Suppose that I have already calculated the two-point correlation function for a Lagrangian with no interations using the path integral formulation.(adsbygoogle = window.adsbygoogle || []).push({});

[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]?

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# Two-point correlation function in path integral formulation

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