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In these notes concerning correlation functions in quantum field theory available at http://www.phas.ubc.ca/~gordonws/526/Propagators.pdf is stated, after equation $(3)$, that the correlation functions for QED vanishes unless the number of Dirac fields involved is equal to the number of adjoint fields. Even if this statement seems quite reasonable to me I can't find a general proof for it. Does anyone have a hint about it? I suppose it is in some way related to the fact that the Dirac field appears always quadratically in the QED Lagrangian, but how are the two facts actually linked?

Does the same argument apply to a field theory described by a Lagrangian like $ L= L_D + L_{KG} +L_{int}$, where the first term is the Dirac Lagrangian, the second is the Klein-Gordon Lagrangian and the interaction term is $L_{int} = -g \phi \bar \psi \psi$ ?

Finally, is a similir argument used to prove that correlation functions with an odd number of photon fields are vanishing?

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    $\begingroup$ Hint: consider the unitary operator $U(\theta)$ acting on $\psi$ as a $\mathrm U(1)$ phase rotation, $U(\theta)\psi U^\dagger(\theta)\equiv\mathrm e^{iq_\psi\theta}\psi$, and which leaves the vacuum invariant, $U(\theta)|0\rangle=|0\rangle$. $\endgroup$ – AccidentalFourierTransform Sep 27 '17 at 9:32
  • $\begingroup$ do you mean that using that operator and its adjoint I can find that correlation functions with a different number of fields and adjoint fields are equal to themselves times a phase factor different from 1 and hence must vanish? $\endgroup$ – L.R. Sep 27 '17 at 22:01
  • $\begingroup$ zacly.${}{}{}{}$ $\endgroup$ – AccidentalFourierTransform Sep 27 '17 at 22:04
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You can see that from the Feynman rules. The Fermion propagator is such that you always contract a $\psi$ with a $\overline{\psi}$. Also, the vertex in $L_{\rm int}$ has the same number (i.e., 1) of $\psi$'s and $\overline{\psi}$'s. Therefore if the external legs (i.e., the fields involved in your correlation) don't have this property of having the same number of $\psi$'s and $\overline{\psi}$'s, you simply can't construct your Feynman diagram and the sum over the latter is zero. This is the diagrammatic counterpart of the more conceptual explanation in the comment by AccidentalFourierTransform.

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  • $\begingroup$ +1 diagrammatic proofs are always preferred IMHO, because you don't have to worry about symmetry breaking, anomalies, etc. In this case, it is almost obvious that no regulator will break the symmetry I mentioned in my comment. $\endgroup$ – AccidentalFourierTransform Sep 27 '17 at 22:10

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