# Fermi's Golden Rule: non-sinusoidal $t\to\infty$

1. Mar 12, 2014

### MisterX

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 $a \in \mathbb{R}, a > 0$ ?

2. Mar 13, 2014

### andrien

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 $c_n^1$.

3. Mar 13, 2014

### Bill_K

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.

4. Mar 13, 2014

### Avodyne

5. Mar 13, 2014

### Bill_K

Of course you can. But that was not the question. The question, I believe, was about Fermi's Golden Rule.

6. Mar 13, 2014

### Avodyne

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."

7. Mar 13, 2014

### Bill_K

Take a look at the title of the thread.