Timeline for Do we need an orthonormal basis in Quantum Mechanics?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2015 at 17:28 | history | edited | Qmechanic♦ |
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| Jan 23, 2015 at 17:27 | comment | added | Qmechanic♦ | Possible duplicates: physics.stackexchange.com/q/39602/2451 and links therein. | |
| Jan 23, 2015 at 16:37 | vote | accept | Xin Wang | ||
| Jan 23, 2015 at 16:34 | answer | added | alanf | timeline score: 0 | |
| Jan 23, 2015 at 8:17 | vote | accept | Xin Wang | ||
| Jan 23, 2015 at 8:17 | |||||
| Jan 23, 2015 at 1:23 | answer | added | Nogueira | timeline score: 4 | |
| Jan 23, 2015 at 0:06 | vote | accept | Xin Wang | ||
| Jan 23, 2015 at 8:17 | |||||
| Jan 23, 2015 at 0:04 | answer | added | By Symmetry | timeline score: 13 | |
| Jan 22, 2015 at 23:58 | comment | added | Xin Wang | @MarkMitchison I suspected something like that, thank you. | |
| Jan 22, 2015 at 23:57 | comment | added | Xin Wang | @Phoenix87 could you explain this relationship maybe in an answer? | |
| Jan 22, 2015 at 23:57 | comment | added | Mark Mitchison | @XinWang Physically, orthogonality of the eigenvectors means that the states are distinguishable. Therefore, if one has an "observable" with non-orthogonal eigenvectors, it means there does not exist even in principle a measurement that allows you to determine the value of that observable with certainty. This is a very strange property for an observable to have, although I can't see why this should actually be forbidden. | |
| Jan 22, 2015 at 23:56 | comment | added | Phoenix87 | @XinWang functional calculus is also physics. Without it you are not allowed to regauge your instruments... | |
| Jan 22, 2015 at 23:55 | comment | added | Xin Wang | @ACuriousMind sorry, I don't understand what PT symmetric means. | |
| Jan 22, 2015 at 23:54 | comment | added | Xin Wang | @Phoenix87 the functional calculus is rather a mathematical construct, I was wondering more about why this is necessary from a physical perspective. | |
| Jan 22, 2015 at 23:53 | comment | added | Phoenix87 | non-normal operators don't have functional calculus, and this is something you want most of the times. | |
| Jan 22, 2015 at 23:47 | comment | added | ACuriousMind♦ | No. PT symmetric, but not Hermitian Hamiltonians are one example. | |
| Jan 22, 2015 at 23:36 | history | asked | Xin Wang | CC BY-SA 3.0 |