Timeline for Properties of spectrum of a self-adjoint operator on a separable Hilbert space
Current License: CC BY-SA 3.0
13 events
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| Jun 23, 2016 at 15:44 | comment | added | Valter Moretti | You see that the spectrum fills the complete interval $[0,1]$ but not every point of the spectrum is an improper eigenvalue, the points in $\mathbb Q$ are proper eigenvalues whose eigenvectors are just the (normalizable) vectors $\psi_k$. The irrational points of $[0,1]$ are instead elements of the continuous spectrum, but there is no associated non-normalizable eigenfunctions associated with each of them. | |
| Jun 23, 2016 at 15:42 | comment | added | Valter Moretti | I think an explicit example is helpful where on the one hand it is evident that the statement is false, on the other hand it is also clear that the example has no much physical sense. Consider, in $H = L^2(\mathbb R)$ (thus separable), a Hilbert basis $\{\psi_k\}_{k \in {\mathbb Q} \cap [0,1]}$, $\mathbb Q$ being the set of rational numbers and define the bounded self-adjoint operator $A := \sum_{k \in {\mathbb Q} \cap [0,1]} k |\psi_k\rangle \langle \psi_k|$. It is an easy exercise to prove that $D(A)=H$, $\sigma(A)= [0,1]$, but $ \sigma_p(A)= {\mathbb Q} \cap [0,1]$. | |
| Jun 23, 2016 at 14:49 | comment | added | Valter Moretti | Indeed the second statement is false rigorously speaking. It is more or less correct if the Hilbert space is separable... | |
| Jun 23, 2016 at 12:18 | comment | added | Wildcat | However, the next sentence on the very same page reads: "If the spectrum is continuous (i.e., the eigenvalues fill out an entire range) then the eigenfunctions are not normalizable, and they do not represent possible wave functions [...]" How this can be right then, if, as you said, point spectrum can be continuous? For me it look like that given a self-adjoint operator with a continuous spectrum I could not actually tell is a particular spectral value an eigenvalue or not. Besides, how do I normalize eigenvectors that corresponds to eigenvalues which form a continuous point spectrum? | |
| Jun 23, 2016 at 12:13 | comment | added | Wildcat | So, if I understand thing correctly, if a self-adjoint operator has a discrete spectrum (all spectral values are isolated points) then each spectral value is an eigenvalue, consequently, the corresponding vector is an eigenvector that is normalizable and could represent a possible state of a system. And this is in line with what Griffiths says in his Introduction to Quantum Mechanics (p. 100): "If the spectrum is discrete (i.e., the eigenvalues are separated from one another) then the eigenfunctions lie in Hilbert space and they constitute physically realizable states." | |
| Jun 22, 2016 at 9:08 | vote | accept | Wildcat | ||
| Jun 20, 2016 at 13:50 | comment | added | Wildcat | Oh, yes, I knew the definition. But I wrongly thought that countability implies discreteness, so to speak. Now, after reading your and Martin's responses, I know that this notions are unrelated: a countable set is not necessarily "discrete", i.e. there might be no "gaps" between its elements. | |
| Jun 20, 2016 at 11:48 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 20, 2016 at 11:41 | comment | added | Valter Moretti | A set $S$ is said to be countable if there is an injective map $f : S \to \mathbb N$... | |
| Jun 20, 2016 at 11:04 | comment | added | Wildcat | Looks like I have a wrong notion of countability in my mind... | |
| Jun 19, 2016 at 16:37 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 19, 2016 at 16:20 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 19, 2016 at 16:14 | history | answered | Valter Moretti | CC BY-SA 3.0 |