# 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 $E_i \leftrightarrow E_f$.

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I'm afraid I don't quite see this - could you expand on your argument perhaps? –  Edward Hughes Dec 17 '12 at 15:09
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

I) Well, OP evidently knows that it is the density $\rho_f(E_f)$ of final (rather than initial) states that appear in Fermi's golden rule

$$\tag{1} \Gamma_{fi} ~=~ \rho_f(E_f) \frac{2\pi}{\hbar} |W_{fi}|^2.$$

Here we adorn the density $\rho_f$ with a subscript $f$, to make that point clear, following a suggestion by MarkWayne. Instead it seems that OP's actual question is:

Must the energy $E_f$ [which here denotes a pertinent average of final states that we summed over in a sufficiently small energy interval, and which appears inside $\rho_f(E_f)$ in eq. (1)] approximately match the energy $E_i$ of the initial state $i$, or not?

II) A crucial role is played by the time-dependence of the interaction potential $V(t)$ in the Hamiltonian

$$\tag{2} H~=~H_0+V(t).$$

For instance, in the harmonic perturbation [1], the interaction potential reads

$$\tag{3} V(t)~=~\sum_{\pm}W^{\pm} e^{\pm\mathrm{i}\Omega t},$$

where $\Omega$ is the angular frequency of absorption/stimulated emission. (We need at least two terms in the potential (3) to make the interaction operator $V(t)$ Hermitian.) One may show that this favors transitions of the form

$$\tag{4} E_f~\approx~E_i\pm\hbar\Omega.$$

So in the harmonic perturbation, $\rho_f(E_f)$ and $\rho_f(E_i)$ are in general different.

III) However, in the derivations of Fermi's golden rule in many elementary textbooks (which always use time-dependent perturbation theory), the interaction term $V(t)$ is often treated as time-independent (corresponding to $\Omega=0$). This means that the initial and final state in such time-independent treatments must have approximately the same energy, cf. also a comment by Lubos Motl.