Timeline for Spectrum of operator involving ladder operators
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 19, 2023 at 21:41 | history | became hot network question | |||
| Jan 19, 2023 at 21:25 | answer | added | Nik Weaver | timeline score: 4 | |
| Jan 19, 2023 at 14:44 | vote | accept | António Borges Santos | ||
| Jan 19, 2023 at 14:38 | answer | added | Carlo Beenakker | timeline score: 6 | |
| Jan 19, 2023 at 14:36 | comment | added | fedja | Just think of why and in what sense the eigenvalues of the finite approximations do converge to those of the full operator and you'll see (unless you are much more inventive than I) that not only all proofs will fail for the eigenvalue $0$, but it is even impossible to make a not obviously false general statement about how one should recover its multiplicity from the values of the eigenvalues of finite approximations. | |
| Jan 19, 2023 at 14:19 | comment | added | fedja | Unless "smart work-around" means determining the presence/multiplicity of $0$ by alternative means having nothing to do with the main truncation scheme, I do not see one but I'll be happy to be proved wrong :-) | |
| Jan 19, 2023 at 14:14 | comment | added | António Borges Santos | @fedja but maybe there is a smart work-around here? i suppose that is what makes it a numerics question ;) | |
| Jan 19, 2023 at 14:12 | comment | added | fedja | However my main comment on the impossibility to say anything about the presence of the eigenvalue $0$ remains valid. | |
| Jan 19, 2023 at 14:07 | comment | added | fedja | Ah, right. I misunderstood the notation. I somehow confuser it with $a+a^*$. Stupid me! | |
| Jan 19, 2023 at 14:05 | comment | added | António Borges Santos | @fedja maybe you are talking about a different object but what I see in this case is $$ H_N = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} $$ which seems to have rank $2$. | |
| Jan 19, 2023 at 14:01 | comment | added | fedja | Wrong. Consider $N=2$, really. It takes 1 minute at most ;) | |
| Jan 19, 2023 at 14:00 | comment | added | António Borges Santos | @JochenGlueck thank you, that was embarassing... | |
| Jan 19, 2023 at 14:00 | comment | added | António Borges Santos | @fedja, the truncated Hamiltonian has two degenerate matrices at the off-diagonal blocks, right? So the truncated Hamilltonian has a zero eigenvalue of geometric multiplicity $2$? | |
| Jan 19, 2023 at 13:59 | history | edited | António Borges Santos | CC BY-SA 4.0 |
edited body; edited title
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| Jan 19, 2023 at 13:55 | comment | added | fedja | I don't understand the comment about $0$ of multiplicity $2$ for truncations: the sum of two degenerate matrices does not need to be even degenerate, forget about eigenvalue $0$ of multiplicity $2$ (just look at $N=2$). In general, truncation to large size and taking the limit for compact self-adjoint matrices allows you to faithfully recover all non-zero eigenvalues but can tell nothing about whether the eigenvalue $0$ is there (and I doubt if any finite size approximation is capable of that). Your case may be special, but it won't be pure "numerics" then anyway. | |
| Jan 19, 2023 at 13:41 | history | asked | António Borges Santos | CC BY-SA 4.0 |