Fermi's Golden Rule - Inelastic collision probability

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SUMMARY

Fermi's Golden Rule (FGR) for inelastic transitions provides the probability of a transition due to a perturbation over a given time interval. The probability is influenced by the energy difference between two states (B - A) and the density of states (ρ(E)), which accounts for the number of final states per energy interval. The discussion clarifies that the transition probability is time-independent as it represents a first-order approximation, despite the initial state occupation decreasing over time due to interactions. The equation for the probability is defined as P = 2E |V|² D / (B - A)², where D is the density of states and V is the matrix element.

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  • Understanding of Fermi's Golden Rule and its applications in quantum mechanics
  • Familiarity with the concepts of inelastic collisions and energy states
  • Knowledge of the density of states (DOS) in quantum systems
  • Basic grasp of perturbation theory in quantum mechanics
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Master J
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Fermi's Golden Rule for an inelastic transition states gives the probability of a transition, due to a perturbation (which is constant in a given time interval).

For 2 energy states A and B, it is proportional, among others quantities, to what I'll call E. This E is the small range of energy states around the final state B ---> ie. the uncertainty, since we can't know 100% the final energy state (uncertainty principle).

Obviously, for an inelastic collision, B - A > E, otherwise we would have conservation of energy.

So, I think I have all that right! But here is my question...the above mentioned probability, is it time INdependent? I have been told it is. However, it doesn't seem so to me:

B - A > E homework > E therefore, 2pih/t > E where h is h-bar and w the Bohr frequency between the 2 states.
So, as time goes on, it would seem to that the probability of finding, for example, an electron in a higher energy state would decrease, ie. it would have most likely relaxed by say photon or phonon emission.
Any insight in this? Am I right?

Perhaps...it is not explicitly time dependent??
 
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What's contained in Fermi's Golden Rule is not a "E, a range of energies" Rather it's ρ(E), the density of states (number of final states per energy interval.) And the reason it's in there has nothing to do with the Heisenberg uncertainty principle, or because energy is being slightly unconserved.

It's in there because energy *is* precisely conserved, and the transition probability contains a delta function, δ(EA - EB). In order to get something finite, one sums over all final states having an energy in the immediate vicinity of EB and defines a transition probability to this entire group of states.

The answer is time-independent because it's a first-order approximation. Does not take into account the fact that the occupation of the initial state will slowly decrease due to the interaction.
 
Ah, I have just realized that what I am looking at is in fact a precursor to FGR.

Sorry, I am not talking about FGR, which is a probability per unit time, i am talking about the probability itself. FGR is derived from the elastic consideration of this, and the equation I am talking about is from the inelastic. It is:

P = 2E. |V|^2. D / (B-A)^2 where D is the DOS, V is the matrix element.
 

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