Timeline for How does De Broglie–Bohm theory or pilot wave theory explain the results of the Stern–Gerlach experiment?
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| Sep 15, 2021 at 13:21 | comment | added | David Spector | Part 2: As to the transfer of momentum from one particle to another, that is a classical phenomenon like temperature or pressure. The behavior of multiple particles is completely determined by their initial positions (not spins) and the Schrödinger equation, which describes the entire geometry of the experiment, including the measurement device. | |
| Sep 15, 2021 at 13:21 | comment | added | David Spector | Part 1: I do not see the stated weakness in BI. In the BI, particles exist and have deterministic trajectories, which are wholly determined by the kinetic component of the Schrödinger equation. Since intrinsic particle spin is contained in the Schrödinger equation, the BI explains the Stern–Gerlach experiment. In BI, the measurement apparatus is included in the overall experiment, not considered part of an "environment", so there is no wave function collapse. | |
| Apr 25, 2015 at 11:17 | comment | added | Sofia | @BT About the inconsistency mentioned in my comment, I had a talk with prof. D. Durr, an expert in BI, but what he answered me was that this is what the equations say. I wasn't pleased by this answer, I see this as a conceptual problem in BI. Still, I won't refute BI for this weakness, as the concept of collapse in the SQT is also very unpleasant. | |
| Apr 25, 2015 at 11:07 | comment | added | Sofia | @BT Please read attentively my answer and comments, I told you, my time is very short. When two quantum objects interact, you write the Schrodinger equation. See what I answered to your question 2). | |
| Apr 25, 2015 at 11:06 | history | edited | Sofia | CC BY-SA 3.0 |
italics, boldface, and numbering the OP's questions
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| Apr 25, 2015 at 10:58 | comment | added | Sofia | @BT The supporters of the BI are not very clear about the mechanism of impressing a detector. They only say that the detector is impressed if and only if the Bohmian particle impinges on it. Of course it seems inconsistent to claim that the physical properties, mass, charge, etc., are carried by the guide-wave (at every point and point of it), however the detector makes a recording only at the contact with the Bohmian particle, that possesses only position and Bohmian velocity. About transfer of momenta, why don't you read attentively my answer? (see the next comment) | |
| Apr 24, 2015 at 20:55 | comment | added | B T | Thanks for answering my questions, this is very helpful. So I do understand that its not the bohmian particle that directly interacts with the detector (or more precisely, th wave functions of the bohmian particles that comprise the detector). But surely the bohmian particle interacts with the detector indirectly, right? So What I meant to ask is, how does momentum get transferred from one bohmian particle to another? Does the bohmian particle's position/velocity affect its wave-function, which then in turn interacts with the wave functin of the detector? | |
| Apr 24, 2015 at 20:35 | history | answered | Sofia | CC BY-SA 3.0 |