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,
(H0+H′−iℏ∂∂t)∑nan(t)|n⟩e−itEn/ℏ=0,
where En and |n⟩ are the stationary eigenvalues and eigenfunctions of H0 .
Rewrite this equation as that of the time evolution of the coefficients a an(t),
iℏdak(t)dt=∑n⟨k|H′|n⟩an(t)eit(Ek−En)/ℏ.
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 an(t)=δn,i, which simply says that the system stays in the initial state i.
For states k≠i, ak(t) becomes non-zero due to H′≠0 and these are assumed to be small due to the weak perturbation. Hence, one can plug in the zeroth order form a an(t)=δn,i into the above equation to get the first correction for the amplitudes ak(t),
iℏdak(t)dt=⟨k|H′|i⟩eit(Ek−Ei)/ℏ,
which integrates to
iℏak(t)=2⟨k|H′|i⟩eiωt/2sinωt/2ω
for ω≡(Ek−Ei)/ℏ, for a state with ai(0)=1,ak(0)=0, transitioning to a state with ak(t) (again, k≠i).
The transition rate is then
Γi→k=ddt|ak(t)|2=2|⟨k|H′|i⟩|2ℏ2sinωtω,
a sinc function peaking sharply for small ω. At ω=0, sin(ωt)/ω=t, so the transition rate varies linearly with t for an isolated state |k⟩ !
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 ρ(E), they must be integrated over their energies, and whence the corresponding ωs,
Γi→f=2ℏ∫∞−∞dωρ(ω)|⟨f|H′|i⟩|2sinωtω.
For large t, the sinc function is sharply peaked at ω≈0, and negligible outside [−2π/t,2π/t]; the density and transition element can be taken out of the integral, so that the rate
Γi→f=2ρ|⟨f|H′|i⟩|2ℏ∫∞−∞dωsinωtω
is now merely proportional to a constant Dirichlet integral, π.
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 ak(t)s invalidates lowest-order perturbation theory, which requires ak<<ai.)
If it helps, try to have a look to this article http://onlinelibrary.wiley.com/doi/10.1002/9783527665709.app6/pdf