Problem on calculating decay rates/lifetimes

[AuraCrystal]

Homework Statement

Given the Lagrangian

[itex]\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,[/itex]

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

Homework Equations

Dyson expansion, etc.

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 p_i and the final momenta be p₁ and p₂):

[itex]<f|S|i>=<p_1, p_2|S|p_i>[/itex]

To the lowest order in [itex]\mu[/itex], the S-matrix is

[itex]S=-i\mu \int d^4 x \mathcal{H}_{interaction}=-i\mu \int d^4 x \Phi \phi \phi[/itex]

so to lowest order, the transition amplitude is

[itex]<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>[/itex]
Considering just the last bit, namely
[itex]\phi \phi \sqrt(2E_i) a_{\Phi}^{\dagger}(p_i) |0> [/itex]
We can expand the [itex]\phi[/itex] in terms of creation and annihilation operators, right? For the annihilation operator, it'll commute with the [itex]a_{\Phi}^{\dagger}(p_i)[/itex] so that term will give 0. For the other[itex]\phi[/itex], we get the same thing. So you're going to get something like

[itex]\int a_{\phi}^{\dagger}(p_i) a_{\phi}^{\dagger}(p_i) a_{\Phi}^{\dagger}(p_i) |0>[/itex]

Can't you just commute [itex]a_{\Phi}^{\dagger}(p_i)[/itex] to the left? When bracketed with the final state which has no [itex]\Phi[/itex] 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?

 

[TSny]

Can't you just commute [itex]a_{\Phi}^{\dagger}(p_i)[/itex] to the left? When bracketed with the final state which has no [itex]\Phi[/itex] 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?

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

 

[AuraCrystal]

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