Calculating the decay rate for h -> phi phi process

[Peeter]

Homework Statement

Given a coupling $h \; \partial_\mu \phi^a \partial^\mu \phi^a$, meant to model the first order interaction of the Higgs field $h$ to boson fields $\phi^a$, compute the width $\Gamma(h \rightarrow \phi^3 \phi^3)$ of the Higgs particle to decay to two longitudinal (say) $Z$-bosons (hence the index $3$).

We've been told that we only need to consider first order perturbative terms.

Homework Equations

$$\begin{aligned}{\left\langle {\Omega} \right\rvert} T( h(x_1) \phi^3(x_2) \phi^3(x_3) e^{-i \int dt H_{\text{int}} }) {\left\lvert {\Omega} \right\rangle}&=\frac{{\left\langle {0} \right\rvert} T( h(x_1) \phi^3(x_2) \phi^3(x_3) e^{-i \int dt H_{\text{int}} }) {\left\lvert {0} \right\rangle}}{{\left\langle {0} \right\rvert} T( e^{-i \int dt H_{\text{int}} }) {\left\lvert {0} \right\rangle}} \\ &=\text{some of all connected amputated diagrams with three external points,}\end{aligned}$$
where $T$ designates the time ordering operator.

The full Lagrangian (calculated in a previous problem set) was found to be:

$$\begin{align}\mathcal{L} =\frac{\left\lvert {m} \right\rvert^2}{2 \lambda} \partial_\mu h \partial^\mu h+\frac{\left\lvert {m} \right\rvert^2}{4 \lambda} ( 1 + h )^2 \text{Tr}{{\left( { \partial_\mu \Omega^\dagger \partial^\mu \Omega } \right)}}+ \left\lvert {m} \right\rvert^2\frac{\left\lvert {m} \right\rvert^2}{2 \lambda} \left( { 1 + h } \right)^2- \lambda\left( {\frac{\left\lvert {m} \right\rvert^2}{2 \lambda}} \right)^2\left( { 1 + h } \right)^4=\frac{\left\lvert {m} \right\rvert^2}{\lambda} \mathcal{L}',\end{align}$$
where $\mathcal{L}'$ is the ``canonically normalized (I assumed that to mean that there's a one-half factor on the kinetic terms) Lagrangian
$$\begin{align}\mathcal{L}' = \frac{1}{{2}} \partial_\mu h \partial^\mu h+\frac{1}{{4}}( 1 + h )^2 \text{Tr}{{\left( { \partial_\mu \Omega^\dagger \partial^\mu \Omega } \right)}}+\frac{1}{{2}}\left\lvert {m} \right\rvert^2\left( { 1 + h } \right)^2-\frac{\left\lvert {m} \right\rvert^2}{4}\left( { 1 + h } \right)^4,\end{align}$$
where
$$\begin{align}\Omega = e^{i \sigma^a \phi^a}.\end{align}$$

The Attempt at a Solution

The conjugate momenta are
$$\begin{align}\pi^a = \frac{\partial {L_{\text{int}}}}{\partial {(\partial_0 \phi^a)}} = 2 h \partial_0 \phi^a,\end{align}$$
so the interaction Hamiltonian is
$$\begin{aligned}H_{\text{int}} &= \int d^3 x\left( {\pi^a \partial_0 \phi^a - L_{\text{int}}} \right) \\ &=\int d^3 x \left( { 2 h (\partial_0 \phi^a)^2 - h (\partial_0 \phi^a)^2 + h (\partial_k \phi^a)^2} \right) \\ &=\int d^3 x h(x) \left( { (\partial_0 \phi^a)^2 + (\partial_k \phi^a)^2} \right),\end{aligned}$$

This is a somewhat strange seeming Hamiltonian (to me, at least given my rudimentary knowledge), as the only interactions I've seen in class have involved powers of the fields, whereas this one has derivatives of those too.

Based on the scalar theory example $\phi \phi \rightarrow \phi \phi$ example we worked in class, I figured that I need to calculate the first order diagrams of
$$\begin{aligned}{\left\langle {0} \right\rvert} &T( h(x_1) \phi^3(x_2) \phi^3(x_3) e^{-i \int d^4 z H_{\text{int}}(z) }) {\left\lvert {0} \right\rangle} \\ &={\left\langle {0} \right\rvert} T( h(x) \phi^3(y) \phi^3(w) ( 1 - i \int d^4 z h(z) \left( { \partial_0 \phi^3(z) \partial_0 \phi^3 z + \partial_k \phi^3(z) \partial_k \phi^3(z) } \right) ){\left\lvert {0} \right\rangle},\end{aligned}$$
Because $h, \phi^a$ commute, as well as any $\phi^a, \phi^b, a \ne b$, it seems to me that the contraction of $h(x), h(z)$ results in a Feynman propagator
$$\begin{align}D_F(x - z)={\left\langle {0} \right\rvert} T( h(x) h(z)) {\left\lvert {0} \right\rangle}.\end{align}$$
We can also have contractions between $\phi^3(y) \phi^3(w)$ and the derivatives of the $\phi(z)$'s in the interaction integral, or contractions between the $\phi^3(y), \phi^3(w)$'s and the derivative terms, but all the $h, \phi^a$ contractions should be zero since those fields commute. I thought the interpretation of the contractions of the derivatives fields would be
$$\begin{aligned}{\left\langle {0} \right\rvert} T( \phi^3(x_i) \partial_\mu \phi^a(z)) {\left\lvert {0} \right\rangle}&={\left\langle {0} \right\rvert} T( \phi^3(x_i) \partial_\mu \phi^a(z)) {\left\lvert {0} \right\rangle} \\ &={\left\langle {0} \right\rvert} T( \phi^3(x_i) \partial_\mu \phi^3(z)) {\left\lvert {0} \right\rangle} \\ &=\partial_\mu^{(z)} D_F( x_i - z ),\end{aligned}$$
the derivative of a Feynman propagator for $\phi^3$ between the $x_i$ and $z$ vertices.

It seems to me that there's only two diagrams (sketch attached), but the derivatives in the interaction are still bothering me, and I suspect that I've messed up the starting point for this problem significantly.

 

[member 553083]

Any insight or guidance would be greatly appreciated.A:You are correct in your approach so far. The only thing to note is that the contraction of the derivatives in the interaction Hamiltonian should be done with the usual propagator for scalar fields, i.e.\begin{align*} \langle 0|T(\phi(x_1) \partial_\mu \phi(z))|0\rangle = \partial_\mu^z \Delta (x_1-z)\end{align*}where $\Delta (x_1-z)$ is the scalar propagator.The two diagrams you mention are the only ones that contribute at leading order. The only thing left to do is calculate their value and take the imaginary part.