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Vanishing commutator for spacelike-separated operators?

  1. Sep 2, 2010 #1
    In David Tong's QFT notes (http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf p. 43, eqn. 2.89) he shows how the commutator of a scalar field [tex]\phi(x)[/tex] and [tex]\phi(y)[/tex] vanishes for spacelike-separated 4-vectors x and y, establishing that the theory is causal. For equal time, [tex]x^0=y^0[/tex], the commutator is given by:

    [tex] [\phi(x),\phi(y)] = \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{\vec{p}^2+m^2}} (e^{i \vec{p}\cdot(\vec{x}-\vec{y})} - e^{-i \vec{p}\cdot(\vec{x}-\vec{y})}) [/tex]

    He says that this vanishes because "we can flip the sign of [tex]\vec{p}[/tex] in the last exponent as it is an integration variable." What does he mean here?

    I think I see how something equivalent can be done by Lorentz transforming the (x-y) in the second exponent to -(x-y), since it's just a change of coordinates, but I'm not sure why you can do that in one exponent without doing it in the other, and why Tong says you can also flip the sign of p because it's being integrated over.

    Any help would be appreciated!
     
  2. jcsd
  3. Sep 2, 2010 #2
    Split the integral into (integral 1) - (integral 2). To see that you can flip the sign of p in the second integral think about the one dimensional case first:

    [tex]\int_{- \infty}^{\infty} f(x) dx = \int_{- \infty}^{\infty} f(-x) dx[/tex]
     
  4. Sep 2, 2010 #3
    Gah. That's ridiculously simple. I don't know why I didn't think of that.

    Well, just to make sure I'm getting it right... the trick is in the fact that the integrals are both taken over -infinity to infinity, so you can flip the sign of the integrated variable without changing the value of the integral, yeah?

    Thanks a ton.
     
  5. Sep 2, 2010 #4
    Yeah. It doesn't matter "which way" you integrate along the axis. In general of course you have

    [tex]\int_a^b f(x) dx = \int_{-b}^{-a} f(-x) dx[/tex]
     
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