Inflaton Decay Rate After inflation, (p)reheating is supposed as the mechanism which is responsible for restoring hot Big Bang. Always the resulting decay rate of inflatons into bosons and fermions are mentioned without derivation in literature (Perhaps its assumed to be very simple!). I couldn't find derivations anywhere. Can anyone help me? 
$$
\mathcal L = ... +g_1\phi^2\chi^2+g_2\phi\bar{\psi}\psi+\sigma\phi\chi^2
$$
The last term needs symmetry breaking to appear of course. $\phi$ is the inflaton field and $\chi$ & $\psi$ are arbitrary bosonic and fermionic fields. The remaining parameters are coupling constants.
 A: As the general details of reheating are quite complicated, I won't say much about it, instead I'll just focus on answering the simpler part: given a Lagrangian, how to compute the decay rate (or, in the case of the first term, the cross section)?
Now, the decay rate for a general reaction in which one initial particle decays into $n$ final ones is given by:
$$d\Gamma=\frac{1}{2E}|\mathcal{M}_{fi}|^2d\Phi^{(n)}$$
where $E$ is the energy of the initial particle, $d\Phi^{(n)}=(2\pi)^4\delta^4(P_i-P_f)\prod\limits_{i=1}^n\frac{d^3p_i}{(2\pi)^3\, 2E_i}$ is the phase space element in momentum space, and of course $\mathcal{M}$ is the probability amplitude of the transition. The calculation can be done in three steps, employing the Feynman rules:
1) $\mathcal{L}_1=g_2\phi\bar\psi\psi$
This is a standard 3-field coupling at a single point, or in other words a decay of a scalar particle into a fermion-antifermion pair, and so the transition amplitude is:
$$i\mathcal{M}^{(1)}=ig_2\bar{u}^{s'}(p')v^{s}(p)$$
Here $p$ is the momentum of the antifermion, and $p'$ is that of the fermion. So the square of the modulus is simply:
$$|\mathcal{M}^{(1)}|^2=g_2^2 \, Tr\left[\bar{v}^{s}(p) u^{s'}(p') \bar{u}^{s'}(p')v^{s}(p)\right]$$
In order to compute this we of course sum over the final polarizations, to get:
$$|\mathcal{M}^{(1)}|^2=g_2^2 Tr\left[(\gamma\cdot{p}'+m)(\gamma\cdot{p}-m)\right]$$
We have just two terms that don't vanish due to trace identities, and so:
$$|\mathcal{M}^{(1)}|^2=g_2^2\, (4p\cdot p'-4m^2)=4g_2^2\,(p\cdot p'-m^2)$$
As the reaction involves two final particles, we have:
$$d\Phi^{(2)}=\frac{d^3p}{(2\pi)^3\, 2E}\frac{d^3p'}{(2\pi)^3\, 2E'}(2\pi)^4\delta^4(P_I-p-p')$$
we will evaluate the transition element in the rest frame of the initial particle, so $E_I=M_I$, the inflaton mass, and $\textbf{p}+\textbf{p}'=0$. In that case the 3D dirac delta goes away, and we have just:
$$d\Phi^{(2)}=\frac{1}{(2\pi)^2}\frac{1}{4EE'}\delta(M_I-E-E')d^3p$$
We then write $d^3p=p^2\,dp\,d\Omega$, and since $E=E'$ since they have the same mass, the integral is:
$$d\Phi^{(2)}=\frac{1}{32\pi^2}\sqrt{1-\frac{4m^2}{M_I^2}}d\Omega$$
Note that all of this was done just employing the delta functions, so while $\mathcal{M}$ depends on momenta, the above result is independent of the scattering element.
The matrix element evaluated in the COM frame gives:
$$|\mathcal{M}^{(1)}|^2=4g_2^2\,(E^2+\textbf{p}^2-m^2)=8g_2^2\, \textbf{p}^2=8g_2^2\,\left(\frac{M_I^2}{4}-m^2\right)$$
The last equality is a consequence of energy conservation. Finally we have:
$$d\Gamma=\frac{1}{2M_I}8g_2^2\,\left(\frac{M_I^2}{4}-m^2\right)\frac{1}{32\pi^2}\sqrt{1-\frac{4m^2}{M_I^2}}d\Omega$$
The integral over the angular variables is just $4\pi$, and thus:
$$\Gamma=\frac{M_I}{8\pi}g_2^2\,\left(1-\frac{4m_\psi^2}{M_I^2}\right)^{\frac{3}{2}}$$
Since the inflaton is usually considered to be a lot more massive than any known fermion, in the limit $m_\psi\ll M_I$ we get $\Gamma=\frac{M_I g_2^2}{8\pi}$.
2) $\mathcal{L}_2=\sigma\phi\chi^2$
The only difference between this process and the previous one is in the matrix element (it's again a 3-field interaction at a point), which now reads:
$$i\mathcal{M}^{(2)}=i\sigma\Rightarrow |\mathcal{M}^{(2)}|^2=\sigma^2$$
Just plugging this into the previous equations gives:
$$d\Gamma=\frac{1}{2M_I}\sigma^2\frac{1}{32\pi^2}\sqrt{1-\frac{4m_\chi^2}{M_I^2}}d\Omega\Rightarrow \Gamma=\frac{\sigma^2}{16\pi M_I}\sqrt{1-\frac{4m_\chi^2}{M_I^2}}$$
Note that $\sigma$, unlike $g_2$ which is dimensionless, has dimensions of mass.
3) $\mathcal{L}_3=g_1\phi^2\chi^2$
This is a scattering term, and luckily it only involves scalar fields, and no propagators, so the matrix element is just:
$$i\mathcal{M}^{(3)}=ig_1\Rightarrow |\mathcal{M}^{(3)}|^2=g_1^2$$
The differential cross section for a $2\rightarrow 2$ processes in the COM frame for elastic scattering is:
$$\frac{d\sigma}{d\Omega}=\frac{1}{64\pi^2 s}|\mathcal{M}^{(3)}|^2$$
Here $s$ is the square of the total energy in the COM frame, which is $s=(E_1+E_2)^2=4E^2$. Thus:
$$\sigma=\frac{g_1^2}{4\pi E^2}$$
The general formulas for the differential cross section and the decay rate as well as the Feynman rules used here can be found in pretty much any QFT textbook (Peskin & Schroeder, Maggiore, etc.). Note that I may have missed some combinatorical factor(ial)s in the last two processes since the the final particles (and initial ones in the third part) are identical, hence other diagrams which are permutations of the standard ones probably contribute, which multiplies the scattering element, but it also affects the size of the phase space (the phase space generally decreases by a factorial).
