In Fermi's Golden Rule, does the transition probability increase linearly with time or quadratically with time?

Question

When deriving Fermi's Golden rule, we get that the probability of a quantum system transitioning from an initial state $|i\rangle$ to a final state $|f\rangle$ is $$P_{i\rightarrow f}(t)=\frac{|V_{fi}|^2}{\hbar^2}\frac{\sin^2\left(\frac{\omega_{fi}-\omega}{2}\cdot t\right)}{\left(\frac{\omega_{fi}-\omega}{2}\right)^2}.\tag{1}$$ If we wish to examine the behaviour of this probability when the resonance condition $\omega_{fi}=\omega$ is met, we can do the following $$P_{i\rightarrow f}(t)=\frac{|V_{fi}|^2}{\hbar^2}\frac{\sin^2\left(\frac{\omega_{fi}-\omega}{2}\cdot t\right)}{\left(\frac{\omega_{fi}-\omega}{2}\cdot t\right)^2}\cdot t^2\\ \Rightarrow P_{i\rightarrow f}(t)=\frac{|V_{fi}|^2}{\hbar^2}\text{sinc}^2\left(\frac{\omega_{fi}-\omega}{2}\cdot t\right)\cdot t^2 $$ Now since $sinc(0)=1$, we have that when $\omega_{fi}=\omega$, the probability of transition increases quadratically with time: $$P_{i\rightarrow f}(t)=\frac{|V_{fi}|^2}{\hbar^2}t^2 \tag{2}$$ This quadratic increase in time is actually not unique to the resonance condition. It occurs for all the secondary peaks in the $P_{i\rightarrow f}(t)$ function. So seemingly, the probability for a transition should increase quadratically with time. However, all the books I have read (McIntyre, Shankar, Griffiths etc) as well as all the responses on this site pertaining to Fermi's Golden Rule that I have seen then go on to say that $$\lim_{t\rightarrow \infty}P_{i\rightarrow f}(t)=\lim_{t\rightarrow \infty}\frac{|V_{fi}|^2}{\hbar^2}\frac{\sin^2\left(\frac{\omega_{fi}-\omega}{2}\cdot t\right)}{\left(\frac{\omega_{fi}-\omega}{2}\right)^2}$$ $$\therefore \lim_{t\rightarrow \infty}P_{i\rightarrow f}(t)=\frac{2\pi}{\hbar^2}|V_{fi}|^2\delta(\omega_{fi}-\omega)\cdot t \tag{3}$$ This result now seems to indicate that the probability of transition increases linearly with respect to time provided we are looking at $P_{i\rightarrow f}(t)$ for large enought times. We no longer have the quadratic time dependence that equation (2) suggests we should. So what has happened here? Why do equation (1) and (2) imply that the probability should increase quadratically with time (regardless of how large $t$ is) but when we take the limit as $t\rightarrow \infty$, all of a sudden we now only have a linear increase with time?

I have seen the post over here (Transition probability derivation: How to prove $\lim_{\alpha\rightarrow\infty} \frac{\sin^2\alpha x}{\alpha x^2} ~=~\pi\delta(x)$?) which proves the limit but the answer does not explain how the transition probability $P_{i\rightarrow f}(t)$ somehow seems to be both increasing quadratically with time according to equations (1) and (2) but only linearly according to equation (3). Any help on this issue would be greatly appreciated!

Answer 1

1. OP is considering an argument $x$ for the ${\rm sinc}^2(x)$ function with $|x|\ll 1$, where the probability $P \propto t^2$ is indeed quadratic in $t$.

2. But Fermi's golden rule applies to the region $|x|\gtrsim 1$, where $P \propto t$ is linear in $t$, cf. e.g. this, this and this Phys.SE posts.

Answer 2

The transition from a $t^2$ dependence to a $t$ dependence happens when the evolution time $t$ becomes larger than the coherence time of the drive field. To see this explicitly, we can write down the amplitude of the final state as an integral from the first order perturbation theory: \begin{equation} a_f(t)=-\frac{i}{\hbar}\int_0^t e^{i\omega_{fi}t'}V_{fi}(t')dt' \end{equation} The final state probability is then given by $P_f(t)=|a_f(t)|^2$. If the drive field is purely monochromatic, then $V_{fi}(t)=V_{fi}e^{-i\omega t}$, and for short time you obtain the $t^2$ dependence of the final state probability. This is justified if the integration time $t$ is short compared to the coherence time of the drive field (or the inverse of the spectral bandwidth of the drive field). If, however, the drive field is broadband, and you're interested in the probability of the final state at time that is long compared to the coherence time, then the Fermi's Golden rule applies. If the different frequency of the drive fields are not correlated, then we have: \begin{equation} P_f(t)=\frac{1}{\hbar^2}|\int_0^t e^{i\omega_{fi}t'}V_{fi}(t')dt'|^2 \end{equation} This integral is a windowed Fourier transform of a stochastic drive field. Thus, for the integration time $t$ much larger than the coherence time, we will have: \begin{equation} |\int_0^t e^{i\omega_{fi}t'}V_{fi}(t')dt'|^2=2\pi t S(\omega_{fi}) \end{equation} where $S(\omega)$ is the power spectral density of the field $V_{fi}(t)$, in the sense that $\int_{-\infty}^{\infty} S(\omega)d\omega=<|V_{fi}(t)|^2>$. Then, we obtain a linear dependence of the final state probability: \begin{equation} P_f(t)=\frac{2\pi t}{\hbar^2}S(\omega_{fi}), \end{equation} which is the result of Fermi's Golden rule. And also, this derivation focuses on the situation where the drive field has a spectral width. The same method applies when the final state is an ensemble of continuum states.