Suppose we have a lagrangian $\mathcal{L}$ made by different fields, i.e. \begin{equation} \mathcal{L}= \mathcal{L_0} + g\phi\partial_\mu\phi A^\mu, \end{equation} where $\mathcal{L_0}$ is the free lagrangian for the fields $A^\mu$ and $\phi$.

Why is the 2-point correlation function $$\langle\Omega|T(\phi A^\mu)|\Omega\rangle = 0~?$$ In my lecture notes, my professor just assume this. Is it true in general for all types of contractions between different fields? Why? I get the intuitive explanation: if in the free lagrangian there are no terms like $\phi A^\mu$ we cannot turn a $\phi$ particle into an $A^\mu$ particle, but how can I see this more rigorously?

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    $\begingroup$ Your action has a $\phi \to - \phi$ symmetry so your correlators must satisfy the same thing. It follows that your correlator vanishes. $\endgroup$
    – Prahar
    Commented Dec 1, 2023 at 10:36

1 Answer 1


The story for a (connected) 2-point function of 2 different fields is very similar to what goes on for a 1-point function, cf. e.g. this & this related Phys.SE posts:

  1. Either a symmetry ensures that the 2-point function vanishes automatically (this is what happens in OP's example, cf. the above comment by Prahar),

  2. Or we add a quadratic interaction counterterm proportional to the 2 fields in the action in such a way that the 2-point function vanishes, i.e. the vanishing of the 2-point function is imposed as a renormalization condition.


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