I am currently studying quantum mechanics and I was introduced to Dirac notation (bra-ket notation). I have read multiple times that quantum states are represented by vectors called kets ($|\psi\rangle$) which belong to an "abstract Hilbert space".

However, I wasn't satisfied with this approach, since it doesn't give us information about which Hilbert space we are talking about.

Reading the Wikipedia page for the "position operator", I learned that this operator is usually defined on the space of tempered distributions $\mathcal{S}'(\mathbb{R}^N)$, which is the continuous dual space of $\mathcal{S}(\mathbb{R}^N)$, the space of Schwartz functions.

I also found an article written by David Carfi, which says that we can indeed use tempered distributions as our quantum state vectors, but I do have some questions about this space :

  • is there an inner product $(.,.)$ we can define on $\mathcal{S}'(\mathbb{R}^N)$ ?
  • is there an inner product $(.,.)$ we can define on $\mathcal{S}'(\mathbb{R}^N)$ such as $\big(\mathcal{S}'(\mathbb{R}^N), \, (.,.)\big)$ is a Hilbert space ?
  • if not, how can we define the bra corresponding to a $|\psi\rangle$ ket ?

Thank you.

  • 2
    $\begingroup$ There is no sensible inner product between a tempered distribution and another tempered distribution. Tempered distributions act on Schwartz class functions; this is an asymmetric dual pairing, in contrast to the situation of Hilbert spaces. $\endgroup$
    – Ian
    Aug 2, 2021 at 16:03
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    $\begingroup$ If I recall correctly, the way to deal with this is through rigged hilbert spaces, en.m.wikipedia.org/wiki/Rigged_Hilbert_space $\endgroup$
    – Nick Alger
    Aug 2, 2021 at 16:16
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    $\begingroup$ What is fundamental to the development of the subject? It seems to me that observables and measurements are fundamental, at least in terms of the Physics. And rules of how the observables relate. Out of these, you should be able to build Hilbert spaces on which to represent the observables and measurements as operators on vector spaces. Constructs such as tempered distributions may be useful in leading up this model. $\endgroup$ Nov 29, 2021 at 15:37

2 Answers 2


For a particle which moves in a space of $N$ dimensions, one has $N$ position operators $\hat{q}_1,\ldots,\hat{q}_N$ and also $N$ momentum operators $\hat{p}_1,\ldots,\hat{p}_N$. These are unbounded operators (i.e., not everywhere defined) on the Hilbert space $\mathcal{H}=L^2(\mathbb{R}^N)$. For a function $\psi(q)=\psi(q_1,\ldots,q_N)$ in this space, the action of these operators is given by $$ \hat{q}_j(\psi)(q)=q_j\psi(q) $$ and $$ \hat{p}_j(\psi)(q)=-i\frac{\partial\psi}{\partial q_j}(q)\ . $$ Here $j$ is an index while $i$ is the square root of $-1$, and I set Plank's constant equal to $1$ for simplicity. A good domain on which to define these operators is the space of test functions $\mathscr{S}(\mathbb{R}^N)$ which is inside $L^2$. Now if one wants eigenvectors for these operators, unfortunately they are not to be found in $L^2$, but in the bigger space of distributions $\mathscr{S}'(\mathbb{R}^N)$. As mentioned in a comment, there is a theory for that called the theory of rigged Hilbert spaces, but I would not recommend learning it.

So the answer to the question is: no, the physical Hilbert space is not the space of temperate distributions. Note however that for a scalar Bosonic quantum field theory in $d$-dimensional spacetime, the physical Hilbert space is $\mathcal{H}=L^2(\mathscr{S}'(\mathbb{R}^{d-1}),d\nu)$ for some Borel probability measure $\nu$ on a $\mathscr{S}'(\mathbb{R}^{d-1})$. So there, distributions play a role in defining the Hilbert space, albeit with one extra layer of abstraction since $\mathscr{S}'(\mathbb{R}^{d-1})$ replaces what before was $\mathbb{R}^N$.

