Compatibility Thm HW: Can We Find More Orthon Eigenstates?

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SUMMARY

The discussion revolves around the Compatibility Theorem in quantum mechanics, specifically regarding the uniqueness of eigenstates associated with operators \(\hat{A}\) and \(\hat{B}\). It is established that if the eigenvalues \{A_i, B_j\} correspond uniquely to a vector in the basis, then \{\hat{A}, \hat{B}\} constitutes a Complete Set of Commuting Observables (CSCO). The confusion arises from the assertion that \(\tilde{u_1}\) and \(\tilde{u_2}\) are the only eigenstates of \(\hat{B}\) in the specified plane, despite the potential for additional orthonormal eigenstates within the span of degenerate states. The discussion clarifies that the eigenvectors of \(\hat{B}\) are unique when the eigenvalues are not degenerate.

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davon806
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Homework Statement


Please see the following,I am confused by the word "only".
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Homework Equations

The Attempt at a Solution


I understand that the Compatibility theorem ensures we can find a basis of common eigenfunctions of \hat{A} ,\hat{B}.If each pair of eigenvalues {A_i,B_j} identifies uniquely one vector of the basis,then the set {\hat{A} ,\hat{B}} forms a CSCO.
Here they are asserting \tilde{u_1} , \tilde{u_2} are the only eigenstates of B in the plane.I don't see the reason for that.Could we find more orthonormal eigenstates of B in the plane spanned by the degenerate states?
 
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davon806 said:
Could we find more orthonormal eigenstates of B in the plane spanned by the degenerate states?
They are not degenerate with respect to B. That's exactly the case (1) discussed there, the two eigenvectors with respect to B are unique. If both eigenvalues of B are the same, see case (2).
 
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