# Homework Help: Problem on calculating decay rates/lifetimes

1. Nov 27, 2012

### AuraCrystal

1. The problem statement, all variables and given/known data
Given the Lagrangian

$\mathcal{L}=\frac{1}{2} ( \partial_{\mu} \Phi)^2-\frac{1}{2}M^2 \Phi ^2 + \frac{1}{2} ( \partial_{\mu} \phi)^2-\frac{1}{2}M^2 \phi ^2-\mu \Phi \phi \phi,$

[The last term, the interaction term allows a $\Phi$ particle to decay into 2 $\phi$ particles, assuming of course that M>2m.] Calculate the decay rate of that process to lowest order in $\mu$

2. Relevant equations

Dyson expansion, etc.

3. The attempt at a solution
OK, so I'm stuck on computing the transition amplitude here. Following Aitchison and Hey, the transition amplitude is (letting the initial momentum be pi and the final momenta be p1 and p2):

$<f|S|i>=<p_1, p_2|S|p_i>$

To the lowest order in $\mu$, the S-matrix is

$S=-i\mu \int d^4 x \mathcal{H}_{interaction}=-i\mu \int d^4 x \Phi \phi \phi$

so to lowest order, the transition amplitude is

$<p_1, p_2|S|p_i>=-i\mu<0|a_{\phi}(p_1) a_{\phi}(p_2) \sqrt(2E_1) \sqrt(2E_2) \int d^4 x \Phi \phi \phi \sqrt(2E_i) a_{\Phi}^{\dagger}(p_i) |0>$
Considering just the last bit, namely
$\phi \phi \sqrt(2E_i) a_{\Phi}^{\dagger}(p_i) |0>$
We can expand the $\phi$ in terms of creation and annihilation operators, right? For the annihilation operator, it'll commute with the $a_{\Phi}^{\dagger}(p_i)$ so that term will give 0. For the other$\phi$, we get the same thing. So you're gonna get something like

$\int a_{\phi}^{\dagger}(p_i) a_{\phi}^{\dagger}(p_i) a_{\Phi}^{\dagger}(p_i) |0>$

Can't you just commute $a_{\Phi}^{\dagger}(p_i)$ to the left? When bracketed with the final state which has no $\Phi$ particles, this'll give 0. Did I make a mistake? Or do I just have to consider a higher order term in the Dyson expansion?

2. Nov 27, 2012

### TSny

When moving $a_{\Phi}^{\dagger}(p_i)$ to the left, wouldn't you need to consider what happens when moving it past $\Phi$?

3. Nov 29, 2012

### AuraCrystal

^Yeah, I forgot about that! Thanks! :)