Close-to-resonance driving frequency in time dependent perturbation theory

Question

In a time-dependent perturbation theory problem, the Hamiltonian is given $H = H_0 + V$, where $V$ is a perturbation that varies sinusoidally with time. $V = V_0 \sin \omega t$. Supposed that the original Hamiltonian $H_0$ is solved, and the (non-degenerate) eigenstates are given by

$$H_0 | n \rangle = E_n |n \rangle$$

Suppose the system is initially prepared in the state $|m \rangle$. The perturbation is switched on for a duration of $T$. I would like to consider the transition probability $P_{m \to n}$ after the perturbation is switched off. If the driving frequency is close to the resonance frequency, one term in the perturbation can be ignored (rotating wave approximation). If I did not make any mistake, the result using first-order perturbation theory is

$$P_{m \to n} = \left|\frac{\langle m |V_0| n \rangle}{E_n - E_m - \hbar \omega} \right|^2 \sin^2 \left( \frac{E_n - E_m - \hbar \omega}{2 \hbar} T \right)$$

This means that the maximum transition probability is proportional to $\left(\frac{1}{E_n - E_m - \hbar \omega}\right)^2$. This quantity can be made arbitrarily large by using driving frequency close to resonance. But the $\sin^2$ term is small, so the small-angle approximation can be used.

\begin{align} P_{m \to n} &\approx \left|\frac{\langle m |V_0| n \rangle}{E_n - E_m - \hbar \omega} \right|^2 \left( \frac{E_n - E_m - \hbar \omega}{2 \hbar} T \right)^2 \\ &= \frac{|\langle m |V_0| n \rangle|^2}{4\hbar^2} T^2 \end{align}

So it suggests that the transition probability is larger when the system is exposed to the perturbation for a longer time. However, the probability should never exceed one. So where did the perturbation theory fail? Or did I make a mistake somewhere?

Answer 1

There are two situations in which your approximate calculation is going to fail. One situation is at long times. You made use of the approximation that argument of the sine function was small, so that you could approximate $$\sin\left(\frac{E_{n}-E_{m}-\hbar\omega}{\hbar}T\right)\approx\frac{E_{n}-E_{m}-\hbar\omega}{\hbar}T.$$ However, it should be clear that, no matter how close to resonance (that is, how close to $E_{n}-E_{m}=\hbar\omega$) you get, there will always be times $T$ large enough that the argument of the sine is no longer small. At times for which $$T\gtrsim|\omega_{mn}-\omega|^{-1}=|\delta|^{-1}$$ [where $\omega_{mn}\equiv(E_{n}-E_{m})/\hbar$, and $\delta\equiv\omega-\omega_{mn}$ is called the "detuning"], you need to use the evaluate the trigonometric function fully. Since $|\sin^{2}x|\leq 1$, the transition probability is never going to exceed $$P_{m\rightarrow n}\leq\left|\frac{\langle m |V_0| n \rangle}{E_n - E_m - \hbar \omega} \right|^{2}.$$

The other situation where your expression has a problem is if the driving perturbation is too strong. If the matrix element $\langle m|V_{0}|n\rangle=\hbar\omega_{R}/2$ is too large, the expressions in your question can also grow larger than unity. ($\omega_{R}$ is known as the "Rabi frequency," after the inventor of magnetic resonance.) This problem occurs because you have only considered the transition perturbatively; the amplitude for the system to be in the upper state $|n\rangle$ only receives a contribution at linear order in $V_{0}$, corresponding to the absorption of a single photon from the driving field. Physically, as the probability of being in the higher-energy $|n\rangle$ state grows, the oscillating applied field begins to stimulate downward transitions back to the lower state $|m\rangle$, but this is not captured by the lowest-order perturbative calculation.

However, it turns out that in the rotating wave approximation, the Hamiltonian for the two-level system is actually exactly solvable. You can find the solution in section 1.7 of these course notes. (Peculiarly, those course notes are hosted on a server from Cal Tech, but they were written by David Pritchard and Wolfgang Ketterle for the 8.421 Atomic and Optical Physics I class at MIT.) The "dressed atom" solution for the transition probability is $$P_{m\rightarrow n}=\left(\frac{\omega_{R}^{2}}{\omega_{R}^{2}+\delta^{2}}\right) \sin^{2}\left(\frac{\sqrt{\omega_{R}^{2}+\delta^{2}}}{2}T\right).$$ This expression is clearly between zero and one, as it should be. When $\omega_{R}$ is small compared with $|\delta|$, this reduces to the perturbative expression. However, when the system is strongly driven, near to resonance, so that $\omega_{R}\gg|\delta|$, the system oscillates back and forth between the lower and upper states with frequency $\omega_{R}$. (The presence of the Rabi frequency in the denominator of $P_{m\rightarrow n}$ also has other interesting effects, including changing the resonant line width. This is known as "power broadening," and it has a natural explanation in the fact that when $\omega_{R}$ is large, the lifetime of the upper state is not set primarily by spontaneous decays but by the Rabi oscillations back from $|n\rangle$ to $|m\rangle$.)

Answer 2

First-order perturbation does not guarantee probability conservation. And, if the perturbation is too strong as in your case, the first-order approximation is invalid.