Timeline for Double slit experiment from first principles of QM
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13 events
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| Aug 27, 2014 at 12:39 | answer | added | gherardo | timeline score: 2 | |
| Jul 10, 2014 at 8:26 | comment | added | Trimok | @Sbaniala : For the most simple approach, the mathematics are quite the same as the Classical wave-optics formulation. However, the Quantum interpretation is that these are probability amplitudes, and not real light wavefunctions. You sum the probability amplitudes with the correct phase difference, and then you take the square of the modulus to have the probability. | |
| Jul 9, 2014 at 22:54 | comment | added | Sbaniala | @ACuriousMind, It makes sense to me what you are trying to say. Could you please elaborate it, with connection to Born rule, as a complete answer? It would be good for others as well. | |
| Jul 9, 2014 at 22:01 | comment | added | ACuriousMind♦ | You have to add @ before my name, else I don't get notified to your reply. I'm not sure I fully understand you, but are you asking why the interpretation of scalar products, known as the Born rule, holds? Then the answer is: Within the standard Copenhagen framework, it is an axiom, and alternative frameworks might have it as a derivation, but then they will have some other axioms instead. | |
| Jul 9, 2014 at 18:58 | comment | added | Sbaniala | Dear ACuriousMind, What I am interested to know is how to establish the probability amplitude of the expressions in the Copenhagen interpretation. The probability amplitude from the two slits are only different in phase but have the same wave number. I could not understand how does that happen? I am asking this question to understand the Copenhagen interpretation in relation to the double slit experiment. | |
| Jul 9, 2014 at 16:47 | comment | added | ACuriousMind♦ | " What causes the probability amplitude to be identical with the prepared states?" You must mean something else, as the amplitude is a number, not a state. Also, what about Feynman path integral disturbs you? It is from first principles, is it not? Just as some problems are hideously complicated in Newtonian mechanics but crystal clear in Lagrangian mechanics, so it can occur in QM that the path integral solves elegantly what is otherwise very ugly. | |
| Jul 9, 2014 at 14:45 | comment | added | Sbaniala | Thank you for your insight but I have difficulty with performing the numerical simulations. It that the only way to explain it ? I am basically interested to understand why the two probability amplitudes due to the two slits are what they are? Basically the prepared states are definite momentum states along a direction perpendicular to the plane of the slits and the screen. What causes the probability amplitude to be identical with the prepared states? | |
| Jul 9, 2014 at 14:34 | comment | added | Trimok | If you accept Schrodinger equation, you just have to consider infinite potential in the plan $z=0$ of the slits (except for the $2$ slits where the potential is zero); of course, you have to do numerical simulations. | |
| Jul 9, 2014 at 13:46 | comment | added | Sbaniala | Please make sure that I have read all the answers related to this question in the PSE. I am looking for a rigorous description of electron double slit interference patters only from the first principles of quantum mechanics (but not the Feynman Integral approach) | |
| Jul 9, 2014 at 13:41 | review | First posts | |||
| Jul 9, 2014 at 13:45 | |||||
| Jul 9, 2014 at 13:35 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
deleted 4 characters in body; edited tags
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| Jul 9, 2014 at 13:26 | history | edited | Earth is a Spoon | CC BY-SA 3.0 |
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| Jul 9, 2014 at 13:24 | history | asked | Sbaniala | CC BY-SA 3.0 |