The casimir operator $\textbf{L}^2$ commutates with the elements $L_i$ of the angular momentum operator $\textbf{L}$:

$$ [\textbf{L}^2, L_i] = 0. $$

However, the $L_i$ do not commute among themselves:

$$ [L_i, L_j] = i\hbar\epsilon_{ijk}L_k. $$

This makes sense so far, but it leaves me wondering how their eigenspaces relate to each other. I remember some theorem that diagonalizable, commuting matrices share their eigenspaces. If those operators could be expressed as complex matrices (in the finite-dimensional case), they surely are diagonalizable. So it follows that $\textbf{L}^2$ has the same eigenspaces as the three $L_i$, but that would imply that they commute among themselves, which is not the case.

What am I missing? What is the relation between the eigenspaces of these operators?

  • 1
    $\begingroup$ Commuting diagonalisable matrices can be simultaneously diagonalised (i.e., with a common basis of eigenvectors). This does not mean they share their eigenspaces (two basis vectors may have the same eigenvalue for one of the matrices, but not for the other). And $A$ may share a common basis of eigenvectors both with $B$ and with $C$ individually, without $B$ and $C$ sharing a common basis of eigenvectors (think $A=0$ or $A=I$). $\endgroup$ Aug 22, 2015 at 7:59

5 Answers 5


OP is essentially pondering if commutativity is a transitive relation, ie. if three normal operators$^{1}$ $A$, $B$, and $C$ satisfies

$$ [A,B]~=~0\quad \wedge\quad [B,C]~=~0 \quad\stackrel{?}{\Rightarrow} \quad [A,C]~=~0 .\tag{T}$$

The answer is No, but OP argues via the existence of a common basis of eigenvectors for two commuting normal operators that eq. (T) should hold.

To most clearly expose the flaw in OP's argument, pick $B$ to be proportional to the identity. Then $B$ commutes with everything. Clearly we can then find two non-commuting normal operators $A$ and $C$, so that eq. (T) is violated! And a basis of eigenvectors for $A$ cannot be a basis of eigenvectors for $C$, and vice versa.


$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

  • $\begingroup$ "The flaw" is exactly OP's question, as $L^2$ is proportional to the identity. $\endgroup$
    – Omry
    Aug 22, 2015 at 8:12

Two operators may be simultaneously diagonalized if and only if they commute.

As you can see, $L_z$ commutes neither with $L_x$ nor with $L_y$ – and not with any other linear combinations different from a multiple of $L_z$ – so there's no way to diagonalize two different components of $L_i$ at all.

However, $L_z$ (and similarly other components) commutes with $L^2$, so $L_z$ and $L^2$ may be simultaneously diagonalized.

It means that whatever basis you have and $L_z,L^2$ expressed with respect to this basis, one may find a matrix $U$ on the Hilbert space such that both $UL^2 U^{-1}$ as well as $U L_z U^{-1}$ are diagonal matrices. No other component etc. may be added. We say that $L^2,L_z$ form a "complete set of commuting observables" describing the angular part of the wave function of one particle (or all the internal angular momentum, spin degrees of freedom of any particle).

You reach something like a "paradox" by discussing eigenspaces. The problem with your reasoning is that the eigenspaces are multi-dimensional in most cases but they are not subspaces of each other.

$L^2$ has different eigenvalues $\ell(\ell+1)\hbar^2$ for $\ell=0,1,2,3,\dots $ Also, half-integer values are possible for the general "spin".

But if you consider the full Hilbert space, the eigenspace corresponding to the eigenvalue with $\ell$ is not one-dimensional. Instead, it is at least $(2\ell+1)$-dimensional. It is exactly this-dimensional if there are no other degrees of freedom. If there are other degrees of freedom, the dimension of the eigenspace is a multiple of $(2\ell+1)$.

On the other hand, the eigenspace of $L_z$ is associated with the eigenvalues of this operator, $m$. You may get the eigenvalue $L_z=m$ for $\ell = |m|$ but you may also get it for $\ell=|m|+1$, $|m|+2$, or any other number greater than $|m|$ by a positive integer. So the eigenspace of $L_z$ is the linear envelope of the union of one-dimensional subspaces of eigenspaces of $L^2$.

The eigenspace of $L_x$ corresponding to the eigenvalue of $m$ picks different one-dimensional subspaces from those. Their basis vectors are neither parallel nor orthogonal to those associated with $L_z$. Similarly for $L_y$.

The multi-dimensional eigenspace of $L^2$ with the eigenvalue $\ell(\ell+1)\hbar^2$ and the multi-dimensional eigenspace of $L_z$ with the eigenvalue of $m$ have an intersection – in the simplest case of wave functions on the sphere, one-dimensional intersection – that corresponds to all states with the eigenvalues given by the quantum numbers $(\ell,m)$.

  • $\begingroup$ The first sentence needs an additional hypothesis (unless you have a definition of "operator" that makes them automatically diagonalisable). $\endgroup$ Aug 22, 2015 at 8:04

The eigenspaces of the quadratic Casimir $L^2 = L_x^2+L_y^2+L_z^2$ of the Lie algebra of infinitesimal rotations $\mathfrak{so}(3)$ are precisely the irreducible representations of $\mathfrak{so}(3)$ - we usually label a representation by its highest weight $l$, which is in this case just a number telling you what the largest possible value for any of the $L_i$ is. The representations labeled by $l$ have dimension $2l+1$, and all vectors of the same irreducible representation have $l(l+1)$ as the eigenvalue of $L^2$, so $L^2$ has degenerate eigenspaces (as it has to, since it commutes with every element of the Lie algebra, so it must be a multiple of the identity on irreducible complex representations by Schur's lemma).

We can get non-degenerate labels for the states by choosing any of the three directions (commonly one takes $L_z$) and labeling states with $l$ additionally by their eigenvalue $m_l$ for $L_z$. Now, every eigenspace with eigenvalue $l(l+1)$ for $L^2$ is spanned by the $2l+1$ states with eigenvalues $-l,-l+1,\dots,l-1,l$ for $L_z$. The operators $L_y$ and $L_x$ also can lift this degeneracy, but since the $L_i$ do not commute with each other, their eigenvectors are a different choice of basis for the eigenspaces.

In the context of atomic orbitals, $l$ is usually called azimuthal quantum number and $m_l$ is called magnetic quantum number.



The $L_i$ has many eigenspaces corresponding to many eigenvalues. Each of those eigenspaces is also an eigenspace of the Casimir operator.

So they share common eigenspaces in the sense that there are eigenspaces that are eigen to both. But they don't share them in the sense that they are the same.

Look at the hydrogen atom. There are energy eigenspaces and there Casimir eigenspaces. A Casimir eigenspace of eigenvalue $0$ contains vectors of every possible energy. And an energy eigenvector with eigenvalue $E$ c contains vectors of lots of different angular momentums. But there are common eigenvectors that have a fixed Casimir eigenvalue and a fixed energy.

There exist common eigenvectors for commuting operators but that doesn't mean a random eigenvector of one is an eigenvector of the other.


When I was asking this question, I didn't understand the relation between the commutativity of two operators and their eigenspaces:

If an operator $A$ commutates with another operator $B$, then $A$ leaves the eigenspaces of $B$ invariant:

$$ B\psi = \epsilon\psi \implies BA\psi = AB\psi = \epsilon A\psi $$

But this does not imply that $\psi$ is an eigenstate of $A$.

Maybe I mistook "leave B's eigenspaces invariant" for "B's eigenvectors are A's eigenvectors".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.