Timeline for Does the commutator of anything with itself not vanish?
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Oct 9, 2014 at 14:43 | comment | added | user121330 | Forgive stupid old me... Perhaps you can explain how non-commuting numbers would allow for a non-self-commuting operator? | |
| Oct 9, 2014 at 8:32 | comment | added | leftaroundabout | @user121330: what do you mean? Sure, composition of linear operators doesn't commute in general... otherwise there would be no reason to ever consider the commutator. But how is this in dissonance with this answer? | |
| Oct 8, 2014 at 20:04 | comment | added | user121330 | Anyhow, I'd love to up-vote your answer, but like I said, vector spaces are non-commutative as a rule. | |
| Oct 8, 2014 at 17:37 | comment | added | user121330 | Linear vector spaces as a rule are non-commuting. The easiest axiom to get rid of for a non-vanishing self-commutator would be the existence of the additive inverse. That said, the heart of your argument is gold - Qmechanic's profs are being pedantic - the Grassmann-odd operator's commutator (non-super!) with itself vanishes trivially. | |
| Oct 8, 2014 at 16:11 | history | answered | user10851 | CC BY-SA 3.0 |