Timeline for How should I understand the completeness relation of the form $\sum_{n} \phi_n(x) \overline{\phi_n}(y)=\delta(x-y)$?
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| Jun 24, 2022 at 23:46 | comment | added | André Henriques | The fact that you start with an operator $A$ is irrelevant to the question. You simply have a basis $\{\phi_n(x)\}_{n\in\mathbb N}$ of $L^2(\mathbb R)$, and you're asking whether $\sum_{n}\phi_n(x) \overline{\phi_n}(y)=\delta(x-y)$ in the sense of distributions on $\mathbb R^2$ (I think that the answer should be yes, but I'm not sure). The "completeness" that you're referring to is the statement that $\{\phi_n(x)\}_{n\in\mathbb N}$ forms a genuine orthonormal basis of $L^2(\mathbb R)$, and not just a set of orthonormal linearly independent vectors. | |
| Jun 24, 2022 at 19:53 | comment | added | Isaac | Could you explain more in detail..? | |
| Jun 24, 2022 at 19:43 | comment | added | Isaac | How about the second question of mine? Is it possible for the operator $e^{-A}$ to go inside the sum $\sum_n$? If so, how do we interpret it? | |
| Jun 24, 2022 at 19:41 | comment | added | Christian Remling | The formal "justification" is to disentangle the relation $f(x)=\sum_n \langle \phi_n, f \rangle \phi_n(x)$, which is actually correct if interpreted properly (the series converges in $L^2$), and that's really all the identity is saying if you want to give it a rigorous interpretation. | |
| Jun 24, 2022 at 19:25 | comment | added | Isaac | OK, I edited my question. | |
| Jun 24, 2022 at 19:19 | history | edited | Isaac | CC BY-SA 4.0 |
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| Jun 24, 2022 at 19:04 | comment | added | paul garrett | No, self-adjointness alone is not enough, either for bounded or unbounded (self-adjoint) operators. For example, multiplication by $1/(1+x^2)$ is bounded self-adjoint, but has no eigenvectors. Likewise, multiplication by $x^2$ is unbounded, and has self-adjoint extensions... but has no eigenvectors. But I guess this is just part of the hypotheses you want. Or have "generalized eigenvectors", doing distributional stuff, but then the sum is not a literal sum, etc. | |
| Jun 24, 2022 at 19:03 | history | edited | Isaac | CC BY-SA 4.0 |
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| Jun 24, 2022 at 18:59 | comment | added | Isaac | Ok I will asume that as well. | |
| Jun 24, 2022 at 18:39 | comment | added | paul garrett | You might want to be assuming also that $A$ has compact resolvent, to assure existence of an orthonormal basis of eigenvectors. | |
| Jun 24, 2022 at 18:18 | history | asked | Isaac | CC BY-SA 4.0 |