Momentum conserving delta-function in the transfer matrix of quantum-field-theoretic scattering theory

Question

The $S$-matrix vanishes unless the initial and final states have the same total $4$-momentum, so it is helpful to factor an overall momentum-conserving $\delta$-function:

$$\mathcal{T}=(2\pi)^{4}\delta^{4}(\sum p)\mathcal{M}.$$

Here, $\delta^{4}(\sum p)$ is shorthand for $\delta^{4}(\sum p^{\mu}_{i} - \sum p^{\mu}_{f})$, where $p^{\mu}_{i}$ are the initial particles' momenta and $p^{\mu}_{f}$ are the final particles' momenta. In this way, we can focus on computing the non-trivial part of the $S$-matrix, $\mathcal{M}$. Thus we have

$$\langle f|S-\mathbb{1}|i\rangle = i(2\pi)^{4}\delta^{4}(\sum p)\langle f|\mathcal{M}|i\rangle.$$

My question:

Since the only way to implement the $4$-momentum conservation is by integrating over the delta-function, does this mean that $\langle f|S-\mathbb{1}|i\rangle$ is integrated over to find the probability? What is the integration variable in this case?

Answer 1

Yes, there is an integral, which comes from the LSZ reduction formula, $$ \langle f|i\rangle\sim \int \mathrm dx\ \mathrm e^{ikx}\square_x G(x) $$ where $x=(x_1,x_2\cdots,x_n)$, $k=(k_1,k_2,\cdots,k_n)$ and $G$ is the $n$-point function. If you go to momentum space you'll get that integrand depends on $x$ only through exponentials, and therefore there is a global delta function.

For more details, see here or chapter 10 in Srednicki's book.

Answer 2

There is no need of integrating over anything to obtain the 4-momentum conservation. Indeed, if you think in terms of perturbation theory (Feynman diagrams), each vertex conserve momentum, so that the diagram itself automatically conserves momentum.