Quantum Field Theory is a branch of mathematical physics which is begging for a better understanding. In fact there are no rigorous constructions of interacting QFT in four dimensions. By a rigorous construction I mean a construction of the quadruple $(\mathcal{H},U,\Omega,\phi)$ where $\mathcal{H}$ is the Hilbert space of states, $\Omega$ the vacuum vector, $U$ a unitary representation and $\phi$ an operator valued distribution. These data have to satisfy certain axioms which are called the Wightman axioms. However this program of explicitly constructing QFT turned out to be too difficult and the other strategies emerged: one of them is concerned with the construction of the functional integral, i.e. the problem boils down to the construction of certain measure on the space of distributions. However any interacting QFT is governed by a Lagrangian: I don't see where exactly this Lagrangian enters in the above reasoning. So to be slightly more precise:

Question 1. Suppose that we want to construct QFT following the „original approach” (i.e. constructing it directly). Which one of the Wightman axioms tells us which QFT we have really constructed (i.e. what is the form of the interaction part of a Lagrangian)?

And concerning the functional integral approach:

Question 2. Is there a precise form/conditions for a desired measure (which can be read from the Lagrangian) or is it given only at the heuristic level via some density function with respect to a Gaussian measure (but in the end this measure may turn out to be singular to a Gaussian measure)? If it is given only on a heuristic level how it is possible to know whether we have succeed in our construction?

  • 4
    $\begingroup$ “However any interacting QFT is governed by a Lagrangian” — this is not a true statement. $\endgroup$ Jul 20, 2020 at 0:26
  • $\begingroup$ @Aaron Do you have an interesting and/or toy example of a QFT without a lagrangian in mind? Truebaran: Not an expert, but my impression is that there are theorems (perhaps due to Osterwalder-Schraeder) that allow one to pass between the formulation in terms of a hilbert space to a formulation in terms of correlators-- these might be helpful to you. $\endgroup$ Jul 20, 2020 at 0:47
  • $\begingroup$ @Phil Tosteson I'm aware of this connections therefore I formulated two separate questions: for the functional integral approach I can see some data from Lagrangian floating around (but still I would like to know how this is made rigorous) but suppose that I would like to construct QFT directly as a quadruple $(\mathcal{H},\Omega,U,\phi)$-I don't see which Wightman axioms adresses the question ,,which QFT actually was constructed'' (corresponding to what interaction?)... $\endgroup$
    – truebaran
    Jul 20, 2020 at 1:17
  • 5
    $\begingroup$ Many CFTs can be constructed directly without a Lagrangian. Self-dual fields also usually (always?) don’t have a Lagrangian description. $\endgroup$ Jul 20, 2020 at 1:19
  • 2
    $\begingroup$ Non-lagrangian supersymmetric gauge theories are also quite fashionable, e.g. the 6d N= (2,0) theory. $\endgroup$ Aug 22, 2020 at 6:48

2 Answers 2


The axioms don't tell you what theory you constructed. For that you need to go beyond the construction of correlation functions of the elementary field $\phi$ (the basic chapter on renormalization in QFT textbooks) and produce, e.g., by a point-splitting procedure, correlations with insertion of composite fields like $\phi^3$. You should then identify your theory via the equation of motion, e.g., $-\Delta\phi+m^2\phi=-\lambda\phi^3$ holding inside correlations. To see how this is done rigorously, see the article by Feldman and Rączka in Ann. Phys. 1977 or the recent article by Gubinelli and Hofmanová.

Also, an interesting example is the following. Let ${\rm Conf}_n(\mathbb{R}^2)$ denote the configuration space of $n$ points in $\mathbb{R}^2$, i.e., the set of tuples $(x_1,\ldots,x_n)$ made of $n$ distinct points in $\mathbb{R}^2$. Consider the functions $S_n:{\rm Conf}_n(\mathbb{R}^2)\rightarrow\mathbb{R}$ given by $$ S_n(x_1,\ldots,x_n)=\sqrt{\sum_q\ \prod_{1\le i<j\le n}|x_i-x_j|^{\frac{q_iq_j}{2}}} $$ where the sum is over "neutral configurations of charges" $q=(q_1,\ldots,q_n)\in\{-1,1\}^n$ such that $\sum_i q_i=0$.

One can show that $\forall n,\exists K_n>0,\forall (x_1,\ldots,x_n)\in{\rm Conf}_n(\mathbb{R}^2)$, $$ S_n(x_1,\ldots,x_n)\le K_n\prod_{i=1}^{n}\left(\min_{j\neq i}|x_j-x_i|\right)^{-\frac{1}{8}}\ . $$ The quickest proof I know for this inequality is following the optimal matching argument in Appendix A of "Complex Gaussian multiplicative chaos" by Lacoin, Rhodes and Vargas. Then it is not hard to show that for every Schwartz function $f\in\mathscr{S}(\mathbb{R}^{2n})$, the integral $$ \int_{{\rm Conf}_n(\mathbb{R}^2)}S_n(x_1,\ldots,x_n)\ f(x_1,\ldots,x_n)\ d^2x_1\cdots d^2x_n $$ converges and defines a temperate distribution in $\mathscr{S}'(\mathbb{R}^{2n})$. This is explained in Section 2 of my CMP article "A Second-Quantized Kolmogorov-Chentsov Theorem via the Operator Product Expansion".

Now it is a fact that the resulting distributions $S_n\in \mathscr{S}'(\mathbb{R}^{2n})$ satisfy the Osterwalder-Schrader Axioms, and therefore can be analytically continued into Wightman distributions satisfying the Wightman Axioms and thus via a GNS type construction, give in the end a quadruple $(\mathcal{H},U,\Omega,\phi)$ obeying the Gårding-Wightman Axioms. The $S_n$ are also the moments of a probability measure on $\mathscr{S}'(\mathbb{R}^{2})$.

Quiz: What is the Lagrangian of this theory?

I will come back later with an answer, but regarding the construction via probability measures, I explained this already so I will not repeat myself and refer to

Reformulation - Construction of thermodynamic limit for GFF

A set of questions on continuous Gaussian Free Fields (GFF)

A roadmap to Hairer's theory for taming infinities

Quiz answer: It is the Ising CFT. Note that I tried to see if one has an equation of motion of $\phi^4$ type but my computations got out of hand rather quickly when looking for an explicit $\phi^3$.

  • $\begingroup$ Thank you very much @Abdelmalek Abdesselam, great detailed answer (as always!) $\endgroup$
    – truebaran
    Jul 21, 2020 at 0:24

It's not necessarily possible to uniquely identify a quantum field theory from its Wightman data.

If you've done the construction yourself, you'll know which theory it is, because you chose the coordinates on the space of fields, constructed the measure and the observables, and showed they obey Osterwalder-Schrader. But the Wightman data you construct this way could have been reached starting with some other set of coordinates on a different space of fields. There might be a dual QFT which gives the same Wightman data, starting with a different set of generators.

The canonical example here is the sigma model to a torus $\mathbb{R}/R\mathbb{Z}$ of radius $R$. (I'm not sure if this thing has been built via constructive methods, but it's easy to build by canonical quantization.) This QFT gives the same Wightman data as the sigma model to the torus of radius $1/R$. The only difference is the choice of generators and the underlying classical limits.

  • 1
    $\begingroup$ Just off my memory, I think Gawedzki in his CFT lectures in the IAS volumes on QFT and strings, gives a rigorous account of the $R$ to $1/R$ duality for a free field. But I don't remember if he uses the functional integral or the canonical quantization approach. $\endgroup$ Feb 15, 2021 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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