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If I have a Lagrangian in momentum space of the form

$$ \mathcal{L} = W_\mu^{ \dagger}(p)f(p)^{\mu \nu}W_\nu(p) $$

how is the propagator for the field related to the function $f(p)$ (e.g. is it just given by $(f(p)^{-1})_{\mu \nu}$ or some other relation)?

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  • $\begingroup$ Am I right to assume that you mean $F^{\mu \nu}$ is a function of $p$ rather than containing fields? I.e. this is a kind of mass term? $\endgroup$
    – Orca
    Apr 4, 2016 at 19:13
  • $\begingroup$ Yes. Sorry -- poor notation on my part. $F^{\mu \nu}$ is just some function of $p$, not a strength tensor. Have edited the question. $\endgroup$
    – Hans Schmuber
    Apr 4, 2016 at 19:20
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    $\begingroup$ Interesting question! My first guess would be that in the propagator you would just replace m with f(p) like you said, and integrate over the fields in the path integral as usual, but I'm not sure. I wonder what the physical implications would be? $\endgroup$
    – Orca
    Apr 4, 2016 at 20:22

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The form of the propagator is simple - in momentum space it is simply the inverse of the term coupling your two bosons, as you pointed out: $$G \propto f(p^2)^{-1}$$ provided the $f(p^2)$ is indeed invertible.

To get the Lorentz indices right is not as trivial. As I understand it, if these are vector bosons, the more correct form would be to write $$\mathcal{L} = P_T^{\mu\nu} f(p^2) W_\mu W_\nu$$ where $P_T$ is a transverse projection operator, $P_T^{\mu\nu}f(p^2) \sim \left\langle J^\mu_a J^\nu_a \right\rangle \sim (p^2 \eta^{\mu\nu} - p^\mu p^\nu)f(p^2)$.

Then the propagator will be $$G_{\mu\nu} = (P_T)_{\mu\nu} f(p^2)^{-1}$$ since $P^{\mu\nu}_T (P_T)_{\mu\nu} = 1$.

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