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

The usual definition of the path integral in QM 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 $$\mathfrak{F}_E[x(t)]=e^{-iS_E[x(t)])}$$ 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.