In time-dependent perturbation theory where $H=H_0+V$ and $V$ is considered small and has no explicit time dependence, the standard text-book treatment of the leading order probability amplitude for the system to make a transition from $|i\rangle$ to $|f\rangle$ is

$$ P_{f\leftarrow i}(\Delta t)=\big|\langle f|V|i \rangle\big|^2\frac{4\sin^2(\omega_{fi}\Delta t/2)}{\hbar^2\omega^2_{fi}}. $$

If I consider transitions between two states of the same energy, I take the $\omega_{fi}\rightarrow0$ limit, giving me

$$ P_{f\leftarrow i}(\Delta t)=\big|\langle f|V|i \rangle\big|^2(\Delta t)^2, $$

which grows without bound in time, as $\Delta t^2$. I would take this to mean that perturbation theory fails at long times. Why then am I allowed to take the large time limit to derive Fermi's Golden rule, without risking failure of perturbation theory?

  • $\begingroup$ Sakurai's Modern Quantum Mechanics (p.323 in the revised edition) offers a treatment of this where he first gets it down to be linear in time by considering a continuous range of final states and then calculates the rate per unit time. However, I also don’t see how one would interpret $P_{f \leftarrow i} (\delta t) > 1$. $\endgroup$
    – Claudius
    Dec 6, 2012 at 21:38

1 Answer 1


First, a correction. The first formula is the probability, not probability amplitude.

And it's computed at the leading order only, "linearized" in a sense, so of course it is only a good approximation for $P_{f\leftarrow i}\ll 1$. When the probability becomes comparable to one, subleading and higher-order corrections become important because one must also study how the newly created coefficients in front of other states – states absent in the initial state – change by the time evolution.

The perturbation theory always becomes inadequate when the perturbation, in this case the matrix element $\langle f |V|i\rangle$, is too large. But one must properly understand what "too large" means. And it means $P_{f\leftarrow i} \geq O(1)$ which is equivalent to $\langle f |V|i\rangle \cdot \Delta t \geq O(\hbar)$. For transitions at $\omega_{fi}\to 0$, the requirement for "how small the perturbation matrix element has to be" simply gets tougher, the upper limit becomes smaller. One more equivalent way to say it: for the perturbation theory to be OK, you need to have $\Delta t\ll \hbar / \langle f |V|i\rangle$.

However, your treatment has one more problem. Well, one of two problems. If you consider the transition to a discrete final state that just happens to have a finite energy, you are dealing with degenerate perturbation theory and you should first rediagonalize $H_0+V$ in this Hilbert subspace, to find out that the actual energy eigenstates differ from the original initial state and their energies actually differ.

If you consider a transition to a final state that belongs to a continuum, then you're interested in the integrated probability over $\omega_f$, anyway, and in that case, $\sin^2 Y / Y^2$ may be approximated by a multiple of the delta-function which imposes the "naive" energy conservation law. See e.g. this document for some intro to the method. My inequality appears as (11.40) on page 104.


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