Consider the evaluation by discretization of the path integral $$\int e^{iS[x(t)]}\mathfrak{D}x(t),\quad S[x(t)]=\int_{t}^{t'}\left[\frac{m}{2}\dot{x}(\tau)^2-V(x(\tau))\right]d\tau.$$

One discretizes time with a partition $t_0,\dots, t_N$ with $t_k = t_0 + k\epsilon$ and $t_0=t$ and $t_N=t'$. A path $x(t)$ is discretized by taking the points $x_k = x(t_k)$ and connecting by straight lines.

Finally one has to discretize the action. My impression from some notes and texts is that one follows one "educated guess" procedure:

  1. Replace derivatives by differences of neighboring points divided by lattice spacing $\epsilon$;

  2. Replace integrals by Riemann sums;

So $S[x(t)]$ would be discretized as: $$S_N(x_0,\dots, x_N)=\sum_{k=1}^N \left[\frac{m}{2}\left(\dfrac{x_k-x_{k-1}}{\epsilon}\right)^2-V(x_k^\ast)\right]\epsilon$$

Where $x_k^\ast$ is some point along the straight line joining $x_{k-1}$ and $x_k$.

My question is: is this actually one educated guess or can the discretization be derived by applying the action functional $S$ to the polygonal path which linearly interpolates the points $(t_k,x_k)$?

For the free part it indeed works. If we apply $S$ to the linear interpolation we get the same as above. But for the potential part we end up with something like

$$\sum_{k=1}^N \int_{t_{k-1}}^{t_k} V\left(x_{k-1}+\frac{x_{k}-x_{k-1}}{\epsilon}(\tau-t_{k-1})\right)d\tau.$$

If one Taylor expands this, one gets something similar to the Riemann sum, but there are higher order corrections so that the "guessed" discretization would just be equal to first order.

So can the guessed discretizations be in some other way derived by applying the action functional to the linear interpolation path? If not, is there some other systematic procedure which gives it, or it is really just a guess?


1 Answer 1


In the end we take $\lim_{\epsilon\rightarrow 0}$. So first order matching is sufficient.
(Just a small comment but I don't have much reputation :))


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