and the 4-position $x_\mu = (\vec{x}, t)$ (up to whatever index and sign convention they're using). $\endgroup$
    – Michael
    Mar 3, 2013 at 8:38
  • $\begingroup$ The previous poster is correct, but PS also mentions this a few pages earlier in the their text. I don't have it on me right now but have a look at it. $\endgroup$
    – levitopher
    Mar 3, 2013 at 16:21
  • $\begingroup$ @Blackie Eq.(2.40) on page 23 shows it's Lorentz invariant. $\endgroup$
    – luyuwuli
    Dec 16, 2013 at 8:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.