Compatibility of qunatum measurements

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The discussion centers on the non-commutation of two operators A and B in quantum mechanics and whether the commutator [A,B]psi can equal zero for any state psi. It is noted that the only apparent case where this occurs is when psi equals zero, indicating no particle. Participants are encouraged to consider specific examples, such as the commutator [x^2, p], to identify states that might yield zero despite the commutator not being identically zero. The conversation highlights the importance of position eigenstates in this context. Ultimately, the exploration of quantum states and their interactions with operators reveals complexities in the behavior of quantum measurements.
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The question is "two operators A,B do not commute". Is it true that
([A,B]psi) does not equal zero for any state psi?"

I'm not sure if there are any cases in which a state psi could produce this result (except for the obvious one where psi = 0 and there is no particle).

Thanks
 
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Here is a hint: think about the commutator [x^2 , p]. Can you find a state that gives zero when you hit it with the commutator even though commutator isn't identically zero?
 
I've tried applying the commutator to an arbitrary psi and i get i*h bar * 2x* psi. The only case where I can think of this equalling zero is when psi = 0, which is the obvious case of having no particle.
 
You can't think of any state that gives zero when you hit it with the position operator? Hint: think about position eigenstates.
 
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