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I know hardly anything about quantum field theory (QFT) but I'm giving a try to understand some ideas of it. As far as I understand, in QFT one is interested in studying measures such as: \begin{eqnarray} d\mu(\varphi) \propto e^{-S(\varphi)}d\varphi \tag{1}\label{1} \end{eqnarray} where $S$ is a given action and $d\varphi$ an 'a priori' measure. In particular, one of the most interesting actions is given by: \begin{eqnarray} S(\varphi) = \int_{\mathbb{R}^{d}}\bigg{[}\frac{1}{2}\langle \varphi, (-\Delta+m^{2})\varphi \rangle + g\varphi(x)^{4} dx\bigg{]}. \tag{2}\label{2} \end{eqnarray} In view of (\ref{2}), expressions such as: \begin{eqnarray} d\mu(\varphi) \propto e^{-S_{0}(\varphi)}e^{-\int_{\mathbb{R}^{d}}g\varphi(x)^{4}}d\varphi \equiv e^{-g\int_{\mathbb{R}^{d}}g\varphi(x)^{4}dx}d\mu_{G}(\varphi) \tag{3}\label{3} \end{eqnarray} are very common in the theory, where the term $e^{-g\int_{\mathbb{R}^{d}}\varphi(x)^{4}dx}$ is usually viewed as a perturbation of a Gaussian measure. This being said, I'd like to clarify something. As far as I understand, the measure $d\mu_{G}$ in (\ref{3}) is a Gaussian measure on $\mathcal{S}'(\mathbb{R}^{d})$ induced by a positive-definite function $e^{-\frac{1}{2}B(f,Gf)}$ where $B$ is a quadratic form induced by a bilinear and continuous map $B$ defined on $\mathcal{S}(\mathbb{R}^{d})\times \mathcal{S}(\mathbb{R}^{d})$ in terms of the Green's function $G$ of the massive Laplacian $-\Delta + m^{2}$. Thus, the representation $-g\int_{\mathbb{R}^{d}}\varphi(x)^{4}dx$ is only formal (what does it mean?), since $\varphi$ is supposed to be a function on $\mathcal{S}'(\mathbb{R}^{d})$. It turns out that expressions such as (\ref{3}) end up being very confusing to non-experienced students like me, once we're often misguided to wrong conclusions or get wrong understandings of the subject. I think, however, that there is a deeper reason for such notations than just being a simpler way of writting things down. I think this has to be with the fact that one is interested in seeing $\varphi(x)$ as 'tempered distributional'-valued random variables, but I'm not sure. So I'd like to clarify these concepts. Why do quantum field theorists use pointwise notation to express distributions and what can one benefit from it?

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    $\begingroup$ QFT per se doesn't reference random variables. However, one observes that, formally, QFT path integrals, continued to the Euclidean, look like partition functions in Statistical Mechanics, which does reference random variables. This mapping both allows one to use intuition from Statistical Mechanics to think about QFT, as well as provides paths towards computational methods. To do physics, it's enough to be quite informal about this mapping, but by all means flesh it out for yourself in whatever way seems most natural to you. $\endgroup$ – Michael Engelhardt Apr 16 at 0:53
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    $\begingroup$ I’m having trouble understanding the question, but the separation of the $\phi^4$ is not natural from the physics and really is just the statement that you’re doing perturbation theory. The whole thing is really a measure on ... something. Maybe looking at lattice regularizations will help your intuition? $\endgroup$ – Aaron Bergman Apr 16 at 1:19
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    $\begingroup$ If you’re trying to think of the path integral rigorously, you’re going to have a lot of trouble. The case above, for example, likely does not exist as a theory. Unless you want to pursue it as a field of research, you’ll probably have better luck of thinking of it as an integral over some inchoate ‘space’ of ‘functions’ and assuming that the lattice has a continuum limit when the theory makes sense, even if it’s been too hard to prove in most cases as yet. $\endgroup$ – Aaron Bergman Apr 16 at 1:49
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    $\begingroup$ The $\phi^{4} $ is motivated by the physics. The real world is not described by a quadratic action. The infinities you note associated with the Gaussian case are quite innocuous compared to what you have to contend with once you introduce interactions such as the $\phi^{4} $. $\endgroup$ – Michael Engelhardt Apr 16 at 6:57
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    $\begingroup$ I think the issue is that you are taking the method of making the Gaussian integral rigorous and trying to extend that to other QFTs. This hasn’t worked, and, as Michael said above, it’s far from obvious that it’s the right direction to go to. The action is a classical quantity, with honest functions that can be point wise multiplied. The path integral is a not rigorous thing based on that classical action that has a plausible path to definition via the lattice using exactly that classical action. $\endgroup$ – Aaron Bergman Apr 16 at 14:30
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The quick answer is that (3) written by physicists is not to be taken too seriously by mathematicians. However, it is a statement of a goal or research problem which is to find a rigorous definition/construction of what (3) is trying to say.

