A while ago I asked the following question: Understanding Measure in Path integrals and got to the conclusion that path integral measures are infinite products of $d\phi(x_i)$ for some scalar field $\phi$.

What does this differential mean? How would one evaluate it?


1 Answer 1


The differential in a path integral means the same thing as in an ordinary integral, such as $$ \int_{-\infty}^\infty d\phi\ f(\phi). $$ A "path" integral, defined on a lattice as described in Qmechanic's answer to your earlier question, is just an ordinary multi-variable integral with one integration variable $\phi(x)$ for each point $x$ in the lattice. Think of $x$ as an index labeling the different variables. The lattice can be very large and very fine, as long as the total number $N$ of sites is finite. We can have $N=10^{100000}$ if we like. It's still just an ordinary multi-variable integral, and the integrand is just an ordinary function of all of those variables.

If the lattice had only one point ($N=1$), then the "path" integral would reduce to an ordinary single-variable integral, written like this: $$ \int_{-\infty}^\infty d\phi(x)\ f\big(\phi(x)\big). $$ Again, $\phi(x)$ is just an unusual notation for a single variable. On a lattice with a large number $N\gg 1$ of points (as usual), this becomes $$ \int_{-\infty}^\infty d\phi(x_1)\ \int_{-\infty}^\infty d\phi(x_2)\ \cdots \int_{-\infty}^\infty d\phi(x_N)\ f\big(\phi(x_1),\phi(x_2),...,\phi(x_N)\big). $$ The collection of variables $\phi(x_n)$ is written to look like a function of $x$, because the intuition is that for a very large and very fine lattice, we might as well be working with a continuum of variables... but to define what that means, we use a lattice. (It also relies on the integrand having a special form.) The continuum limit isn't taken until after evaluating the integrals, if it's ever actually taken at all.

  • $\begingroup$ Ohhh! This makes perfet sense now. Thank you so much $\endgroup$
    – Craig
    Dec 17, 2018 at 4:35

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