Simultaneous eigenstates of two commutating operators When two operators switch have a complete set of simultaneous eigenstates, are the simultaneous eigenstates that are part of this complete set all the simultaneous eigenstates of the two existing operators?
Could there be other simultaneous eigenstates not belonging to this set?
For example, when we're searching for the simultaneous eigenstates of the commuting operators $H$, $L^2$ and $L_z$ in the problem of a particle inside a central field, we look for them among the spherical harmonics.
How can we be sure that simultaneous eigenstates of $H$, $L^2$ and $L_z$ are among simultaneous eigenstates of $L^2$ and $L_z$, namely spherical harmonics?
Is it not possible for there to be simultaneous eigenstates of $H$, $L^2$, $L_z$ outside the set of simultaneous eigenstates of $L^2$, $L_z$?
 A: This question can be answered without anything quantum: the set of things that have properties $A$, $B$, and $C$ is always contained within the set of things that have properties $A$ and $B$, because anything with $A$, $B$, and $C$ must (by definition) have $A$ and $B$.
The converse is not true: something with properties $A$ and $B$ doesn't necessarily have all three properties  $A$, $B$, and $C$.
To apply to your problem: anything that is a simultaneous eigenstate of $H$, $L^2$, and $L_z$ must by definition be an eigenstate of $L^2$ and $L_z$. Therefore it is impossible to have a simultaneous eigenstate of $H$, $L^2$, and $L_z$ outside the set of simultaneous eigenstates of $L^2$ and $L_z$.
A: Lets say $\psi$ denotes the fictional missing eigenstate. We consider 2 cases: (1) $\psi$ is orthogonal to the complete set of the current eigenfunctions $\phi_{i}$ and (2) $\psi$ is not orthogonal to the set of the current eigenfunctions. For case (1) since the set is complete:
$$\psi=\sum_{i}c_{i}\phi_{i}=\sum_{i}\langle\psi\cdot\phi_{i}\rangle\phi_{i}$$ but due to our assumption $\langle\psi\cdot\phi_{i}\rangle=0$. Thus, $\psi=0$.
In case (2) we can express $\psi$ as a linear combination of the complete set members:
$$\psi=\sum_{i}\langle\psi\cdot\phi_{j}\rangle\phi_{i}$$
Now if $\psi$ is the eigenvector corresponding to a nondegenerate eigenvalue then $\langle\psi\cdot\phi_{j}\rangle=0$ since we want $\psi\neq\phi_{j}$ for all members of the set (based on our assumption). On the other hand if $\psi$ corresponds to a degenerate eigenvalue then there is a subset of the complete set of which $\psi$ may be expressed as a linear combination. Note that while this subset is degenerate, we use the Gram-Schmidt process to create orthogonal members for the set. For the sake of simplicity lets say that the set is $\{\phi_{1},\phi_{2}\}$ where $\langle\phi_{1}\cdot\phi_{2}\rangle=0$. Using the Gram-Schmidt process we can define a substitute pair of orthogonal eigenvectors with $\psi_{1}=\psi$ and $\psi_{2}$. Due to degeneracy they will have the same eigenvalue and will be linear combinations of $\phi_{1}$ and $\phi_{2}$. More importantly, this implies that the degenerate subset of the complete set may be replaced by a degenerate subset which $\psi$ becomes a member of and this contradicts the assumption that $\psi$ is missing from the complete set.
