Dirac-delta function in Derivation of Fermi's Golden Rule

I was following along Mark Thomson's Modern Particle Physics, and got stuck on this book's derivation of Fermi's Golden Rule (On page 53):

"... If there are d$$n$$ accessible final states in the energy range $$E_f \rightarrow E_f +dE_f$$, then the total transition rate $$\Gamma_{fi}$$ is given by $$\begin{equation} \Gamma_{fi} = 2\pi \int|T_{fi}|^2 \frac{dn}{dE_f} \lim_{T\rightarrow \infty} \left\{ \frac{1}{T} \int_{-\frac{T}{2}}^{-\frac{T}{2}} e^{i(E_f - E_i)t} \delta(E_f-E_i) dt \right\}dE_f \tag{A}. \end{equation}$$ The delta-function in the integral implies that $$E_f=E_i$$ and therefore $$(\text{A})$$ can be written $$\begin{equation} \Gamma_{fi} = 2\pi \int|T_{fi}|^2 \frac{dn}{dE_f} \delta(E_f-E_i) \lim_{T\rightarrow \infty} \left\{ \frac{1}{T} \int_{-\frac{T}{2}}^{-\frac{T}{2}} dt \right\}dE_f \tag{B}. \end{equation}$$ ... (and so on) "

Based on the explanation between the steps, I don't understand why $$E_f$$ and $$E_i$$ should be the same. I do know that in order for $$E_f$$ to be the same as $$E_i$$, the Dirac-delta functions supposed to be integrated with the exponent, but this is not the case. Any explanation for this would be appreciated.

• Are you asking why $E_f$ is set equal to $E_i$ in the $t$-integral, despite it not being integrated over? Dec 1 '20 at 9:53
• @NiharKarve yes, that's what am i wondering Dec 1 '20 at 10:14
• FWIW, the Fermi Golden rule is derived in my Phys.SE answer here. Dec 1 '20 at 10:29

This is just an application of the rule \begin{align} \int dx\;f(x)g(x) \delta(x-y) = f(y)g(y) &= f(y)\int dx\; g(x)\delta(x-y)\\ &= \int dx\; f(y)g(x) \delta(x-y) \end{align} where we have used the definition of the delta function. Intuitively the Dirac delta is zero except where its argument vanishes, so the value $$f$$ only matters at that point, so we can make the substitution and get the same answer.

Here $$f$$ is the seemingly complicated function $$f(E_f) = \lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2} dt \;e^{\imath (E_f-E_i) t}$$ and $$g(E_f) = |T_{if}|^2 \frac{d n}{dE_f}$$

• Just wanna make sure, do you mean $f(E_f)$, $g(E_f)$, and $E_f-E_i$ on the exponent part in $f(E)$? Dec 1 '20 at 12:04
• @Andrijauhari yes. I will edit to be clearer Dec 1 '20 at 12:13

I think this can be quickly resolved with a reference to the wikipedia page on the dirac delta:

Approach 1) as T goes to infinity, the integral becomes delta of the energy difference, so there may be a square of dirac deltas which is the same as a dirac delta. However, I'm not sure why the division by T can be absorbed in the dirac delta... As commented below, this is physically unreasonable.

Approach 2): The answer is easier than possibility 1 - the definition of the Dirac delta means that the argument, here $$E_i-E_f$$, is zero. This is used to set the argument of the exponential to zero.

Approach 2 seems to be what was used here. Reviewing the wikipedia page on the Dirac delta might help you get a handle on why it is this way.

• The square of deltas is ill-defined. But it could be that the first delta should not be there and then comes in as a consequence of the math as you explained. Dec 1 '20 at 10:28
• While I agree that the square of deltas is ill-defined, I think that most physicists would just square the definition involving infinity if the argument is 0 and 0 otherwise by squaring the components in this piecewise-defined definition. In that case, there is a good chance that the author (as a member of "most physicists") would use that the square of a dirac delta is the same dirac delta. As such, I made such an assumption in my answer. Dec 1 '20 at 10:36
• I know that Physicists always like to be mathematically sloppy (after all, I am one myself), but this is not one such case. A squared dirac delta almost certainly means that there has been an error in the computation as it would imply a physically infinite result. Written in physics style: $\int dx \delta(x) \delta(x) = \delta(0) = \infty$. Dec 1 '20 at 10:45
• Thanks, I never thought of the square like that. Dec 1 '20 at 10:54
• @PrawwarP Thank you so much, Approach 2 is what I'm trying to convince to myself. I'd love to vote your answer, but I'm still new here, so I can't. Dec 1 '20 at 12:02