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

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 ?

• A quantum physical sysem is defined by its Hamiltonian, so the two operators cannot form a CSCO, if the Hamiltonian is excluded. Dec 3, 2020 at 13:41
• @DanielC There's no reason why a CSCO has to include the Hamiltonian of the system. I can't imagine many situations in which one wouldn't include the Hamiltonian, given its obvious importance, but in principle it isn't necessary. $(P^2+X^2, L^2,L_z)$ is a CSCO for $L^2(\mathbb R^3)$, regardless of what operator you choose to generate time evolution. Dec 3, 2020 at 14:47
• Yes, but those operators do not define the system. Only the Hamiltonian does. Dec 3, 2020 at 15:07
• @DanielC That's true, but I don't see how it's relevant here. Dec 3, 2020 at 16:58
• @DanielC That's not what complete means in this context. Complete does not mean exhaustive, but rather that there is no degeneracy left in the basis of common eigenfunctions. Dec 3, 2020 at 18:34

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.

• Very claryfing answer, thank you. Dec 3, 2020 at 14:39
• Just a little question. 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 ? Dec 4, 2020 at 10:45
• @NunzioDamino I've edited your follow-up questions and my response into the end of my answer. Dec 4, 2020 at 16:20

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.

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).

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 $$L^2$$.

• The Runge-Lenz vector commutes with $H$, but not with $L_z$. $H$, $L^2$, and $L_z$ (and the spin operators, if we're talking about those too) are sufficient to distinguish between the states that have degenerate energies. Dec 3, 2020 at 14:34
• @ JoshuaTS. Yes. I mispoke. My second error today. I should wait until my coffee has kicked in before reading stack exchange! Dec 3, 2020 at 14:35