Problem on calculating decay rates/lifetimes

[AuraCrystal]

Homework Statement

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

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):

<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 going to 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?

 

[TSny]

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?

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

 

[AuraCrystal]

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