Timeline for Transformation of Observable Operators
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 15, 2020 at 5:11 | comment | added | Himanshu | When you didn't change the operator that's what you call the active transformation that's the case for which I have given explanation. But you can do the same with passive transformation where the transformation will change. I hinted on last paragraph. | |
| Nov 15, 2020 at 4:19 | comment | added | Jeff | I think the key thing here is that Ballentine wants to perform a transformation on a state vector but not change any of the observables. I think that is what you were alluding to with keeping the eigenvectors on the same line under the rotation. | |
| Nov 15, 2020 at 0:22 | comment | added | Jeff | It still isn't wholly satisfying for me. The position observable, $X$, has eigenvectors for every point in space. So if you do a space translation by $\mathbf{a}$, of $|x\rangle$, $|x'\rangle = |x+a\rangle$. But, Ballentine is insisting that one also translates the position operator $X$. | |
| Nov 14, 2020 at 21:11 | vote | accept | Jeff | ||
| Nov 14, 2020 at 21:11 | |||||
| Nov 14, 2020 at 20:26 | history | answered | Himanshu | CC BY-SA 4.0 |