Suppose we are given the simple expression

$$ F(k) = \frac{1}{E^2-E(k)^2} $$

which has a pole when $E^2 = E(k)^2$ and where $E, E(k)$ are real numbers. When working with this expression (e.g. inside an integral) we can add some infinitesimal constant $\epsilon >0$ to take care of the poles and let this $\epsilon$ go to zero in the end.

What I want to show now is that the imaginary part of the square of $F$ is proportional to a delta function $\delta\left( E^2 - E(k)^2 \right)$. What I tried was simply to calculate

$$ F^2 = \lim_{\epsilon\to 0}\frac{1}{(E^2+E(k)^2+i \epsilon)^2} = \lim_{\eta\to 0} \frac{1}{(E^2-E(k)^2)^2 + i\eta(E^2-E(k)^2)} $$

where $\eta = 2\epsilon$ and where I have set $\epsilon^2 = 0$. My idea was to use the representation of the delta function

$$ \pi\delta(x) = \lim_{\eta \to 0} \frac{\eta}{\eta^2+x^2} $$

but I failed to bring $F^2$ to this form. Is there any way to do this anyway? For example by introducing an infinitesimal parameter somewhere else in the fraction? Or maybe by doing something completely different?

  • $\begingroup$ I would like to know if there is any particular for which you need to consider the square of $F$. As a matter of fact, $F$ itself can be analytically continued as the (Fantappié) indicatrix of the distribution $\delta\big(E-E(k)\big)$: this seems to rule out the possibility to do the same thing for its square $F^2$. $\endgroup$ – Daniele Tampieri Aug 14 '18 at 17:55

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