Timeline for Mathematical probabilistic interepretation of probability amplitude
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2012 at 17:39 | comment | added | N. Virgo | @JustinSolomon learnt a lot of what I know from Leonard Susskind's quantum entanglements course, available on YouTube. It assumes less of a maths background than you have, but it's pretty good for concentrating on the mathematical structure rather than the physics, and might be what you're looking for. | |
| Oct 5, 2012 at 16:52 | comment | added | Justin Solomon | Interesting! This makes it quite hard for us applied math types to evaluate whether it's useful for our own research :-) . I wonder if there's anyone who provides an introduction to quantum physics purely probablistically, sort of in the spirit of this document: scottaaronson.com/democritus/lec9.html | |
| Oct 5, 2012 at 12:18 | comment | added | N. Virgo | Not everybody agrees with one of the above positions (I certainly don't), and the interpretation of QM is a small but very active field today. However, it's really hard, and there's nothing like a consensus. | |
| Oct 5, 2012 at 12:16 | comment | added | N. Virgo | @JustinSolomon in terms of an explanation, the problem is that the historical development of QM mostly took the form of people making wild guesses about how to perform the calculations, which somehow rapidly converged to a very successful formalism that nobody knew how to correctly interpret. The attitude of most physicists today is either "shut up and calculate" (i.e. it's not a physicist's job to worry about what the equations actually mean) or words to the effect that no interpretation is needed and the formalism of QM already contains everything you need to know about the physics. | |
| Oct 4, 2012 at 19:23 | comment | added | N. Virgo | @JustinSolomon yes, I guess really the first and second expressions should be for $p(D|E,M)$ and the third for $p(D|E,\neg M)$, where $M$ is a Boolean that's true if you made the measurement of which slit the photon passed through, and false if not. Then the third expression is actually $$p(D|E,\neg M)= |a(S_1|E,M)a(D|S_1,M) +a(S_2|E,M)a(D|S_2, M)|^2,$$ with the r.h.s. conditioned on $M$ rather than $\neg M$ to indicate that the $a$'s themselves don't depend on whether the measurement was made. | |
| Oct 4, 2012 at 13:48 | comment | added | Justin Solomon | But hopefully there is at least an explanation for relative phase? | |
| Oct 4, 2012 at 13:42 | comment | added | dmckee --- ex-moderator kitten | Note that relative phases have meaning, but the absolute phases do not. This is not the only place in physics where one gets to choose an arbitrary starting point, however, and people don't generally bother themselves with the "Why?" of it. | |
| Oct 4, 2012 at 13:26 | comment | added | Justin Solomon | This is a very clear explanation of some of the phenomena I was having trouble understanding -- thanks! As you mention, the second and third equations for $p(D|E)$ yield different values, the third being the result of not trying to detect the slit. Is there a probabilistic explanation that's hiding here? E.g. that the second equation actually is somehow conditioned on having made an observation, which adds or removes some degree of probabilistic independence? | |
| Oct 4, 2012 at 12:19 | history | answered | N. Virgo | CC BY-SA 3.0 |