Let's say you have two operators A and B such that when they act on an eigenstate they yield a measurement of an observable quantity (so they're Hermitian). A and B do not commute, so they can't be measured simultaneously. My question is this: You have a matrix representation of A and B and you find their eigenvalues and corresponding eigenstates (eigenvectors), let's say they each have a basis of 3 eigenstates. My understanding is that if their eigenstate basis is not identical, they are not simultaneously measurable quantities and thus do not commute. What if A and B, which don't commute share a 1 eigenstate in common out of their 3 states? Does that mean that the two observables are simultaneously measurable when acting on that eigenstate? What if you had a mixed state, not a singular state, would there then be a probability of the observables being simultaneously measurable and a probability that they won't be? This wouldn't make sense to me in terms of non-commuting operators, but I'm not sure.(adsbygoogle = window.adsbygoogle || []).push({});

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# I Eigenstate of two observable operators

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