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Pauli Jordan Green's function

  1. Sep 22, 2013 #1
    Hi everyone,

    I'm going through some lecture notes on Quantum Field Theory and I came across a derivation of an explicit form of the Pauli Jordan Green's function for the Klein-Gordon field.

    The equations used in my lecture notes are equivalent to the ones in http://www.physics.byu.edu/faculty/berrondo/wt752/Invariant%20Functions.pdf

    My question is actually quite simple: how does equation 4.144 follow from 4.143 in the above pdf? (the same equations are in my lecture notes as well).
  2. jcsd
  3. Sep 22, 2013 #2


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    It's just due to the Euler identity

    $$e^{i\theta} = \cos \theta + i \sin\theta.$$

    This identity shows up in a lot of places, so it is a good one to remember!
  4. Sep 22, 2013 #3


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    There's a bit more to it than that, isn't there? Looks to me like he pulled a switcheroo in notation. In Eq 4.143, p·x is a 4-dimensional dot product, but in Eq. 4.144, the same p·x is 3-dimensional.
  5. Sep 22, 2013 #4
    I see that I'd need to use [itex]\sin(x) = 1/2i (e^{ix}-e^{-ix})[/itex], but it seems like the factor [itex]\exp(i\vec{p}\cdot \vec{x})[/itex] is factored out, which can't be true. They way I see it is

    [itex]\exp(-ip\cdot x)-\exp(ip\cdot x)=\exp(-iEt+i\vec{p}\cdot\vec{x})-\exp(iEt-i\vec{p}\cdot\vec{x})[/itex], in which it is not possible to just factor [itex]\exp(i\vec{p}\cdot \vec{x})[/itex] out and be left with [itex]\sin(Et)[/itex] (with some factors)
    Last edited: Sep 22, 2013
  6. Sep 22, 2013 #5
    Yes, the p and x on equation 4.144 should be bold to make that point clear
  7. Sep 22, 2013 #6
    why not?
  8. Sep 22, 2013 #7
    I see your problem. You're forgetting that the integral is over a symmetric region. Make the change of variables [itex] \vec{p}' = - \vec{p} [/itex] in one of the integrals and all is well
  9. Sep 22, 2013 #8
    the spatial momentum goes from -infinity to infinity.so you can change the sign in one of them without any effect
  10. Sep 22, 2013 #9
    But wouldn't that introduce a factor -1 (Jacobian) as well, so that the term left over would be a cosine rather than a sine?
  11. Sep 22, 2013 #10
    There is another -1 factor from the reversal of the integration limits
  12. Sep 22, 2013 #11
    Ah of course! Thanks!
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