Let us assume we live on Euclidean $\mathbb{R}^d$ and consider the normalized partition function

\begin{equation} \begin{aligned} Z : D &\to \mathbb{R} \\ J &\mapsto \frac{\int \exp \left[ -S \left( \psi \right) + \left \langle J, \psi \right \rangle\right] \mathcal{D} \psi}{\int \exp \left[ -S \left( \psi \right) \right] \mathcal{D} \psi} \end{aligned} \end{equation}

where $D$ is taken to be the space of test functions on $\mathbb{R}^d$, $\left \langle \cdot, \cdot \right \rangle$ is the standard $L^2$ inner product and $S : D \to \mathbb{R}$ denotes some suitable classical action.

Some naive ideas to make sense of the above expressions might be:

  • (M1) From a mathematical point of view, there is no Lebesgue-like measure on infinite-dimensional vector spaces. However, from my (totally inadequate) understanding of the Wiener measure, there exists a perfectly fine Gaußian measure $\mu$ (on a larger space than $D$) such that in most relevant cases $\exp \left[ - S \left( \psi \right) \right] \mathcal{D} \psi := \exp \left[ - S_{\mathrm{int}} \left( \psi \right) \right] \mathcal{D} \mu\left( \psi \right)$ where the right hand side now is well-defined.
  • (M2) One may consider Hilbert space completions $H$ of $D$ with respect to some inner product and define the above quotient as a limit of integrations over finite-dimensional exhaustions of $H$ that preferably lie within $D$.

But somehow such integrals are widely considered mathematically ill-defined. What precisely is it that goes wrong? Some ideas of mine are:

  • (P1) There is no unique way to embed $D$ into a larger space on which the Wiener measure might be defined
  • (P2) The numerator/denominator in the above expression are infinite when considered as Lebesgue integrals with respect to the Wiener measure and taking the quotient requires some nonunique choice of limit.
  • (P3) The approximation by finite-dimensional integrals is dependent on the choice of exhausting subspaces
  • (P4) The approximation by finite-dimensional integrals is dependent on the choice of inner product on $D$
  • (P5) The quotient in the definition of $Z$ is dependent on how the limits of exhausting subspaces is taken, i.e in analogy to $\lim_{n \to \infty} \int_{-n}^n \sin \left( n \pi x \right) \mathrm{d} x \neq \lim_{n \to \infty} \int_{-2n}^{2n + 1} \sin \left( n \pi x \right) \mathrm{d} x$.

1 Answer 1


I took the liberty to put some labels (M1), (M2) for the two suggested methods for making sense of the integral, as well as (P1)-(P5) for list of problems that could arise. This allows me to refer to specific items in the question.

If one replaces $J$ by $iJ$ with $J$ real-valued then the functional $Z$ is the characteristic function of the probability measure (within the totally standard framework of Lebesgue integration theory) one would like to define rigorously. The most convenient space $D$ for the test functions $J$ is Schwartz space $\mathscr{S}(\mathbb{R}^d)$. In this case the space where the wanted probability measure would live is the dual space $\mathscr{S}'(\mathbb{R}^d)$ of temperate Schwartz distributions.

About (M1): This does not work. Also, avoid the incorrect (and error prone) terminology of "Wiener measure". The latter is about $d=1$ where the measure $\mu$ is supported on a space of continuous functions and where multiplying by the in Radon-Nikodym weight $\exp(-S_{\rm int}(\psi))$ makes sense, at least in finite volume. For $d\ge 2$, the measure $\mu$, seen as a probability law for a random distribution $\psi$, is such that with probability 1, the quantity $S_{\rm int}(\psi)$ does not make sense, e.g., because it contains terms like $\psi(x)^4$. So in this sense (M1) is not a solution to the problem. However one can correct this attempt by introducing a Fourier cutoff and thus modifying $\mu$. This gives a situation similar to (M2) where one has a well defined sequence of measures on $\mathscr{S}'(\mathbb{R}^d)$ indexed by the cutoffs (frequency and volume) and the question becomes: does this sequence converge in the a suitably topology in the space of probability measures on $\mathscr{S}'(\mathbb{R}^d)$. The most natural topology here is weak convergence which corresponds to pointwise (in $J$) convergence of the corresponding functionals $Z$. This same question of convergence of the sequence of approximations (using finitely many lattice points) is the main question to be addressed by method (M2).

