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

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SUMMARY

Fermi's Golden Rule is traditionally applied to constant or sinusoidal time-varying potentials; however, the discussion explores its application to non-sinusoidal perturbations of the form V_0(𝑥)f(t). The specific case of f(t) = e^{-at} with a > 0 is examined, emphasizing the need for Fourier decomposition of f(t) to relate it back to transition probabilities. The density of states ρ(E) at the final energy E is crucial for defining the transition from discrete initial states to a continuum of final states. The general formalism of time-dependent perturbation theory remains applicable in these scenarios.

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MisterX
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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 V_0\left( \mathbf{x}\right)f\left(t\right), where f(t) 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 f(t) in time, but it's not clear to me how that might be related back to transition probabilities. In particular, what if f(t) = e^{-a t} with a \in \mathbb{R}, a > 0 ?
 
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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##.
 
MisterX said:
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 V_0\left( \mathbf{x}\right)f\left(t\right), where f(t) 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 f(t) in time, but it's not clear to me how that might be related back to transition probabilities. In particular, what if f(t) = e^{-a t} with a \in \mathbb{R}, a > 0 ?
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.
 
Avodyne said:
You can still apply the general formalism of time-dependent perturbation theory.
Of course you can. But that was not the question. The question, I believe, was about Fermi's Golden Rule.
 
The question was "what if the perturbation is of the form V_0\left( \mathbf{x}\right)f\left(t\right), where f(t) is not a constant or sinusoidal?" And the answer is, "apply the general formalism of time-dependent perturbation theory."
 
Avodyne said:
The question was "what if the perturbation is of the form V_0\left( \mathbf{x}\right)f\left(t\right), where f(t) is not a constant or sinusoidal?" And the answer is, "apply the general formalism of time-dependent perturbation theory."
Take a look at the title of the thread.
 

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