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The compatibility theorem states that, if the operators $\hat A$ and $\hat B$ representing the observables $A$ and $B$ do commute, then there exists a common eigenbasis for their eigenstates, and those observables are compatible.

I wonder if this theorem is biconditional: would it be possible to find a common eigenbasis of eigenstates for a pair if $A$ and $B$ do not commute? Or is the if of the theorem actually an if and only if, and the condition that they commute is both necessary and sufficient?

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    $\begingroup$ Hint: To see why the reverse statement holds true, expand a generic state in a common eigenbasis of $A$ and $B$ and compute the commutator. $\endgroup$ – Jakob May 24 at 12:09
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Yes it is if and only if. Given two observables $X$ and $Y$:

  1. $[X,Y]=0$

  2. $X$ and $Y$ have a common eigenbasis

are equivalent statements.

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