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?