[Physics] Mutual or same set of eigenfunctions if two Hermitian operators commute

commutatorhilbert-spacelinear algebraoperatorsquantum mechanics

  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!

Best Answer

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.

Related Question