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 Answers 4


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

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

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

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

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

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

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

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

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 my Phys.SE answer here.


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

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? $\endgroup$ Commented Dec 17, 2012 at 15:09
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    $\begingroup$ 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. $\endgroup$ Commented Dec 17, 2012 at 15:15
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    $\begingroup$ 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
    Commented Dec 17, 2012 at 16:58
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    $\begingroup$ @MarkWayne - that makes a huge amount of sense. I was thinking that $\rho(E_i)$ was the density of initial states. In fact it's not. It's the density of final states at energy $E_i$! So $\rho(E_i)=\rho(E_f)$ makes perfect sense, and there's no difference between the two formulae I've written in the question. Is that a fair summary? Thanks so much for your help, everyone! $\endgroup$ Commented Dec 17, 2012 at 17:04
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    $\begingroup$ @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
    Commented Apr 20, 2017 at 12:57

The key is understanding the transition from summation to integration over energy. We start with $$\Gamma_{fi} ~=~ \sum_{f}\frac{2\pi}{\hbar}\delta (E_f - E_i)|M_{fi}|^2,$$ where summation over $f$ mean summation over all quantum numbers of the final states. We now assume that the matrix element is dependent only on the energies, i.e., it can be written as $$M_{fi}=M(E_f,E_i),$$ and we replace summation by the integration over the energy of the final states: $$\Gamma_{fi} ~=~ \int \rho(E_f) \frac{2\pi}{\hbar}\delta (E_f - E_i) |M(E_f, E_i)|^2\textrm{d}E_f.$$

The second important element is using the basic property of the delta function, already mentioned in the comments and other answers: $$ \int f(x)\delta(x-x_0)dx=\int f(x_0)\delta(x-x_0)dx=f(x_0)$$ (provided that the integration interval contains $x_0$.) Thus $$\Gamma_{fi} = \rho(E_f) \frac{2\pi}{\hbar} |M(E_f, E_f)|^2=\rho(E_i) \frac{2\pi}{\hbar} |M(E_i, E_i)|^2,$$ where it is understood that the initial and the final energies are the same. This is however less transparent than writing $M_{fi}$, which explicitly points out at the nature of the matrix element.

Remark: Personally, I find this textbook method of introducing the density-of-states as overcomplicating the issue. A quite general definition is $$ \rho(E)=\sum_f\delta(E-E_f),$$ in which case the final equation is readily obtained as $$\Gamma_{fi} = \sum_{f}\frac{2\pi}{\hbar}\delta (E_f - E_i)|M(E_f, E_i)|^2= \left[\sum_{f}\delta (E_f - E_i)\right]\frac{2\pi}{\hbar}|M(E_i, E_i)|^2= \rho(E_i)\frac{2\pi}{\hbar}|M(E_i, E_i)|^2 $$


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.


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