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I am stuck at the beginning of a problem where I am given an interaction term that modifies the regular QED Lagrangian. It involves the interaction between a photon field and a massive vector boson: $\mathcal{L}_{int_1} = \frac{1}{4}g_1G_{\mu \nu}F^{\mu \nu}$. Here, $F^{\mu \nu}$ is the electromagnetic field tensor. Similarly, $G_{\mu \nu} = \partial_{\mu}B_{\nu}-\partial_{\nu}B_{\mu}$ where $B_{\mu}$ is the massive vector field.

I am trying to derive the Feynman rule for this vertex. I am not sure what to do. Usually, I just bring down a 4-momentum when I have a field with derivatives in the Lagrangian, but in this case, it looks like two of the 4-momenta would end up forming a dot product together, while the other two would not, and I would be left with a term without any indices and two terms with indices, which doesn't make sense. Does anyone know what the Feynman rule for this vertex is and why it is so?

EDIT: I think the Feynman rule might be $i\mathcal{M} = -i g_1 (g_{\mu \nu}-\frac{k_\mu a_\nu}{k \cdot q})$. I obtained this by an analogous computation for finding the photon propagator. Is doing this allowed?

The main problem that I am trying to solve involves the diagram $f\bar{f} \to \phi\phi^*$. From comments, I realize that I have two options: 1) I can diagonalize the mass matrix and use the new mass as the mass in the propagator between the two end vertices or 2) I can consider the infinite series of $A$ propagator > $B$ propagator > ... > $A$ propagator > $B$ propagator and obtain the same answer. The problem I am having with diagonalizing the mass matrix is that in previous problems where I have diagonalized a mass matrix, my interaction term did not contain derivative terms. So my attempt at a solution (which I am trying now) is to just bring down a 4-momentum and go from there. But then wouldn't that make my mass momentum dependent? With the second method, I think I would need to still know the feynman rule for the interaction vertex, which circles back to my original problem.

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Is this a HW problem? –  DJBunk Dec 28 '12 at 2:31
    
No its not. I am doing extra problems after the completion of this past semester, and this is one of the ones I came across. –  Joe Joe Dec 28 '12 at 6:27
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1 Answer

up vote 1 down vote accepted

Your first interaction term is bilinear in 2 gauge fields. Terms of this form indicate that you need to diagonalize your mass matrix. So you could work out the Feynman rules for your first interaction, but your question is sort of a moot point since you won't have to do perturbation theory for this term, it will get assimilated into the propagators in the diagonal basis.

EDIT:

Just so you get your orginial question answered, the Feynman rule for your first interaction is obtained by Fourier transforming the Action:

$S = \frac{g_1}{4}\int d^4 x F^{\mu} G_{\mu \nu} = \frac{g_1}{4}\int d^4 x \left( \partial_\mu F_\nu -\partial_\nu F_\mu \right) \left( \partial^\mu G^\nu - \partial^\nu G^\mu \right) $

$=\frac{g_1}{2}\int d^4 x \left( F_\nu \partial_\mu \partial^\nu G_\mu - F_\nu \Box G_\nu \right) = \int d^4 p \tilde{F}_{\nu} (-p) \left[ \frac{g_1}{2} \left( p^2 \eta^{\nu \mu} - p^\nu p^\mu \right) \right] \tilde{G}_\mu (p) $

where the Feynman verte rule is in the square brackets now.

Now, to rephrase my original point, the Lagrangian

$\mathcal{L} = \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \frac{1}{4} G^{\mu \nu} G_{\mu \nu} +m^2 G^\mu G_\mu+ g_1 \frac{1}{4} F^{\mu \nu} G_{\mu \nu}$

is trivial, in that it is free. You can see this by doing perturbation theory, or you can make things easy on yourself and diagonalize the quadratic portion of the action.

Let me know if you want further clarification.

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So, overall, I have a process at tree level where $f + \bar{f} \to \phi + \phi^*$, where $f$ is a fermion, I would have two propagators: the photon is released from the $f \bar{f}$ annihilation, then there is a vertex where the photon transitions over to the massive vector boson, then the massive vector boson transitions to the scalar anti-scalar pair. Are you saying that I can diagonalize the mass matrix and then write the product of two propagators as one propagator? –  Joe Joe Dec 28 '12 at 14:15
    
@Joe: you're almost there. The 'propagating' d.o.f. will not be the photon and the $B$-field, but linear combinations of them, and their masses will depend on $g_1.$ But in terms of these new fields, the interaction vertices look different! If you choose to ignore this mass matrix issue (you can), the full photon propagator will contain terms of the form << photon line > B line > photon line > ...> B line > photon line >>. Summing over all these terms gives rise to a mass (dep. on $g_1$). Similarly, the full $B$-propagator would contain terms with photon line insertions. –  Vibert Dec 28 '12 at 14:27
    
So, if I were to circumvent diagonalizing the mass matrix, wouldn't I need a Feynman rule for the interaction vertex? I have done a problem before where I did the second method which you talked about, but I believe that I needed to know what the feynman rule was in order to do that. –  Joe Joe Dec 28 '12 at 15:01
    
@JoeJoe-if you don't diagonalize the mass matrix, then yes, you would need an addl vertex, but the vertex would just have a single $ F_\mu $ coming in and $G_\mu$ field coming out. So its sort of a weird vertex and you are just making things harder on yourself. Diagonalizing the mass matrix takes all these interactions into account once and for all. As a simpler example, you could always treat the mass term of a fermion as a perturbation with separate left and right propagators, but why bother when you can just treat all the mass insertions once and for all treating it as a massive fermion. –  DJBunk Dec 28 '12 at 17:09
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