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If I have a Lagrangian of the form:

$$ \mathcal{L} = k \bar{\psi} \varepsilon^{\mu \nu} \lambda^a \phi G^a_{\mu \nu} + h.c. $$

[where $\phi, \psi$ are fermions, $\lambda^a$ are Gellmann matrices, $\varepsilon^{\mu \nu} $ is some antisymmetric tensor and $G^a_{\mu \nu}$ is the gluon field stength tensor.]

And I want to have the Feynman rule for the vertex and it's conjugate I would usually compute the following:

$$ \frac{\delta S}{\delta \psi \delta \phi \delta A_\mu^a} $$ [where $A_\mu$ is a gluon field].

Following that prescription, I have $$ S = k \int d^4 x \bar{\psi} \varepsilon^{\mu \nu} \lambda^a \phi G^a_{\mu \nu} $$ ignoring color for a moment, $$ \implies \frac{\delta S}{\delta \psi \delta \phi } = 2k \int d^4 x \gamma^0 \varepsilon^{\mu \nu} \left( \partial_\mu A_\nu + i g_s A_\mu A_\nu \right) $$

using the convention that $G_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + i g_s \left[ A_\mu, A_\nu \right] $

Taking the Fourier transform to do the last variation I get a leftover $A_\mu$ in the expression though, don't I?

Something seems off here.

With a nicer field strength tensor, like $F^{\mu \nu}$ I get a nice expression at the end only in terms of the momentum of the photon I have a leftover term involving an $A_\mu$.

The initial variation shouldn't be with respect to the field strength tensor, should it?

I'm messing up something trivial, could someone point me in the right direction?

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up vote 2 down vote accepted

There's a vertex with one gluon, and also a vertex with two gluons.

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I am an idiot. Thank you so much! – user2053414 May 27 '13 at 4:09

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