Imagine the following path integral

$$\int_{x(0)=x_i}^{x(T)=x_f} \mathcal{D}x \, e^{\frac{i}{\hbar}S[x]}.$$

This integral is defined over the space of all paths that satisfy the boundary conditions $x(0)=x_i$ and $x(T)=x_f$. I am interested in defining a metric on this space in order to be able to quantify how similar two different paths are. I appreciate any suggestions on how to proceed.

  • $\begingroup$ The path integral measure is often defined by first considering a lattice with finite spacing $\epsilon$, integrating over the value at every lattice point separately. Then one takes the limit $\epsilon \to 0$ to end up with the path integral. Maybe a similar approach could be used to define how "close" two functions are to each other. $\endgroup$ – scaphys Apr 21 at 22:58
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    $\begingroup$ Alternatively, have a look at $L^p$-distance. $\endgroup$ – scaphys Apr 21 at 23:07

In $\mathbb{R}^n$ the typical notion of distance is $|x| = \sqrt{\sum_i x_i^2}$. This naturally generalizes to square-integrable functions on $[a,b]$, $|f| = \sqrt{\int_a^b |f(x)|^2 dx}.$ Thus, one notion of the distance between $f,g$ on $[a,b]$ is $$|f-g|=\sqrt{\int_a^b |f(x)-g(x)|^2 dx}.$$

  • $\begingroup$ Thank you so much. In my case $f,g$ are complex functions; then I guess I have to modify this to $$|f-g|= \sqrt{\int_a^b (f(x)-g(x))(f(x)-g(x))^{\ast} \, dx.$$ Correct? $\endgroup$ – B. T. Apr 22 at 7:14
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    $\begingroup$ Yes, but you can use the same notation nevertheless, since for $z \in \mathbb{C}$, the typical norm is defined as $|z| \equiv\sqrt{zz^*}$. $\endgroup$ – scaphys Apr 22 at 10:23
  • $\begingroup$ I am glad to help. That is correct, see the comment by @scaphys. $\endgroup$ – user26872 Apr 22 at 21:25

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