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Two-point correlation function in path integral formulation

  1. Sep 6, 2014 #1
    Suppose that I have already calculated the two-point correlation function for a Lagrangian with no interations using the path integral formulation.
    [tex]\langle \Omega | T[\phi(x)\phi(y)] | \Omega \rangle = \frac{ \int \mathcal{D}\phi \phi(x)\phi(y) \exp[iS_0] }{ \int \mathcal{D}\phi \exp[iS_0] }.[/tex]
    If I now add an interaction, such that the new action may be written as [itex]S = S_0 + S_I,[/itex] the new two-point correlation function is obviously
    [tex]\langle \Omega | T[\phi(x)\phi(y)] | \Omega \rangle = \frac{ \int \mathcal{D}\phi \phi(x)\phi(y) \exp[i(S_0+S_I)] }{ \int \mathcal{D}\phi \exp[i(S_0+S_I)] }.[/tex]
    My question is:
    Would I have to do the calculations again for the new expression? Or is there some short cut, such as factoring out [itex]\exp[iS_I][/itex] so that the new expression is a product of the old expression and [itex]\exp[iS_I][/itex]?
  2. jcsd
  3. Sep 6, 2014 #2


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    Sure, the new 2-point correlation functions will be some nonlinear function of the 'free one'. Therefore you need a series expansion for it, in the same way you need it for the generating functional(s).
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