This is a question which has plagued me for a long time.

In the path Integral, we insert$\int\mathcal{D}\Phi(x)\vert\Phi\rangle\langle\Phi\vert=\mathcal{I}$ and $\int\mathcal{D}\Pi(x)\vert\Pi\rangle\langle\Pi\vert=\mathcal{I}$

where $\hat{\phi}(x)\vert\Phi\rangle=\Phi(x)\vert\Phi\rangle$, $\hat{\pi}(x)\vert\Pi\rangle=\Pi(x)\vert\Pi\rangle$ and $\hat{\phi}(x)$,$\hat{\Pi}(x)$ are the field operator and its canonical conjugate operators.

In principle, we should and could verify these completeness relations. But it appears to be tedious work. Meanwhile, I'm not even sure about what the integration measure means. What I have so far is the field operator eigenstates expanded in creation and annihilation operators. See the answer to this problem https://physics.stackexchange.com/q/706613.

The next step should be properly defining the integration measure and calculation by brute force. The measure is continuous limit of discretization of spatial point in practice. I'm not looking to do it myself. There must have been someone who proved this rigorously. I'll appreciate it if someone can post an answer or refer me to relevant resources.

I can think of a possibly unrigorous argument: field operators are Hermitian so their eigenstates must form a complete basis of the Hilbert space. But I'm not convinced that such an argument should close the case. Since the field theory grows out of Fock space, we should be able to verify it explicitly. Moreover, a proper definition of integration measure is essential.


1 Answer 1


"There must have been someone who proved this rigorously."

No, there is no rigorous formulation of path integration or its measure for general quantum field theory. The most comprehensive work in this direction is as far as I know still Glimm and Jaffe's work in 2 and 3 spacetime dimensions, cf. their book "Quantum Physics - A Functional Integral Point of View".

The construction of a rigorous quantum field theory equivalent to the Standard Model is still an open problem, known as the Yang-Mills millenium problem.

  • $\begingroup$ There is a rigorous formulation of path integral in Euclidean signature. Kind of the point of the book, if I recall correctly... $\endgroup$ Oct 7, 2023 at 9:09
  • $\begingroup$ @PeterKravchuk I'm not sure what you mean. Yes, all the rigorous path integrals I'm talking about here are Euclidean (and then make statements about the Minkowski version via the rigorous version of Wick rotation, the Osterwalder-Schrader reconstruction theorem) because the Minkowski version has just far worse convergence behavior. But there still is no such Euclidean path integral in arbitrary spacetime dimension or for arbitrary QFTs that I know of. The book by Glimm and Jaffe I reference just constructs these integrals for polynomially interacting scalar fields in 2d and 3d. $\endgroup$
    – ACuriousMind
    Oct 7, 2023 at 9:18
  • $\begingroup$ I guess I was confused by the use of "general" in your answer. However, I do think that what you're saying is somewhat like saying "there's no rigorous construction for a general probability measure". That is, the general concept is axiomatized, there are known examples, and there are conjectural examples. It is not clear what it would even mean to have a rigorous construction for "a general qft." $\endgroup$ Oct 7, 2023 at 11:11
  • $\begingroup$ @PeterKravchuk By a "general" construction I mean one that well-defines at least a large part of the QFTs we usually deal with in physics, i.e. those with polynomial Lagrangians and non-Abelian gauge fields in arbitrary (or at least 4) spacetime dimensions. The point is that physics acts as if the Lagrangian suffices to make the path integral measure $\mathcal{D}\phi\mathrm{e}^{-S_E[\phi]}$ exist, so "rigorous general QFT" would need to either actually show this is the case or specify exactly for which Lagrangians this works (and the SM Lagrangian should be included in this set). $\endgroup$
    – ACuriousMind
    Oct 7, 2023 at 11:49
  • $\begingroup$ There are so many QFT known today, most of which are non-Lagrangian, that I don't think one can reasonably expect any result that can be called "general" in a useful way. Also, my understanding was the SM is not expected to be UV-complete (in a mathematical sense). Millennium problem is about YM, which is expected to be UV-complete on is own. $\endgroup$ Oct 7, 2023 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.