Timeline for Quantum mechanics: Probability current density in terms of velocity vs. in terms of continuity equation
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2023 at 4:13 | answer | added | Amberd0g | timeline score: 0 | |
| Feb 9, 2021 at 5:15 | comment | added | pglpm | Just curious: what does the exclamation mark on the equality sign mean? It's a surprising equality? | |
| Feb 9, 2021 at 5:13 | comment | added | pglpm | I'm not sure whether for time-dependent Hamiltonians, for example, the probability current is still conserved. From a more general point of view, a probability current is associated with an observable/POVM rather than a system. It seems you're interested in that of the position observable. | |
| Feb 9, 2021 at 5:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 4, 2020 at 22:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 20, 2020 at 7:44 | comment | added | LLang | That is an interesting idea. I assume that it is also possible to add operators that depend on spin, since it commutes with position. Having a look at operators from the Breit-Pauli Hamiltonian (en.wikipedia.org/wiki/Breit_equation), which are possible candidates to be added to the nonrelativistic Hamiltonian, it seems they can all be written in terms of these basic ingredients (position, momentum, spin). | |
| Mar 19, 2020 at 17:58 | answer | added | Charles Francis | timeline score: 1 | |
| Mar 19, 2020 at 15:21 | comment | added | don't train ai on me | Sorry, didn't realize you meant even more general than that. It is an interesting question whether this is possible to show. Maybe it could be shown for a hamiltonian which is an arbitrary function of the position and momentum operators (written as a taylor series in powers of those variables). I think that's as general as it gets, until fields are quantized. | |
| Mar 19, 2020 at 6:08 | comment | added | LLang | As I said, I am able to do the derivation for specific forms of the Hamiltonian, like the nonrelativistic one you wrote down. However, I am looking for a general proof. The reason is that the derivation of the current density from the continuity equation is quite cumbersome, while the commutator of the Hamiltonian and position is easily calculated. If I knew that both lead to the same results, I would always follow the latter approach. | |
| Mar 18, 2020 at 20:47 | comment | added | don't train ai on me | If you write $H=p^2/2m + V(\vec{r})$ you can find the form of the velocity operator explicitly. Position commutes with the potential. Then it should be doable. | |
| Mar 18, 2020 at 19:35 | history | asked | LLang | CC BY-SA 4.0 |