Is $(L^2, L_z)$ a complete set of commuting observables?

Question

According to the main definition we define a (C.S.C.O.) complete set of commuting observables $(A,B,C, \dots)$ if:

1. Every commutator between the operators of the list is $0$
2. If we fix the eigenvalues of the operators there exists a unique eigenvector with these eigenvalues.

(Anyway, there is a reference for the exact formal definition of this concept ? In every textbook I have this concept is introduced with just a brief discussion on the subject.)

If I follow blindly this definition I conclude that ($L^2$, $L_z$) is a CSCO, because if a fix a value of $l$ and a value of $m$ there exists a unique eigenvector (namely a unique spherical harmonic for every fixed value of $l$ and $m$). But if this set is complete, why in the study of the Hydrogen atom I can add to the set the Hamiltonian $H$ ?

For myself the set must not be complete, because if I fix just one value of $l$ or either $m$, I can clearly notice the degeneracy. I even think that the latter reasoning may serve as a method to find that the set of observables is not complete, but I haven't found any reference in the literature.

So, what parts of my reasoning are wrong ?

Answer 1

It depends on the Hilbert space.

If I follow blindly this definition I conclude that $(L^2, L_z)$ is a CSCO, because if a fix a value of $l$ and a value of $m$ there exist a unique eigenvector (namely a unique spherical harmonic for every fixed value of $l$ and $m$)

This is true if you're considering $L^2(S^2)$, the natural Hilbert space for particles confined to the surface of a sphere. However, the Hydrogen atom lives in $L^2(\mathbb R^3) \simeq L^2(\mathbb R \times S^2)$. In the latter space, the eigenstates of fixed $l$ and $m$ are degenerate; since the hydrogen atom wavefunctions can be written $\psi_{nlm}$, clearly for a fixed $l$ and $m$ we can have many different states corresponding to an infinity of possible $n$'s.

The addition of the hydrogen atom Hamiltonian as a third commuting observable breaks this degeneracy, and so $(H,L^2,L_z)$ are a complete set of commuting observables for $L^2(\mathbb R^3)$.

Note also that if we consider the spin of the electron as well, our Hilbert space becomes $L^2(\mathbb R^3) \otimes \mathbb C^2$, and the states of fixed $n,l,m$ are now doubly degenerate. To break this degeneracy, we need to add another mutually-commuting observable such as $S_z$.

In the latter case if I add the $S_z$ operator, now the states with $(n,l,m,s_z)$ are degenerate and I can lift this degeneracy by adding $S$ resulting at the end with a C.S.C.O ? And In general I can state that the degeneracy is equal to the dimension of the Hilbert space minus the number of operators ?

The answer to both questions is no. If your Hilbert space is $L^2(\mathbb R^3)\otimes \mathbb C^2$ and you consider the observables $(H,L^2,L_z)$, then the eigenspace corresponding to some $(n,l,m)$ is two-dimensional, because a general eigenstate of $H,L^2,$ and $L_z$ would be of the form

$$\Psi_{nlm} = \psi_{nlm}(\mathbf x) \otimes\pmatrix{\alpha \\ \beta}$$

for some arbitrary $\alpha,\beta\in \mathbb C$. To lift this degeneracy, we add $S_z$ to the set. Now the most general state corresponding to e.g. $(n,l,m,+1/2)$ would be of the form

$$\Psi_{nlm\uparrow} = \psi_{nlm}(\mathbf x) \otimes \pmatrix{\alpha \\ 0 }$$

for arbitrary $\alpha\in\mathbb C$, so the corresponding eigenspace is one-dimensional. This is what we mean by non-degeneracy in this context.

The answer to your second follow-up question is also no. There's no connection between the number of operators and the dimensionality of the Hilbert space. A simple example would be the infinite dimensional Hilbert space $L^2(\mathbb R)$ equipped with harmonic oscillator Hamiltonian $H_{QHO}$. Because $H_{QHO}$ has no degeneracy, it comprises a CSCO all by itself.

Answer 2

What you’re missing is to account for “complete”. In practice this means: do you have enough observables to uniquely label quantum states? In the case of $L^2$ and $L_z$, it will not be enough to uniquely label hydrogen states, or the states of a 3d harmonic oscillator, or for that matter the states in any 3d central potential.

Answer 3

Regarding the Hydrogen atom, being in an eigenspace of both $L^2$ and $L_z$ means knowing the type of orbital the electron is in ($s$, $p$, $d$, etc.) - this gives the $l$ label - and also which specific orbital it is in ($p_x$, $p_y$, $d_{x^2-y^2}$, etc.) - this gives the $m$ label - see here.

However, every shell (labelled by $n$) has an $s$-orbital, every shell with $n\geq 2$ has a $p_x$, $p_y$ and $p_z$ orbital, etc. In other words, knowing that the electron "is in a $p_x$ orbital" doesn't give complete information, the remaining information is given by specifying which eigenspace of the Hamiltonian we are in (this decides the $n$ label).

Answer 4

The hydrogen atom, or more precisely the quantum Kepler problem, has another variable that commutes with $L^2, L_z$ and $H$. This is the Runge-Lenz vector and it accounts for the extra degeneracy with $E$ depending only on $n$ and not on $L^2$.