Magnetic moment in four-fermion theory

Question

I'm trying to calculate the neutrino magnetic moment in the theory with this additional term in the Lagrangian: $\frac{a}{M^2}(\bar{\nu}\sigma_{\mu\nu}\nu)(\bar{e}\sigma^{\mu\nu}e)$, where $\sigma^{\mu\nu}=\frac{i}{2}[\gamma^{\mu},\gamma^{\nu}]$. This is the interaction of neutrino with electron (for simplicity I'm considering only electrons and electron neutrinos) As far as I understand, I should calculate a vertex $\Gamma^{\mu}$, corresponding to this diagram.

Then I should split it up to obtain a term proportional to $\frac{i\sigma^{\mu\nu}q_{\nu}}{2m}$ ($q = \bar{p}-p$), which would be magnetic moment.

But I don't quite understand how to write down an integral for this diagram. I'm thinking of something like this ($p$ for neutrino, $\bar{p}$ for antinutrino): $$\int\frac{d^4k}{(2\pi)^4}\frac{\gamma_{\mu}}{\gamma (p+k)-m+i\epsilon}\frac{\sigma^{\mu\nu}}{\gamma (\bar{p}+k)-m+i\epsilon}$$ Howewer I'm not sure about upper and lower indices, since in usual cases there is a boson propogator proportional to $g_{\mu\nu}$, which provides the correct contraction, but not in this case. On the other hand two indices must be contracted because $\Gamma^{\mu}$ has one upper index.

P.S. I know, that the integral is divergent, but nevertheless I'd like to know how to write it down.

Answer 1

The Feynman-rule for this 4-fermion-vertex is $\frac{ia}{M^2} (\sigma_{\mu\nu})_{ij} (\sigma^{\mu\nu})_{lm}$, where $i,j$ are the Spinor-indices of the neutrinos, while $l,m$ are the spinor-indices of the electrons. You see, that this vertex has no free Lorentz-indices and four free spinor-indices, as it should be.

For the amplitude you find

$$i\Gamma^\mu = i \frac{a}{M^2}\bar{u}(\bar p)\sigma_{\rho\lambda}u(p) \int \frac{d^4 k}{(2\pi)^4} \frac{\text{Tr}\left[\sigma^{\rho\lambda}(\not p + \not k + m) \gamma^\mu (\bar{\not p}+ \not k + m)\right]}{((p+k)^2-m^2 + i\epsilon)((\bar p + k)^2 - m^2 + i\epsilon)}, $$ where you now need to find terms of the form $\bar u(\bar p) \sigma_{\mu \nu} q^\nu u(p)$ because the diagram has a free Lorentz-index $\mu$. The trace appears, because you have a closed fermion loop.

I'm sure, that you can work the rest out on your own - if not, feel free to ask a follow-up question.