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Fermi golden rule

  1. Jul 31, 2011 #1
    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,
  2. jcsd
  3. Aug 1, 2011 #2


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    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.
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