In many body quantum theory, Feynman diagrams are commonly used to calculate green's function. My question is: does the diagrammatic method works for all kinds of green's function? For casual green's function, $$ G(x_1, x_2)=-i\langle T{\psi(x_1)\psi^{\dagger}(x_2)}\rangle $$ it can be represented as a line from $x_2$ to $x_1$ and interactions can be represented by some vertices. However, we can have this diagrammatic technique because we have time order product in the bracket and can use Wick's theorems to reduce multi-points functions into products of several two point functions.

Now consider Greater Green's function: $$ G^{>}(x_1, x_2) = -i\langle\psi(x_1)\psi^{\dagger}(x_2)\rangle. $$ In this case, we don't have time ordering and Wick's theorem cannot be applied. Do we still have similar diagrammatic techniques DIRECTLY for calculating $G^{>}(x_1, x_2)$, i.e. we can still represent $G^{>}(x_1, x_2)$ and interaction as some lines and vertices.

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    $\begingroup$ Look up Keldysh approach (a parallel development by Kadanoff&Baym might be closer to Schwinger's own ideas) $\endgroup$
    – Roger V.
    May 10 at 19:19

1 Answer 1


You have to be way more careful with your notation. For the case that you refer to equilibrium-Green-function-techniques for systems that are described within a grand-canonical-ensemble (as its being done in the book from Altland or Negele and Orland) then the short answer to your question is:

No, there is no way of determining $G^{>}$ directly.

I am not certain how this answer changes when one considers Green-functions within the Keldysh-formalism. However... Within this formalism that I mentioned you can determine $G^{>}$ indirectly and quite easily with diagrammatic techniques; You first determine the imaginary-time-Green-function with diagrammatic techniques, then analytically continue this to get the retarded Green-function, and then use one of the identities that you obtain from the Lehmann-representation to obtain the greater Green-function.

Depending on your level of knowledge I can also elaborate this with concrete formulas.


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