Timeline for The Heisenberg Uncertainty Principle and the 1D particle in a box
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2021 at 16:17 | comment | added | Gert | physics.stackexchange.com/questions/335184/… "A simple way would be to realise that when n=0, the magnitude of the momentum is 0, and thus there are no 'positive' and 'negative' values it could take: it most definitely has a momentum of exactly zero, with no uncertainty. This would be allowed, if you were not in a box. However, placing yourself in a box, meaning that Δx<∞, means that you necessarily have a minimal non-zero momentum, using the argument you mentioned earlier." Seems to confirm my point about $p=0$. | |
| Mar 14, 2021 at 4:30 | vote | accept | Gert | ||
| Mar 13, 2021 at 23:44 | comment | added | DanielC | It is trivial to show that the ground state energy can't be $0$, but this has nothing to do with the HUP, which could make statements only on the spectrum and eigenfunctions of the coordinate and momentum, if these two operators and their commutator is properly defined. In the well, as in general for no interaction, $H=\frac{P^2}{2m}$, but this squaring is tricky, because it involves unbounded operators, so if one draws conclusions from the HUP for $x$ and $p$, has to be careful to exend them to $H$ (spectrum and eigenfunctions). | |
| Mar 13, 2021 at 23:24 | comment | added | user87745 | Relating to the original question of OP, the HUP is not what makes the ground state energy non-zero, right? Because the energy eigenstates are not even in the domain of definition of the commutator $[x,p]$. | |
| Mar 13, 2021 at 23:14 | history | answered | DanielC | CC BY-SA 4.0 |