Timeline for Adjoint of an operator
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2018 at 12:39 | answer | added | DisintegratingByParts | timeline score: 1 | |
| May 22, 2018 at 13:37 | history | tweeted | twitter.com/StackPhysics/status/998920712625811457 | ||
| May 22, 2018 at 8:04 | history | protected | Qmechanic♦ | ||
| May 22, 2018 at 8:04 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 50 characters in body; edited tags
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| May 22, 2018 at 7:19 | answer | added | yuggib | timeline score: 2 | |
| May 22, 2018 at 6:49 | answer | added | Luke | timeline score: 0 | |
| May 22, 2018 at 6:07 | answer | added | the boy who believed | timeline score: -1 | |
| May 22, 2018 at 4:51 | comment | added | Triatticus | You have partially answered your own question by stating the relationship $\hat{x}$ and $\hat{p}$ have to their adjoints. Now think about what the adjoint of an object like $i\hat{p}$ is | |
| May 22, 2018 at 4:19 | comment | added | quirkyquark | Also, as we know in Q-Mech, $\widehat{x}$ and $\widehat{p}$ are the two basic operators, once we find that they are self-adjoint, how do we find the adjint of operators like $a$, as I have mentioned in the question? | |
| May 22, 2018 at 4:15 | comment | added | quirkyquark | @SRS Yes. For the cases when we express the operators in terms of matrices (finite or infinite), the adjoint is defined as the transpose of complex conjugate of the given matrix. But, if we haven't expressed it in the form of a matrix, then how do we take the adjoint of the operator, that is my question. | |
| May 22, 2018 at 4:01 | comment | added | SRS | Have you looked at the definition of the adjoint of an operator? | |
| May 22, 2018 at 3:47 | history | asked | quirkyquark | CC BY-SA 4.0 |