Density of states in Fermi's golden rule

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SUMMARY

The discussion centers on the density of states, denoted as ##\rho(E_{final})##, in the context of Fermi's golden rule. It clarifies that in scenarios without time-dependent potentials, ##\rho(E_{final})## is equivalent to ##\rho(E_{initial})##. The density of states is defined as the number of states within an energy interval ##[E_{f},E_{f}+dE]## divided by ##dE##, resulting in ##dN/dE##. The conversation addresses the challenge of infinite states within this interval and suggests using box quantization to achieve discrete states, ensuring that the final rate remains finite despite the infinite possibilities.

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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?
 
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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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