# Derivative coupling of neutrinos to massless Goldstone boson - calculation of decay width

I have a theory with a derivative coupling of neutrinos $\nu_{i,j}$ to a massless Goldstone boson $\phi$: \begin{equation} g_{ij}\partial^\mu \phi_\mu\bar{\nu_i}\nu_j. \end{equation} Now I want to calculate the decay width of the decay of $\nu_i$ into lighter $\nu_{j,k}$, either via \begin{equation} \nu_i \rightarrow \nu_{j,k} + \phi, \end{equation} or via \begin{equation} \nu_i \rightarrow \nu_{j,k} + \nu_{j,k} +\bar{\nu}_{j,k}, \end{equation} where $\phi$ is an intermediate state.

How do I do that and infer the lifetime of the neutrino $\nu_i$?

## 1 Answer

To calculate the decay width, you will need to first compute the scattering amplitudes for all processes contributing to the decay, to your desired order in perturbation theory. To compute the width for a two-body phase space, one has to evaluate the integral,

$$\Gamma = \int \frac{d\Omega_{\mathrm{cm}}}{4\pi} \frac{1}{8\pi} \left( \frac{2|p|}{E_{\mathrm{cm}}} \right) \frac{1}{2m} |\mathcal{M}|^2$$

where $\mathcal M$ denotes the amplitude with the appropriate sum over spins and so forth, $m$ is the mass of the initial particle, $E_{\mathrm{cm}}$ the centre of mass energy and $p$ denotes the 3-momenta of either resultant particle in the centre of mass frame as well.

The three-body phase space is more complicated, and not treated in all textbooks. Evaluating the integration measure over the phase space, one finds one must integrate the differential,

$$\frac{d\Gamma}{dE_1 dE_2} = \frac{|\mathcal{M}|^2}{64\pi^3m}$$

where $E_1$ and $E_2$ are the energies in the rest frame of the initial particle, of two of the three resultant particles. Rather than sketch the details here (which is just tedious manipulation), you can watch the derivation by Prof. Mark Wise (Caltech) in this lecture video.

From the width $\Gamma$, you can also compute the half-life $\tau = 1/\Gamma$. I assume you know how to derive the Feynman rule of the vertex needed for the calculation; a derivative coupling gives rise to factors of momenta in the Feynman rule.

• Thank you very much for your answer. I don't know how to calculate the amplitude if I have not four but only three in/out-going particles, and if there is no intermediate state. Can you help me with that? – Thomas Jan 25 '16 at 18:00
• @LCF If you're not able to do this yourself, it seems you may be unfamiliar with basic principles in quantum field theory to apply Feynman rules to Feynman diagrams of processes. I'd suggest revising QFT. – JamalS Jan 25 '16 at 18:28
• I had many lectures on QFT but we never treated derivative couplings. – Thomas Jan 25 '16 at 19:15
• @LCF Well, instead of just constants in a vertex rule, with derivatives in coupling terms, you also pick up factors of momenta flowing into the vertex. Look up some notes on the Feynman rules for QCD (Yang-Mills), it should make it clear. – JamalS Jan 25 '16 at 21:44