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?

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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!).

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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.

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  • $\begingroup$ 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 '13 at 5:59
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    $\begingroup$ 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 Brown Mar 3 '13 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 '13 at 16:21
  • $\begingroup$ @Blackie Eq.(2.40) on page 23 shows it's Lorentz invariant. $\endgroup$ – luyuwuli Dec 16 '13 at 8:48

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