Timeline for Eigenvalues and spectrum of the adjoint
Current License: CC BY-SA 4.0
27 events
| when toggle format | what | by | license | comment | |
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| Jan 3, 2020 at 12:18 | comment | added | Arnold Neumaier | @lcv: Yes, I corrected my statement after having reading your comment! Thanks for pointing this out. | |
| Jan 2, 2020 at 10:24 | comment | added | lcv | @RobertFurber I was talking to the OP. In the original post the spectrum was defined as the set of points for which $x - A$ is not invertible. | |
| Jan 2, 2020 at 8:34 | comment | added | Robert Furber | @lcv To whom are you addressing your comment? Use the @ symbol before the user name. The definition of spectrum for an unbounded is given in the Wikipedia article that András Bátkai linked to, and doesn't make any difference to the discussion (the spectrum of $T$ is still the same, and I already had to be using the unbounded definition for $a^*$ because it's an unbounded operator). | |
| Jan 1, 2020 at 14:46 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
corrected definition of the spectrum
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| Jan 1, 2020 at 14:27 | comment | added | lcv | Note also that your definition of spectrum is the one for (works only for) bounded operators. Check out the general one for the unbounded case. | |
| Jan 1, 2020 at 14:12 | comment | added | Robert Furber | @ArnoldNeumaier I don't know. | |
| Jan 1, 2020 at 13:03 | comment | added | Arnold Neumaier | @RobertFurber: Do generalized eigenvectors always exist in the interior of the continuous spectrum of a self-adjoint operator? | |
| Jan 1, 2020 at 13:02 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
corrected statement about generalized eigenvectors
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| Jan 1, 2020 at 13:01 | comment | added | Robert Furber | @ArnoldNeumaier Yes. | |
| Jan 1, 2020 at 12:07 | comment | added | Arnold Neumaier | @RobertFurber: So the continuous spectrum of $T$ is the discrete set $\{0\}$? | |
| Jan 1, 2020 at 1:19 | comment | added | Robert Furber | @ArnoldNeumaier The spectrum of the creation operator $a^*$ is all "residual spectrum", i.e. the kernel of $a^* - \lambda$ is $\{0\}$ for all $\lambda \in \mathbb{C}$, but the range of $a^* - \lambda$ is not dense, essentially because the bra $\sum_{n=0}^\infty \frac{\lambda^n}{\sqrt{n!}} \langle n |$ vanishes on the image of $a^{*}$, so it doesn't have closed range. As a result, I think it would be difficult to find any kind of eigenvector for $\lambda \in \mathbb{C}$ in this case, even if you allow "eigenvectors" outside the Hilbert space. | |
| Dec 30, 2019 at 20:49 | comment | added | Robert Furber | @ArnoldNeumaier But there is no corresponding generalized eigenvector (when $T$ is considered as an operator on the nuclear space $s$ of sequences that rapidly converge to $0$) | |
| Dec 30, 2019 at 20:49 | comment | added | Robert Furber | @ArnoldNeumaier There must be something in the air, because I was just mentioning the fact that that isn't true elsewhere. Consider the operator $T|n\rangle = \frac{1}{n+1}|n\rangle$. It is self-adjoint, its eigenvalues are $\frac{1}{n+1}$ for each $n = 0,1,2,\ldots$, but additionally it is not invertible, so $0$ is a spectral value (and part of the continuous spectrum). | |
| Dec 30, 2019 at 14:07 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
made language more precise
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| Dec 30, 2019 at 13:46 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
corrected misprint
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| Dec 30, 2019 at 11:27 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
made the question more precise in the light of the comments so far
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| Dec 30, 2019 at 11:08 | comment | added | Jochen Glueck | @ArnoldNeumaier: Unfortunately, I don't know. | |
| Dec 30, 2019 at 9:17 | comment | added | Arnold Neumaier | @JochenGlueck: For self-adjoint operators, for each point in the continuous spectrum, there are unnormalizable eigenstates in the dual of a dense nuclear subspace. Does this generalize to general closed operators? What would be these generalized eigenstates in the case of $a^*$? | |
| Dec 29, 2019 at 21:07 | history | edited | YCor |
edited tags
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| Dec 29, 2019 at 19:41 | comment | added | Jochen Glueck | Thank you for the clarification! To expand on @MichaelRenardy's comment: It is true that, in your example, $a^*$ has no eigenvalues, so its eigenspectrum (i.e. the set of its eigenvalues; many people also call it the point spectrum) is empty. However, the spectrum of $a^*$ (which is, in general, a larger set than the point spectrum), is not empty but coincides with the complex conjugate of the spectrum of $a$, i.e. with the complex plane. | |
| Dec 29, 2019 at 18:05 | comment | added | András Bátkai | @ArnoldNeumaier: see here: en.wikipedia.org/wiki/… | |
| Dec 29, 2019 at 16:15 | comment | added | Arnold Neumaier | @MichaelRenardy: What is their difference, and how are they related to each other? | |
| Dec 29, 2019 at 15:51 | comment | added | Michael Renardy | I suspect that you are confusing the spectrum with the eigenspectrum. Indeed, eigenvalues of an operator need not be eigenvalues of the adjoint. | |
| Dec 29, 2019 at 15:26 | comment | added | Arnold Neumaier | @JochenGlueck: I defined the operators and gave a link. | |
| Dec 29, 2019 at 15:25 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
added definitions of the operators in my example
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| Dec 29, 2019 at 14:32 | comment | added | Jochen Glueck | For a closed densely defined linear operator $A$ on a Hilbert space, the spectrum of $A^*$ is in fact always the complex conjugate of the spectrum of $A$, just as in the finite-dimensional case (see for instance Theorem III.6.22 in "T. Kato: Perturbation theory for linear operators" (1980)). To better understand your example, a reference or a precise definition of $a$ would probably be helpful (since the notions "annihilation operator" and "creation operator" seem to be used in various contexts in quantum physics). | |
| Dec 29, 2019 at 10:56 | history | asked | Arnold Neumaier | CC BY-SA 4.0 |