It is instructive to do (M2) in detail even for constructing the massive free field, i.e., the measure $\mu$. This is done in the two MathOverflow posts



What can go wrong: Let me now consider the problems that could arise and discuss (P1)-(P5). Proposing the latter as a list of potential problems, in fact amounts to putting the cart before the horse, because this tacitly presumes that existence of a limit or some outcome of method (M1) or (M2) is unproblematic, and all one has to do is worry about uniqueness of this outcome. The main problem that could occur is that there may be no limit, even if one restricts to a subsequence (loss of tightness or probability mass escaping to infinity). The other problem is that the limit obtained may be Gaussian, i.e., not interesting. See


for more on this but to make a long story short, one has to give oneself more flexibility, by allowing the couplings to depend on the cutoff, in order to get interesting limits. How to do that hinges on a renormalization group analysis. See

What is the Wilsonian definition of renormalizability?

Finally, it expected that when the limit exists, it will be unique, i.e., independent of the arbitrary choices in setting up the approximation sequence, but only few results of this type have been established rigorously. For the $\phi_3^4$ model, a recent construction is in the article by Gubinelli and Hofmanová https://arxiv.org/abs/1810.01700

  • $\begingroup$ Thanks for the elaborations! It is hard to see why the seemingly huge freedom of renormalization schemes boils down to a finite-dimensional variety. Do you happen to have a mathematical intuitive guide as to why this should be the case? From a physics point of view it is not too surprising. Also, maybe naive, but why do we need to go all the way to probability measures on distribution spaces? There are much simpler completions of test function spaces that one may envisage being sufficient (e.g Sobolev spaces). Related: Why would physics care about the topology on the space of test functions? $\endgroup$
    – iolo
    Commented Sep 7, 2020 at 15:50
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    $\begingroup$ I need several comments, one for each question. Q1 about finite dim variety: this is a bit of a long story but it is explained in the link "what is the Wilsonian definition of renormalizability" in my answer. The short summary is that this is because unstable manifolds of typical RG fixed points are finite dimensional, i.e., there are finitely many relevant couplings. There are exceptions though, e.g., the Gaussian fixed point in 2d. $\endgroup$ Commented Sep 7, 2020 at 18:24
  • $\begingroup$ Q2: The question does not make sense, since it compares apples to oranges. The typical space of test functions $D$ is already complete. So this is not about completion but rather taking the dual $D'$. It is possible to pick for $D$ a Sobolev space of positive regularity (see the GH arXiv article I linked to), but then $D'$ will be of negative regularity and made of "functions" $\psi$ which are not functions. The classical fields $\psi$ being integrated over in the functional integral belong to $D'$ and even if you want $E'$ to be a Sobolev space, the question is still about constructing... $\endgroup$ Commented Sep 7, 2020 at 18:30
  • $\begingroup$ ...a probability measure on $E'$. Moreover this measure is not obtained directly but as a limit. So you need to talk about the space of probability measures on $D'$ and tell me what convergence there means to you. $\endgroup$ Commented Sep 7, 2020 at 18:32
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    $\begingroup$ Unfortunately, this does not work or so to speak Murphy's law applies here maximally with a vengeance. In dimension $d\ge 2$, the integral cannot be over a space of functions like $L^2$. It has to be over a space of generalized functions or Schwartz distributions. This is also related to the need for UV renormalization. The need for integrating over a space of distributions is already the case for the free field. $\endgroup$ Commented Sep 8, 2020 at 16:02

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