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Jul 1, 2022 at 19:30 comment added bdzh Yes, but that $\Lambda$ i found before was the bandwith of the continuous spectrum, and the result you found for the the total transition rate gives the same thing. Is there any relation?
Jul 1, 2022 at 19:05 comment added user200143 One way to derive Fermi's golden rule is is to calculate the rate as the time derivative of the transition probability. $\frac{d}{dt} |\langle f|e^{-iHt}|i\rangle|^2$. The probability of staying in the initial state is 1 minus the sum of all transition probabilities. In lowest order this is an exponential whose Fourier transform gives a Lorentzian. So the exponential factor which gives the total transition rate gives the Lorentzian bandwidth.
Jul 1, 2022 at 3:02 comment added bdzh So the bandwith of the continuous spectrum is the same as the transition rate from the discrete state to the continuum?
Jul 1, 2022 at 3:01 vote accept bdzh
Jul 1, 2022 at 1:32 history answered user200143 CC BY-SA 4.0