Heavy sterile neutrino decay into Z bosons. (Feynman Diagram)

HypnoFunk
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Hello there,

I am looking at the decays of Heavy sterile neutrinos in the their mass states to Z bosons.

Using the feynman rules, how would I go about calculating the decay amplitude. I have only ever seen cases where the Z boson is a propagator. Would my expression look something like this:

<br /> M = \kappa\frac{g}{2\cos\theta_{W}}\overline{\nu_{1}}(2)\gamma^{\mu}P_{L}N_{m}(1)\epsilon_{\mu}^{*}(3)<br />

where \kappa is some constant. Essentially the Feynman diagram looks like a standard weak vertex between two neutrinos. The only difference is that one of them is Heavy (GeV-TeV) scale. However my question really is about whether I just take the standard vertex and include the \epsilon^{*\mu} vector for the outgoing Z-boson.

The standard vertex factor reads
<br /> -i\frac{g}{2\cos\theta_{W}}\kappa\gamma^{\mu}P_{L}<br />

Is it enough for me to just add \epsilon^{\mu *} for the outgoing Z boson?

Thank you to anyone who can help.
 
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i think the part of z (ϵμ∗ )boson is correct but there is a problem in vertex. Because sterile neutrino does not interact in weak interaction so it will not couple to z boson. so you need to do something different here. You can check Mahapotro's " Massive neutrinos in physics and astrophysics " . check out the chapter on Left right symmetric model. there check sections like FCNC and heavy neutrino decay. I think you can find your answer there. So check if possible.
 
I think that's right - if it's sterile, the vertex is 0, right?

More realistically, you will have to have to define your couplings and decide how sterile is sterile.
 
Vanadium, in my experience, sterile neutrino models (well, interesting ones at least) generally have small mixings between the sterile and active neutrinos. I think that's what Hypnofunk's \kappa is intended to quantify.
 
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