I wonder if there is a detailed derivation of the quasi-electron lifetime:
\begin{equation} \frac{1}{\tau_k}=\frac{2\pi}{\hbar}\frac{1}{V^2}\sum_{k', q}\sum_{\sigma}|V_q|^2f_{k'}(1-f_{k-q})(1-f_{k'+q})\delta(\epsilon_{k-q}-\epsilon_{k}+\epsilon_{k'+q}-\epsilon_{k'}) \end{equation}
from Fermi golden rule. Although the result is stated in many literature and textbooks, I did not find an explicit derivation from Fermi golden rule anywhere so far.
The following argument follow basically this proof.
The golden rule is a straightforward consequence of the Schrödinger equation, solved to lowest order in the perturbation $H'$ of the Hamiltonian,
$$\left(H_{0}+H'-\mathrm{i} \hbar \frac{\partial }{\partial t}\right)\sum _{n}a_{n}(t)|n\rangle \mathrm {e} ^{-\mathrm {i} tE_{n}/\hbar }=0,$$
where $E_n$ and $|n⟩$ are the stationary eigenvalues and eigenfunctions of $H_0$ .
Rewrite this equation as that of the time evolution of the coefficients a $a_{n}(t)$,
$$\mathrm{i} \hbar {\frac {\mathrm {d} a_{k}(t)}{\mathrm {d} t}}=\sum _{n}\langle k|H'|n\rangle a_{n}(t)\mathrm {e} ^{\mathrm {i} t(E_{k}-E_{n})/\hbar }.$$
This equation is exact but normally cannot be solved in practice.
For a weak constant perturbation $H'$ which turns on at $t=0$, we can use perturbation theory. Namely, if $H ′ = 0$ it is evident that a $ a_{n}(t)=\delta_{n,i}$, which simply says that the system stays in the initial state $i$.
For states $ k\neq i$, $a_{k}(t)$ becomes non-zero due to $H'\neq 0$ and these are assumed to be small due to the weak perturbation. Hence, one can plug in the zeroth order form a $a_{n}(t)=\delta_{n,i}$ into the above equation to get the first correction for the amplitudes $a_{k}(t)$,
$$\mathrm {i} \hbar {\frac {\mathrm {d} a_{k}(t)}{\mathrm {d} t}}=\langle k|H'|i\rangle \mathrm {e} ^{\mathrm {i} t(E_{k}-E_{i})/\hbar },$$
which integrates to $$\mathrm {i} \hbar a_{k}(t)=2\langle k|H'|i\rangle \mathrm {e} ^{\mathrm {i} \omega t/2}{\frac {\sin \omega t/2}{\omega }}$$
for $\omega \equiv (E_{k}-E_{i})/\hbar$, for a state with $a_i(0) =1, a_k(0)=0$, transitioning to a state with $a_k(t)$ (again, $k\neq i$).
The transition rate is then
$$\Gamma_{i\rightarrow k}={\frac {\mathrm {d} }{\mathrm {d} t}}\left|a_{k}(t)\right|^{2}={\frac {2|\langle k|H'|i\rangle |^{2}}{\hbar ^{2}}}{\frac {\sin \omega t}{\omega }},$$
a $sinc$ function peaking sharply for small $\omega$. At $\omega =0$, $\sin(\omega t)/\omega =t$, so the transition rate varies linearly with $t$ for an isolated state $|k\rangle$ !
By dramatic contrast, for states of energy $E$ embedded in a continuum, they must be all accounted for collectively. For a density of states per unit energy interval $\rho(E)$, they must be integrated over their energies, and whence the corresponding $\omega$s,
$$\Gamma_{i\rightarrow f}={\frac {2}{\hbar }}\int _{-\infty }^{\infty }\mathrm {d} \omega \rho (\omega )|\langle f|H'|i\rangle |^{2}\,{\frac {\sin \omega t}{\omega }}.$$
For large $t$, the $sinc$ function is sharply peaked at $\omega\approx 0$, and negligible outside $[-2\pi/t,2\pi/t]$; the density and transition element can be taken out of the integral, so that the rate $$\Gamma_{i\rightarrow f}={\frac {2\rho |\langle f|H'|i\rangle |^{2}}{\hbar }}\int _{-\infty }^{\infty }\mathrm {d} \omega {\frac {\sin \omega t}{\omega }}$$ is now merely proportional to a constant Dirichlet integral, $\pi$.
The time dependence has vanished, and the constant decay rate of the golden rule follows.
As a constant, it underlies the exponential particle decay laws of radioactivity. (For excessively long times, however, the secular growth of the $a_k(t)$s invalidates lowest-order perturbation theory, which requires $a_k<< a_i$.)
If it helps, try to have a look to this article http://onlinelibrary.wiley.com/doi/10.1002/9783527665709.app6/pdf
