# Limit of the $\sin^2$ function in the derivation of Fermi's golden rule

In the derivation of Fermi's golden rule one typically arrives at an expression of the form $$\frac{\sin^2(\omega t)}{\omega^2}$$ which is then converted to $$\pi t\delta(\omega).$$ I cannot follow this step. I know the following identity $$\delta (\omega) = \lim_{t\rightarrow \infty}\frac{\sin^2(\omega t)}{\pi |t|\omega^2}$$ from which i would assume that one extends the first expression by $$\frac{t}{t}$$ and then does the limit. But how can you pull the $$t$$ out of the limit ? Is this rigorous or is this an approximation ?

I think it should be like this $$\lim_{t\rightarrow\infty} \frac{t}{t}\frac{\sin^2(\omega t)}{\omega^2} \not = \pi t\delta(\omega)$$ and the equation should be $$\lim_{t\rightarrow\infty} \frac{t}{t}\frac{\sin^2(\omega t)}{\omega^2} = \pi \delta(\omega) \lim_{t\rightarrow \infty}t.$$ Am i wrong in the above equations ? Otherwise i don't see how Fermi's goldene rule could ever work since we assume at one time that $$t$$ is so large that we can approximate a function in the limit that $$t$$ goes to infinity while on the other hand $$t$$ has to be small such that pertubation theory of first order is accurate. These conditions seem to contradict each other but in every book i find this step. I haven't found any satisfactory answer so far regarding this step. I know the general conditins for pertubation theory but i find the form with the dirac delta function nonsensical. I assume that i go wrong at some point since no one ever brings this point up, please point out my error if i did something wrong.

• @Aaron Stevens the The book by Cohen-Tannoudji, Quantum Mechanics, vol 2 contains this step. He gives a definition of the dirac delta function as a limit in the apendix and then arrives at the expression which stumps me. The expression should still contain the limit as i wrote it in my opinion. Commented Sep 17, 2019 at 19:07
• My problem is mostly of mathematical nature, what happens when you have something like $\lim_{t \rightarrow \infty } (f(t)g(x,t) )$, where $g(x,t)$ turns into $\delta (x)$ in this limit. I expect it to be $\delta(x) \lim_{t \rightarrow \infty} f(t)$ and not be $f(t) \delta (x)$. The validity of Fermi's golden rule is not my main concern. Commented Sep 17, 2019 at 19:13
• Hi @Hans Wurst: Which page/section/equation in Cohen-Tannoudji? Link. Commented Sep 19, 2019 at 2:10
• @Qmechanic I checked your link and my problem is also contained in the english version. In cohen-tannoudjii_v2.djvu, page 1300 equation C-32. I do not understand how he can pull out a factor of $t$ from the expression $\lim_{t\infty }F(t, \frac{E-E_i}{\hbar})$. In the limit this factor should be evaluated at the limit. I am not sure if he is right or wrong, i would have expected the use of the approximate symbols and not an equality sign. This is why i ask, is it really an equality or just an approximation what is given in equation C-32. Commented Sep 25, 2019 at 8:31
• Commented Sep 25, 2019 at 22:18

1. In the comments it became clear that OP is trying to understand the formula $$\lim_{t\to\infty} F\left(t,\frac{E-E_i}{\hbar}\right)~=~\pi t~\delta\left(\frac{E-E_i}{2\hbar}\right)~=~2\pi\hbar t~\delta(E-E_i) , \tag{C-32}$$ taken from Ref. 1, where $$F(t,\omega)~=~\left[\frac{\sin(\omega t/2)}{\omega/2}\right]^2. \tag{C-7}$$
2. It is clear that eq. (C-32) does not make sense as an ordinary limit, since the rhs. should then be independent of $$t$$. (It makes sense as an asymptotic series, though.)
3. However, if the goal is just to derive Fermi's golden rule, there is a short-cut: It turns out that the $$F$$-function (C-7) is proportional to the probability $$P(t)$$, which grows with time $$t$$. The trick is to instead consider the quotient $$\frac{P(t)}{t}$$. Then it becomes mathematically well-defined to consider the limit $$\lim_{t\to\infty}\frac{P(t)}{t}$$, i.e. $$\lim_{t\to\infty} \frac{F\left(t,\frac{E-E_i}{\hbar}\right)}{t}~=~\pi ~\delta\left(\frac{E-E_i}{2\hbar}\right)~=~2\pi\hbar ~\delta(E-E_i) , \tag{C-32'}$$ cf. this Phys.SE post. This in turn implies that for large enough times $$t$$ the probability $$P(t)$$ grows proportionally with $$t$$, and that the transition rate $$\frac{dP(t)}{dt}~=~\frac{P(t)}{t} \tag{*}$$ is given by the proportionality factor.