# Biconditionality of the compatibility theorem for commuting operators

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?

• 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. – Jakob May 24 at 12:09

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