[proofplan] The proof expands the commutator of $J_i$ with the quadratic operator $J^2$ using the product rule for commutators. The angular momentum commutation relations convert each first-order commutator into a linear combination of the components $J_k$. The resulting double sum contains a factor symmetric in the indices $j,k$ multiplied by the Levi-Civita symbol, which is antisymmetric in those same indices; pairing the terms indexed by $(j,k)$ and $(k,j)$ gives cancellation. The argument is purely algebraic on the common invariant domain $D$. [/proofplan]

[step:Expand the commutator with the quadratic operator $J^2$]Fix $i \in \{1,2,3\}$ and $\psi \in D$. Since each $J_a$ maps $D$ into $D$, every product of the operators $J_1,J_2,J_3$ appearing below is defined on $\psi$. Recall that $J^2:D\to D$ is the [linear map](/page/Linear%20Map) defined by \begin{align*} J^2\phi=\sum_{j=1}^{3}J_j^2\phi \end{align*} for every $\phi\in D$. For linear maps $A,B,C : D \to D$, the commutator product identity on $D$ is \begin{align*} [AB,C]\psi = A[B,C]\psi + [A,C]B\psi . \end{align*} Indeed, \begin{align*} A[B,C]\psi + [A,C]B\psi = A(BC\psi-CB\psi)+(ACB\psi-CAB\psi). \end{align*} Cancelling the middle two terms gives \begin{align*} A[B,C]\psi + [A,C]B\psi = ABC\psi-CAB\psi = [AB,C]\psi . \end{align*} Applying this identity with $A=B=J_j$ and $C=J_i$, and using linearity of the commutator in its first argument, gives \begin{align*} [J^2,J_i]\psi = \sum_{j=1}^{3} [J_j^2,J_i]\psi . \end{align*} For each $j$, \begin{align*} [J_j^2,J_i]\psi = J_j[J_j,J_i]\psi + [J_j,J_i]J_j\psi . \end{align*} Therefore \begin{align*} [J^2,J_i]\psi = \sum_{j=1}^{3}\left(J_j[J_j,J_i]\psi + [J_j,J_i]J_j\psi\right). \end{align*}[/step]

[guided]Fix one component index $i \in \{1,2,3\}$ and one vector $\psi \in D$. The point of assuming $J_a : D \to D$ for each $a$ is that all iterated expressions such as $J_jJ_k\psi$, $J_jJ_kJ_i\psi$, and $J_iJ_j^2\psi$ remain meaningful as vectors in $D \subset H$. The quadratic angular momentum operator $J^2:D\to D$ is defined by \begin{align*} J^2\phi=\sum_{j=1}^{3}J_j^2\phi \end{align*} for every $\phi\in D$. We first prove the algebraic identity that lets us differentiate a product inside a commutator. Let $A,B,C : D \to D$ be linear maps. By the definition of the commutator, \begin{align*} [AB,C]\psi = AB(C\psi)-C(AB\psi). \end{align*} We insert and subtract the same middle term $ACB\psi$: \begin{align*} AB(C\psi)-C(AB\psi) = ABC\psi-ACB\psi+ACB\psi-CAB\psi . \end{align*} Grouping the first two terms and the last two terms gives \begin{align*} ABC\psi-ACB\psi+ACB\psi-CAB\psi = A(BC\psi-CB\psi)+(ACB\psi-CAB\psi). \end{align*} Since $ACB\psi-CAB\psi=(AC-CA)B\psi=[A,C]B\psi$, this becomes the commutator product identity: \begin{align*} [AB,C]\psi = A[B,C]\psi + [A,C]B\psi . \end{align*} Now apply this identity to the product $J_j^2 = J_jJ_j$ and the operator $J_i$. For each $j \in \{1,2,3\}$, take $A=J_j$, $B=J_j$, and $C=J_i$. This gives \begin{align*} [J_j^2,J_i]\psi = J_j[J_j,J_i]\psi + [J_j,J_i]J_j\psi . \end{align*} Since $J^2$ is defined by \begin{align*} J^2\psi = \sum_{j=1}^{3} J_j^2\psi , \end{align*} linearity of the commutator in its first entry gives \begin{align*} [J^2,J_i]\psi = \sum_{j=1}^{3}\left(J_j[J_j,J_i]\psi + [J_j,J_i]J_j\psi\right). \end{align*} This is the desired expansion: the commutator with the quadratic operator has been reduced to commutators between individual angular momentum components.[/guided]

[step:Substitute the angular momentum commutation relations]Let $\hbar\in\mathbb{R}$ denote the reduced Planck constant, and let $\varepsilon_{abc}$ denote the Levi-Civita symbol on $\{1,2,3\}$, so $\varepsilon_{abc}=1$ for even permutations of $(1,2,3)$, $\varepsilon_{abc}=-1$ for odd permutations, and $\varepsilon_{abc}=0$ when two indices coincide. By the defining angular momentum commutation relations assumed in the theorem statement, for each $j \in \{1,2,3\}$ the relation with indices $j,i$ gives \begin{align*} [J_j,J_i]\psi = i\hbar \sum_{k=1}^{3} \varepsilon_{jik}J_k\psi . \end{align*} Because $J_j$ is linear and maps $D$ into $D$, \begin{align*} J_j[J_j,J_i]\psi = i\hbar \sum_{k=1}^{3} \varepsilon_{jik}J_jJ_k\psi . \end{align*} Also, since $J_j\psi \in D$, applying the same commutation relation to the vector $J_j\psi$ gives \begin{align*} [J_j,J_i]J_j\psi = i\hbar \sum_{k=1}^{3} \varepsilon_{jik}J_kJ_j\psi . \end{align*} Substituting both identities into the expansion from the previous step yields \begin{align*} [J^2,J_i]\psi = i\hbar \sum_{j=1}^{3}\sum_{k=1}^{3}\varepsilon_{jik}(J_jJ_k+J_kJ_j)\psi . \end{align*}[/step]

