Square of the Dirac Delta Shenanigans

Question

The following is a purely mathematical problem, but it often arises when dealing with transition probabilities in Quantum Field Theory.

In QED the transition probability amplitudes often contain a Dirac delta of the form:

$$\delta (E_f-E_i)$$

but since we have to deal with Born's rule we are interested in the square of this quantity, and so we must ask ourself what it means to take the square of a Dirac delta. Usually we deal with the problem (to be precise in the context of box quantisation) in the following way:

We write the Dirac delta in Fourier form¹:

$$\delta (E_f - E_i)=\lim _{T \to +\infty}\frac{1}{2\pi}\int _{-T/2}^{+T/2}e^{i(E_f-E_i)t}dt$$

obtaining:

$$2\pi\delta (E_f - E_i)=\lim _{T \to +\infty}\int _{-T/2}^{+T/2}e^{i(E_f-E_i)t}dt=\lim _{T \to +\infty}\left[\frac{e^{i(E_f-E_i)t}}{i(E_f-E_i)}\right] _{-T/2}^{+T/2}=\lim _{T \to +\infty}\frac{2 \sin(\Delta E T/2)}{\Delta E} \tag{1}$$

From $(1)$ we operate the following magic trick (i.e. abuse): since we want to find the square of the Dirac delta we can think of writing

$$(2\pi)^2\delta ^2(\Delta E)=\lim _{T \to +\infty}\frac{4 \sin^2 (\Delta E T/2)}{(\Delta E)^2} \tag{2}$$

As we were saying the operation carried out between $(1)$ and $(2)$ is not formally correct, but even accepting $(2)$ as granted the following last step remains for me a mystery: for some reason unknown to me we can now state that

$$\lim _{T \to +\infty}\frac{4 \sin^2 (\Delta E T/2)}{(\Delta E)^2}=(2\pi)T\delta (\Delta E)\tag{3}$$

and so finally

$$\delta ^2(\Delta E)=\frac{T}{2\pi}\delta (\Delta E) \tag{4}$$

My question of course is: how do we derive (3)? Why is it valid?

In answering please keep in mind that this question is asked by somebody with limited to non existing knowledge on distribution theory, and I also suspect that an answer to this question may be particularly useful to people with such a limited mathematical background; so please don't take to much as granted.

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[1] If you want more context on this you can check out this previous question on Math S.E.

Answer 1

Pick any absolutely integrable function $\eta(x)$ such that $\int_{-\infty}^\infty dx \ \eta(x)=1 $. We use $\eta$ to 'construct' a Dirac delta (a so called nascent delta or approximation to the identity )

$$\tag{1} \frac{1}{\epsilon}\eta\left(\frac{x}{\epsilon}\right) \to \delta (x) \qquad ,\qquad \epsilon \to 0 $$

Where in the first expression $\to$ means in the sense of distributions. Concretely this means

$$\tag{2} \lim_{\epsilon\to0}\ \epsilon^{-1}\int\limits_{-\infty}^\infty dx \ \eta(x/\epsilon) f(x)=f(0) $$

For any suitable test function $f(x)$. Note that (2) is exactly the action of the delta: $\int_{-\infty}^\infty dx \ \delta(x)f(x)=f(0)$. To prove (1) you can look at theorem 9.8 in this answer.

Now we apply (1) to your question. Let $T=1/\epsilon$ then (1) reads

$$\tag{3} T \eta(xT)\to\delta(x) \qquad , \qquad T\to\infty $$

In your case we pick $\eta(x)=\frac{2}{\pi}\frac{\sin^2(x/2)}{x^2}$ so that

$$\tag{4} T\eta(xT)=\frac{2}{\pi}\frac{\sin^2(xT/2)}{x^2T}\to\delta(x) \qquad,\qquad T\to\infty $$

So we have the equivalency

$$\tag{5} \frac{4\sin^2(xT/2)}{x^2}\to2\pi T\delta(x) \qquad ,\qquad T\to\infty $$

Just like (2), the expression in (5) means concretely that

$$\tag{6} \lim_{T\to\infty} T^{-1} \int\limits_{-\infty}^\infty dx \ \left[\frac{4\sin^2(xT/2)}{x^2} \right]f(x) = 2\pi f(0) $$

Note that I moved the $T$ to LHS so that the limit expression makes sense. The freedom to choose any suitable $\eta$ is very useful for 'picking out' Dirac deltas in various complicated expressions.