In quantum mechanics, people introduce the notion of "continuous basis" (I actually don't know the mathematical denomination of it). It is not a Schauder basis. I would like to know what could be a good definition of it, and what are the possible difficulties of defining it.

We have a Hilbert space $\mathcal{H}$, and a family of "vectors", $ \left\lbrace|x \rangle \right\rbrace_{x\in \mathbb{R}} $ such that any vector $|\psi \rangle $ in $\mathcal{H}$ can be written

" $|\psi \rangle = \int_{\mathbb{R}} \psi(x)|x \rangle $ "


$\int_{\mathbb{R}} |x \rangle \langle x | dx = Id$

  • We don't say where those $|x \rangle $ live, and in general they are not in $\mathcal{H}$. Usual example from physics $\mathcal{H}=L^2(\mathbb{R})$, and where we think of $|x \rangle$ as the delta distribution.

  • It looks a little bit like the spectral theorem that allows to write self adjoint operators as integral with a projector valued measure on the spectrum. I also came across that word, "direct integral" that may have a link.

  • we usually also take "orthonormal basis", i.e $\langle x |y \rangle = \delta(x-y)$ (what is the meaning of seing it as a distribution in 2 variables)

Of course the decomposition of a vector has to be unique. In the example $\mathcal{H}=L^2(\mathbb{R})$ the coefficients is itself a function $\psi\in L^2(\mathbb{R})$, then unique in the sense "equal almost everywhere"

Among all the questions such presentation raises:

  • is the hilbert space structure playing any role? example in defining $\langle x |$ as the "dual basis". Because since those vectors are not even in the hilbert space, their scalar product is not defined.
  • is there any difficulty to define an abstract vector space generated by the family $ \left\lbrace|x \rangle \right\rbrace_{x\in \mathbb{R}} $. (Then of course there is still the problem of why a vector from our starting hibert space is equal to some construction in another abstract space.)
  • If this whole business is actually well defined, we can see fourier transform as a particular case of writing a vector in the following generalized basis (also from physics)

$|p\rangle:= \left\lbrace x\rightarrow \frac{1}{\sqrt{2\pi}} e^{ipx}\right\rbrace$

and the theorems of Fourier transform being an isometry and being invertible would just say that $|p\rangle $ is an "orthonormale" basis.

  • 1
    $\begingroup$ “We don’t say where those $|x\rangle$ live—who is “we”? Theoretical physicists don’t—mathematical physicists do. ;) $\endgroup$
    – The User
    Jun 8, 2013 at 22:23
  • $\begingroup$ One good reference for rigged Hilbert spaces, from a physics point of view but relatively rigorous, is Quantum Mechanics I by Galindo and Pascual. $\endgroup$ Jul 22, 2013 at 15:15
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    $\begingroup$ My question here is related to your question: math.stackexchange.com/q/3171921/71829 $\endgroup$ Apr 7, 2019 at 13:27

2 Answers 2


You may want to have a look at Gelfand triples. In the example of the impulse operator and the Fourier transform you consider the triple consisting of the Schwartz space, the Hilbert space $L^2(\mathbb{R}^n)$ and the dual of the Schwartz space. The momentum operator is defined on the Schwartz space. The Fourier transform is well-defined on this triple: All the three spaces get mapped to themselves. The waves $e^{ipx}$ are elements of the dual of the Schwartz space and are a complete set of generalised eigen values of the momentum operator.

Let me sketch the general situation: You have a Hilbert space $H$ and a nuclear topological vector space $S$ linearly embedded into the Hilbert space such that the scalar product is continuous with respect to the topology of $S$. We consider self-adjoint operators $T$ which are defined on $S$. Then generalised eigenvectors as elements of the dual space of $S$ can be defined in the obvious way. Let $\sigma$ be the spectrum of $T$. We can decompose the Hilbert space as a direct integral using a measure $\mu$

$H=\int^\oplus H(\lambda)\mathrm{d}\mu(\lambda)$

such that $T$ acts as a multiplication by $\lambda$ on each space $H(\lambda)$. There exists a unitary operator $U$ mapping $H$ to some space $L^2(X)$ where $T$ acts as a multiplication by a $\sigma$-valued function $a$. It can be proven that for each $x$ in $X$ there exists a $\phi_x$ in the dual space of $S$ such that for every $f\in S$ the functions $Uf\colon X\to\mathbb{C}$ and $x\mapsto\phi_x(f)$ are equal almost everywhere. The functional $\phi_x$ is a generalised eigenvector corresponding to the eigenvalue $a(x)$, that means, given a test function $f\in S$:


The space $H(\lambda)$ corresponds, informally spoken, to a space of functions defined on $a^{-1}(\left\{\lambda\right\})$. Thus, by taking linear combinatons, for every element $x\in H(\lambda)$ there is a generalised eigenvector $\phi$ for the eigenvalue $\lambda$. If you choose an orthonormal basis of each space $H(\lambda)$ the corresponding generalised eigenvectors are “complete” in a sense related to the direct integral decomposition.

In the book Introduction to Axiomatic Quantum Field Theory by Bogolubov, Logunov and Todorov you can read a short description, but they do not prove it. They refer to Generalized Functions, Vol. 4: Applications of Harmonic Analysis by Gelfand and Vilenkin.


I believe the situation you asked about is addressed by the notion of "rigged Hilbert space"; see for example http://en.wikipedia.org/wiki/Rigged_Hilbert_space . (If I remember correctly, I came across this notion in the book of Bogolyubov and Shirkov on quantum field theory.)

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    $\begingroup$ I agree with this answer but I'd like to add that the standard way to deal with this issue is to appeal to spectral theory rather than trying to interpret the notation $|x\rangle$ literally as a vector in some auxiliary space. Don't interpret everything physicists say literally ... that way madness lies. $\endgroup$
    – Nik Weaver
    Jun 8, 2013 at 19:39
  • $\begingroup$ @Nik Understanding the spectral theorem without references to notions of generalised eigenvectors is certainly more fundamental (from a mathematical point of view). But that it is possible to get these eigenvectors as distributions is in fact non-trivial and an addition insight. (Similar point: You do not only study the Fourier transform on $L^1$ and $L^2$, but studying it on continuous functions, the Schwartz space or distributions is insightful, too) $\endgroup$
    – The User
    Jun 8, 2013 at 19:51

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