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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.

Qmechanic
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Andri jauhari
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2 Answers2

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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 $$ g(E_f) = |T_{if}|^2 \frac{d n}{dE_f} $$

By Symmetry
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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.

PrawwarP
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