Fermi golden rule

1. Jul 31, 2011

mahblah

A way to write Fermi golden rule is
$W_{fi} = \sum{\frac{d P_{fi}}{dt}} = \frac{2 \pi}{\hbar} \sum_{f} |V_{fi}|^2 \delta(\varepsilon_f - \varepsilon_i)$

where "i" is the initial unperturbed state and "f" is the final state of an ensemble of final states (i sum over them).

But because of $\delta( \varepsilon_f - \varepsilon_i)$ i'm asking $( \varepsilon_f - \varepsilon_i) =0$, so the inital and final state are the same??

they say that because of conservation energy must be $( \varepsilon_f - \varepsilon_i) =0$, but the external potential $V$ (i turn it on at time t=0) does not change the energy of the system (so it should be $\varepsilon_f \neq \varepsilon_i$)?

thanks all and sorry for my english,
MahBlah.

2. Aug 1, 2011

Bill_K

Take a look at Messiah, Ch XVII, Sect I.4 where he has a good explanation of this. Energy does not have to be conserved, since you turned V on at t=0. However the energy disconserving transitions (away from the peak of the delta function) have a finite probability. Whereas the conserving ones (near the peak) have a probability that grows linearly with time, hence the transition rate w = dW/dt is finite.