Eigenfunctions of compatible observables that are not shared

Question

I'm using D.J. Griffiths's Introduction to Quantum Mechanics (3rd. ed), reading about the angular momentum operators $\mathbf L=(L_x,L_y,L_z)$ and $L^2$ in chapter 4. The author discusses eigenfunctions $f$ that are specifically eigenfunctions of both $L^2$ and $L_z$:

$$L^2f = \lambda f\qquad\qquad\qquad L_zf = \mu f$$

The fact that $f$ can be an eigenfunction of both follows from $[L^2,L_z]=0$: $L^2$ and $L_z$ commute and are thus compatible, meaning that measuring the one brings the system into an eigenstate of itself and also the other.

After a discussion involving the transformation of the angular momentum operators to spherical coordinates, the author writes at the end:

"Conclusion: Spherical harmonics are the eigenfunctions of $L^2$ and $L_z$." (emphasis his)

Now, I'm worried that this is only partially true, and that it should really be "Spherical harmonics are shared eigenfunctions of $L^2$ and $L_z$", not blatantly "the eigenfunctions", since we assumed during the derivation that $f$ was an eigenfunction of both operators. What about the other eigenfunctions?

That's not too bad, I thought at first: now that we neatly know the properties of the shared eigenfunctions, can't we just write the remaining eigenfunctions as a linear combination of the shared eigenfunctions to analyse their properties as well? The answer seems to be no, on second thought, since in chapter 3 of the same textbook, the following was posed as a derivable theorem:

$$\textrm{incompatible observables $A$ and $B$ do not have a complete set of shared eigenfunctions}$$

equivalent to either of the following statements:

$$[A,B]\neq0 \Rightarrow \textrm{$A$ and $B$ do not have a complete set of shared eigenfunctions}$$ $$\textrm{$A$ and $B$ have a complete set of shared eigenfunctions} \Rightarrow [A,B]=0$$

That means that we can't necessarily write the remaining eigenfunctions of $L^2$ and $L_z$ as linear combinations of the shared ones, because the arrow points the wrong way for it to definitely be possible.

Researching the theorem's content, I came across two threads that stated something about this:

• This thread says: "Let's start with only 2: operators $A$ and $B$. If $[A,B]=0$, there is at least one orthonormal basis of common eigenvectors."

• This thread seems to make a stronger assertion: "(...) compatible operators are guaranteed only to have the same eigenvectors, not the same eigenvalues."

So, after all, maybe we can analyse the eigenfunctions of both that aren't shared, but I have no proof that this is possible. To converge onto a question, I'm wondering:

1. Is the $\Rightarrow$ in the given theorem generalisable to a $\Leftrightarrow$?
   • This would justify that the author only discuss the shared eigenfunctions.
2. In a stronger fashion, do compatible observables even have eigenfunctions that they don't share in the first place? If that's true, what if a measurement is made and such an eigenfunction is gained - are the observables then suddenly incompatible?
   • If this is not true, then I feel the author is justified in claiming that the spherical harmonics are indeed the eigenfunctions of $L^2$ and $L_z$, since they are shared.

Answer 1

Claim: $[\hat A,\hat B]=0$ $\iff$ $\hat{A}$ and $\hat{B}$ have a complete set of common eigenfunctions.

Proof

If $\hat{A}$ and $\hat{B}$ have a complete set of common eigenfunctions $|\psi_n\rangle$, then $\hat{A}\hat{B}|\psi_n\rangle=\hat{A}B_n|\psi_n\rangle=B_n\hat{A}|\psi_n\rangle=B_nA_n|\psi_n\rangle=\hat{B}A_n|\psi_n\rangle=\hat{B}\hat{A}|\psi_n\rangle$. Thus $[\hat{A},\hat{B}]|\psi\rangle=0$ for any state $|\psi\rangle$, thus $[\hat{A},\hat{B}]=0$.

On the other hand, say $[\hat{A},\hat{B}]=0$. Let $|\psi_n\rangle$ be a complete set of eigenstates of $\hat A$. Let's focus on all the eigenstates of $\hat{A}$ with a given eigenvalue $\lambda$. Then $\hat{A}\hat{B}|\psi_n\rangle=\hat{B}\hat{A}|\psi_n\rangle=\hat{B}\lambda|\psi_n\rangle=\lambda\hat{B}|\psi_n\rangle$. In other words, $\hat{B}$ takes an eigenstate of $\hat{A}$ with eigenvalue $\lambda$ to another eigenstate of $\hat{A}$ with eigenvalue $\lambda$. Thus, if you write $\hat{B}$ in the basis of eigenstates of $\hat{A}$, it will take a block-diagonal form with nonzero elements only between eigenstates with the same eigenvalue of $\hat{A}$.

Thus, in the basis of eigenstates of $A$, if we arrange the eigenstates in order of increasing eigenvalue, we have

$$ \hat{B} = \left[\begin{matrix} \begin{matrix}b_{11} & b_{12}\\b_{21}&b_{22}\end{matrix}& 0 & \cdots & 0&\\0&\begin{matrix}b_{33} & b_{34} & b_{35}\\b_{43}&b_{44}&b_{45}\\b_{53}&b_{54}&b_{55}\end{matrix}& \cdots & 0\\ \vdots & &\ddots\\0&\cdots&&b_{nn} \end{matrix}\right] $$

Here, I've assumed that the first two eigenstates of $\hat{A}$ are degenerate, the next three are degenerate, etc. You should convince yourself that you can diagonalize this matrix by only combining eigenstates of $\hat{A}$ with the same eigenvalue. Once you've done this, you have a complete basis that consists of mutual eigenstates.

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As a side note, I glanced through Griffiths and was shocked to discover he did not prove this. I know Griffiths is considered a "lighter" textbook but I always thought it had a reasonable presentation of the basics. Consider reading chapter 1 of Shankar for a more complete presentation of the linear algebra. In particular, on page 43 he provides this proof.

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To answer your second question, consider the $Y_{\ell m}$ functions, which I'll denote by $|\ell, m\rangle$. These are eigenfunctions of both $L^2$ and $L_z$, with eigenvalues $\hbar^2\ell(\ell+1)$ and $\hbar m$, respectively. On the other hand, you should be able to convince yourself that the state $|1,-1\rangle+|1,1\rangle$ is an eigenstate of $L^2$ but not $L_z$, while the state $|1,1\rangle+|2,1\rangle$ is an eigenstate of $L_z$ but not $L^2$. If you want some intuition about how these states interact with measuring, I'll give an example. Say the angular momentum state of a particle is $\frac{1}{\sqrt{3}}[|1,1\rangle+|1,-1\rangle+|2,1\rangle]$. If we measure $L^2$, we have a $2/3$ chance of measuring $\ell=1$, and a 1/3 chance of measuring $\ell=2$. If we measure $\ell=1$, then the resulting state is $\frac{1}{\sqrt{2}}[|1,1\rangle+|1,-1\rangle]$, which still doesn't have a definite $L_z$. Then if we measure $L_z$, we'll get $m=\pm 1$ with probability $1/2$. The final state will be $|1,1\rangle$ or $|1,-1\rangle$. The "compatibility" of $L_z$ and $L^2$ doesn't mean that you have a definite $L_z$ value iff you have a definite $L^2$ value. On the other hand, it DOES mean that if you have a definite $L^2$ value and measure $L_z$, you still have a definite $L^2$ value.