I'm trying to learn some quantum mechanics by myself, and because of my mathematics background, I'm trying to understand it in a rigorous way. Since then, I've been intrigued by the use of rigged Hilbert spaces in this context. The topic caught my attention one more time after my discussion on physics stack exchange about the Dirac formalism.

Repeating a little bit what I've said in the linked question, physicists use $|x\rangle$ as eigenvectors of the position, and write every element $|\psi\rangle$ of the underlying Hilbert space as: \begin{eqnarray} |\psi\rangle = \int dx \,\psi(x) |x\rangle \tag{1}\label{1} \end{eqnarray} This is analogous to what is usually found in the theory of Hilbert spaces, in which for every $x\in \mathscr{H}$ and $\{e_{n}\}_{n\in \mathbb{N}}$ Hilbert basis: \begin{eqnarray} x = \sum_{n\in \mathbb{N}}\langle e_{n},x\rangle e_{n} \tag{2}\label{2} \end{eqnarray}

As it turns out, the space of vectors $|\psi\rangle$ are usually taken to be $L^{2}(\mathbb{R}^{d})$, so that the position operator $\hat{x}$ makes sense as a multiplication operator on a dense subspace. However, it is a self-adjoint operator with continuous spectrum, so it has no eigenvectors at all in $L^{2}(\mathbb{R}^{d})$ and, thus, $|x\rangle$ does not make mathematical sense in this context. In any case, the use of (\ref{1}) demand that the coefficients $\psi(x)$ to be elements of $L^{2}(\mathbb{R}^{d})$ themselves, just as the coefficients $\alpha_{n} = \langle e_{n},x\rangle$ are elements of $\ell^{2}(\mathbb{N})$ in (\ref{2}), so we end up using the components $\psi(x)$ to do actual calculations most of the time (at least, explicit calculations such as solving the Schrödinger equation).

This is what intrigues me. The above scenario is present at the very beginning of every discussion of quantum mechanics and, at least to me, these very few words are more than enough to justify the need of "some bigger space", which seems to be a rigged Hilbert space. However, a quick search on the internet and you find only a few things about rigged Hilbert spaces, in general papers discussing its connections with quantum mechanics. I've checked also my books on rigorous quantum mechanics and none of them seem to address the problem. However, all of them discuss spectral theory, which I've heard is another way to approach all this.

Question: Why does it seem that rigged Hilbert spaces are still a paper subject, not fully incorporated into the usual expositions of rigorous quantum mechanics, while spectral theory is deeply rooted? Is the spectral theory formalism preferable to treat the problems mentioned above instead of rigged Hilbert spaces by any reason? And why is this?

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    $\begingroup$ Taking the Hilbert space to be $L^2(\mathbb{R}^d)$ is of course already a bit problematic in view of the Laplacian occurring in the Schrödinger equation; the more appropriate setting would appear to be a Sobolev space $W^{k,2}(\mathbb{R}^d)$ with $k\ge 1$ in order to at least allow a weak solution. $\endgroup$
    – gmvh
    Commented Feb 14, 2021 at 14:25
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    $\begingroup$ @gmvh indeed. But most books use $L^{2}(\mathbb{R}^{d})$. It is a really difficult process if you are studying it for the first time, as I am, and most of the discussions are not "quite there". $\endgroup$
    – MathMath
    Commented Feb 14, 2021 at 14:47
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    $\begingroup$ You should learn first John von Neumann's foundations auf quantum mechanics, especially unbounded operators. $\endgroup$
    – jjcale
    Commented Feb 14, 2021 at 15:48
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    $\begingroup$ @gmvh: I'm not sure I follow your comment. The common point of view is to consider the Laplace operator $\Delta$ as an unbounded but closed operator on $L^2(\mathbb{R}^d)$. In other words, $\Delta$ maps from the domain $D(\Delta) := W^{2,2}(\mathbb{R}^d) \subseteq L^2(\mathbb{R}^d)$ to $L^2(\mathbb{R}^d)$, and its graph is closed as a subset of $L^2(\mathbb{R}^d) \times L^2(\mathbb{R}^d)$. This is also the setting in which it makes sense to consider $\Delta$ as a self-adjoint operator. [to be continued]. $\endgroup$ Commented Feb 14, 2021 at 21:30
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    $\begingroup$ [continuation] Solvability of the Schrödinger equation is not a problem in this setting: it follows from Stone's theorem and the theory of $C_0$-groups that, for every initial value $u_0 \in L^2(\mathbb{R}^d)$, the equation $\dot \psi = i\Delta \psi$ has a unique mild solution in $L^2(\mathbb{R}^d)$ $\endgroup$ Commented Feb 14, 2021 at 21:30

4 Answers 4


Why are rigged Hilbert spaces a paper subject, not usually treated in rigorous textbooks: this is a good question. If you learn QM the way I did, you start by understanding that the physicists' Dirac delta is fictional (i.e., doesn't live in $L^2(\mathbb{R}^d)$), then you learn von Neumann's approach via spectral theory, which makes everything rigorous, and then sometime later you hear about rigged Hilbert spaces as a way to make the Dirac delta function rigorous. At that point your reaction may be that this just seems like extra work for the purpose of giving literal meaning to the things physicists say, and why bother doing this?

