# Fermi's Golden Rule and Density of States

I know Fermi's Golden Rule in the form

$$\Gamma_{fi} ~=~ \sum_{f}\frac{2\pi}{\hbar}\delta (E_f - E_i)|M_{fi}|^2$$

where $\Gamma_{fi}$ is the probability transition rate, $M_{fi}$ are the transition matrix elements.

I'm struggling to do a derivation based on the density of states. I know that under certain circumstances it's a good approximation to replace $\sum_f$ with $\int_F \rho(E_f) \textrm{d}E_f$ to calculate the transition probability, for some energy range $F$.

Doing this calculation I obtain

$$\Gamma_{fi} ~=~ \int \rho(E_f) \frac{2\pi}{\hbar}\delta (E_f - E_i) |M_{fi}|^2\textrm{d}E_f.$$

Now assuming that the $M_{fi}$ are constant in the energy range under the integral we get

$$\Gamma_{fi} ~=~ \rho(E_i) \frac{2\pi}{\hbar} |M_{fi}|^2.$$

Now this is absolutely not what is written anywhere else. Other sources pull the $\rho(E_f)$ out of the integral to obtain Fermi's Golden Rule of the form

$$\Gamma_{fi} ~=~ \rho(E_f) \frac{2\pi}{\hbar} |M_{fi}|^2$$

for any $f$ with $E_f$ in $F$ which makes much more physical sense. But why is what I've done wrong? If anything it should be more precise, because I have actually done the integral! Where have I missed something?

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It's the same thing because $E_i=E_f$ in this treatment, isn't it? – Luboš Motl Dec 17 '12 at 14:32

As proposed by Lubos, the delta function you started with $\delta(E_i-E_f)$ forces the final result to be invariant by $E_i \leftrightarrow E_f$.
Well, are you familiar with identity:$$\delta(x-x_0)f(x) = \delta(x-x_0)f(x_0)$$ true for distributions, it implies quite directly that you can change $\rho(E_f)$ for $\rho(E_i)$ in your second equation. – Learning is a mess Dec 17 '12 at 15:15
Oh of course - apologies for missing that. But surely in general $\rho(E_i)$ and $\rho(E_f)$ are different even if $E_i = E_f$? For example the decay of one particle into two gives you an extra degree of freedom in $\rho(E_f)$ that you didn't have in $\rho(E_i)$. Or is this logic wrong? – Edward Hughes Dec 17 '12 at 15:18
Dear Edward, you're summing or integrating over $f$ and the integrand depends on $M_{fi}$ and you decided you may eliminate the summation/integral completely. This implicitly means that for each energy $E_f$, the state must actually be unique. Otherwise you would have to keep the sum over the other quantum numbers that commute with the energy. – Luboš Motl Dec 17 '12 at 16:17
I think the point of confusion here is that $\rho(E)$ is the density of final states. Perhaps the notation would be more clear if $\rho_f(E)$ were written instead. Now it should be clear that since energy is conserved $\rho_f(E_f)=\rho_f(E_i)$. Note that the density of initial states, which you might write as $\rho_i(E)$ is not equal to $\rho_f(E)$, as your comment, "But surely..." seems to suggest. – MarkWayne Dec 17 '12 at 16:58