Timeline for Is there a spectral theory approach to non-explicit Plancherel-type theorems?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Aug 9, 2013 at 23:38 | answer | added | paul garrett | timeline score: 2 | |
| Aug 9, 2013 at 21:21 | answer | added | user80744 | timeline score: 1 | |
| Aug 18, 2012 at 3:16 | comment | added | Greg Kuperberg | @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. | |
| Aug 15, 2012 at 18:53 | comment | added | Helge | I just saw this playing around on meta.... Are you asking a question beyond that spectrally almost every solution is polynomially bounded? | |
| May 2, 2012 at 13:54 | comment | added | Bazin | More subtle is the compactness of the resolvent of the 2D $$ -\Delta_{\mathbb R^2}+x^2y^2. $$ | |
| May 2, 2012 at 13:53 | comment | added | Bazin | 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. | |
| Nov 10, 2011 at 19:53 | history | asked | Greg Kuperberg | CC BY-SA 3.0 |