# Does this vertex equal 0?

If I have an interaction term in my Lagrangian that looks like:

$\mathcal{L}_{int} = (\partial_\mu B_\nu)(A^\mu B^\nu - A^\nu B^\mu)$ where B is a massive spin-1 field.

Am I correct in thinking that the vertex rule associated with this term is 0? Since the vertex rule would be proportional to $k_B^\mu (g^{\nu \rho} - g^{\rho \nu})$ = 0

• related: Feynman Rules for massive vector boson interactions (note that $\mathcal L=G_{\mu\nu}A^\mu B^\nu$, using their notation) Commented Jan 16, 2016 at 22:38
• @AccidentalFourierTransform in what way? Commented Jan 16, 2016 at 22:40
• Well, I thought the other post might be helpful for you: both in your case and in the other post, the intention is to find the vertex function for a bilinear interaction of a massive vector field and the electromagnetic field. I think the posts are related. Commented Jan 16, 2016 at 22:43
• If anyone still has an answer to my question that would be great! Commented Jan 16, 2016 at 23:06

Suppose you have, by definition, that $$\tag 1 \frac{\delta A^{\nu}}{\delta A_{\mu}} \equiv g^{\mu \nu}$$ In Fourier space the 3-vertex generated by your largangian is obtained by "amputation" of fields by using $(1)$, $$\Gamma^{\alpha \beta \delta} \sim \frac{\delta^{2}}{\delta B_{\alpha}\delta B_{\beta}}\frac{\delta}{\delta A_{\gamma}}\left(k_{\mu}B_{\nu}(A^{\nu}B^{\mu} - A^{\mu}B^{\nu})\right) =$$ $$=\frac{\delta^{2}}{\delta B_{\alpha}\delta B_{\beta}}k_{\mu}B_{\nu}\left(g^{\gamma \nu}B^{\mu} - g^{\gamma \mu}B^{\nu} \right) =$$ $$=2k_{\mu}\left( g^{\beta \gamma}g^{\alpha \mu} - g^{\gamma \mu}g^{\alpha \beta} - g^{\alpha \gamma}g^{\beta \mu} + g^{\gamma \mu}g^{\alpha \beta}\right) =$$ $$= 2(k^{\alpha}g^{\beta \gamma} - k^{\beta}g^{\alpha \gamma})$$