Fermi's golden rule and Probabilities in QM

Question

In Fermi's golden rule

$$P_{ab}(t)=2\pi t/\hbar \left|\langle\psi_b|V|\psi_a\rangle\right|^2 \delta(E_f-E_i)$$

for transition probability from state $a$ to $b$, how can the probability grow with time to above 1, how does one interpret a probability above 1? How does one use this formula?

And in this formula for the amplitude to propagate from an position eigenstate a to $b$ in time $T$ $\langle a|e^{-iHT}|b\rangle $, how does one interpret this physically? We cant have a particle at exact position a, let alone how would we measure it at the exact time $T$? Is there some integral over $T$ or something to get a physical probability?

What would we measure in an experiment to check that the amplitude is $\langle a|e^{-iHT}|b\rangle$ for a free particle?

Answer 1

Fermi's golden rule originates from first-order perturbation theory, which means that it can be derived from Schrödinger's equation under the condition that the perturbing hamiltonian be weak enough that there isn't a significant depletion of the initial state (for a proof see e.g. this link). As such, it only holds when the transition probabilities are on the order of a few percent (maybe even 10-50%, depending on the situation), and after that you need to take account of second-order effects.

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Regarding your second question, the formula $$⟨\phi|e^{-iHT}|\psi⟩$$ for a transition amplitude makes direct, measurable physical sense only when both the states in question are physically preparable, which excludes position eigenstates. For that kind of state, the amplitude $$⟨x_2|e^{-iHT}|x_1⟩$$ should be regarded as the integral kernel which gives the amplitude of transition between an initial wavefunction $\psi(x_1)$ and a final one $\phi(x_2)$ as $$\int\text dx_1\text dx_2\phi(x_2)^*⟨x_2|e^{-iHT}|x_1⟩\psi(x_1).$$