The answer one is looking for is some clearly important, basic topic in QM which can be treated more easily with rigged Hilbert spaces. I don't know of such a thing, but that is likely because I don't know enough physics. Hopefully someone more knowledgeable can give a good example.

The thesis Quantum Mechanics in Rigged Hilbert Space Language by de la Madrid looks like a good resource. Similar questions on mathoverflow and the physics stack exchange are here and here.

  • $\begingroup$ Actually isnt the rigged Hilbert space depending on the Hamiltonian? I remember when reading Böhm (rigged HS in QM) that he somehow needed the Hamiltonian for defining the subspace of the Hilbert spacel, he used the harmonic oszillator. It went over my head, but for me it doesnt feel right to depend on the hamiltonian (also it is unclear how this construction can be done explicitely for non-integrabel system (at least to me)). $\endgroup$
    – lalala
    Commented Feb 15, 2021 at 15:43
  • $\begingroup$ @lalala Did you mean to comment on my answer? I'm not sure what you're referring to in what I said. $\endgroup$
    – Nik Weaver
    Commented Feb 15, 2021 at 18:18

@Nik Weaver: The answer one is looking for is some clearly important, basic topic in QM which can be treated more easily with rigged Hilbert spaces.

The foremost topic is the theory of resonances.

A resonance is a complex eigenvalue $E-i\Gamma$, with $\Gamma>0$, of a nonself-adjoint analytic continuation of the Hamiltonian. Resonances describe a decaying system such as a heavy nucleus, which will have quasi-bound states at energy $E$ with a life time $\hbar/\Gamma$. The corresponding generalized eigenfunctions (scattering states) live in a rigged Hilbert space, see Description of resonances within the rigged Hilbert space (2006).

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    $\begingroup$ Fantastic! That is just the sort of thing I wanted. I suppose ordinary spectral theory doesn't apply because the operator isn't self-adjoint. $\endgroup$
    – Nik Weaver
    Commented Feb 14, 2021 at 17:12
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    $\begingroup$ thank you; the non-self-adjointness may be an issue, but what I understand is that the eigenfunctions are plane-wave-like scattering states, not bound states, so their orthonormality involves a delta function. $\endgroup$ Commented Feb 14, 2021 at 18:57

You ask 'why spectral theory instead of rigged Hilbert spaces?'.

There are some practical/pedagogical reasons: One is that you'll need basic spectral theory in quantum mechanics anyways, so might as well get as much mileage as you can out of it. Another is that spectral theory can be largely explained by analogy with finite-dimensional linear algebra. Rigged Hilbert spaces require nuclear vector spaces, which while quite nice are not particularly intuitive. (One can cut corners and define them via nested Hilbert spaces, but this is also clearly more complicated than just having one Hilbert space.)

There are also sociological reasons: Much of the work on rigorous quantum physics follows von Neumann, and he used spectral theory. He wasn't wrong to do so; the most important thing to get right at the time was the shocking behavior of atomic spectra. Rigged Hilbert spaces came from Gel'fand and his collaborators thirty years later. By that point, much of the work in rigorous quantum physics was being done by people who'd already set everything up in terms of C*-algebras.

One might also blame Reed & Simon, whose influential and otherwise fabulous analysis book rather sassily only recommends "the abstract rigged approach to readers with an emotional attachment to the Dirac formalism".

Lastly, I'd argue that spectral theory and rigged Hilbert spaces aren't really different things. The theory of rigged Hilbert spaces is just a nice way of setting up spectral theory for unbounded operators. I prefer rigged Hilbert spaces for quantum physics, because they let you make explicit use of the operators you actually want to talk about, instead of inventing circumlocutions to only talk of bounded operators. But it must be acknowledged that the C*-algebra language has allowed a lot of progress to be made, and that the distinctions between these things are invisible to experiment.


The book by Leslie E. Ballentine called "Quantum mechanics - a modern development" explains why we need rigged Hilbert spaces, and proceeds to spend the first chapter building the mathematical machinery necessary.


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