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Im reading Modern Particle Physics from Mark Thomson and I have problems to follow the derivation of Fermi's golden rule (section 2.3.6). In particular equation 2.49:

$$\Gamma_{fi} = 2\pi \int |T_{fi}|^2 \frac{dn}{dE_f}\delta(E_f - E_i) \displaystyle \lim_{T \to \infty} \left\{ \frac{1}{T} \int_{-T/2}^{+T/2} dt \right\} dE_f$$

$$= 2\pi \int |T_{fi}|^2 \frac{dn}{dE_f}\delta(E_f - E_i)dE_f$$

$$= 2\pi |T_{fi}|^2 \left| \frac{dn}{dE_f} \right|_{E_i}.\tag{2.49}$$

Im completely clueless on how to go from the second to the third line. Any idea?

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    $\begingroup$ Isn't this just applying the definition of the Dirac delta? $\endgroup$
    – Anders Sandberg
    Commented Oct 31 at 16:20

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$\delta(x)$ is the Dirac delta function (a.k.a. Dirac delta distribution). It attains the following value: $$\delta(x)=\begin{cases}0&x\neq0\\\infty&x=0\end{cases}$$ Now, this is a strange definition for a function. (Indeed, it is not a function, but a distribution.) But its normalization is set by the following property: $$\int_{-\infty}^\infty f(x)\delta(x)\mathrm{d}x=f(0)$$ You can think of the $\delta(x)$ function as "picking out" the value of $f(0)$ when integrated over $0$. In your case, the $\delta(E_f-E_i)$ function "picks out" the value $E_f=E_i$.

The Dirac delta is ubiquitous in physics; you'll definitely be seeing it around.

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  • $\begingroup$ Thank you very much, now I got it. $\endgroup$
    – Africanus
    Commented Nov 2 at 0:09

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