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I don't understand how eigenfunctions produce possible observable quantities. It just feels like mathematical trickery, and it probably is, but why does the trickery work? It seems unbelievable that the same wavefunction can hand you the possible observed quantities for whatever observable you wish just by "asking" it in the right way.

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    $\begingroup$ Does this link help? $\endgroup$ – dval98 Oct 15 '20 at 1:59
  • $\begingroup$ @dval98 Ok, here's where I'm at. It makes sense that we can find some basis where the basis vectors correspond to states with definite values for some desired observable. I also see how we can find some operator that produces that basis as its set of eigenfunctions. Where I am lost is how the eigenvalues come into play. Wouldn't it make more sense that the eigenvalue would correspond to the probability of measuring the particle to be in that particular eigenstate? $\endgroup$ – Jeff Bass Oct 15 '20 at 3:40
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I think I see where the confusion is (hopefully). The eigenvalue you obtain when you operate on some wave-function (assuming this function is an eigenfunction) is the value of the observable in that exact eigenstate. Every eigenstate is associated with an eigenvalue with zero variance (i.e. 10/10 times you make the measurement, you will get that eigenvalue for that state).

For example, we known the exact value for the electron in the lowest energy level of the hydrogen atom is -13.6 eV (1).

However, as these systems become more complex, we cannot obtain the exact solution for the .

(Note: I am not suggesting they do not exist, just that we do not have finite solutions to them)

So instead, we obtain all the possible eigenvalues with the probabilities that the system will have that value. If the system was in some exact state (x), it is associated with some exact eigenvalue. There isn't any variance in the eigenvalue for that exact state. The probability describes the chance that we will find that system existing in state (x) with that corresponding eigenvalue.

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  • $\begingroup$ This definitely makes more sense now. I think the question became why these eigenvalues had anything to do with observable values, which I asked in a separate question. $\endgroup$ – Jeff Bass Oct 15 '20 at 16:35

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