Square of a finite deltafunction

[center o bass]

Hi. I'm reading "Quantum Field Theory - Mandl and Shaw" about how to derive the cross-section and in the derivation the authors make the following argument

"For large values of T and V, we can then take

\delta_{TV}(\sum p_f' - \sum p_i) = (2\pi)^4 \delta^{(4)}(\sum p'_f - \sum p_i)
and
(**) (\delta_{TV}(\sum p_f' - \sum p_i))^2 = TV(2\pi)^4 \delta^{(4)}(\sum p'_f - \sum p_i)"

where they earlier have defined

(2\pi)^4 \delta^{(4)}(\sum p'_f - \sum p_i) = \lim_{T,V \to \infty} \delta_{TV}(\sum p_f' - \sum p_i) = \lim_{T,V \to \infty} \int_{-T/2}^{T/2}dt\int_V d^3x \exp ( i x( \sum p_f' - \sum p_i)).

My question is how they get in a TV factor when they take the square of that finite delta function in equation (**)? The argument is found in full detail in

http://books.google.no/books?id=Ef4...no&source=gbs_toc_r&cad=4#v=onepage&q&f=false

at page 129. And the equation that I'm wondering about it (8.5).

 

[Jano L.]

Their derivation can be understood in this way:

Let \delta_T (\omega) = \frac{1}{2\pi} \int_{-T/2}^{T/2} e^{-i\omega t} ~dt. It can be shown that approximately

<br /> \int_R \delta_T (\omega) f(\omega) ~d\omega \approx f(0),<br />

so this function behaves similarly as the δ distribution.

Now, what is ( \delta_T(\omega) ) ^2? This can be attributed meaning of a distribution. The integral

<br /> I = \int_R \delta_T(\omega) ^2 f(\omega)~d\omega<br />

can be calculated approximately in this way:

<br /> I = \int_R \delta_T(\omega) [ \delta_T(\omega) f(\omega) ]~d\omega \approx \delta_T(0) f(0)<br />

Since \delta_T(0) = T, we obtain

<br /> I \approx Tf(0).<br />

From this we infer that the expression \delta_T ^2 behaves, under the integral, almost as the function T\delta_T(\omega).

But I would like to say that these are just approximate formulae and one should be careful about using them.