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Aug 1, 2020 at 13:21 vote accept Mr. Palomar
Aug 1, 2020 at 10:35 answer added Vercassivelaunos timeline score: 3
Aug 1, 2020 at 10:12 comment added Mr. Palomar Thanks @Vercassivelaunos. I was under the impression that this would only be true when considering eigenvalues of a single operator, but it seems that I am mistaken.
Aug 1, 2020 at 10:10 answer added Philip timeline score: 1
Aug 1, 2020 at 10:08 history edited Qmechanic CC BY-SA 4.0
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Aug 1, 2020 at 10:08 comment added Vercassivelaunos The physical state of a hydrogen atom doesn't have to be an eigenstate of all these operators. But it can always be written as a linear combination of those eigenstates. In quantum mechanics, we're always looking for a basis of the Hilbert space which makes calculations convenient. The set of eigenstates of the given operators is such a convenient basis. But there's nothing preventing you from finding an atom in a state which is not one of those eigenstates. It will simply take more symbols to write it down, for instance $\frac{1}{\sqrt2}(\vert n=1\rangle+\vert n=2\rangle)$.
Aug 1, 2020 at 10:07 history edited Qmechanic
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Aug 1, 2020 at 10:06 answer added my2cts timeline score: -2
Aug 1, 2020 at 9:54 answer added FGSUZ timeline score: 1
Aug 1, 2020 at 9:43 answer added Leiba Goldstein timeline score: 0
S Aug 1, 2020 at 9:39 history suggested A. Bordg CC BY-SA 4.0
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Aug 1, 2020 at 9:38 review Suggested edits
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Aug 1, 2020 at 9:22 history asked Mr. Palomar CC BY-SA 4.0