As I understand, Fermi's golden rule is a result from first order perturbation, which says that the transition rate of an initial state $|i\rangle$ to a final state $|f\rangle$ is $$ \Gamma_{i\rightarrow f} =2\pi \rho|\langle f |H'| i\rangle|^2 $$ where $H'$ is the perturbative term in the Hamiltonian and $\rho$ is the density of states (of dimension $[E]^{-1}$) near the final state $|f\rangle$. However, in quantum field theory when people apply Fermi's golden rule to compute decay rate or cross section, they often replace $\langle f|H'|i\rangle$ by the scattering amplitude $\mathcal{M}_{fi}$, which is (up to some constant factor) a matrix element of the $S$-matrix and is typically computed using a time-ordered exponential $$\left\langle f\left|T\exp\left(-i\int_{-\infty}^\infty dt H'_I(t)\right)\right|i\right\rangle.$$ What is the justification of such replacement?


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Your second expression is really the more fundamental one, in the sense that Fermi's Golden Rule (FGR) is just an approximation. You can't really go from FGR to the field theoretic result, but you can see why they're equivalent in a certain limit:

As you said, we are using first-order perturbation theory to derive FGR, which means the timescales involved are short compared to the transition rate $\Gamma$. In addition, our perturbation is constant. So, if we turn on our perturbation at some time and wait a short time $\Delta t$, we just expand the operator you wrote to get $$1-iH'\Delta t,$$ throwing away higher-order terms. Now, compare this with $$S \sim 1 + i \mathcal{M},$$ which makes sense if we think of $\mathcal{M}$ as corresponding to something non-trivial to happen. The probability is proportional to $|\mathcal{M}|^2$, by the normal Born rule, and so putting these two expressions together we find ourselves back with FGR.

To actually see why your second expression is even valid in the first place is a bit more involved, but you essentially just repeatedly evolve your state infinitesimally according to the normal Schroedinger equation, and then stick it all together in the correct order (hence the $T$ and integral). This is just the Dyson series.

Aaaand I just saw this was asked 2 years ago; oh well.


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