# $S$-matrix elements in path integral approach

How to calculate $$S$$-matrix elements of quantum electrodynamics using path integral formalism?

Long story short, using the path integral we can evaluate vacuum expectation values of time ordered products of operators. Whenever the said operators are field operators, what we find are correlation functions and is with this correlation functions that we can evaluate S-matrix elements. We do it by means of the LSZ reduction formula.

As you can see from the wikipedia article, the formula is a nasty beast even for the simplest case of a $$\lambda\phi^4$$-theory. The formula for QED is even more cumbersome since there are three possible fields instead of only one. For example, only for the fermion part, the LSZ formula reads: given an initial $$N$$-fermion (non interacting) state $$|\alpha, in\rangle$$ and a final $$N^\prime$$-fermion (non interacting) state $$\langle \beta,out|$$, where $$\alpha,\beta$$ are two collective indices for all the particles in the states, the transition probability from the initial state to the final state which is in fact the S-matrix element, is given by $$\langle\beta,out|\alpha,in\rangle = \int \prod_{i=1}^{N} d^{4} x_{j} \frac{i \mathrm{e}^{i p_{j} x_{j}}}{\sqrt{Z}}\left[\left(i\not \partial_{x_{j}}+m\right) u_{\mathrm{p}_{j}}^{s_{j}}\right]_{\alpha_{j}} \prod_{l=1}^{N^{\prime}} \mathrm{d}^{4} y_{l} \frac{i \mathrm{e}^{-i k_{i} y_{l}}}{\sqrt{Z}}\left[\bar{u}_{\mathrm{k}_{l}}^{\sigma_{l}}\left(-i \not \partial_{y_{l}}+m\right)\right]_{\beta_{l}}\color{red}{\left\langle 0\left|\mathrm{T}\left[\Psi_{\beta_{1}}\left(y_{1}\right) \ldots \bar{\Psi}_{\beta_{n^{\prime}}}\left(y_{n^{\prime}}\right) \bar{\Psi}_{\alpha_{1}}\left(x_{1}\right) \ldots \bar{\Psi}_{\alpha_{n}}\left(x_{n}\right)\right]\right| 0\right\rangle}$$

The red bit is the one which we can evaluate using the path integral.

This formula is non-perturbative, which means that is one could evaluate it, we would get an exact result. But as you can see it would be pretty difficult to evaluate an exact result, and I don't know if there are any cases in which is possible. So at the end we come back to our beloved perturbation theory.