Timeline for Common eigenvector of commuting unbounded operators
Current License: CC BY-SA 4.0
23 events
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| Aug 23 at 19:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 25 at 18:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 27, 2023 at 17:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 20, 2023 at 8:43 | history | edited | YCor | CC BY-SA 4.0 |
fixed English
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| Nov 20, 2023 at 8:31 | answer | added | Apoorv Potnis | timeline score: 0 | |
| Aug 24, 2018 at 14:08 | comment | added | Christian Remling | @MatthewDaws: I don't think there's any ambiguity in the terminology. Of course, as you point out, "pure point spectrum" and "purely discrete spectrum" are two different things, but it doesn't seem plausible to assume that the OP must be meaning the latter when he said the former. | |
| Aug 24, 2018 at 6:06 | comment | added | Matthew Daws | @ChristianRemling: Ah. I think this comes down to what exactly we mean by "pure point spectrum". I am following Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space and assuming (maybe wrongly) that what the OP means is the same as "purely discrete spectrum" (meaning $\sigma(T)$ consists only of eigenvalues of finite multiplicity with no finite accumulation point). Proposition 5.12 characterises such operators. | |
| Aug 23, 2018 at 21:41 | comment | added | Tim S | Mhm, okay. I know that theorem you are talking about, thanks for the help. Hope i will get some results out of this | |
| Aug 23, 2018 at 20:15 | comment | added | Christian Remling | You can not conclude that $\phi\in D(S)$ from the fact that $\phi$ is an eigenvector of $T$; for example, if $T=1$, then this assumption says literally nothing on $\phi$. (Similarly, even if you knew that $\phi\in D(S)$, I'm not sure if your calculation is getting you anywhere. You could conclude that $S\phi\in N(T-\lambda)$, but what next?) I think a more abstract approach would work better; for example, both operators can be made multiplication operators by the same unitary transformation. | |
| Aug 23, 2018 at 20:07 | comment | added | Christian Remling | @MatthewDaws: You certainly can't conclude that $|\lambda_n|\to\infty$. For example, the set of eigenvalues could be $\mathbb Q$. | |
| Aug 22, 2018 at 15:56 | comment | added | Tim S | Okay. Then i will try solving this, on that way. Thanks! | |
| Aug 22, 2018 at 15:48 | comment | added | Matthew Daws | Okay. Similarly $e^{itT}\phi = \sum_i e^{it\lambda_i} (\phi|\phi_i) \phi_i$. If this commutes with $e^{itS}$, which has a similar form, for all $t$, then you can perform calculations with bounded diagonal operators which will lead to the result you want. (Also $\lim_{n\rightarrow\infty} |\lambda_n| = \infty$). | |
| Aug 22, 2018 at 15:45 | comment | added | Tim S | So $E(\{\lambda \})$ is the projection on the Eigenspace to $\lambda$. And we can write $T\phi=\sum_i \lambda_i\langle\phi,\phi_i \rangle \phi_i$ where $\{\phi_i\}_i$ is an orthonormal basis composed of eigenvector to eigenvalues $\lambda_i$. | |
| Aug 22, 2018 at 15:37 | comment | added | Matthew Daws | Okay, so you are happy with resolutions of the identity. If $T$ has pure point spectrum, what special form does $E$ take, and hence what special form does $T$ take? | |
| Aug 22, 2018 at 15:26 | comment | added | Tim S | With the general Spectral Theorem: $$ e^{itT}=\int_{\sigma(T)} e^{it\lambda} dE(\lambda)$$ | |
| Aug 22, 2018 at 14:59 | comment | added | Matthew Daws | So another question: how do you define $e^{itT}$? | |
| Aug 22, 2018 at 12:11 | comment | added | Tim S | The spectral Measure of these Operators commute, or $[e^{itT},e^{isS}]=0$ for all $t,s\in \mathbb{R}$, or $TS=ST$ including the equality of the domains. As far as I know, they are equivalent. | |
| Aug 22, 2018 at 11:55 | comment | added | Matthew Daws | What is your definition of when unbounded (self-adjoint) operators commute? | |
| Aug 22, 2018 at 9:35 | comment | added | Tim S | The Domain of the Operator $S$ | |
| Aug 22, 2018 at 9:33 | history | edited | Tim S | CC BY-SA 4.0 |
added 8 characters in body
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| Aug 22, 2018 at 9:29 | comment | added | Yossi Lonke | What is $D(S)$? | |
| Aug 22, 2018 at 9:10 | review | First posts | |||
| Aug 22, 2018 at 10:17 | |||||
| Aug 22, 2018 at 9:07 | history | asked | Tim S | CC BY-SA 4.0 |