The problem asks to identify which operator does NOT commute with the dot product of orbital angular momentum ($ \vec{L}$) and spin angular momentum ($ \vec{S}$). Two operators commute if their commutator is zero. The commutator is defined as $ [A, B] = AB - BA $. We are interested in the operator $ \vec{L} \cdot \vec{S} = L_x S_x + L_y S_y + L_z S_z $.
We need to check the commutation relation $ [\vec{L} \cdot \vec{S}, O] $ for each option $ O $. We use the standard commutation relations for angular momentum: $ [L_i, L_j] = i \hbar \epsilon_{ijk} L_k $, $ [S_i, S_j] = i \hbar \epsilon_{ijk} S_k $, and $ [L_i, S_j] = 0 $.
Let's calculate $ [\vec{L} \cdot \vec{S}, S_z] $:
$ [\vec{L} \cdot \vec{S}, S_z] = [L_x S_x + L_y S_y + L_z S_z, S_z] $Using linearity and the properties of commutators:
$ = [L_x S_x, S_z] + [L_y S_y, S_z] + [L_z S_z, S_z] $We know $ [L_z S_z, S_z] = 0 $ because $ [S_z, S_z] = 0 $.
For the other terms:
$ [L_x S_x, S_z] = L_x [S_x, S_z] + [L_x, S_z] S_x $Since $ [L_x, S_z] = 0 $ and $ [S_x, S_z] = -i \hbar S_y $:
$ = L_x (-i \hbar S_y) + (0) S_x = -i \hbar L_x S_y $Similarly:
$ [L_y S_y, S_z] = L_y [S_y, S_z] + [L_y, S_z] S_y $Since $ [L_y, S_z] = 0 $ and $ [S_y, S_z] = i \hbar S_x $:
$ = L_y (i \hbar S_x) + (0) S_y = i \hbar L_y S_x $Summing the terms:
$ [\vec{L} \cdot \vec{S}, S_z] = -i \hbar L_x S_y + i \hbar L_y S_x = i \hbar (L_y S_x - L_x S_y) $Since this result is non-zero, $ \vec{L} \cdot \vec{S} $ does NOT commute with $ S_z $.
Commutation with $ L^2 $: $ [\vec{L} \cdot \vec{S}, L^2] = [\sum_i L_i S_i, \sum_j L_j^2] $. Each term $ [L_i S_i, L_j^2] $ is zero because $ L_i $ commutes with $ L^2 $, and $ S_i $ acts on a different space and commutes with $ L^2 $. Thus, $ [\vec{L} \cdot \vec{S}, L^2] = 0 $.
Commutation with $ S^2 $: By symmetry, $ [\vec{L} \cdot \vec{S}, S^2] = 0 $. $ S_i $ commutes with $ S^2 $, and $ L_i $ commutes with $ S^2 $.
Commutation with $ (\vec{L} + \vec{S})^2 $: $ (\vec{L} + \vec{S})^2 = L^2 + S^2 + 2 \vec{L} \cdot \vec{S} $. The commutator is $ [\vec{L} \cdot \vec{S}, L^2 + S^2 + 2 \vec{L} \cdot \vec{S}] $. Since $ [\vec{L} \cdot \vec{S}, L^2] = 0 $, $ [\vec{L} \cdot \vec{S}, S^2] = 0 $, and $ [\vec{L} \cdot \vec{S}, 2 \vec{L} \cdot \vec{S}] = 0 $, the total commutator is $ 0 $.
The operator $ \vec{L} \cdot \vec{S} $ commutes with $ L^2 $, $ S^2 $, and $ (\vec{L} + \vec{S})^2 $. It does NOT commute with $ S_z $.