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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 $L^2$, $L_z$?

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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 $L_z$.

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  • $\begingroup$ I agree with you when you say "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 $L_z$" but maybe it is possible to have a simultaneous eigenstate of $H$, $L^2$, and $L_z$ outside the complete set of simultaneous eigenstates of $L^2$ and $L_z$. $\endgroup$
    – Salmon
    Jan 17, 2022 at 16:12
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    $\begingroup$ Won't the completeness ensure that the new eigenstate may be expressed in terms of the spherical harmonics? $\endgroup$
    – Newbie
    Jan 17, 2022 at 16:23
  • $\begingroup$ @Newbie Yes, for sure. But the new eigenstate maybe does not coincide with the eigestates inside the "complete set". This is my question. $\endgroup$
    – Salmon
    Jan 17, 2022 at 17:10
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    $\begingroup$ @Salmone Think of it this way: Either your fictional eigenstate $\psi$ is orthogonal to the complete set $\phi_{i}$ which means then the set is not complete, or its not orthogonal to the complete set which means it should be expressed in terms of a linear combination of the eigenfunctions: $\psi=\sum_{i}c_{i}\phi_{i}$. Since $\psi$ is an eigenfunction $c_{i}=0$ for all but one of the terms, i.e., $\psi$ is already in the set. $\endgroup$
    – Newbie
    Jan 17, 2022 at 17:47
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    $\begingroup$ @Salmone I'm trying to get you to my conclusion without asking anything about Hilbert spaces or technical definitions (even though they answer the question). Instead, just remember that nothing inside of a set can be outside of that set. The "complete set" willl be the biggest set so everything else will be inside of it. If you find something else, it should be added to the complete set (because otherwise the set wasn't "complete"). $\endgroup$ Jan 17, 2022 at 20:39
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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.

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  • $\begingroup$ I don't understand one thing, why in the second case $\psi$ and $\phi_j$ are orthogonal? Didn't we say that in the second case $\psi$ is not orthogonal to the complete set? If they were two eigenvectors relative to the same eigenvalue, why should they be orthogonal? $\endgroup$
    – Salmon
    Jan 17, 2022 at 19:46
  • $\begingroup$ @Salmone Yes, in the second case the assumption is that the $\psi$ is not orthogonal to the $\phi_{i}$ but yet we know that it is an eigenvector and eigenvectors are going to be orthogonal. What I wrote is the math for this contradiction. I will get back to you for the second question. $\endgroup$
    – Newbie
    Jan 17, 2022 at 20:01
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    $\begingroup$ @Salmone I added the degeneracy case. $\endgroup$
    – Newbie
    Jan 18, 2022 at 0:22
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    $\begingroup$ Thank you so much. $\endgroup$
    – Salmon
    Jan 18, 2022 at 0:55
  • $\begingroup$ Sorry, I read your reply again and I still don't understand it. My doubt is as follows: I have, for example, a complete set of simultaneous eigenfunctions of two operators and I wonder: in the complete set of simultaneous eigenfunctions I will have $\phi_1$, $\phi_2$, $\phi_3$....$\phi_n$. If, for example, one of the eigenfunctions within the set corresponds to a degenerate eigenvalue $\lambda_1$, could there be eigenfunctions relating to that same eigenvalue outside the complete set? $\endgroup$
    – Salmon
    Jan 19, 2022 at 16:48

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