Square of a finite deltafunction

  • #1
center o bass
560
2
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

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

where they earlier have defined

[tex] (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)).[/tex]

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).
 
Physics news on Phys.org
  • #2
Their derivation can be understood in this way:

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

[tex]
\int_R \delta_T (\omega) f(\omega) ~d\omega \approx f(0),
[/tex]

so this function behaves similarly as the δ distribution.

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

[tex]
I = \int_R \delta_T(\omega) ^2 f(\omega)~d\omega
[/tex]

can be calculated approximately in this way:

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

Since [itex]\delta_T(0) = T[/itex], we obtain

[tex]
I \approx Tf(0).
[/tex]

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

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