Why is it intuitively unreasonable for this transition probability to grow quadratically in $t$?

Question

In Sakurai's "Modern Quantum Mechanics" section 5.6, there is a seemingly simple statement made that I do not understand the logic of. The author is considering a physical situation in which we "turn-on" a time-independent potential at $t=0$, and ask what the relevant transition probabilities are.

$$\hat{V}(t)=\begin{cases}0 &,t<0 \\ \hat{V} &, t\geq 0\end{cases}$$

If the system under consideration begins in state $|i\rangle$, then to first-order (in the Dyson series), the transition amplitude between that initial state and a final state $|n\rangle$ as a function of time is:

$$c^{(1)}_f(t)=-\frac{i}{\hbar}\int_0^tdt'\,e^{i\omega_{ni}t}V_{fi}=\frac{2V_{ni}}{\hbar\omega_{ni}}e^{\frac{1}{2}i\omega_{ni}t}\sin\left(\frac{\omega_{ni}t}{2}\right)$$

or, more usefully, the transition probability as a function of the energy difference $\Delta E\equiv \hbar \omega_{ni}$:

$$P(t)\approx |c^{(1)}_n(t)|^2=\frac{4|V_{ni}|^2}{(\Delta E)^2}\sin^2\left(\frac{\Delta E t}{2\hbar}\right)$$

Here is the part I don't understand. The author now considers a final state degenerate in energy with the initial state, $\Delta E=0$ - i.e. an energy-conserving transition. In this case, the transition probability to leading order goes like:

$$P(t)=\frac{|V_{fi}|^2}{\hbar^2}t^2$$

The author then makes the following statement:

The probability of $|n\rangle$ after a time interval $t$ is quadratic, not linear, in the time interval during $V$ has been on. This may appear intuitively unreasonable. There is no cause for alarm however,...

I do not understand why one would intuitively expect the transition probability to be linear in $t$, or equivalently why we would expect the transition rate to be constant. Of course if we were dealing with a multi-level/state system in a dynamic steady-state, then I would intuitively expect all transition/flow rates to be constant. But here I cannot see a precise parallel. I have no intuition for this system. Could you help clarify why this might be intuitive?

Answer 1

I do not understand why one would intuitively expect the transition probability to be linear in $t$, equivalently why we would expect the transition rate to be constant.

Sakurai is writing for an audience familiar with Fermi's golden rule, wherein the probability of a quantum system to transition into a particular state (or set of states) with the same energy as the initial state does in fact go linearly in time. However, the golden rule applies to the case of transitions into a continuum of available final states, which is not the case here.

It is possible to show that when you have a continuum, all of the quadratic transition probabilities sum up to a linear overall one. Suppose we have a time dependent perturbation $$H_\text{perturbation} = \cos(\omega t) V$$ for some operator $V$. For a single final state $\left \lvert f \right \rangle$, the transition probability from initial state $\left \lvert i \right \rangle$ is $$P_{fi} = \frac{|V_{fi}|^2}{\hbar^2} \left( \frac{\sin((\omega_{fi} - \omega)t/2)}{(\omega_{fi} - \omega)/2} \right)^2$$ where $\omega_{fi} \equiv (E_f - E_i)/\hbar$. If we have a continuum of final states with density (as a function of energy) given by $\rho(E)$, then the total transition probability to any state of energy $E$ is \begin{align} P &= \int_\text{accessible energies} P_{fi}(t) \rho(E) dE \\ (\text{see note below}) \quad &= \int_{-\infty}^\infty P_{fi}(t) \rho(E=\hbar \omega) \, \hbar \, d\omega \\ &= \frac{2 \pi}{\hbar} |V_{fi}|^2 \rho(E_{fi}) \, t \end{align} which is linear in time. In the second line, we did two things:

1. Convert the integral from energy to frequency using $E=\hbar \omega$.

2. Send the limits of integration to infinity. This is reasonable because the integrand $P_{fi}$ is peaked near $\omega_{fi}$, so adding in more integration range doesn't change the total integral much.

In the third line we assumed $\rho(E)$ is constant over the range of the integral where the integrand is large, i.e. when $E \approx E_{fi}$, and then just evaluated the remaining integral over $P_{fi}$.

Of course if we were dealing with a multi-level/state system in a dynamic steady-state, then I would intuitively expect all transition/flow rates to be constant. But here I cannot see a precise parallel.

Sure, there's no parallel because dynamical steady state is very different from the perturbation problem with an out-of-equilibrium initial state that Sakurai is analyzing here.