Timeline for Reason for the discreteness arising in quantum mechanics?
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11 events
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| Oct 6, 2012 at 13:36 | comment | added | user7757 | Thanks a lot for the reply. It sure made things clearer. | |
| Oct 6, 2012 at 13:34 | comment | added | A.O.Tell | The mathematics of operator spectra is not trivial, and I certainly cannot give you an explanation for discreteness here beyond what I already wrote about the states being spatially constrained in a certain way. Maybe this will help you to study the question: en.wikipedia.org/wiki/Spectrum_(functional_analysis) | |
| Oct 6, 2012 at 13:32 | comment | added | A.O.Tell | You're confusing the mathematics we use to describe the measurement with the actual measurement. Of course we chose a description which matches what we observe a measurement to be. But that does not explain why a measurement does what a measurement does. | |
| Oct 6, 2012 at 13:30 | comment | added | user7757 | OK. How do you mathematically explain the discrete spectrum of eigenvalues. About your second statement: 'why' a measurement forces a system.....But obviously, shoudnt the measurement force the system to be in one of the eigenstates..because if it didnt, we would have more than one eigenvalues, i.e. more than one value for the same observable, which is absurd. | |
| Oct 6, 2012 at 13:28 | comment | added | A.O.Tell | Regarding the unitary representations of symmetry, that issue has mostly been raised by Weyl and Wigner who studied the representation of groups in quantum theory. I'm afraid I cannot explain the details in a comment, but the basic idea is that physically observable symmetries can be described as groups, and the representations of these groups must be contained in the describing mathematics. And it turns out that for quantum theory the only option is a unitary representation. | |
| Oct 6, 2012 at 13:25 | comment | added | A.O.Tell | No, the discrete spectrum of eigenvalues is mathematically well understood. What is not understood is why a measurement forces a system to be in one of the eigenstates associated with the measured eigenvalue after the measurement. This is essentially known as the "quantum measurement problem". You can surely find a lot of information about it if you look for it. | |
| Oct 6, 2012 at 13:22 | comment | added | A.O.Tell | What I meant is that the outcome is picked from a discrete set. Some measurements allow outcomes from a continuous set, like a position measurement for example. | |
| Oct 6, 2012 at 13:22 | comment | added | user7757 | (contd) Do you mean to say that there is no accepted explanation for the discrete spectrum of the eigenvalues? And therefore, the foundations of QM are basically empirical? Help will be much appreciated on this matters. | |
| Oct 6, 2012 at 13:21 | history | edited | A.O.Tell | CC BY-SA 3.0 |
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| Oct 6, 2012 at 13:21 | comment | added | user7757 | Thanks for the answer. Firstly I would like to know which postulate/property of quantum mechanics entails a unitary representation of the symmetry group (is it because of unitarity - the fact that the sum of the probabilities should be 1?) Secondly, I didnt understand your statement: he measurement outcome results in exactly one (often discrete) result. What do you mean by a discrete result, when we say that QM is discrete we mean that the possible outcomes can take only specific values, but how can you talk about the discreteness of a specific result. | |
| Oct 6, 2012 at 13:10 | history | answered | A.O.Tell | CC BY-SA 3.0 |