To motivate my question, please consider a system with total angular momentum $j$. The fact that the largest eigenvalue of $J_z$ is $j$, while $J^2 = J_x^2 + J_y^2 + J_z^2$ has all eigenvalues equal to $j(j+1)$ is often ascribed to the uncertainty principle.

For example, quoting page $51$ of the textbook "Models of Quantum Matter" by Hans-Peter Eckle,

"...However, the eigenvalue of $L^2$ is $l(l+1)$, larger than $l^2$. This implies that the angular momentum operator $\bf{L}$ can never align with certainty with $L_3$ and the uncertainty principle is satisfied. If $\bf{L}$ could be aligned with $L_3$, then $L_1=L_2=0$ and we would have simultaneously sharp values of all three components of the angular momentum operator, in contradiction to Heisenberg's uncertainty relations..."

To what extent is this reasoning true in general? It is hard for me to formulate my question more precisely, but I will attempt to do so:

Consider some set of Hermitian operators $K_i$ which all pairwise do not commute but instead each commute with the sum of their squares, $K^2 = \sum_i K_i^2$. Is it guaranteed that $K^2$'s largest eigenvalue is strictly greater than any of the eigenvalues of the individual $K_i^2$? The strictly greater is key, as it is greater than or equal by this answer. I hope to see that the failure of the individual $K_i$ to commute amongst themselves imposes a stronger statement.

  • 3
    $\begingroup$ I think you also need to assume that $K^2$ commutes with all the $K_i$ in order to have a potential problem with the uncertainty principle (otherwise $K^2$ itself is not observable at the same time as the $K_i$). I think that could potentially be an important point, since $K^2$ being an invariant (ie, commuting with each $K_i$) points to having a rotational symmetry. $\endgroup$
    – Andrew
    Mar 9, 2022 at 6:29
  • $\begingroup$ @Andrew That's a good point about needing $K^2$ to be commuting with the $K_i$ in order to bump up against the uncertainty principle. I've edited my question accordingly. $\endgroup$
    – user196574
    Mar 9, 2022 at 18:57

1 Answer 1


The following two operators $A$ and $B$ have the properties that:

  • $A$ and $B$ do not commute;
  • $A^2 + B^2$ commutes with both $A$ and $B$; and
  • the largest eigenvalue of $A^2$ equals the largest eigenvalue of $A^2 + B^2$:

$$ A = \begin{bmatrix} \lambda & 0 & 0 \\ 0 & 0 & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 0 \end{bmatrix} \qquad B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & i/\sqrt{2} \\ 0 & -i/\sqrt{2} & 0 \end{bmatrix} $$ for $\lambda \geq 1$. We have $$ [A, B] = \begin{bmatrix} 0 & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & i \end{bmatrix} \neq 0 $$ so the two operators do not commute. However,
$$ A^2 = \begin{bmatrix} \lambda^2 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1/2 \end{bmatrix} \quad \text{ and } \quad A^2 + B^2 = \begin{bmatrix} \lambda^2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. $$ and we can see that for any $\lambda > 1$, the largest eigenvalues of both $A^2$ and $A^2 + B^2$ are $\lambda^2$. It can also be easily shown that $A^2 + B^2$ commutes with both $A$ and $B$.

The "loophole" being exploited here is that $[A, B]$ has a non-trivial null space even though the commutator does not itself vanish. Since $A$ and $B$ are simultaneously diagonalizable on this subspace, we're allowed to have a state with zero uncertainty of $A$ and $B$ lying within this subspace.

  • $\begingroup$ This is very nice. Is the converse true that if $[A,B]$ has a trivial null space then one must get strictly greater than? $\endgroup$
    – user196574
    Mar 9, 2022 at 23:24
  • $\begingroup$ @user196574: That would be the obvious question, but I don't immediately see a way to prove it. I'll have to think on it further. $\endgroup$ Mar 10, 2022 at 13:06

Your Answer

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

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