How to properly define the "discretization of a functional"?

Question

In the derivation of the path integral formulation of quantum mechanics, most Physics books end up finding the following (or similar) expression:

$$K(q',t';q,t)=\lim_{N\to \infty}\int\left[\prod_{k=1}^N dq_{k}\right]\left[\prod_{k=0}^N \dfrac{dp_{k}}{2\pi}\right]\\ \qquad \qquad \qquad \times \exp\left[i\sum_{k=1}^{N+1}\left\{\sum_a (q_{k,a}-q_{k-1,a}))p_{k-1,a}-H(q_k,p_{k-1})\epsilon\right\}\right]$$

Here one has sliced the interval $[t,t']$ into subintervals of equal length $\epsilon$.

One now pick $q(t)$ and $p(t)$ functions which satisfy $q(t_k)=q_{k}$ and $p(t_k)=p_{k}$ and notice that the $N\to \infty$ limit of the exponent is

$$S[p,q]=\int_{t}^{t'} \dot{q}(\tau)p(\tau)-H(q(\tau),p(\tau)) d\tau$$

Because of that they say that $K(q',t';q,t)$ is one evaluation by discretization of $$\int e^{iS[p,q]}\mathfrak{D}q\mathfrak{D}p.$$

But I have a problem here. They don't seem to give a definition of what it means to discretize a functional. Instead, they give just one example.

My question is: what is the precise definition of discretization of a functional? In other words, what is the definition of discretization of $$\int \mathfrak{F}[p(\tau),q(\tau)]\mathfrak{D}q(\tau)\mathfrak{D}p(\tau)$$

that when applied to the functional $e^{iS[p,q]}$ gives $K(q',t';q,t)$?

Answer 1

There is no such definition. Derivations you refer to have little to do with rigorous mathematics. If sense can actually be made of the path integral (which happens to be true for some path integrals), questions of proper discretizations of various actions and other functionals have to be studied case by case. In many examples this is a research-level, often partially open problem.

It is also worth noting that this discretization process is typically highly ambiguous. This is also related to the do called operator ordering problem. Cf. Weyl ordering and normal ordering, for instance.