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A Fermis golden rule and transition rates

  1. Mar 30, 2016 #1
    When calculating the mean lifetime of an electron in a solid under the influence of a perturbation (for example electron-phonon interaction) we often apply Fermi's golden rule but the rate always has to be weighted by appropriate distribution functions (for example fermi functions). These take into account wether a given state is occupied of unoccupied and hence a candidate for the transition. Why do we need an "additional" probability for an unoccupied state? I mean, in the context of the Hilbert space the transition of a many-particle state to another isn't different from transitions between one particle states fundamentally. Isn't the probability already contained in the transition rate? In other words: The golden rule states: Given that the electron was in the state A (probability p(A)=1) what is the probability to find it in a continuum of states close to B at a time t? The rate is the time derivative of that probability. Isn't it redundant then to introduce a probability (1-P(B))? I hope I expressed myself clearly.
     
  2. jcsd
  3. Mar 30, 2016 #2
    well it looks very simple that now an electron has to transit between states A and B .... so the probability of the combined states A and B should be 1 . rather than probability of finding the electron in state A is to be 1.

    The transition probability is proportional to the density of final states . It is reasonably common for the final state to be composed of several states with the same energy - such states are said to be "degenerate" states. In many cases there will be a continuum of final states, so that this density of final states is expressed as a function of energy.
    i think one has to analyze the different terms appearing in the transition probability expression (as in the golden rule) which is a workable approximation.
     
  4. Apr 1, 2016 #3
    If you'd calculate the transitions of one particle you would only use the golden rule. The probability to find the electron in the continuum with energy close to Ea would then be an integral of the rate -no distribution function needed. When you use many particle systems you have to use distribution functions additionally. But fundamentally, a transition from a particular one-particle state to another is not different from the transition of a many particle state to another. The underlying math is the same. But why are they treated as if they're different?
     
  5. Apr 1, 2016 #4
    *integral over time
     
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