Let the Euclidean Lagrangian be $L_e(\phi,\dot{\phi})=\frac{1}{2}\dot{\phi}^2+\frac{1}{2}m^2{\phi}^2+\frac{\lambda}{4!}{\phi}^4$,for interacting case.Then,


$=\frac{\int D\phi e^{-\int_{-\infty}^{+\infty}dt L_e}\phi(t_a)\phi(t_b)....}{\int D\phi e^{-\int_{-\infty}^{+\infty}dt L_e}}$

$=\frac{\int D\phi e^{-\int_{-\infty}^{+\infty}dt {L^0}_e}\phi(t_a)\phi(t_b)....\sum_{n=0}^{\infty}(-\frac{\lambda}{4!})^{n}\frac{1}{n!}[\int_{-\infty}^{+\infty}dt {\phi}^4]^n}{\int D\phi e^{-\int_{-\infty}^{+\infty}dt {L^0}_e}\sum_{n=0}^{\infty}(-\frac{\lambda}{4!})^{n}\frac{1}{n!}[\int_{-\infty}^{+\infty}dt {\phi}^4]^n}$

$=\frac{[\int D\phi e^{-\int_{-\infty}^{+\infty}dt {L^0}_e}(\phi strings)]/[\int D\phi e^{-\int_{-\infty}^{+\infty}dt {L^0}_e}]}{[\int D\phi e^{-\int_{-\infty}^{+\infty}dt {L^0}_e}differentstrings of \phi]/[\int D\phi e^{-\int_{-\infty}^{+\infty}dt {L^0}_e}]}$

Then,we apply Wick's Theorem to the numerator and Denominator.This is what is normally done.My question is: We know that this is not a free theory,i.e. $\hat{\phi}$ can not be decomposed into creation,anhiliation operators.These are purely interacting fields.Although from the second step only functions appear and no operators appear,these functions are different from that of the free case.Can we apply Wick's theorem in such case to the numerator and denominator?

  • $\begingroup$ 1. Your formulae are very hard to read. 2. Please make clearer what exactly you mean by "Wick's theorem" and what about applying it here is uncelar to you. There is more than one way to set up perturbation theory in the path integral formalism and what exactly the answer is there depends on which one you're trying to do. $\endgroup$
    – ACuriousMind
    Commented Sep 5, 2016 at 17:02
  • $\begingroup$ Wick's theorem:The vacuum expectation value of a time ordered product of 'free' fields is the sum of all possible contractions.For a two point function the answer is $t_a----t_b$ + tadpole $\endgroup$
    – Annie
    Commented Sep 5, 2016 at 17:09
  • $\begingroup$ $l_e$ has been separated into free and interacting parts.And the exponential over interacting part has been expanded in terms of taylor series.Then in the final step,divided the numeratora and denominator by the same thing,so that I can bring it into the form where I can use Wick's theorem. $\endgroup$
    – Annie
    Commented Sep 5, 2016 at 17:12
  • $\begingroup$ Wick's theorem does not rely on the fields being free; in path integral language, Wick's theorem is nothing but pairing rules for Gaussian integral. On the other hand, in the canonical quantization picture, in spirit of perturbation theory, $\phi$ can be treated as free fields, and interaction effects build upon this to give corrections. $\endgroup$ Commented Sep 5, 2016 at 18:33

1 Answer 1


In order to apply Wick's theorem to interacting QFT in operator formalism you have to pass to the interaction picture of quantum mechanics. In this formalism, the Hamiltonian is split in two (unphysical) parts. The free part is responsible for the Heisenberg evolution of field operators (to which Wick's theorem is perfectly applicable), and the interaction part is responsible for the Schroedinger-like evolution of quantum states.

Take a look at this question and at my notes on interaction picture and deriving Feynman rules.

I hope this answers your question.


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