- #1
mahblah
- 21
- 2
A way to write Fermi golden rule is
[itex] W_{fi} = \sum{\frac{d P_{fi}}{dt}} = \frac{2 \pi}{\hbar} \sum_{f} |V_{fi}|^2 \delta(\varepsilon_f - \varepsilon_i) [/itex]
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 [itex] \delta( \varepsilon_f - \varepsilon_i) [/itex] I'm asking [itex] ( \varepsilon_f - \varepsilon_i) =0 [/itex], so the inital and final state are the same??
they say that because of conservation energy must be [itex] ( \varepsilon_f - \varepsilon_i) =0 [/itex], but the external potential [itex]V[/itex] (i turn it on at time t=0) does not change the energy of the system (so it should be [itex] \varepsilon_f \neq \varepsilon_i[/itex])?
thanks all and sorry for my english,
MahBlah.
[itex] W_{fi} = \sum{\frac{d P_{fi}}{dt}} = \frac{2 \pi}{\hbar} \sum_{f} |V_{fi}|^2 \delta(\varepsilon_f - \varepsilon_i) [/itex]
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 [itex] \delta( \varepsilon_f - \varepsilon_i) [/itex] I'm asking [itex] ( \varepsilon_f - \varepsilon_i) =0 [/itex], so the inital and final state are the same??
they say that because of conservation energy must be [itex] ( \varepsilon_f - \varepsilon_i) =0 [/itex], but the external potential [itex]V[/itex] (i turn it on at time t=0) does not change the energy of the system (so it should be [itex] \varepsilon_f \neq \varepsilon_i[/itex])?
thanks all and sorry for my english,
MahBlah.