Timeline for Essential self-adjointness of differential operators on compact manifolds
Current License: CC BY-SA 3.0
11 events
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| Dec 18, 2011 at 1:21 | comment | added | Daniel Tausk | Let me add another comment to set the record straight: if $L$ is $\frac{d}{dx}\sin(x)\frac{d}{dx}$ defined in $C^\infty(S^1)$ and $L'$ is one of the several self-adjoint extensions of $L$ then, while the unitary flow $e^{itL'}$ does not preserve $C^\infty(S^1)$ (otherwise $L$ would be essentially self-adjoint), the map $\psi(t)=e^{itL'}\psi(0)$ solves the associated PDE $\frac{d}{dt}\psi=iL\psi$ in a perfectly reasonable sense, because it solves $\frac{d}{dt}\psi=iL'\psi$ and $L'$ is just the restriction to some subspace of $L^2(S^1)$ of the extension of $L$ to distributions. | |
| Dec 17, 2011 at 14:05 | vote | accept | Daniel Tausk | ||
| Dec 17, 2011 at 14:05 | comment | added | Daniel Tausk | Double checked. $L=\frac{d}{dx}\sin(x)\frac{d}{dx}$ is not essentially self-adjoint in $C^\infty(S^1)$. Thanks. | |
| Dec 16, 2011 at 9:17 | comment | added | Daniel Tausk | I think that in the Coulomb potential example you have an open set (not a zero measure set) of initial conditions (electrons that start too slow) that end up in the singularity in finite time. In any case, it seems that you are right about the original issue. $L=\frac{d}{dx}\sin(x)\frac{d}{dx}$ is not essentially self-adjoint in $C^\infty(S^1)$. Boring computations with Fourier series yield a non zero solution of $(L^*+i)\psi=0$. So you solved the problem. But I should recheck my computations before I assert that more confidently. | |
| Dec 15, 2011 at 21:29 | comment | added | Terry Tao | True, but in that case only a zero measure set of trajectories blow up. Looking at the energy curves for $\sin(x) \xi^2$ it seems to me that almost all trajectories will blow up either forwards or backwards in time, so I would still put my money on quantum incompleteness for this system (or some variant thereof, for instance, one could play with raising the power of $\sin(x)$ here). A model case to play with might be the operators $\frac{d}{dx} x^m \frac{d}{dx}$ on the real line; of course this isn't a compact manifold, but it might still be informative. | |
| Dec 15, 2011 at 20:22 | comment | added | Daniel Tausk | Right. Thanks. Though, thinking a bit more about this, there are very simple examples in which classical trajectories die in finite time, while the quantum problem is complete. Say, one electron in Coulomb potential: several classical trajectories die in finite time with the electron falling into the singularity (say, if the electron starts at rest), while the corresponding quantum Hamiltonian is essentially self-adjoint in $C^{\infty}_c(\mathbb{R}^3)$. | |
| Dec 15, 2011 at 19:41 | comment | added | Terry Tao | It's true that there will be an abstract self-adjoint extension of real differential operators such as $\frac{d}{dx} \sin(x) \frac{d}{dx}$ (because the deficiency indices must match), but the flow maps $e^{itL}$ associated to such extensions have no reason to preserve the initial domain of the operator, and thus need not solve the associated PDE in any reasonable sense (e.g. distributional sense). As such I would consider these abstract propagators to be "unphysical". | |
| Dec 15, 2011 at 19:33 | comment | added | Terry Tao | Yes, this is what I mean by classical vs. quantum. Intuition from both physics and microlocal analysis (e.g. Egorov's theorem on pseudodifferential operators) suggests that $e^{itL}$ should approximately propagate in phase space along the Hamiltonian flow of the symbol (i.e. the quantum flow approximates the classical flow). If the classical flow is incomplete, this suggests (though does not quite prove) that one has quantum incompleteness also. | |
| Dec 15, 2011 at 18:53 | comment | added | Daniel Tausk | By the way, if you take $L=\frac{d}{dx}\sin(x)\frac{d}{dx}$ (which is the simplest example of a symmetric operator with principal part $\sin(x)\frac{d^2}{dx^2}$) then $L$ surely admits one self-adjoint extension (because its coefficients are real functions) and such a self-adjoint extension does generate a unitary propagator $e^{itL}$. I don't know, though, whether $L$ is essentially self-adjoint. Since $L$ is quite simple I maybe able to figure that out "by hand". | |
| Dec 15, 2011 at 18:45 | comment | added | Daniel Tausk | Just to make sure I understand what you are talking about (and maybe I don't), when you say "corresponding classical problem" you mean "classical" as in "classical versus quantum", with a quantization rule like replacing $\xi$ with $-i\frac{d}{dx}$ in the classical Hamiltonian? If that's the case, is there a mathematical reason for believing that the two problems (the "classical" and the "quantum") are related or are you just relying on insights coming from physics? | |
| Dec 15, 2011 at 6:00 | history | answered | Terry Tao | CC BY-SA 3.0 |