# Square of a finite deltafunction

1. Oct 17, 2012

### 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

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

2. Oct 17, 2012

### 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

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

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

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

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 $\delta_T(0) = T$, we obtain

$$I \approx Tf(0).$$

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.