Why multiply by volume when integrating over final momenta in scattering amplitude calculations? I am learning about calculating decay rates from quantum field theory amplitudes from David Tong's lecture notes (page 74 in the notes, 24 in document). However, I have some doubts:


*

*When he says the ﬁrst is to integrate over all possible momenta of the ﬁnal particles, he uses the measure $V \int \frac{d^{3} p_{i}}{(2 \pi)^{3}}$. I'm not sure why that factor of V is there. I have come up with some explanations, the best of which is that the delta function $\delta^{(4)} (p_{F}-p_{I})$, due to considering only a finite space does not actually multiply by infinity, but by V. However, I need several factors of V, which I don't see how to get.

*The next question is about the meaning of the probability he calculates. He gets a Probability which is something finite times a $t$, so that it grows infinitely in time. I don't get how does he then get from this the half life, as I don't even know how to make sense of a diverging probability like this (normalizing it in the normal way I know, would just give all the probability way at infinity).
So clearly, I must be missing some important things here.
 A: Typically in this kind of scattering calculation one works within a fictional finite volume as a regularisation technique, to ensure that your momentum states are normalisable. This is what Tong is doing here. In a finite box of volume $L^3$ with periodic boundary conditions, each component of the momentum of a plane-wave eigenstate can only take discrete values $p_{i} = \frac{2\pi n_i}{L}$, where $n_i$ is an integer. Therefore each state occupies a box of volume $(\frac{2\pi}{L})^3$ in momentum space.
In order to get the observable decay rate, one really needs to perform a discrete sum over final momentum states. However, this sum is much easier to perform by approximating it by an integral. (This approximation is good when $L$ is very large, so the momentum states are closely packed in $p$-space.) The momentum integral calculates the total volume in $p$-space occupied by the final states. What you really want is the number of final states within this total $p$-space volume. Therefore you must divide the integral by the volume occupied by each state:
$$ \sum\limits_{\mathbf{p}} \approx \frac{L^3}{(2\pi)^3}\int\mathrm{d}^3\mathbf{p}.$$
The factor $\frac{L^3}{(2\pi)^3}$ is also referred to as the density of states in $p$-space, for obvious reasons. Another way of seeing why this factor must be there is for dimensional consistency: the factor $L^3$ cancels the $[L]^{-3} $ dimension coming from the momentum integration.
About the second question, see the comment by Michael Brown below the OP's post. A first-order perturbation calculation based on the Golden Rule will only give the correct transition probability for short times. You cannot extrapolate transition probabilities thus calculated up to very long times, since higher order processes start to become important.
