Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
3  
It's the same thing because $E_i=E_f$ in this treatment, isn't it? –  Luboš Motl Dec 17 '12 at 14:32
add comment

2 Answers

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

share|improve this answer
    
I'm afraid I don't quite see this - could you expand on your argument perhaps? –  Edward Hughes Dec 17 '12 at 15:09
2  
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
4  
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
show 2 more comments

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.

References:

  1. J.J. Sakurai, Modern Quantum Mechanics, 1994, Section 5.6.
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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