Approximation for the square of a Dirac delta function

Question

I am working in the appendix to Section II.6 in Zee's QFT book, 2nd Ed. I am trying to compute the cross section for a meson to decay to two mesons as $\varphi\to\eta+\xi$ with three respective momenta $k,p,q$. To simplify the problem, I have imposed the big box condition which quantizes the $k,p,q$. I have computed the probability amplitude

$$ \mathcal{A}\propto \big(2\pi)^4\delta^{(4)}(p+q-k) ~~, $$

where "$\propto $" means "is proportional to." Now I need to square $\mathcal{A}$ to get the probability. I substitute the integral form of the $\delta$-function so

$$ |\mathcal{A}|^2\propto \left[\big(2\pi)^4\delta^{(4)}(p+q-k)\right ]^2=\big(2\pi)^4\delta^{(4)}(p+q-k)\int\!d^4x\,e^{ix(p+q-k )}~~. $$

Zee simplifies this to

$$ |\mathcal{A}|^2\propto\big(2\pi)^4\delta^{(4)}(p+q-k)\int\!d^4x =\big(2\pi)^4\delta^{(4)}(p+q-k)VT~~,$$

where $VT$ is the 4-volume of spacetime in the box we have chosen. Zee cites as motivation for the approximation $e^{ix(p+q-k )}=1$ the fact we are using the big box normalization instead of considering all of spacetime. I do not see why the big box normalization which imposes quantization on the momenta $k,p,q$ allows us to approximate the integrand as being equal to one. How does he do this? We have used the box to quantize momenta so I do not see what it does for the present integral over $d^4x$.

Answer 1

The box normalization isn't motivation for setting $e^{ix(p+q-k)}=1$; Zee is just comforting those readers who (correctly) balk at the idea of squaring a delta function by reminding them that all of these calculations are being performed in a finite box, which means e.g. that momenta are quantized, not continuous, and that the Dirac deltas would be replaced by Kronecker deltas if we were sufficiently motivated to go back and perform this calculation while explicitly keeping $V$ finite.

The reason the exponential is set to one is simply because it appears next to $\delta^{(4)}(p+q-k)$; if $p+q-k\neq 0$, then that delta function "evaluates" to zero, so you may as well make the replacement $e^{ix(p+q-k)}\rightarrow e^{ix(0)}=1$.