Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    Bill_K

    User Avatar
    Science Advisor

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pauli Jordan Green's function
  1. Green function for SHO (Replies: 2)

Loading...