  • 1
    $\begingroup$ Yes, and/but an expression of the Schwartz space of tempered distributions as the limit of Hilbert spaces related to the quantum harmonic oscillator, to show that it's nuclear Frechet... and then to see that its dual is a colimit of Hilbert spaces, is useful. I have the impression that Dirac operated in a fashion that somehow took these things for granted... $\endgroup$ Aug 2, 2021 at 21:36
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    $\begingroup$ The way I see it, Gelfand's rigged Hilbert space approach is just a general machine for proving isomorphisms of spaces of test functions/temperate distributions with much simpler sequence spaces of rapid decay/at most polynomial growth, see for example galaxy.cs.lamar.edu/~rafaelm/webdis.pdf Now I find these isomorphisms extremely useful, but I only need that for $\mathscr{S}'(\mathbb{R}^N)$ where I can do it by hand. If working with distributions on more general Nash manifolds, then I guess rigged Hilbert spaces might be really needed. $\endgroup$ Aug 2, 2021 at 21:48
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    $\begingroup$ It is certainly true, in this situation, as in many, that the thing that "must be done", is in fact correct, although the certification of its correctness may only be possible at a more sophisticated level... I myself am a bit amused by this sort of disconnect between "factual correctness" and "absolute truth". :) $\endgroup$ Aug 2, 2021 at 22:02
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    $\begingroup$ Thank you for your answer. The issue I have when using $L^2$ as our physical Hilbert space is that operators such as the position operator have no eigenvectors in this space. Therefore, it seems to me that there is a contradiction with one of the postulates of quantum physics, which says that the set of eigenvectors of all observables form a basis of $\mathcal{H}$. $\endgroup$
    – Loris
    Aug 3, 2021 at 8:09
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    $\begingroup$ The postulates of quantum physics should not be taken too literally. In particular, the notion of "basis" it calls for is not that of elementary linear algebra courses. $\endgroup$ Aug 3, 2021 at 17:33
  1. In the space of tempered distributions, we cannot define everywhere-defined inner products, but we can define suitable and useful partial inner products on some convenient linear subspaces. In particular, for example, on the subspace of tempered distributions induced by square-integrable functions (or equivalent classes of them), we can define the usual $L^2$ inner product. Moreover, we have infinitely many other interesting partial inner products, in tempered distribution space, for Quantum Mechanics; for instance, one associated with any “Dirac orthonormal basis” $e$ of tempered distribution space itself. One important example of such products is the natural partial inner product $(.|.)_\delta$ defined on the linear span of the family $\delta$ of all Dirac deltas (which are eigenvectors of the position operator and constitute a continuous Dirac-orthonormal basis of the entire tempered distribution space). The inner product of two finite linear combinations of Dirac deltas $\sum_x a_x \delta_x $ and $\sum_x b_x \delta_x$, in such inner product, is defined (naturally) as the complex number $$\biggl\langle \sum_x a_x \delta_x \ \bigg| \ \sum_x b_x \delta_x \biggr\rangle_\delta \ := \ \sum_x a_x^* b_x$$ (all the summations are indeed finite, even if they are indexed by the Euclidean space $E=\mathbb R ^n$). In a perfectly analogous way, we can define a partial inner product $(.|.)_e$, for each continuous Dirac orthonormal basis $e$ of tempered distribution space.

  2. In order to obtain, for each continuous Dirac-orthonormal basis $e$, a natural Hilbert space contained into (not equal to) the space of tempered distributions, we can take the standard completion $H_e$ of the above pre-Hilbert spaces $$(\rm{span}(e),(.|.)_e).$$ Observe that those Hilbert spaces are strictly contained into the space of tempered distributions, their topologies are strictly stronger than the natural topologies of the distribution space and they are non-separable Hilbert spaces.

  3. Inside the above “partial” Hilbert spaces of tempered distributions, we can define (in the usual way) a bra $\langle u |$, for every ket $| u \rangle$. For instance, in the Hilbert space $H_\delta$, the bra of a finite linear combination $\sum_x a_x \delta_x $ is the linear functional $$\sum_{x \in E} a^*_x \langle \delta_x |$$ sending every finite linear combination $$\sum_x b_x \delta_x$$ to the complex number $$\sum_x a^*_x b_x.$$ The bra associated with every delta $\delta_x$ is the functional associating with each $\sum_y a_y \delta_y $ the complex number $a_x$.

  4. We can put together the $L^2$ Hilbert subspace and the spaces $H_e$ by orthogonal sum. For instance, a more comprehensive Hilbert subspace associated with the position operator is the orthogonal sum of $L^2$ and $H_\delta$.

  5. To claim that the right spaces for the foundations of Quantum Mechanics are the spaces of tempered distributions (instead of the naked $L^2$ spaces, which are indeed just contained in some $\mathcal S’$) doesn’t mean that such spaces are Hilbert spaces or that they are the only possible spaces adoptable in theoretical physics… it does mean that they are the minimal required spaces offering enough structures (algebraic, topological, metric, unitary, … ) inside them to satisfy usual basic necessities of the theory: sufficient richness of observables; continuity of fundamental operators; enough versatile superposition operations; continuous basis clearly and efficiently defined; Dirac orthogonality; enough Dirac orthonormal eigenbases to cover a certain variety of reasonable continuous observables; a powerful and effective functional calculus; normalizability of some states that cannot belong to $L^2$, but having (nevertheless) a obvious and straightforward probability interpretation in quantum mechanics, for example the Dirac deltas, which are (simply) states concentrated at one point; enough partial inner products to reproduce efficiently the bra-ket calculus in the continuous case; enough extension to cover, in one unique coherent realm, square-integrable functions, Dirac deltas, De Broglie waves, different kind of normalizability and relative versions of the uncertainty inequalities outside $L^2$.


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