Shared eigenstates of angular momentum

Question

The angular momentum operators $L_x$, $L_y$ and $L_z$ don't commute with each other but they do all commute with the operator $L^2 = L_x^2+L_y^2+L_z^2$. I know that if two matrices $A$ and $B$ commute and $v$ is an eigenvector of $A$ and $A$ has no identical eigenvalues then $v$ must also be an eigenvector of $B$.

I am wondering how to interpret that. Are for example all eigenstates of $L_z$ also eigenstates of $L^2$? If this is the case, does this mean that $L^2$ has more eigenstates than $L_z$ since it also shares eigenstates with $L_x$ and $L_y$? Do the all eigenstates of $L^2$ consist of the union of the eigenstates of $L_x$, $L_y$ and $L_z$?

Answer 1

Are for example all eigenstates of $L_z$ also eigenstates of $L^2$?

No. There are plenty of situations where you can have eigenstates of $L_z$ that are not eigenstates of $L^2$.

The simplest example is probably the one you have in mind and the first one you were introduced to (if your route into quantum mechanical angular momentum followed the normal steps), namely, the hydrogen atom. (Or, by extension, any particle in 3D in a central potential.)

Here, the eigenstates are normally denoted as $|n,\ell,m⟩$, where $n$ is the principal quantum number, $\ell$ labels the eigenvalue of $L^2$ and $m$ labels the eigenvalue of $L_z$.

The important thing to note here is that for any given value of $m$, there are multiple different $\ell$ subspaces that contain states with that eigenvalue of $L_z$. This means that you can form superpositions with the chosen $m$ but with different $\ell$: say, something like $$ |3,1,1⟩+|3,2,1⟩ = |3p_1⟩ + |3d_1⟩, $$ which has a well defined $L_z=1$, but which does not have a well-defined $L^2$.

---

Now, as pointed out in the existing answer on this thread, once you get more into the details of quantum mechanics and things get correspondingly more abstract (and, particularly, once you learn to contextualize the quantum mechanical theory of angular momentum as one application of group representation theory), then it becomes more common to consider Hilbert spaces that contain a single eigenvalue of $L^2$ (in technical language, "irreducible representations", often abbreviated as "irrep"s).

If you are within that setting, then all eigenstates of $L_z$ are also eigenstates of $L^2$, simply because all the states in the space under consideration are eigenstates of $L^2$, and they all have the same eigenvalue. However, from the way you phrased the question, I think it's unlikely that you're thinking about a setting like this one.

---

Regarding the second half of your question,

If this is the case, does this mean that $L^2$ has more eigenstates than $L_z$ since it also shares eigenstates with $L_x$ and $L_y$? Do the all eigenstates of $L^2$ consist of the union of the eigenstates of $L_x$, $L_y$ and $L_z$?

there is largely no objective way of comparing which operator has "more" eigenstates. In any nontrivial situation (i.e. when you have more than one eigenspace of $L^2$ in your Hilbert space), there is an infinity of eigenstates of $L^2$ that are not eigenstates of $L_z$ (and not eigenstates of $L_x$ or $L_y$ either), and there is also an infinity of eigenstates of $L_z$ which are not eigenstates of $L^2$.

There are two trivial cases in this regard,

• the case of $\ell=0$, where there is a single state involved; and
• the case of $j=1/2$, where every eigenstate of $J^2$ is an eigenstate of a component $\hat{\mathbf n}\cdot\mathbf J$ of $\mathbf J$ along some direction $\hat{\mathbf n}$ (but not necessarily the restricted set of Cartesian coordinates along some pre-chosen set of axes);

but for anything bigger than that, the sets in question are just different enough that you cannot really compare them.

Answer 2

There's a slight detail you are missing.

I know that if two matrices 𝐴 and 𝐵 commute and 𝑣 is an eigenvector of 𝐴 and 𝐴 has no identical eigenvalues then 𝑣 must also be an eigenvector of 𝐵

It's more delicate than that! What you have is that, given two matrices A and B which commute, then you can find a basis of common eigenvectors. So in fact, you can calculate an eigenvector of A in such a way that it is not an eigenvector of B. The important thing is that you can find eigenvectors which are common to both. But given an eigenvector of A it is not necessarily eigenvector of B and vice-versa.

The same goes for angular momentum. The eigenvectors of $L^2$ are not necessarily eigenvectors of $L_z$, but you can always find a set that is common to $L^2$ and $L_z$.

Answer 3

Thanks to Emilio and Henrique and this post I found the answer to my questions, and I would like tu sum it up here :

Two commuting matrices share at least one common eigenbasis, but if there are some degeneracies, they won't share all their eigenvectors. $L^2$ and $L_z$ can have some degeneracies so in general not all eigenstates of $L_z$ are eigenstates of $L^2$ and vice-versa. A good examples of matrices having such properties would be the identity matrix and some non commuting matrices. The identity commutes with all the other matrices and since its eigenvalues are all the vector it share an eigenbasis with the other matrices, but not all its eigenvectors.

Answer 4

Are for example all eigenstates of $L_z$ also eigenstates of $L^2$?

Yes: that's what you generic rule dictates, no? What holds you back? A is $L_z$ here, assuming you are in an irreducible representation, so $L_z$ is not degenerate.

If this is the case, does this mean that $L^2$ has more eigenstates than $𝐿_z$ since it also shares eigenstates with $𝐿_𝑥$ and $𝐿_𝑦$?

Of course. What holds you back?

Do the all eigenstates of $𝐿^2$ consist of the union of the eigenstates of $𝐿_𝑥$, $𝐿_y$, $𝐿_z$?

They include the union of the above, but in general, they can, and do, include many more states which are not eigenstates of all of the above.

For an irreducible representation, i.e. for states with a common eigenvalue for $L^2$, this operator is proportional to the identity, so all states of the corresponding dimensionality are its eigenstates. This is an infinity of them.

Illustrate it with spin 1/2, where all these operators are proportional to the Pauli matrices and the 2d Identity matrix, respectively. So any complex 2-vector is an eigenstate of $L^2$.

---

Edit (geeky) in response to comment

Tip of the hat @Emilio for bringing up the complication of when you are working in a Hilbert space which is a direct sum of more than one irrep, so reducible multiplets as in the famously degenerate one of the Hydrogen atom.

In this case, absorbing radial dependences appropriately, the state for, e.g., n =3, $ \alpha |l=1,m=1\rangle + \beta|l=2,m=1\rangle$, is an eigenstate of $L_z$ with eigenvalue 1, for all β and α; but, naturally, unless β =0 or α =0, this is degenerate, and fails the hypothesis for A. In this case, $L_z$ is degenerate, and does not share all of its eigenstates with $L^2$, since $L^2(\alpha|1,1\rangle + \beta|2,1\rangle)=2\alpha |1,1\rangle +6 \beta|2,1\rangle $.

Mastering the irrep above, you can easily handle this one.