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I am a mathematician who is trying to understand Wightman axioms. I do not understand what an operator-valued distribution is, because in this context people say that operators can be unbounded, and as far as I know, unbounded operators do not have any structure on them. Indeed, if we were only dealing with bounded operators, I would guess operator-valued distributions would just be the tensor product (of vector spaces) of tempered distributions and bounded linear operators, but since this is not the case I am not really sure of what people mean.

Edits based on the received answers and comments: I do not believe the answer "it is just a set-theoretic function between the Schwartz space and unbounded operators" is satisfying because it is missing the point that it should behave in some sense like a distribution (otherwise it would not be called distribution I guess).

The reason why I am asking this is that I need to understand the maths behind it, not its meaning so I am looking for a mathematical answer from the mathematical physics community (in particular every word should be well-defined, which rules out words like abstract object or transforms, unless these words are in turn given a definition).

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    $\begingroup$ It is simply an object that when evaluated on a test function gives an operator. I do not see why the unboundedness should be a problem. Most operators in quantum theories are unbounded, although they are usually closed. $\endgroup$
    – mike stone
    Commented Nov 15, 2022 at 18:12
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    $\begingroup$ Well, my problem is that what you are saying is that basically it is a set-theoretic function from the set of tempered functions to the set of unbounded operators. However, if this is true, I do not see why the word "distribution" is there. I would guess that you should at least require some basic properties that make a quantum field resemble a distribution. What you are saying suggests that if $F$ is a field and $f$ is a tempered function, it might as well happen that $F(2f) \neq 2F(f)$, which I hoped were true. $\endgroup$
    – Nicolò Cavalleri
    Commented Nov 15, 2022 at 20:02
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    $\begingroup$ Possibly relevant: ncatlab.org/nlab/show/operator-valued+distribution#Definition $\endgroup$
    – Andrew
    Commented Nov 15, 2022 at 22:44
  • $\begingroup$ It might be worth considering asking this question at Math SE $\endgroup$
    – Níckolas Alves
    Commented Nov 16, 2022 at 20:22
  • $\begingroup$ @NíckolasAlves I will do it next time, it is just that googling stuff of this kind I saw that this topic was only present in this forum, so I thought it was more likely people would know about it here $\endgroup$
    – Nicolò Cavalleri
    Commented Nov 16, 2022 at 20:39

2 Answers 2

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Ok, I found the answer in Folland's book. An operator-valued distribution is a function $\Phi$ from the Schwartz space to operators on $\mathcal{H}$ that has the following property: for every $\eta, \xi \in \mathcal{H}$, $\langle \eta \mid \Phi \mid \xi \rangle$ is a tempered distribution.

Just a comment for who comes next: this property is a very strong request and mathematically is the whole reason why it makes sense to call it a distribution. Without this property, things get pretty wild.

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    $\begingroup$ Intuitively, well-defined is a recursive concept that means that something is defined in terms of other well-defined things ultimately coming down to axioms and logic in some foundation theory, which you can assume to be ZFC for the purpose of this question although it is really unimportant. This can be made more precise, but I do not think this is the point. I am happy to point out definitions of the terms that I used if anyone does not actually understand what I refer to. $\endgroup$
    – Nicolò Cavalleri
    Commented Nov 16, 2022 at 20:15
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    $\begingroup$ This is up to you of course, but I think it might improve your answer if you elaborated on the comment "Without this property, things get pretty wild." I'm a physicist, not a mathematician, but I'd be interested in learning a little bit about the subtleties. (I upvoted because I always like to see when people are able to answer their own questions) $\endgroup$
    – Andrew
    Commented Nov 17, 2022 at 0:20
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    $\begingroup$ @Andrew I just meant that the kind of objects which are just functions from the Schwartz space to operators are so many and have none of the properties you expect from distributions, like linearity etc. It's like considering all functions from R to R (a set with cardinality even bigger than the one of continuum) instead that linear functions from R to R, which are basically just R $\endgroup$
    – Nicolò Cavalleri
    Commented Nov 17, 2022 at 0:30
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    $\begingroup$ @NicolòCavalleri: Unfortunately, the proposed definition does not work, if the intent is to use it in the context of Wightman axioms. For a test function $f$, the evaluation of the operator $\Phi(f)$ on an abitrary vector $\xi$ in $\mathcal{H}$ will be undefined because you will need $\Phi(f)$ to be an unbounded operator, i.e., a partially defined operator. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Nov 18, 2022 at 20:42
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    $\begingroup$ The key here is to have a common dense subspace $\mathcal{D}\subset\mathcal{H}$, where all the operators $\Phi(f)$, $f$ a test function, are well defined. A good source for this is the second volume of the book by Reed and Simon on mathematical physics. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Nov 18, 2022 at 20:44
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Maybe the link that you are missing is this understanding of the word 'distribution', as used in mathematical analysis.

A (complex-valued) distribution, sometimes also called 'generalized function', is a function $$\varphi: \mathcal S \to \mathbb C, \tag1$$ where $\mathcal S$ is a space of 'well-behaved' functions known as test functions. The precise choice depends on what you're doing, but it is typically something like the set of smooth functions with compact support.

The reason that we call them 'generalized functions' is that if you have a function $f:\mathbb R\to\mathbb C$, you can normally also define a distribution $\varphi_f$, whose value on a test function $g\in\mathcal S$ is given by $$ \varphi_f(g) = \int_{-\infty}^\infty f(x)^*g(x)\mathrm dx. $$ However, this class can also include pathological objects like the Dirac delta, which fit the mold of $(1)$ but for which the use of the language of normal functions can be quite awkward.

In any case: an operator-valued distribution is a linear function that, like in $(1)$ accepts test functions in $\mathcal S$, but which returns operators that act on your Hilbert space $\mathcal H$ of interest.

I will leave the rigorous details to others, but hopefully this will be useful for you, if this was indeed the gap that needed to be filled, and, if not, then for others after you ;-).

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