Timeline for Non-degeneracy of ground state in quantum mechanics
Current License: CC BY-SA 3.0
16 events
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| Jun 13, 2021 at 14:52 | comment | added | Xuda Ye | The proof by Glimm and Jaffe is nice. The idea is to prove that the operator $A = e^{-t H}$ has a positive kernel, that is, $A$ acting on a positive wavefunction always produces a positive one. | |
| Jun 2, 2017 at 5:57 | comment | added | lcv | @fff123123 No it's not :) Check somewhere the definition of spectrum. I guess Wikipedia is a good starting point | |
| Jun 2, 2017 at 4:32 | comment | added | fff123123 | @lcv But $e^{ikx}$ is also not square integrable. We still view it as eigenstate. | |
| Jun 2, 2017 at 4:30 | comment | added | lcv | @fff123123 Me neither. Anyway, without further specifications in this context one would assume that the Hilbert space is $L^2(\mathbb{R})$ the space of square summable functions on the line. A constant function is not in this space unless the constant is zero. Hence your function is not in the (Hilbert) space and $0$ is not an eigenvalue. However $0$ belongs to the spectrum of $-\Delta$. | |
| Jun 1, 2017 at 21:11 | comment | added | fff123123 | @lcv Sorry, I'm not a mathematician. May I ask a naive question? For operator $\Delta$, is $f(x)= constant$ an eigenstate of $\Delta$ with eigenvalue $0$? | |
| Jun 1, 2017 at 21:06 | comment | added | lcv | @Christian Remling it seems I was typing at the same time.. | |
| Jun 1, 2017 at 21:02 | comment | added | lcv | Generally with the term ground state one implicitly requires it to be an eigenvector (i.e. a state). You must enforce a condition such that the infimum of the spectrum is indeed an eigenvalue. A part from this yes, Perron-Feibenius kind of argument apply quite generally | |
| Jun 1, 2017 at 20:59 | comment | added | Christian Remling | Also, by a "ground state" one usually means an eigenfunction with eigenvalue equal to the minimum of the spectrum. In general, there is of course no reason why the bottom of the spectrum should be an eigenvalue, and $V=0$ is a simple example where it isn't. The natural way to understand the question seems to be to assume that $\min\sigma$ indeed is an eigenvalue (which will for example follow from the assumption that $V\to\infty$ from one of the other answers). | |
| Jun 1, 2017 at 20:57 | comment | added | fff123123 | @ChristianRemling Sorry, a typo. Thanks | |
| Jun 1, 2017 at 20:57 | history | edited | fff123123 | CC BY-SA 3.0 |
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| Jun 1, 2017 at 20:29 | history | edited | fff123123 | CC BY-SA 3.0 |
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| Jun 1, 2017 at 20:14 | history | edited | fff123123 | CC BY-SA 3.0 |
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| Jun 1, 2017 at 20:11 | comment | added | fff123123 | @lcv Certainly, we always assumed the potential bounded from below. For detailed requirement and proof, you can consult the book I cite. | |
| Jun 1, 2017 at 20:02 | comment | added | fff123123 | @lcv Even the $V(x)$ is constant, the ground state is still unique. Because the eigenstate of $-\Delta^2$ is $e^{i k x}$ for $k\neq 0$ there is double degenerate. For $k=0$, it's unique up to a phase. | |
| Jun 1, 2017 at 8:49 | comment | added | lcv | What about $V(x)$ constant? | |
| May 31, 2017 at 21:31 | history | answered | fff123123 | CC BY-SA 3.0 |