Timeline for Weird Behaviour of the act of measurement to a quantum system
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2013 at 18:00 | comment | added | QnoP | You're more than welcome :) and thanks for the link, it was very interesting. | |
| Oct 25, 2013 at 17:09 | comment | added | el psy Congroo | Thanks for your clear explanation, I think I have grasped the main idea already. In case you are interested : This is the method of measuring the energy without looking at its momentum when they don't commute nobelprize.org/nobel_prizes/physics/laureates/2012/… | |
| Oct 25, 2013 at 17:08 | vote | accept | el psy Congroo | ||
| Oct 25, 2013 at 12:14 | comment | added | QnoP | I wanted to add that the only situation in which $V(x)$ commutes with momentum is where $V(x) = constant$, which doesn't really have a physical significance. Unless $V$ has different functional dependencies other than $x$ (like t for instance). | |
| Oct 25, 2013 at 10:24 | comment | added | QnoP | In these cases, the potential and kinetic energies can't be separated by measurement, this means measuring the momentum can't give you the kinetic portion of the energy you measure. | |
| Oct 25, 2013 at 10:11 | comment | added | QnoP | In the case of a non-free particle, $[H,p]$ depends on the commutation of the potential ($V(x)$) with $p$. If $[V(x),p]=0$ then $[H,p]=0$ also. But in the case that $H$ doesn't commute with $p$, the eigenfunctions of the operators aren't the same. So, when you measure the energy for example, the wave function will collapse into one of the energy eigenstates, then if you measure the momentum afterwards, the system will collapse into one of the eigenstates of $p$, and you'll lose the information about the energy. Therefor you can't have the measurements of energy and momentum at the same time. | |
| Oct 24, 2013 at 22:20 | comment | added | el psy Congroo | so is it true that for a non-free particle case, [H,p] happens not commuting? What would happen to the wave function when you try to measure the energy of the system? I'm wondering how could one measure the "ENERGY" without refering to momentum and its potential? | |
| Oct 24, 2013 at 22:17 | comment | added | el psy Congroo | In the case of a bound state, we have boundary condition to restrict the value of k. But for a unbound/scattering state, we don't have BC , hence Fourier analysis was used to construct a wavepacket to make the wavefunction normalizable | |
| Oct 24, 2013 at 21:23 | history | edited | QnoP | CC BY-SA 3.0 |
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| Oct 24, 2013 at 21:17 | history | answered | QnoP | CC BY-SA 3.0 |