# Mutual or same set of eigenfunctions if two Hermitian operators commute

1. If two operators commute, do they have "a mutual set of eigenfunctions", or "the same set of eigenfunctions"? My quantum chemistry book uses these as if they are interchangeable, but they do not seem to be the same in a very significant way.

2. One direct consequence of my confusion with respect to this arises when considering angular momentum operators and the fact that: $$[\hat{L}^2, \hat{L}_x] = [\hat{L}^2, \hat{L}_y] = [\hat{L}^2, \hat{L}_z] = 0$$ which implies that $\hat{L}^2$ shares a mutual set of eigenfunctions with $\hat{L}_x, \hat{L}_y,$ and $\hat{L}_z$. However, the spherical harmonics (which I thought were the only eigenfunctions of $\hat{L}^2$) are only eigenfunctions of $\hat{L}_z$ and $\hat{L}^2$ (when considering these 4 operators)! Thus what are the eigenfunctions that $\hat{L}^2$ shares with $\hat{L}_x$ and $\hat{L}_y$ since we know there must be some from the commutation relation!? (I understand that we can redefine which axis is x, y, and z, but my point is that only one of the three axes can have its angular momentum component operator have the spherical harmonics as eigenfunctions regardless of how you define your axes).

3. $\hat{L}_x$, $\hat{L}_y$, and $\hat{L}_z$ do not commute with each other, yet they all three commute with a fourth common operator as already mentioned, $\hat{L}^2$! It doesn't make sense to me how it's possible for $A$ to commute with $B$ and $B$ to commute with $C$ yet $A$ not to commute with $C$.

4. If two operators don't commute, can they still share, for example, 1 eigenfunction, or must they not share any eigenfunctions at all?

Any answer that doesn't presuppose any extensive math background beyond differential equations and basic linear algebra would be greatly appreciated! I'm taking first semester quantum chemistry. Thanks!

• For part 3, replace "B" with "the identity" and see if there's still no sense to be made ;)
– user10851
Nov 28, 2015 at 5:35

Assumptions: I will be talking about Hermitian (more generally self-adjoint) operators only. This means that I will assume that the operators in question have a set of eigenvectors that span the Hilbert space. As mentioned by tomasz in a comment, this is not exactly necessary, since more general statements can be made, but since we are dealing with basic QM, I figure this simplification is reasonable.

Questions 1 and 2

The statement is that if two operators commute, then there exists a basis for the space that is simultaneously an eigenbasis for both operators. However, if (for instance) one of the operators has two eigenvectors with the same eigenvalue, any linear combination of those two eigenvectors is also an eigenvector of that operator, but that linear combination might not be an eigenvector of the second operator.

Case in point: we consider the states $|l,m\rangle = |1,1\rangle$ and $|l,m\rangle = |1,-1\rangle$. These are both eigenvectors of $\hat{L}^2$ and $\hat{L}_z$. The eigenvalues of $\hat{L}_z$ are $\hbar$ and $-\hbar$, respectively, but the state $$|1,1\rangle + |1,-1\rangle$$ is clearly not an eigenvector of $\hat{L}_z$, but it is still an eigenvector of $\hat{L}^2$. This is probably why we use the term "mutual" rather than "same". Finally, we form linear combinations of the $|1,m\rangle$ states to get the $l=1$ states that are eigenvectors of, say, $\hat{L}_y$.

As a straight-forward example, consider the following two matrices: $$L = \left[ {\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}} \right]$$ and $$Z = \left[ {\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}} \right].$$ The vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are eigenvectors of both operators. The latter two are eigenvectors of $L$ with the same eigenvalue (namely $1$), but they are eigenvectors of $Z$ with different eigenvalues (namely, $1$ and $-1$). If we instead use the vectors $(0,1,1)$ and $(0,1,-1)$, these are still eigenvectors of $L$ with eigenvalue $1$, but they are no longer eigenvectors of $Z$, as you can verify.

Question 3

As cleverly pointed out by Chris White in a comment, the identity operator commutes with every single other operator, and yet there are operators that don't commute with each other. This is the simplest of counter-examples to your intuition. Commutativity is not a transitive property.

I don't know whether your intuition about the problem was mathematical or physical. If it was physical, then this is something you have to just get used to, because it is part of nature that things work this way. If it is mathematical, then perhaps the counter-example provided above will help.

As a straight-forward example, add to the above matrices the matrix $$X = \left[ {\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}} \right]$$ This operator commutes with $L$ but not $Z$.

Question 4

It is certainly okay for two operators that don't commute to share an eigenvector. In fact, $\hat{L}_x$, $\hat{L}_y$, and $\hat{L}_z$ all share an eigenvector: the state $|l,m\rangle = |0, 0\rangle$.