[guided]We now replace the abstract commutators in the expansion by the angular momentum relations. Here $\hbar\in\mathbb{R}$ is the reduced Planck constant, and $\varepsilon_{abc}$ is the Levi-Civita symbol on $\{1,2,3\}$: it equals $1$ on even permutations of $(1,2,3)$, equals $-1$ on odd permutations, and equals $0$ when two indices coincide. The theorem statement assumes that these angular momentum commutation relations hold on the common domain $D$. Thus, for each $j \in \{1,2,3\}$, the relation with ordered indices $j,i$ states that, for the fixed vector $\psi \in D$, \begin{align*} [J_j,J_i]\psi = i\hbar \sum_{k=1}^{3} \varepsilon_{jik}J_k\psi . \end{align*} The domain condition is used twice here. First, because each $J_k\psi$ lies in $D$, the finite sum on the right is a vector in $D$. Second, because $J_j:D\to D$ is linear, applying $J_j$ to that finite sum gives \begin{align*} J_j[J_j,J_i]\psi = i\hbar \sum_{k=1}^{3}\varepsilon_{jik}J_jJ_k\psi . \end{align*} For the other term, the input to the commutator is not $\psi$ but $J_j\psi$. This is legitimate because $J_j\psi\in D$. Applying the same angular momentum relation to this vector gives \begin{align*} [J_j,J_i]J_j\psi = i\hbar \sum_{k=1}^{3}\varepsilon_{jik}J_kJ_j\psi . \end{align*} Substituting these two identities into the expansion of $[J^2,J_i]\psi$ gives \begin{align*} [J^2,J_i]\psi = i\hbar \sum_{j=1}^{3}\sum_{k=1}^{3}\varepsilon_{jik}(J_jJ_k+J_kJ_j)\psi . \end{align*} The proof has now reduced the desired commutation statement to a finite algebraic cancellation indexed by the two component labels $j$ and $k$.[/guided]

[step:Cancel the symmetric operator factor against the antisymmetric Levi-Civita symbol]For fixed $j,k \in \{1,2,3\}$, define the operator $S_{jk}:D \to D$ by \begin{align*} S_{jk}(\phi) = (J_jJ_k+J_kJ_j)\phi \end{align*} for every $\phi \in D$. Then $S_{jk}=S_{kj}$ by definition. The Levi-Civita symbol is antisymmetric in the first and third displayed indices here: \begin{align*} \varepsilon_{kij} = -\varepsilon_{jik}. \end{align*} Thus, for every $j,k \in \{1,2,3\}$, \begin{align*} \varepsilon_{jik}S_{jk}\psi+\varepsilon_{kij}S_{kj}\psi = \varepsilon_{jik}S_{jk}\psi-\varepsilon_{jik}S_{jk}\psi = 0 . \end{align*} Terms with $j=k$ also vanish because $\varepsilon_{jij}=0$. Hence the whole double sum is zero: \begin{align*} \sum_{j=1}^{3}\sum_{k=1}^{3}\varepsilon_{jik}(J_jJ_k+J_kJ_j)\psi = 0 . \end{align*} Therefore \begin{align*} [J^2,J_i]\psi = 0 . \end{align*} Since $i \in \{1,2,3\}$ and $\psi \in D$ were arbitrary, $J^2$ commutes with each angular momentum component on $D$.[/step]

[guided]The remaining expression is \begin{align*} i\hbar \sum_{j=1}^{3}\sum_{k=1}^{3}\varepsilon_{jik}(J_jJ_k+J_kJ_j)\psi . \end{align*} The scalar factor $i\hbar$ is irrelevant for vanishing, so we focus on the double sum. For fixed $j,k \in \{1,2,3\}$, define $S_{jk}:D\to D$ by \begin{align*} S_{jk}(\phi) = (J_jJ_k+J_kJ_j)\phi \end{align*} for every $\phi\in D$. This operator is symmetric in $j$ and $k$: exchanging the two labels gives \begin{align*} S_{kj}(\phi) = (J_kJ_j+J_jJ_k)\phi = S_{jk}(\phi) \end{align*} for every $\phi\in D$. The Levi-Civita symbol has the opposite behaviour: it is antisymmetric when two indices are exchanged. With the middle index $i$ fixed, swapping $j$ and $k$ gives \begin{align*} \varepsilon_{kij} = -\varepsilon_{jik}. \end{align*} Therefore the two terms in the double sum indexed by $(j,k)$ and $(k,j)$ cancel: \begin{align*} \varepsilon_{jik}S_{jk}\psi+\varepsilon_{kij}S_{kj}\psi = \varepsilon_{jik}S_{jk}\psi-\varepsilon_{jik}S_{jk}\psi = 0 . \end{align*} When $j=k$, the term vanishes separately because the Levi-Civita symbol is zero whenever two indices coincide: \begin{align*} \varepsilon_{jij}=0. \end{align*} Thus every term in the finite double sum is either paired with an equal and opposite term or is zero on its own. Hence \begin{align*} \sum_{j=1}^{3}\sum_{k=1}^{3}\varepsilon_{jik}(J_jJ_k+J_kJ_j)\psi = 0 . \end{align*} Substituting this into the expression for $[J^2,J_i]\psi$ gives \begin{align*} [J^2,J_i]\psi = 0 . \end{align*} Because the component index $i\in\{1,2,3\}$ and vector $\psi\in D$ were arbitrary, this proves that $J^2$ commutes with each angular momentum component on the common invariant domain $D$.[/guided]