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Consider a 3-dimensional Hilbert space spanned by the normalized eigenstates $|1\rangle,|2\rangle,|3\rangle$ of an operator $A$. Consider a normalized superposition, $|\psi\rangle=c_1|1\rangle+c_2|2\rangle+c_3|3\rangle$. If one measures the operator $A$, the possible final states are $|1\rangle,|2\rangle$ and $|3\rangle$ with probabilities $|c_1|^2,|c_2|^2$ and $|c_3|^2$ respectively.

  1. Suppose we measure an operator $B$ that commute with $A$. Is it always true that the possible final states are again $|1\rangle,|2\rangle$ and $|3\rangle$ with probabilities $|c_1|^2,|c_2|^2$ and $|c_3|^2$ respectively?

  2. What about the special case when $B=f(A)$ (say, $f(A)=A^2$) where $f(A)$ is some function of the operator $A$?

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    $\begingroup$ What if $B$ is the identity? $\endgroup$ – Emilio Pisanty Jun 16 '18 at 13:13
  • $\begingroup$ @EmilioPisanty Nice! Is there a more nontrivial example? I've also added a special case of first bullet point in the second bullet point. $\endgroup$ – SRS Jun 16 '18 at 13:49
  • $\begingroup$ Sure. The point is that your question bumps into trouble whenever B has degenerate eigenspaces. The identity is an extremal case but, as usual in linear algebra, you can just restrict that behaviour to a smaller subspace and impose different behaviour elsewhere. Or, in simpler language, make B degenerate over 1 and 2 but not 3. $\endgroup$ – Emilio Pisanty Jun 16 '18 at 19:08
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  1. The answer is: No, the operator $B$ could have degenerate eigenvalues, cf. Emilio Pisanty's above example. Then the final state could differ from $|1\rangle$, $|2\rangle$ or $|3\rangle$.

  2. This can even occur if $B=f(A)$. E.g. if $f(x)=x^0$. Or e.g. if $f(x)=x^2$ and $A=|1\rangle\langle 1| - |2\rangle\langle 2|$.

See also this related Phys.SE post.

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