As a more trivial example of operators that share an eigenvector that don't commute, consider the following two: $$A = \left[ {\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}} \right]$$ and $$B = \left[ {\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}} \right]$$ They clearly share the eigenvector $(1,0,0)$, but the lower blocks of the matrices A and B are respectively the Pauli-$z$ and Pauli-$x$ matrices, which do not commute.

• About the existence of an eigenbasis: this is provided that the two operators do have eigenbases at all. The general fact there is that two commuting operators preserve one another's eigenspaces (as an obvious consequence of linearity), but it certainly doesn't mean that when two operators commute, they have to necessarily have eigenbases at all. Of course, they do have them if they are hermitian. Nov 28, 2015 at 13:17
• @tomasz. I will add this clarification. Nov 28, 2015 at 15:07
• @tomasz Not all self-adjoint operators have eigenbases (only the ones either compact or with compact resolvent); but they have an associated spectral decomposition. And the property of "commuting" for unbounded self-adjoint operators is slightly more complicated than the usual $[A,B]=0$ (you need to check that the associated spectral families commute, in order to have a common spectral decomposition and other desirable properties). Nov 28, 2015 at 16:09
• @yuggib. This is also true. I figured I'd avoid the complexities of operators acting on infinite-dimensional Hilbert spaces, given this is a question from a student in basic quantum chemistry course (although I did use the term "self-adjoint", so...) Nov 28, 2015 at 16:14
• @JohnDoe. You can define equivalent spherical harmonics that are eigenfunctions of $L_x$ and $L_y$; they are just rotations of the spherical harmonics. For instance, to get the eigenfunctions of $L_x$, rotate the coordinates so that the z-axis becomes the x-axis (or maybe the z-axis becomes the negative x-axis; I'd have to check). These rotated functions are no longer eigenfunctions of $L_z$, but they have the same exact structure, and in fact if you defined $\theta$ and $\phi$ relative to the x-axis, they would be identical in form (but still different functions). Sep 27, 2016 at 17:23
1. We usually say that if two operators, $\hat{A}$ and $\hat{B}$ commute, then they have a simultaneous set of eigenstates. Saying that the eigenstates are the same isn't really correct.

For example, let operator $\hat{A}$ be hermitian and act on elements of the Hilbert Space $\mathcal{H}_A$ and let operator $\hat{B}$ also be hermitian and act on elements of the Hilbert Space $\mathcal{H}_B$ and let $\mathcal{H}_A \ncong \mathcal{H}_B$ so that the two spaces are clearly distinct.

By the spectral theorem, $\hat{A}$ has a set of eigenvectors $\lvert \psi_{a_n} \rangle$ with real eigenvalues $a_n$ that form a basis for $\mathcal{H}_A$ and similarly for $\hat{B}$, $\lvert \phi_{b_n} \rangle$, $b_n$ and $\mathcal{H}_B$. Then, any state of the form $\lvert \psi_{a_n} \rangle \times \lvert \phi_{b_n} \rangle$ is a simultaneous eigenvector of both $\hat{A}$ and $\hat{B}$; however, the sets of states $\lvert \psi_{a_n} \rangle$ and $\lvert \phi_{b_n} \rangle$ are certainly not the same; they aren't even elements of the same space!

1. Recall that in the physicist's spherical coordinate system, $\theta \in [0, \pi]$ is the azimuthal angle measured from the $+ \hat{z}$ axis and $\phi \in [0, 2\pi]$ the polar angle. So, the spherical harmonics that you see on Wikipedia have implicitly chosen a particular axis to be called $\hat{z}$.

As you point out, which axis we call which is arbitrary. So, the simultaneous eigenfunctions of $\hat{L}^2$ and $L_x$ or $L_y$ can be written is the same form as the spherical harmonics, except that we now let $\theta \to \theta_x$ or $\theta \to \theta_y$ be an azimuthal angle measuring the angles with respect to the $x$ or $y$ axis and then let $\phi \to \phi_x$ or $\phi_y$ be the corresponding polar angle. Essentially what we're doing here is just shuffling around which label we put on which axis.

What we have are three different representations of the same set of eigenfunctions of $\hat{L}^2$. Because these are equivalent representations, we can certainly write the simultaneous eigenfunctions of $\hat{L}^2$ and $\hat{L}_x$ as a linear combination of the eigenfunctions of $\hat{L}^2$ and $\hat{L}_z$, which I leave to you as a worthwhile exercise.

1. Chris White's comment hopefully makes it clear that we should not always trust our intuition. If commutativity were transitive, as you suggest, then we would be forced to conclude that all operators commute!

2. Operators can certainly share some simultaneous eigenfunctions even if they don't commute. For example, $Y_0^0$ doesn't contain any $\theta$ or $\phi$ in its usual representations, so it is the same regardless of how you label the axes. $Y_0^0$ is then an eigenfunction of all of $\hat{L}^2$, $\hat{L}_x$, $\hat{L}_y$, and $\hat{L}_z$.

Whether non-commuting operators share some or no eigenfunctions depends exactly on what their commutator is.