Density of states in Fermi's golden rule

1. Dec 17, 2015

Coffee_

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. Dec 18, 2015

DrDu

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.