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Density of states in Fermi's golden rule

  1. Dec 17, 2015 #1
    Fermi's golden rule contains a term that is the density of the final states ##\rho(E_{final})##. For my problem we have no time depending potentials so that's the same as ##\rho(E_{initial})##.

    If I understand the definition of ##\rho## correctly, it's the number of states in an interval ##[E_{f},E{f}+dE]## divided by ##dE## which just gives ##dN/dE##.

    However... what if there are infinite different states in this interval?

    Example:

    A -> B + C + D

    The final wavefunction will be a three particle wavefunction that will be able to distribute any final energy ##E_f## in an infinite number of ways among the different particles ##B,C,D##. This would mean that ##\rho## is infinite here.

    What am I missing?
     
  2. jcsd
  3. Dec 18, 2015 #2

    DrDu

    User Avatar
    Science Advisor

    You are missing nothing. You could do a box quantization, i.e. assume that the particles are trapped in a large box, so that the states become discrete. In the limit that the box becomes infinitely large, you recover the continuum. Note that the probability amplitude of a single state will also scale like 1/sqrt V where V is the box volume. Hence the coupling to the states will also decrease so that the final rate will remain finite.
     
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