Timeline for Abelian Symmetry groups and Degeneracy
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 4, 2017 at 7:41 | vote | accept | Primo-uomo | ||
| Jul 3, 2017 at 15:48 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jul 3, 2017 at 15:43 | comment | added | Valter Moretti | Yes!... I added a comment regarding commutativity | |
| Jul 3, 2017 at 15:42 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jul 3, 2017 at 15:29 | comment | added | Primo-uomo | As a clarification, this argument could be extended to a general Abelian group simply because the key point here, is that one can find a non-commuting operator if n>1, yes? In addition to this, could you please explain how the commutation relations between $A$ and $H$, and $P$ come to be? Thank you so much! | |
| Jul 3, 2017 at 15:24 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jul 3, 2017 at 15:24 | comment | added | Valter Moretti | Right, I forgot a factor $P$... corrected. Actually, it was not necessary since $Ran(B) \subset \cal H_E$, it holds $PB=B$... | |
| Jul 3, 2017 at 15:02 | comment | added | Primo-uomo | I am unable to see why $A'$ is self adjoint, or should the definition include $PBP$ instead of just $BP$ (or is $PB=BP=B$)? | |
| Jul 3, 2017 at 14:58 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jul 3, 2017 at 14:44 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jul 3, 2017 at 14:38 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jul 3, 2017 at 14:30 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jul 3, 2017 at 14:25 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jul 3, 2017 at 13:51 | history | answered | Valter Moretti | CC BY-SA 3.0 |