## Is there a spectral theory approach to non-explicit Plancherel-type theorems?

Teaching graduate analysis has inspired me to think about the completeness theorem for Fourier series and the more difficult Plancherel theorem for the Fourier transform on $\mathbb{R}$. There are several ways to prove that the Fourier basis is complete for $L^2(S^1)$. The approach that I find the most interesting, because it uses general tools with more general consequences, is to use apply the spectral theorem to the Laplace operator on a circle. It is not difficult to show that the Laplace operator is a self-adjoint affiliated operator, i.e., the healthy type of unbounded operator for which the spectral theorem applies. It's easy to explicitly solve for the point eigenstates of the Laplace operator. Then you can use a Fredholm argument, or ultimately the Arzela-Ascoli theorem, to show that the Laplace operator is reciprocal to a compact operator, and therefore has no continuous spectrum. The argument is to integrate by parts. Suppose that $$\langle -\Delta \psi, \psi\rangle = \langle \vec{\nabla} \psi, \vec{\nabla \psi} \rangle \le E$$ for some energy $E$, whether or not $\psi$ is an eigenstate and even whether or not it has unit norm. Then $\psi$ is microscopically controlled and there is only a compact space of such $\psi$ except for adding a constant. The payoff of this abstract proof is the harmonic completeness theorem for the Laplace operator on any compact manifold $M$ with or without boundary. It also works when $\psi$ is a section of a vector bundle with a connection.

My question is whether there is a nice generalization of this approach to obtain a structure theorem for the Laplace operator, or the Schrödinger equation, in non-compact cases. Suppose that $M$ is an infinite complete Riemannian manifold with some kind of controlled geometry. For instance, say that $M$ is quasiisometric to $\mathbb{R}^n$ and has pinched curvature. (Or say that $M$ is amenable and has pinched curvature.) Maybe we also have the Laplace operator plus some sort of controlled potential --- say a smooth, bounded potential with bounded derivatives. Then can you say that the spectrum of the Laplace or Schrödinger operator is completely described by controlled solutions to the PDE, which can be interpreted as "almost normalizable" states?

There is one case of this that is important but too straightforward. If $M$ is the universal cover of a torus $T$, and if its optional potential is likewise periodic, then you can use "Bloch's theorem". In other words you can solve the problem for flat line bundles on $T$, where you always just have a point spectrum, and then lift this to a mixed continuous and point spectrum upstairs. So you can derive the existence of a fancy spectrum that is not really explicit, but the non-compactness is handled using an explicit method. I think that this method yields a cute proof of the Plancherel theorem for $\mathbb{R}$ (and $\mathbb{R}^n$ of course): Parseval's theorem as described above gives you Fourier completeness for both $S^1$ and $\mathbb{Z}$, and you can splice them together using the Bloch picture to get completeness for $\mathbb{R}$.

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 Only a simple remark. In the non-compact case, the paradigmatic example is the harmonic oscillator $$-\Delta_{\mathbb R^d}+\frac{\vert x\vert^2}{4}$$ with spectrum $\frac{d}{2}+\mathbb N$. The eigenvectors are the Hermite functions with an explicit expression from the so-called Maxwellian $\psi_0=(2\pi)^{-d/4}\exp{-\frac{\vert x\vert^2}{4}}$ and the creation operators $(\alpha!)^{-1/2}(\frac{x}{2}-\frac{d}{dx})^\alpha \psi_0$. In one dimension the operator $-\frac{d^2}{dx^2}+x^4$ (quartic oscillator) has also a compact resolvent, but nothing explicit is known about the eigenfunctions. – Bazin May 2 2012 at 13:53 More subtle is the compactness of the resolvent of the 2D $$-\Delta_{\mathbb R^2}+x^2y^2.$$ – Bazin May 2 2012 at 13:54 I just saw this playing around on meta.... Are you asking a question beyond that spectrally almost every solution is polynomially bounded? – Helge Aug 15 at 18:53 @Helge - That's part of the story, but in the ordinary Plancherel theorem, not the hardest part to state or prove. You would also want some statement about the spectral measure (that is, the projection-valued measure produced by the spectral theorem) associated to the Laplace or Schrodinger operator. Again, if you have a Laplace operator on a closed manifold, there is an algorithm to diagonalize it completely. The completeness theorem is considered very important, and not just the fact that you can find eigenfunctions. – Greg Kuperberg Aug 18 at 3:16