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I'm trying to understand how bad things could possibly get without Cauchy completeness as a criterion for Hilbert spaces in quantum mechanics. Obviously, doing calculus on a pre-Hilbert space would be complicated but could there still be some reasonable version of the spectral theorem? Why or why not? Some elaboration and perhaps even some references would be appreciated.

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    $\begingroup$ One example to consider is the Laplacian $\Delta$ on the pre-Hilbert space $E = C_c^\infty(\Omega)$, where $\Omega$ is some nice domain (e.g. a ball). Then $\Delta$ is symmetric, everywhere defined, and negative definite, so you would hope any "reasonable" version of the spectral theorem would apply to it. Now you would also hope any "reasonable" version of the spectral theorem would let you define the semigroup $e^{t\Delta}$ on $E$, but this is impossible since solutions of the heat equation do not stay compactly supported. $\endgroup$
    – Nate Eldredge
    Commented Dec 28, 2019 at 17:18

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Here is a simple example that shows that the idea of spectral theory on pre-Hilbert spaces in the sense you are asking is hopeless. Consider the pre-Hilbert space consisting of the restrictions of all complex polynomials to $[0,1]$, as a dense subspace of $L^2[0,1]$. Then let $A$ be the operator of multiplication by $x$. The spectral projections of this operator are characteristic functions; none of them except $0$ and $1$ are polynomials.

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There is a very nice setup which is suitable for precise mathematical understanding of quantum mechanics including the “delta-function-like eigenvectors”: that of a rigged Hilbert space. It is a Hilbert space $H$ together with a fixed dense continuous inclusion of a locally convex (often assumed nuclear) topological vector space $\Phi\hookrightarrow H$. An example to think of is the inclusion of the Schwartz space $\mathcal S(\mathbb R^n)$ into $L^2(\mathbb R^n)$.

And indeed, there is a very satisfactory spectral theory of selfadjoint operators on rigged Hilbert spaces which gives, for instance, the precise meaning to the statement “the delta functions $\delta_x$, $x\in[0,1]$ form a complete system of generalised eigenvectors for the operator of multiplication by $x$ on $L^2([0,1],\mathrm{Leb})$”.

A systematic treatment can be found in the classical source

I. M. Gelfand and N. J. Vilenkin. Generalized Functions, vol. 4: Some Applications of Harmonic Analysis. Rigged Hilbert Spaces. Academic Press, New York, 1964.

Applications to some classical problems of quantum mechanics can be found in the Ph.D. thesis R. de la Madrid, Quantum Mechanics in Rigged Hilbert Space Language (2001).

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    $\begingroup$ Serious question, is there any substantive advantage to working with rigged Hilbert spaces other than being able to rigorously interpret the physicists' delta functions? My feeling has always been that the "right" way to make delta functions rigorous is via spectral theory, but that could just be ignorance on my part. $\endgroup$
    – Nik Weaver
    Commented Dec 28, 2019 at 23:43
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    $\begingroup$ @NikWeaver, Some aspects of self-adjoint extensions of (restrictions of) self-adjoint operators (e.g., Friedrichs extensions in the semi-bounded case) are nicely explained in terms of $H^1\to L^2 \to H^{\-1}$ and such, in my opinion. $\endgroup$
    – paul garrett
    Commented Dec 29, 2019 at 2:57
  • $\begingroup$ @paulgarrett: thank you, that's a good example. $\endgroup$
    – Nik Weaver
    Commented Dec 29, 2019 at 6:55
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    $\begingroup$ It should be mentioned that Amiel Feinstein, who translated that book into English discovered an error in Gelfand-Vilkenin's proof, involving the treatment of sets of measure zero, which was then corrected by the following paper of G. G. Gould: londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms/… $\endgroup$
    – Robert Furber
    Commented Dec 29, 2019 at 13:39
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    $\begingroup$ I'll also add another oft-unmentioned caveat. People coming from the physics literature often expect that "a complete system of eigenvectors" means that for each spectral value there must exist an eigenvector. This is not the case, as can be seen in the multiplication operator $\frac{1}{1+n}$ on $\ell^2$, which has a complete set of eigenvectors but for which there is no eigenvector, not even in the generalized sense, corresponding to the spectral value $0$. $\endgroup$
    – Robert Furber
    Commented Dec 29, 2019 at 14:05

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