Uncertainty principle manifesting in $j(j+1)$ vs $j^2$

Question

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..."

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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.

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.