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.

1 Answer 1


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


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.

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.

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.

  • $\begingroup$ Thanks for the input! I'm not accepting the answer just yet, as I want my second question to be answered as well. Also, although I'm only taking my first course in QM, I have to say I don't really dig Griffiths all that much. He's kind of all-over-the-place and doesn't refer to rigorous mathematics as often as he should (e.g. Sturm-Liouville theory), and he tends to use problems and examples as theorems. It also takes him just short of 100 pages to get to observables. $\endgroup$
    – Mew
    Commented Apr 26, 2020 at 15:48
  • $\begingroup$ @Mew Edited, maybe it answers both your questions now? $\endgroup$ Commented Apr 26, 2020 at 16:15
  • $\begingroup$ Right. Key take-away, based on your closing sentence: compatibility of two observables does not mean that measuring the first brings the system into an eigenfunction of the second and vice versa. Instead, it means that measuring the second after measuring the first (and vice versa) doesn't invalidate the first of both measurements. $\endgroup$
    – Mew
    Commented Apr 26, 2020 at 16:45
  • $\begingroup$ +1: Excellently thorough answer! @Mew Notice that the fact that measuring one operator does not bring the state to an eigenstate of the second operator has everything to do with the degeneracy of the spectra of the operators. If you have two compatible operators and none of their spectra is degenerate then measuring the one would necessarily bring the state to an eigenstate of both the operators. $\endgroup$
    – user87745
    Commented May 17, 2020 at 0:29
  • 1
    $\begingroup$ @Mew It's not as edgy a case as you might imagine ;) In fact, we rarely care about compatible observables if both of them are non-degenerate--because we'd automatically know about the value of the other by simply measuring one. Think about it, how many examples can you recall where we use two compatible observables with both having non-degenerate spectra? About your other point, yes, but it would only work in one direction, if you measure $m=0$ first, that wouldn't tell you anything about $l$. Also, getting lucky means that you happen to land in a non-degenerate eigen-subspace of an operator. $\endgroup$
    – user87745
    Commented May 17, 2020 at 1:08

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