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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.

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  • $\begingroup$ @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. $\endgroup$ – Hans Wurst Sep 17 '19 at 19:07
  • $\begingroup$ 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. $\endgroup$ – Hans Wurst Sep 17 '19 at 19:13
  • $\begingroup$ Hi @Hans Wurst: Which page/section/equation in Cohen-Tannoudji? Link. $\endgroup$ – Qmechanic Sep 19 '19 at 2:10
  • $\begingroup$ @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. $\endgroup$ – Hans Wurst Sep 25 '19 at 8:31
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  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.

  4. For more details of the proof of Fermi's golden rule, see my Phys.SE answer here.

References:

  1. C. Cohen-Tannoudji, B. Diu & F. Laloe, QM, Vol. 2, 1978; Chapter XIII, p. 1283-1302.
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  • $\begingroup$ What you wrote in point 2 is what i wanted to clarify. Equation C-32 is not sensible if you are approaching it purely from a mathematical point of view. $\endgroup$ – Hans Wurst Sep 26 '19 at 6:51

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