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Locality in the propagator

  1. Jan 9, 2010 #1
    What does it mean to say that the propagator

    [tex]G(x,x')=\int \frac{d^4p}{(2\pi)^4}\left(\frac{e^{ip(x-x')}}{p^2+M^2}\right) [/tex]

    is nonlocal? Does that mean that if x and x' are space-like in separation, this expression is non-zero? If you did have something local represented by a Fourier transform f(p):

    [tex]G(x,x')=\int \frac{d^4p}{(2\pi)^4}f(p)e^{ip(x-x')} [/tex]

    how could you tell just from the form of f(p) that G(x,x') vanished for space-like separations?

    For large M

    [tex]G(x,x')=\left(\frac{1}{M^2}+\frac{\Box}{M^4}+...\right)\delta(x-x') [/tex]

    becomes local because the delta function only allows influences at x=x'. That means the interaction is no longer just local, but instantaneous. Does being local mean being instantaneous, or the broader meaning: being time-like?
  2. jcsd
  3. Jan 12, 2010 #2
    I think I can help out part of the way. However, these are tough integrals to get out.

    For your first question, you haven't chosen a pole prescription (that is the i\epsilon in the denominator); without this the integral is undefined. The four possible choices of pole prescription give: the advanced and retarded propagators, and the Feynman propagator and its complex conjugate. Any of these, or any linear combination, is a green's function of the KG equation in the sense that they satisfy
    (\Box^2 + m^2) G(x) = \delta^4(x)
    (up to normalising by a suitable constant factor). The pole prescription lets you do the p^0 integral, leaving a d^3p integral. If you are good with integrals you can evaluate the 3d integrals to get G(x); each of the propagators can be expressed in closed form, involving Bessel functions.

    Once they're evaluated, the expressions show that the advanced and retarded propagators are local and supported entirely within the forward and backward light cones, while the Feynman propagator is non-local, in the sense that it is non-zero for spacelike x. It is exponentially suppressed for large spacelike separation r, in fact it goes like e^(-mr)/r for large r.

    Not sure whether you can tell the support of a function from the form of its Fourier transform, except by evaluating it!

    I don't think the last equation even makes sense. Does the series even converge?

    Let me know if you want me to expand on the results above. There may be some glitches as I'm recalling them from memory! You might like to check in Bogoliubov & Shirkov if you don't mind cracking a nut with a sledgehammer.


  4. Jan 16, 2010 #3
    Yeah, I'm not sure if the last equation converges either. In fact, I'm not sure how to even derive that last equation from the first. Presumably d^4p is integrated only to a cutoff, so that when M goes to infinity, M^2 can overwhelm p^2. But if d^4p is not integrated to infinity, you don't get the delta function after the binomial expansion.
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