In the Physicists' definition of the path integral, does the result depend on the choice of partitions?

Question

The standard definition of the path integral in Quantum Mechanics usually goes as follows:

1. Let $[a,b]$ be one interval. Let $(P_n)$ be the sequence of partitions of $[a,b]$ given by $$P_n=\{t_0,\dots,t_n\}$$ with $t_k = t_0 + k\epsilon$ where $\epsilon = (b-a)/n$, and $t_0 = a$, $t_n=b$.

2. Let $\mathfrak{F}: C_0([a,b];\mathbb{R}^d)\to \mathbb{C}$ be a functional defined on the space of continuous paths on $[a,b]$. One defines its discretization as the set of functions $\mathfrak{F}_n : \mathbb{R}^{(n+1)d}\to \mathbb{C}$ given by $$\mathfrak{F}_n(x_0,\dots,x_n)=\mathfrak{F}[\xi_n(t)]$$ where $\xi_n(t)$ is the curve defined by taking the partition $P_n$, defining $\xi(t_i)=x_i$ and linearly interpolating between the points - in other words $\xi(t)$ is for $t\in [t_i,t_{i+1}]$ the straight line joining $x_i$ and $x_{i+1}$.

3. One defines the functional integral as the limit $$\int_{C_0([a,b];\mathbb{R}^d)}\mathfrak{F}[x(t)]\mathcal{D}x(t)=\lim_{n\to \infty}\int_{\mathbb{R}^{(n+1)d}} \mathfrak{F}_n(x_0,\dots, x_n) d^dx_0\dots d^dx_n$$

   if it exists.

4. In the case of interest for physics one has $\mathfrak{F}[x(t)]=e^{iS[x(t)]}$ or rather $\mathfrak{F}_E[x(t)]=e^{-S_E[x(t)]}$ the euclidean version.

So by slicing the time axis into equal subintervals, one converts the functional to a sequence of functions, integrates those and takes the limit.

This is the construction outlined for instance in Peskin's book or Sakurai's book, just rewritten in a more "mathematical" form.

Now, if on the very first step we choose another sequence of partitions $(P_n)$ such that the sequence of partition's norms $|P_n|\to 0$ as $n\to \infty$ but which is not the sequence of equal subintervals, would the resulting path integral be different?

I don't see reason why it should be equal. The intervals endpoints are distinct, the interpolations are distinct, hence the maps $\mathfrak{F}_n$ are distinct.

If it is I think this is a big problem. After all, the way we are slicing the time axis is arbitrary and just one trick to make the problem easier to deal with.

Answer 1

The question as such needs a re-formulation, because you are using the wrong notion of integrability.

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With respect to refinement partitions have the structure of a directed set (meaning a partition can be finer than another, and any two partitions admit a common refinement). Thus your intermediate functional $I_P(\mathfrak F)$ is actually a net. The most natural definition of the integrability of $\mathfrak F$ wrt this correspondence is then that $$\mathfrak F \text{ is integrable}:\iff \lim_{P\to\infty}I_P(\mathfrak F)\text{ exists}.$$ This definition works without chosing any particular sequence of partitions $P_n$ that become arbitrarily fine. A consequence of $\mathfrak F$ being integrable is that any sequence $I_{P_n}(\mathfrak F)$ converges (if $P_n$ becomes arbitrarily fine) and the limit does not depend on the sequence. On the other hand if $\lim_n I_{P_n}(\mathfrak F)$ always exists for such sequences $P_n$ and always converges to the same value, the above condition of integrability is satisfied.

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So upon re-formulation the answer is: "Yes, by definition." What is needed now is a discussion of why this reformulation should be necessary, and a further re-formulation of the question so a non-trivial answer becomes possible.

As noted in the comments your situation is practically identical to that of the Riemann integral. For the Riemann integral crappy functions like $d(x)=\begin{cases} 0& x\in\mathbb Q\\ 1& x\notin\mathbb Q\end{cases}$ make it clear that in general that the limit of $I_{P_n}(f)$ will depend on the sequence chosen (let one sequence be supported by rational points, compare with another supported by irrational points). Just like this you can find crappy functionals where it is obvious that the limit depends on the partitions, an example is $$\mathfrak F(\gamma) = \begin{cases}0 &\gamma \text{ is not differentiable at some rational value}\\ 1&\gamma\text{ is differentiable at all rational values and takes on values only in }[0,1]\\0 &\text{else}\end{cases}$$ Here you've got $I_{P_n}(\mathfrak F)=1$ if $P_n$ is supported by irrational points and $I_{P_n}(\mathfrak F)=0$ else.

These examples should make clear why the original definition is an inadequate definition of "integrable".

So as it stands the answer to the question is trivial. How can the question be re-formulated to achieve a non-trivial answer? Compare again to how one can achieve interesting answers in the Riemann integral case.

1. What is a useful class of functions (functionals here) which are integrable?
2. What is a useful class of functions (functionals) in which "$f$ integrable" $\iff$ "$I_{P_n}(f)$ converges for one sequence of partitions"?

For the Riemann integral the usual answer to 1. is functions with finitely many discontinuities. A normal answer to 2. is a functions with a number amount of discontinuities. But far more than these two classes being "usual" answers, these are actually the functions you are interested in. Most people don't actually care about integrating the function $d$ from above, even if you have more sophisticated integrals that can handle it. You don't encounter such a function in physics. As such in the "usual" frame the definition of the Riemann integral that depends on a sequence $P_n$ of partitions is adequate.

However, I don't know any nice answers to the two questions, nor do I have a "usual" class of functionals in mind, on which 2. could be investigated.