# Photon-Massive Vector Boson Vertex Feynman Rule

1. Dec 28, 2012

### stack

1. The Problem
I am trying to find the feynman rule which corresponds to the addition of an interaction term to the QED lagrangian which couples the electromagnetic field to a neutral massive vector boson field. In this problem, $$k^\mu$$ corresponds to the photon 4-momentum and $$q^\mu$$ corresponds to the massive vector boson 4-momentum.

2. Relevant equations
$$\mathcal{L}_{int} = \epsilon F_{\mu \nu}G^{\mu \nu}$$ where $$F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$ and $$A_\mu$$ is the electromagnetic field operator. Similarly, $$G^{\mu \nu}$$ is the same as $$F$$ only now for the massive vector field $$B^\mu$$.

3. The attempt at a solution
Usually, when I find the feynman rule that corresponds to a term that has derivatives in the Lagrangian, I just bring down a momentum from the exponential in the expansion of the field in momentum space. So I thought that the following would work:
$$i\mathcal{M} = i\epsilon (k_\mu q^\mu + k_\nu q^\nu - k_\mu q^\nu - k_\nu q^\mu)$$
However, this does not make sense because two of the terms here would contract and two would not, leaving two terms with indices and two terms without. This just doesn't make sense. So on my second attempt, I figured that there should be a Kronecker delta that makes sure that $\mu$ and $\nu$ are the same:
$$i\mathcal{M} = i\epsilon (k_\mu q^\mu + k_\nu q^\nu - k_\mu q^\nu - k_\nu q^\mu)\delta_\mu^\nu$$
However, it should be easy to see that this is zero. Then I tried to use the fact that the field strength tensors are anti-symmetric to eliminate some terms, but I ran into the same problems. Any help would be greatly appreciated. I am quite lost.

Last edited: Dec 28, 2012
2. Dec 28, 2012

### kevinferreira

Well, your interaction term will have terms like
$$\partial_{\mu}A_{\nu}\partial^{\mu}B^{\nu}$$
and all other possibilities of interchanging $\mu,\nu$, which gives 4 terms.
In momentum space, you have terms coming from each one of these terms that give you
$$k_{\mu}\epsilon_{\nu}q^{\mu}\eta^{\nu}=(k\cdot q)(\epsilon\cdot\eta)$$
where $\epsilon,\eta$ are the polarisations of the photon and the massive vector boson, respectively.
Again, you have all 4 possible scalar products.
So I think the Feynman rule for a vertex will just be given by:
$$i\epsilon(q^{\mu}\eta^{\nu}+q^{\nu}\eta^{\mu}+k^{ \mu} \epsilon^{\nu}+k^{\nu}\epsilon^{\mu})$$
These indices don't need be contracted here on the vertex, they will be contracted with the propagators later on.