Fermi's Golden Rule: Explained

[Master J]

For a perturbation that is constant within a given time, we can use Fermi's Golden Rule.
In developing Fermi's Golden Rule, 2 approximations are made:

(1)

The energy interval E (continuum) to which discrete states are scattered is small enough such that the density of states is constant in this interval.

(2)

The time t is large enough such that the energy interval E is greater than the Bohr frequency. E >> 2 pi h / t
(h is h-bar!)

Now I do not get this 2nd one.

There is a factor involving cosine of (w.t), where w is the Bohr frequency connecting the final and initial states. For small t, this factor oscillates rapidly and has a peak when the 2 states are equal, showing that scattering that preserves the unperturbed energy is dominant. This bit is clear.

Yet why must the energy interval be greater than the Bohr f?? An what does this have to do with time? The Bohr f is really just an energy difference right? I don't see how the inequality from (2) arises, or why it is important?

 

[DrDu]

The spacing of the energy levels has to be much smaller than the energy uncertainty h/t. The energy spacing determines the period of the recurrence time. For long times, a superposition of discrete energy levels will return to the initial state.
In real systems, this condition is usually not so important as the systems are coupled to surroundings which will lead to further decoherence, so that no recurrences occur.

 

[Master J]

Do you mean that the spacing between the initial and scattered to state (final state) must be less that the energy uncertainty, ie., the range of energies to be scattered to?

 

[DrDu]

No, I meant the spacing of the states forming the quasi-contiuum of scattering states.

 

[DrDu]

A great model to understand Fermi's golden rule is the Bixon Jortner model, which treats the constant coupling V of a single initial state s to an infinite ladder equally spaced final states i ( i ranging from -infinity to infinity), which may be solved analytically. On google you certainly find some closer description.

 

[Master J]

Thanks, I'll look that up. I believe its mentioned in one of my QM books.

Ah, so the statement about this energy interval E is just a statement of the Uncertainty Principle:

E > h / t with a factor of 2 pi when using the reduced Plank constant h-bar.

So, t is the time at which we "look" to see which states the particles have been scattered to, the perturbation beginning at t=0 (of course, the most probable is that they have given up that energy from the perturbation and relaxed to their initial state (elastic), as the peak in the cos(w.t) factor shows right?).

Now then, I am thinking that t must be sufficiently large so that the interval only has to be greater than h / t, which will be quite small if t is large! Then, this allows the approximation that the density of states is approximately constant since E is can now be small.

If t wasn't sufficiently large, the range of states the particles could be scattered to would be too large (since E must now be large from UP), making trying to preidict the particles states quite useless.

This is my grasp of it so far now. Am I correct, or at least along the right line?

 

[DrDu]

Let me rephrase the Fermi rule in my words:
The Fermi golden rule holds for time periods $$\Delta t$$ which are
1. so large, that 1the density of final states in the interval $$\Delta E \approx h/\Delta t$$ (energy time uncertainty relation) is approximately constant.
2. so short that the number of final states in the interval $$\Delta E$$ is >>1, so that the final states can be approximated as a continuum.