# Decay width in 2 & 3 body decays, calculating momentum integrals

I'm considering a toy model with two types of scalar particles, one massive $(\Phi)$ and one massless $(\phi)$ with an interaction of the form

$$L_{int}=-\lambda \phi\phi\Phi$$

I'm interested in a decay $\Phi \rightarrow \phi\phi$ and calculating the total decay width for it.

The formula I have for differential decay width is this:

$$\mathrm d\Gamma = \frac{1}{2m_i} \left( \prod_f \frac{\mathrm d^3p_f}{(2\pi)^3} \frac{1}{2E_f} \right) |M|^2 (2\pi)^4 \delta^{(4)}\left(p_i - \sum_f p_f\right)$$

where $i,f$ stands for initial and final.

Now in this simple model the Feynman rules for a vertex contribute only a -$i\lambda$ to the matrix element, so $|M|^2 = \lambda^2$

However I don't know how to actually calculate this integral, don't even know how to start.

Similarly, with a 3-body decay (assuming also a simple interaction with matrix element $\lambda$), I have the formula

$$d\Gamma = \frac{1}{(2\pi)^5}\frac{1}{16m_i}|M|^2\,\mathrm dE_1 \,\mathrm dE_2 \,\mathrm d\alpha\,\mathrm d(\cos\beta)\,\mathrm d \gamma$$

with $E_1,E_2$ the final energies of two out of the three particles, and the angles are the Euler angles.

It would be great if someone could guide me through calculating integrals like that.

For the first case, I got this far (ignoring the constant factors in front):

$$\Gamma = \int \frac{\mathrm d^3 p_1 \mathrm d^3 p_2}{|\mathbf{p_1}||\mathbf{p_2}|}\delta^{(3)}(\mathbf{p_1}+\mathbf{p_2})\delta(m-|\mathbf{p_1}|-|\mathbf{p_2}|)$$

so I get rid of one integration and get

$$\Gamma = \int \frac{\mathrm d^3 p_1 }{|\mathbf{p_1}|^2}\delta(m-2|\mathbf{p_1}|)$$

With polar coordinates,

$$\Gamma = 4\pi \int \frac{\mathrm d |\mathbf{p_1}| }{|\mathbf{p_1}|^2}|\mathbf{p_1}|^2\delta(m-2|\vec{p_1}|) = 4\pi \int_0^\infty \mathrm d |\mathbf{p_1}|\delta(m-2|\mathbf{p_1}|) = 2\pi$$

But how to do the 3 body decay? The amplitude $|M|^2$ is just $\lambda^2$ again, so I can integrate over the Euler angles to get

$$\Gamma = \frac{1}{(2\pi)^3}\frac{\lambda^2}{8m} \int \int dE_1 dE_2$$

but how do I come up with the limits of integration here?

Is this right $\int_0^m dE_1 \int_0^{m-E_1}dE_2$ ?

• Some tips as I don't have too much time until the Weekend: At first the Matrix element is $M=2\lambda$ since there is no $\frac{1}{2!}$ factor in front of the interaction Lagrangian Secondly: You can immediately contract one $d^3p$ with three dimensions of the Delta function. Afterwards you can go to spherical coordinates for the remaining $d^3p$ and immediately execute angular integrals since the Matrix Element has no angular dependency. Finally, use that for massless particles $|\vec{p}| = E$ holds and thus $d|\vec{p}|=dE$ Hope this helps – failtrolol Apr 6 '16 at 21:03
• Thanks, I used your tips but am stuck again (edited the post) – Spine Feast Apr 6 '16 at 22:42
• Just plainly execute the angular integrations for an Independent Matrix element. The Phase space integrations have already been done. The angular factors give you $8\pi^2$. Then think about the correct boundaries for integrating with respect to $dE_1$ and $dE_2$. – failtrolol Apr 7 '16 at 12:16
• Yeah, that's what I got but I don't know how to come up with the bounds. – Spine Feast Apr 7 '16 at 12:18
• Choose one particle, that one can have energy from 0 to $\frac{M_\Phi}{2}$ - These are the boundaries for one Integration. And for the other one? – failtrolol Apr 7 '16 at 12:30