I am currently trying to understand the Fermi's golden rule.
We consider a system with Hamiltonian: $$\hat H = \hat H_0 + \hat Ue^{i \omega t},$$ where the expectation value of $\hat U$ i much smaller than that of $\hat H_0$.
At $t=0$ the systems is in the ground state $\lvert 0 \rangle$ of $\hat H_0$ (we assume that $\hat H_0$ has discrete spectrum). Using first-order, time dependent perturbation theory, we obtain probability of the system being in the state $\lvert n \rangle$ after time $t$: $$\lvert c_n(t) \rvert^2 = \frac{1}{\hbar^2} \lvert \langle n \rvert \hat U \lvert 0 \rangle \rvert^2 \frac{\sin^2 \left(\frac{(\omega_n - \omega_0 - \omega) t}{2} \right)}{(\frac{\omega_n - \omega_0 - \omega}{2})^2}.$$ As far as I understand, this is the probability (not probability density) that we find the system in a state $\lvert n \rangle$ after time $t$ (if we for example turn off the time dependent potential and measure energy of the system). Now, according to my notes, if we take the limit as $t \to \infty$, this probability tends to: $$\lvert c_n(t) \rvert^2 = \frac{2\pi t}{\hbar^2} \lvert \langle n \rvert \hat U \lvert 0 \rangle \rvert^2 \delta(\omega_n - \omega_0 - \omega).$$ This expression is infinite (undefined?) when $\omega_n - \omega_0 - \omega = 0$. However, even after a very long time the transition probability should be finite (and while we're at it, than one).
I would normally expect to see dirac delta when dealing with probability densities , however as I mentioned before we are considering transitions between discrete states, so Kronecker delta is what I would expect to see here.
My question is, what am I missing here?