One has an injective continuous linear map $\iota:\mathscr{S}(\mathbb{R}^d)\rightarrow \mathscr{S}'(\mathbb{R}^d)$ given by $$ \iota(\varphi)=\left(f\longmapsto \int_{\mathbb{R}^d}\varphi(x)f(x)\ d^dx\right) $$ where $f$ is a generic element of $\mathscr{S}(\mathbb{R}^d)$. Via this map it is natural to identify $\mathscr{S}(\mathbb{R}^d)$ with a subset of $\mathscr{S}'(\mathbb{R}^d)$. Moreover, this subset is dense (in fact sequentially dense) in $\mathscr{S}'(\mathbb{R}^d)$. The action $S(\varphi)$ in (2) is perfectly well defined for $\varphi$ in the subset $\mathscr{S}(\mathbb{R}^d)$ but not for $\varphi\in \mathscr{S}'(\mathbb{R}^d)\backslash\mathscr{S}(\mathbb{R}^d)$.

Unfortunately, the Gaussian measure $d\mu_G$ is not supported on $\mathscr{S}(\mathbb{R}^d)$ but on the much bigger space $\mathscr{S}'(\mathbb{R}^d)$. One can be a bit more precise and work in a weighted Sobolev or Besov space of exponent $\alpha$ but the latter would be negative except for $d=1$.

So one has to introduce a regularization in order to remove the regularization. This is explained in my answer to

A roadmap to Hairer's theory for taming infinities

and to

https://physics.stackexchange.com/questions/372306/what-is-the-wilsonian-definition-of-renormalizability

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  • $\begingroup$ Thanks for the answer! Just to clarify, when you say (\ref{2}) is perfectly well defined for $\varphi $ in the subset of $\mathcal{S}(\mathbb{R}^{d})$ you mean $\mathcal{S}'(\mathbb{R}^{d})$ via the injection map you defined? $\endgroup$ – IamWill Apr 16 at 21:31
  • $\begingroup$ your comment does not make sense as written. S without prime is a subset of S' with prime. The action is well defined for phi in the subset S without prime, but not in general for phi's in the complement of S without prime in S' with prime. $\endgroup$ – Abdelmalek Abdesselam Apr 16 at 21:37
  • $\begingroup$ ahh! Sorry I misunderstood it. Now I got it! $\endgroup$ – IamWill Apr 16 at 22:05
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To make an interacting (i.e. not purely quadratic) QFT at all meaningful, you have to impose a regulator. The most transparent regulator in many ways, and the only known regulator that allows to address the full non-perturbative content of the QFT, is a lattice regulator. Replacing $\mathbb{R}^d$ by $\mathbb{Z}^d$ (or a finite subset such as $\Lambda=\mathbb{Z}^d\cap[0;L]^d$ with suitable boundary conditions) makes the path integral measure $\mathrm{d}\varphi=\prod_{x\in\Lambda}\mathrm{d}\varphi_x$ unambiguously defined and removes any problems with pointwise operations on the fields appearing inside $S(\varphi)$. Whether a continuum limit then exists is of course a non-trivial (and generally unsolved) problem.

To answer your explicit question: Quantum field theorists (if such a generalization is permitted), at least when writing expressions such as the action of $\varphi^4$ theory, tend to think not so much in terms of distributions as in terms of fields in the regularized theory, for which pointwise products are unproblematic.

It should also be noted that a QFT is the quantization of a classical field theory, and that is governed by an action that involves pointwise products; I'd consider that sufficient motivation to keep the pointwise notation (which moreover makes the Poincaré invariance of the continuum action apparent, something which a more distribution-centric approach would likely obscure).

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    $\begingroup$ Perhaps it's worth adding that using exactly the classical action is simply abuse of notation. Even on the lattice, it's better to use an appropriately renormalized interaction (in this case, the normally-ordered product $:\phi^4:$). Leaving out the renormalization is also an abuse of notation. I think physicists do it because it works in the d=1 case (quantum mechanics), which was the first case where the path integral was really understood. Not really recommended, as doing this confused their thinking about QFT for decades. $\endgroup$ – user1504 Apr 16 at 15:14
  • $\begingroup$ In the path integral formalism, the fields aren't operators, so there is nothing to normal order at that level. Renormalization is accounted for by writing the lattice action in terms of bare parameters (typically indicated by a subscript 0), which of course aren't the physical renormalized quantities. QFT on the lattice is really pretty rigorous at finite lattice spacing (the devil is in the continuum limit). $\endgroup$ – gmvh Apr 16 at 19:47
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    $\begingroup$ @gmvh: As you know there are two points of view: 1) operators, 2) path integrals. In 1) normal ordering means putting creation ops on the left and annihilation ops on the right. In 2), it is not like there is no normal ordering. There is one but it is expressed differently, i.e., changing monomials like $\phi^4$ to Hermite polynomials $:\phi^4:$. That's what user1504 is referring to. $\endgroup$ – Abdelmalek Abdesselam Apr 16 at 20:31
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    $\begingroup$ To be fair to gmvh: calling the renormalized classical lattice interaction a 'normally ordered product' is itself an abuse of notation. My apologies. :) $\endgroup$ – user1504 Apr 17 at 2:50
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    $\begingroup$ @gmvh: Also, another comment, physicists use evaluation at points instead of the point of view of Schwartz distributions. The reason is not the presence of a regulator which makes sample fields $\varphi(x)$ smooth. It is because correlations $\langle \varphi(x_1)\cdots\varphi(x_n)\rangle$ have singular support on the diagonal. Namely, one can evaluate them pointwise at noncoincident points. Sometimes, however, one needs to consider contact terms. $\endgroup$ – Abdelmalek Abdesselam Apr 17 at 14:02

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