Timeline for Spectral theorem for self-adjoint differential operator on Hilbert space
Current License: CC BY-SA 3.0
19 events
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| Nov 8, 2017 at 18:50 | comment | added | Cameron Zwarich | Unfortunately, the actual proof of the generalized eigenvector expansion given in the 4th volume of Generalized Functions is completely wrong. See Gould's The Spectral Representation of Normal Operators on a Rigged Hilbert Space for a correct proof. While I like the idea of rigged Hilbert spaces myself, the tricky details around the existence of liftings mean that it's easy to make stupid mistakes, apparently even for the originators of the theory. | |
| Nov 8, 2017 at 13:58 | history | edited | Jonathan Gleason | CC BY-SA 3.0 |
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| May 29, 2015 at 19:14 | comment | added | Johannes Hahn | @JonathanGleason Not only can this space be zero. Even if it isn't zero it might still fail to be dense. Also: unbounded operators do not form algebras in general because of the need to keep track of the various domains. If you define the domain of $A+B$ to be $dom(A)\cap dom(B)$ (which is the most obvious choice) then even addition isn't assoziative and doesn't admit negatives. | |
| May 28, 2015 at 13:53 | comment | added | Jonathan Gleason | Question: Let us generalize this definition to the case where a single operator $A$ is replaced by an algebra $\mathcal{A}$: $\Phi :=\left\{ f\in L^2(\mathbb{R}):A(f)\in L^2(\mathbb{R})\text{ for all }A\in \mathcal{A}\right\}$. For $\mathcal{A}$ the algebra generated by the usual position and momentum operators, does this definition give us the Schwartz space? | |
| May 28, 2015 at 13:48 | comment | added | Jonathan Gleason | Suppose we have a partially-defined (un-boudned) operator $A:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ and define $\Phi :=\left\{ f\in L^2(\mathbb{R}):A^n(f)\in L^2(\mathbb{R})\text{ for all }n\in \mathbb{N}\right\}$. Unless I am being incredibly dense, $A$ should restrict to an operator $A|_{\Phi}:\Phi \rightarrow \Phi$. This should always work, though in bad cases we will probably have $\Phi =0$. | |
| May 28, 2015 at 13:37 | comment | added | Johannes Hahn | @JonathanGleason What about differential operators with non-polynomial coefficients? I guess it is possible to find an appropriate space $\Phi$ in many cases, but I don't think there is a general theorem to this end. | |
| May 28, 2015 at 12:38 | comment | added | Jonathan Gleason | I have to admit, I think the Hilbert space itself is a bit superfluous and is only mentioned explicitly in the definition for historical reasons. I usually think of rigged spaces as topological vectors spaces $\Phi$ equipped with an embedding into their conjugate dual $\Phi \hookrightarrow \Phi ^*$. This then gives a sesqui-linear form on $\Phi$, which we can complete to get a Hilbert space if we really want it, but I try to avoid mentioning the Hilbert space directly if I can. From this perspective, considering examples which explicitly mention the Hilbert space feels a bit weird to me. | |
| May 28, 2015 at 12:37 | comment | added | Jonathan Gleason | @JohannesHahn Can you give an example of such an operator? I have to admit, in all the examples I was interested in (all differential operators), you could choose always $\Phi$ so that we had $A:\Phi \rightarrow \Phi$. | |
| May 27, 2015 at 22:33 | comment | added | paul garrett | @JohannesHahn, for "semi-bounded" operators, at least, the somewhat-richer context you observe for (let's say elliptic) differential operators can be set up. So something a bit more than "Gelfand triples" can indeed be formalized... but/and I think in the same spirit. | |
| May 27, 2015 at 19:06 | comment | added | Johannes Hahn | As nice as this theorem might be, there is a catch, isn't there? The operator A has to be $\Phi\to\Phi$ and it does not apply to a general selfadjoint operator $\Phi\to H$, right? For differential operators this might not be as much of a problem since we know that the Schwartz space exists, but what about less familiar operators? | |
| Apr 7, 2013 at 16:42 | comment | added | Mateusz Wasilewski | Jonathan, thank you for clearing it up. I think it is best to always have in mind many equivalent forms of spectral theorem and I agree with you that this one is extremely valuable, because this is how physicists usually think of spectral decomposition. | |
| Apr 7, 2013 at 15:34 | comment | added | Jonathan Gleason | Having explained this in more detail, I'm thinking that "unnatural" isn't the best word for what I meant, because I would likewise argue that a basis-independent formulation is always going to be more natural than a basis-dependent one. In any case, hopefully it is more clear what I meant. | |
| Apr 7, 2013 at 15:31 | comment | added | Jonathan Gleason | @Mateusz Wasilewski I mean that it is unnatural in the sense that the project-valued measured formulation of the theorem is not presented in a way that is actually used in practice (by physicists). Most physicists probably don't even know what a measure is, much less a projection-valued measure, on the other hand, they all know what an eigenvector and eigenvalue is. To be fair, they likewise probably also don't know what the term "generalized" here means, but you could just pass that off as technical machinery required to make what's going on under the hood work; they idea is still the same. | |
| Apr 7, 2013 at 15:26 | comment | added | Jonathan Gleason | @Yemon Choi I've added a verbatim copy of their definition of generalized eigenvalues and eigenvectors. | |
| Apr 7, 2013 at 15:24 | history | edited | Jonathan Gleason | CC BY-SA 3.0 |
added 406 characters in body
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| Apr 7, 2013 at 9:20 | comment | added | Mateusz Wasilewski | I want to add that I do not agree that formulation of spectral theorem with projection-valued measures is unnatural: first, it is more geometric (does not refer to any basis; that is what I like about it even in finite-dimensional case) and it is quite well motivated that projections should correspond to (elementary) observables in quantum mechanics and spectral theorem says then that all self-adjoint operators ought to do so. | |
| Apr 7, 2013 at 9:16 | comment | added | Mateusz Wasilewski | Yemon, a generalized eigenvector with eigenvalue $\lambda$ in your case is just $\delta_{\lambda}$, a tempered distribution. And this what you want to do: extend Hilbert space to some space which includes objects that you would think of as eigenvectors. Maybe a more natural example would be differentiation ($i \frac{d}{dx}$) which obviously have exponentials ("plane waves") as eigenvectors, however they are not square-integrable. | |
| Apr 7, 2013 at 4:57 | comment | added | Yemon Choi | What does "generalized eigenvector" mean in this context? (I am only familiar with the usage from linear algebra.) In particular, what is a generalized eigenvector for the bounded, self-adjoint operator $(M f)(t) = tf(t)$ on $L^2[0,1]$? | |
| Apr 7, 2013 at 4:32 | history | answered | Jonathan Gleason | CC BY-SA 3.0 |