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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?

Roger V.
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Edward Hughes
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4 Answers4

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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.

References:

  1. J.J. Sakurai, Modern Quantum Mechanics, 1994, Section 5.6.
Qmechanic
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As proposed by Lubos, the delta function you started with $\delta(E_i-E_f)$ forces the final result to be invariant by $E_i \leftrightarrow E_f$.

Learning is a mess
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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
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    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
  • Otherwise, Edward, it's hard to understand why you think that $\rho(E_i)$ may be different from $\rho(E_f)$ if $E_i=E_f$. If the argument of a function, $\rho$ in this case, is the same, the value of the function is also the same, isn't it? Note that we're not implying $i=f$: this much isn't the case because the state $|i\rangle$ isn't a function of $E_i$. But things explicitly written to be functions are functions. – Luboš Motl Dec 17 '12 at 16:21
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    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
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    @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! – Edward Hughes Dec 17 '12 at 17:04
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    @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? – nervxxx Apr 20 '17 at 12:57
  • Hi @nervxxx - Sure, you can write the density of states as an operator $\rho(E) = \pi \delta(E-H_0)$ (up to normalization dependent factors), diagonal in the basis of asymptotic states. The discussion here allows for the possibility that there are different particles in the initial and final states. – MarkWayne Apr 24 '17 at 19:15
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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 $$\Gamma_{fi} = \sum_{f}\frac{2\pi}{\hbar}\delta (E_f - E_i)|M(E_f, E_i)|^2= \left[\sum_{f}\delta (E_f - E_i)\right]\frac{2\pi}{\hbar}|M(E_i, E_i)|^2= \rho(E_i)\frac{2\pi}{\hbar}|M(E_i, E_i)|^2 $$

Roger V.
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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.

Jiang-min Zhang
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