Timeline for Why do particles not decohere in their native state?
Current License: CC BY-SA 4.0
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| Nov 25, 2020 at 11:52 | history | edited | anna v | CC BY-SA 4.0 |
fixed link
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| Sep 30, 2016 at 3:56 | comment | added | anna v | of the wavefunction , which is what one can see here physics.stackexchange.com/questions/282843/… calculated as points . A measurement realizes the probability. The constituents exist in the wavefunction, in the mathematical model we have of physics at particle level, one needs a measurement to define interactions. | |
| Sep 30, 2016 at 3:53 | comment | added | anna v | Again it is the difference between a wave function and a probability distribution. If you go to the individual constituents of the atom, and want to create the state it would require second quantization, the electron field and the proton fiel (for a hydrogen atom. The proton is a very complicated entity with gluons and quarks, but let us approximate it with a field). There you could define the wave function as a superposition of all the creation and annihilation paths taken by the electron and proton . These would be in superposition. The probability of finding an electron would be the square | |
| Sep 29, 2016 at 18:37 | comment | added | Cato1974 | Thanks very much Anna. One further question: How would this line of thinking apply to an electron in a probability distribution cloud around a nucleus? Wouldn't we expect the multiple 'states' of the electron to interact with each other causing decoherence? Or is it that we are purely dealing with a mathematical probability distribution rather than the electron truly being in two places at once. | |
| Sep 29, 2016 at 18:35 | vote | accept | Cato1974 | ||
| Sep 29, 2016 at 18:21 | history | answered | anna v | CC BY-SA 3.0 |