Timeline for Must solutions to the time-independent Schrodinger equation that have discrete or negative eigenvalues be square-integrable?
Current License: CC BY-SA 4.0
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| Apr 5, 2023 at 20:11 | comment | added | tparker | @ChristianRemling So do you have an explicit counterexample that satisfies (2) but violates (1') and (3)? | |
| Apr 5, 2023 at 20:09 | comment | added | Christian Remling | @tparker: My answer is still valid (almost) literally even after your edits. The only extra thing one needs is that their are upgrades of the vN-W potential with a dense in $(0,\infty)$ set of eigenvalues and still $V\to 0$. | |
| Apr 5, 2023 at 20:07 | comment | added | tparker | @WillieWong As far as I know, the first equation in my question defines the set of eigenvalues, not the spectrum. (Correct?) My edited question has nothing at all do with the spectrum of the Hamiltonian; I only care about the set of eigenvalues $\lambda$. Any discussion of the spectrum seems extraneous. | |
| Apr 5, 2023 at 20:04 | comment | added | tparker | @ChristianRemling Okay thanks, you may be right that I lack the mathematical background in functional analysis to usefully discuss this question. | |
| Apr 5, 2023 at 20:03 | comment | added | tparker | @ChristianRemling But I said in the OP that the domain of $H$ is the set of smooth functions $\mathbb{R} \to \mathbb{C}$, not $L^2$. At least, I meant to say that; I can edit the question to clarify that. | |
| Apr 5, 2023 at 20:02 | comment | added | Christian Remling | @tparker: Both eigenvalues and spectrum are very relevant in physics (the spectrum is the set of possible energies of the system). May I suggest my functional analysis lecture notes, especially chapters 12, 13, if you want to seriously study some of the background (but it will be a major effort, as these depend on earlier chapters and especially the spectral theorem). Without this background, no meaningful discussion of the precise mathematics involved here is possible. They are available here: math.ou.edu/~cremling/teaching/ln.html | |
| Apr 5, 2023 at 20:02 | comment | added | Willie Wong | @tparker: (2) as you stated cannot imply anything. Just because $\lambda < 0$ doesn't allow you to say anything about whether $\lambda$ belongs to the spectrum or not. Perhaps what you are thinking of is a statement along the lines of "$\lambda < 0$ and in the spectrum implies $\lambda$ is an eigenvalue", which is true in your setting. | |
| Apr 5, 2023 at 19:59 | comment | added | Christian Remling | @tparker: An eigenvalue is, by definition, a $\lambda$ such that $H\psi=\lambda\psi$ (= your first displayed equation) has a solution from the domain of the operator $H$, so in particular $\psi\in L^2$. This will not hold for most $\lambda\in\mathbb R$. For example, if $V=0$, you can easily check that there are no eigenvalues. | |
| Apr 5, 2023 at 19:59 | comment | added | tparker | @WillieWong 1. Yes, I now understand the difference between eigenvalue and spectrum, and I have edited my question to remove any reference to the word "spectrum" to avoid ambiguity. As far as I know, in physics applications the eigenvalues of the Hamiltonian are the relevant quantities, not the spectrum. 2. If $V \equiv 0$, then why isn't any $\lambda > 0$ an eigenvalue with eigenfunctions $\psi(x) = e^{\pm i \sqrt{\lambda} x}$? 3. So we still don't know whether (1) or (2) implies (3), or vice versa? | |
| Apr 5, 2023 at 19:56 | comment | added | Willie Wong | ... the specific example you mentioned has $\lambda$ an embedded eigenvalue: every interval around $\lambda$ intersect other parts of the spectrum, since the essential spectrum for that example includes all of $(0,\infty)$. But if your question is to be interpreted literally, as the statement that $\lambda$ is not a limit point of the set of eigenvalues, I am not sure what the answer should be for the class of Schrodinger operators you wrote down. | |
| Apr 5, 2023 at 19:50 | comment | added | Willie Wong | @tparker "eigenvalue" is not the same as "spectrum". A value $\lambda$ is in the spectrum of the operator $T$ if $T-\lambda I$ doesn't have a bounded inverse; a value $\lambda$ is an eigenvalue if $T-\lambda I$ is not injective. Returning to the Schrodinger operators, in the case $V\equiv 0$, any $\lambda \geq 0$ is in the spectrum but not an eigenvalue. // That (1') implies (2) is false is what you stated yourself in your question. // Regards to (3) it appears that Christian interpreted your statement (3) to mean that $\lambda$ is in the discrete part of the spectrum; ... | |
| Apr 5, 2023 at 19:29 | comment | added | tparker | 1. Why is claim (1) equivalent to the claim that $\lambda$ is an eigenvalue? It seems to me that any $\lambda$ must be an eigenvalue by definition, whether or not $\psi(x)$ is square-integrable. 2. What is "the reason" why (1') $\implies$ (2) is false? I just found one counterexample that satisfies claim (1) but not claim (2); I don't see how this tells us anything about claim (3). 3. Why are the propositions (2) $\implies$ (1') and (2) $\implies$ (3) "manifestly absurd"? Can you give a counterexample of a $V$, $\psi$, and $\lambda$ that satisfy (2) but not (1') or (3)? | |
| Apr 5, 2023 at 17:17 | history | edited | Christian Remling | CC BY-SA 4.0 |
added 2 characters in body
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| Apr 5, 2023 at 16:29 | history | answered | Christian Remling | CC BY-SA 4.0 |