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Square of a finite deltafunction

  1. Oct 17, 2012 #1
    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]
    [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


    at page 129. And the equation that I'm wondering about it (8.5).
  2. jcsd
  3. Oct 17, 2012 #2

    Jano L.

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    Gold Member

    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

    \int_R \delta_T (\omega) f(\omega) ~d\omega \approx f(0),

    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

    I = \int_R \delta_T(\omega) ^2 f(\omega)~d\omega

    can be calculated approximately in this way:

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

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

    I \approx Tf(0).

    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.
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