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Fermi's Golden Rule: non-sinusoidal [itex]t\to\infty[/itex]

  1. Mar 12, 2014 #1
    From what I have seen, Fermi's Golden rule is applied to constant or sinusoidal time varying potentials. But what if the perturbation is of the form [itex]V_0\left( \mathbf{x}\right)f\left(t\right)[/itex], where [itex]f(t)[/itex] is not a constant or sinusoidal? I am not really familiar with the derivation of Fermi's golden rule, and the explanations I was given both seemed very hand-wavy. I know we can Fourier decompose [itex]f(t)[/itex] in time, but it's not clear to me how that might be related back to transition probabilities. In particular, what if [itex]f(t) = e^{-a t}[/itex] with [itex]a \in \mathbb{R}, a > 0[/itex] ?
     
  2. jcsd
  3. Mar 13, 2014 #2
    why don't you take a look at method of variation of constant.To first order of perturbation theory,probability is just given by absolute square of ##c_n^1##.
     
  4. Mar 13, 2014 #3

    Bill_K

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    Fermi's Golden Rule applies specifically to the transition from a discrete initial state to a continuum of final states. One of the factors in it is ρ(E), the density of states at the final energy E. The reason the perturbing potential is restricted to a single frequency is so that E will be well-defined.
     
  5. Mar 13, 2014 #4

    Avodyne

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  6. Mar 13, 2014 #5

    Bill_K

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    Of course you can. But that was not the question. The question, I believe, was about Fermi's Golden Rule.
     
  7. Mar 13, 2014 #6

    Avodyne

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    The question was "what if the perturbation is of the form [itex]V_0\left( \mathbf{x}\right)f\left(t\right)[/itex], where [itex]f(t)[/itex] is not a constant or sinusoidal?" And the answer is, "apply the general formalism of time-dependent perturbation theory."
     
  8. Mar 13, 2014 #7

    Bill_K

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    Take a look at the title of the thread.
     
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