Timeline for The operator equation $AB = \lambda BA$ for self-adjoint operators
Current License: CC BY-SA 4.0
10 events
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| S May 25, 2020 at 19:42 | history | suggested | Matemáticos Chibchas | CC BY-SA 4.0 |
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| May 25, 2020 at 19:16 | review | Suggested edits | |||
| S May 25, 2020 at 19:42 | |||||
| May 25, 2020 at 17:13 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| May 25, 2020 at 15:58 | comment | added | Michael Engelhardt | Certainly, but it seems to me that your argument as stated would therefore need some qualification as to what types of creation operators $A$ and $B$ you're considering - roughly speaking, "local", as you say. If the operators involve some Wilson lines connecting them to some reference point, one gets into trouble. Evidently, the OP's observation extends beyond self-adjoint operators, but there are restrictions - it would be interesting to understand those. | |
| May 25, 2020 at 14:17 | comment | added | Carlo Beenakker | local creation operators are fermionic or bosonic, when you exchange the order only the initial and final position need to be specified, so their statistics is governed by the permutation group; for the exchange of anyons the path by which they are exchanged matters, governed by the braid group. | |
| May 25, 2020 at 13:43 | comment | added | Michael Engelhardt | Sooo ... one can't define creation operators for anyons? | |
| May 25, 2020 at 13:13 | comment | added | Mateusz Wasilewski | For any $q\in (-1,1)$ one can construct operators $A$ and $B$ such that $AB = qBA$, so your last statement is not correct. In the self-adjoint case the condition $AB=\lambda BA$ indeed implies that also $BA = \lambda AB$, hence $AB = \lambda^{2} AB$, but that's not the case in general. | |
| May 25, 2020 at 11:44 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| May 25, 2020 at 11:30 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| May 25, 2020 at 11:25 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |