# Why can't the functional integral be derived in a mathematically rigorous way?

Why can't the functional integral be derived in a mathematically rigorous way? What are the obstacles that we have to overcome in order to achieve that goal?

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–  Qmechanic Jul 5 '12 at 17:35
It depends on what you mean by "mathematically rigorous": do you mean a physicist is satisfied, or a mathematician? –  Alex Nelson Jul 5 '12 at 17:36
The problem is that there is a fundamental difference of opinion in the two communities about measure theory. Physicists live in a world where probability is fundamental and always makes sense, mathematicians live in a universe where the real numbers have a well-order, and there are "non measurable sets". The arguments physicists make are not hard to turn rigorous in the physicist's mathematical universe, Solovay's universe, or "a measurable topos", but in the mathematician's universe you have technical problems, because you have to be consistent with well ordering the distributions. –  Ron Maimon Jul 5 '12 at 19:42
@RonMaimon: mathematicians can choose between a world where reals are well-ordered and one where all sets are measurable. One needs to be consistent with the rules of logic but not with particular set-theoretic foundations. There is currently no way to make logical sense out of the 4D functional integral in Minkowski space; physicists usage is just a guide to heuristic reasoning. –  Arnold Neumaier Jul 6 '12 at 12:19

## 1 Answer

In many cases, the functional integral can be derived in a mathematically rigorous way, see http://physics.stackexchange.com/a/29558/7924

The obstacles are in the lack of sufficiently sharp functional analytic tools in infinite dimensions that would allow one to bound the errors in the approximations that physicists use heuristically.

Overcoming the obstacles probably requires to find a way to perform rigorous nonperturbative renormalization of things that can currently be done rigorous only in the sense of formal power series.

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