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?

  • 4
    $\begingroup$ It's the same thing because $E_i=E_f$ in this treatment, isn't it? $\endgroup$ Dec 17, 2012 at 14:32

3 Answers 3


As proposed by Lubos, the delta function you started with $\delta(E_i-E_f)$ forces the final result to be invariant by $E_i \leftrightarrow E_f$.

  • $\begingroup$ I'm afraid I don't quite see this - could you expand on your argument perhaps? $\endgroup$ Dec 17, 2012 at 15:09
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 17, 2012 at 15:15
  • $\begingroup$ 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? $\endgroup$ Dec 17, 2012 at 15:18
  • 6
    $\begingroup$ 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. $\endgroup$
    – MarkWayne
    Dec 17, 2012 at 16:58
  • 1
    $\begingroup$ @MarkWayne I'm even more confused by what you think was confusing Edward. Why do you say there can be subscripts $i$ and $f$ for the density of states $\rho$? In deriving FGR we treat the unperturbed Hamiltonian $H$ as the dominant piece, which has an associated density of states $\rho(E)$. There is thus only ONE density of states ever, and it is a function of $E$ as measured by $H$. So to me it doesn't make sense to put a $i$ or $f$ as subscripts in $\rho$ to say that there are two functions $\rho_i(E)$ and $\rho_f(E)$. There aren't two 'unperturbed' Hamiltonians! Or am I missing something? $\endgroup$
    – nervxxx
    Apr 20, 2017 at 12:57

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.

For more information, see e.g. also this Phys.SE answer.


  1. J.J. Sakurai, Modern Quantum Mechanics, 1994, Section 5.6.

While the answers above already answered your question, I would like to recommend a paper by myself: Nonsmooth and level-resolved dynamics illustrated with a periodically driven tight binding model.

In this paper, we derived Fermi golden rule as a by-product. Our derivation does not use the delta function.

I believe our derivation is much simpler and more transparent than those in textbooks. It is just a mathematical property of the sinc function.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.