bio | website | ime.usp.br/~tausk |
---|---|---|
location | São Paulo - Brazil | |
age | 39 | |
visits | member for | 4 years, 2 months |
seen | May 1 '12 at 1:18 | |
stats | profile views | 561 |
Associate Professor, Mathematics Department, University of São Paulo, Brazil
Sep 10 |
awarded | Popular Question |
Dec 22 |
awarded | Good Answer |
Dec 19 |
awarded | Nice Question |
Dec 18 |
comment |
Essential self-adjointness of differential operators on compact manifolds
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 |
awarded | Supporter |
Dec 17 |
revised |
Essential self-adjointness of differential operators on compact manifolds
added 705 characters in body |
Dec 17 |
awarded | Scholar |
Dec 17 |
accepted | Essential self-adjointness of differential operators on compact manifolds |
Dec 17 |
comment |
Essential self-adjointness of differential operators on compact manifolds
Double checked. $L=\frac{d}{dx}\sin(x)\frac{d}{dx}$ is not essentially self-adjoint in $C^\infty(S^1)$. Thanks. |
Dec 16 |
awarded | Commentator |
Dec 16 |
comment |
Essential self-adjointness of differential operators on compact manifolds
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 |
comment |
Essential self-adjointness of differential operators on compact manifolds
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 |
comment |
Essential self-adjointness of differential operators on compact manifolds
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 |
comment |
Essential self-adjointness of differential operators on compact manifolds
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? |
Nov 24 |
answered | Analogies between orthogonal/unitary groups and their indefinite counterparts |
Nov 24 |
comment |
Essential self-adjointness of differential operators on compact manifolds
Correcting small imprecision of my previous comment: if $L$ is first-order (symmetric) then $L=-i\big(X+\frac12\mathrm{div}(X)\big)+V$, with $V$ the multiplication operator by a smooth real-valued function. Since $V$ is bounded self-adjoint, the conclusion is the same... |
Nov 23 |
comment |
Essential self-adjointness of differential operators on compact manifolds
($F_t$ denotes the flow of $X$). |
Nov 23 |
comment |
Essential self-adjointness of differential operators on compact manifolds
By the way, the answer is also affirmative if $L$ is first-order. In that case, $L=-i\big(X+\frac12\mathrm{div}(X)\big)$, with $X$ a smooth vector field in $M$. Since $M$ is compact, $X$ is complete and we obtain $L$ as the generator of a one-parameter unitary group $U_t(f)=(f\circ F_t)\sqrt{\det\mathrm{d}F_t}$, which leaves $C^\infty(M)$ invariant. Thus $L$ is essentially self-adjoint on $C^\infty(M)$. |
Nov 23 |
revised |
Essential self-adjointness of differential operators on compact manifolds
edited body |
Nov 23 |
awarded | Student |