For orbital angular momentum

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · The formalism of quantum mechanics · Symmetry, angular momentum and spin

The Structure of Quantum Mechanics, Ch Compatible observables and uncertainty as a theorem

Problem

For orbital angular momentum, the ladder operators act as L^±∣l,m⟩=ℏl(l+1)−m(m±1) ∣l,m±1⟩\hat L_\pm|l,m\rangle=\hbar\sqrt{l(l+1)-m(m\pm1)}\,|l,m\pm1\rangle, with L^x=12(L^++L^−)\hat L_x=\tfrac12(\hat L_++\hat L_-), L^y=12i(L^+−L^−)\hat L_y=\tfrac{1}{2i}(\hat L_+-\hat L_-), and [L^x,L^y]=iℏL^z[\hat L_x,\hat L_y]=i\hbar\hat L_z. A particle is in the state ∣l=1,m=1⟩|l{=}1,m{=}1\rangle, a common eigenstate of L^2\hat L^2 and L^z\hat L_z. ((a)) Find L^x∣1,1⟩\hat L_x|1,1\rangle and L^y∣1,1⟩\hat L_y|1,1\rangle in terms of ∣1,0⟩|1,0\rangle, and hence ⟨L^x⟩\langle\hat L_x\rangle, ⟨L^y⟩\langle\hat L_y\rangle, ΔLx\Delta L_x and ΔLy\Delta L_y in this state. ((b)) Evaluate both sides of (ΔLx)2(ΔLy)2≥14∣⟨[L^x,L^y]⟩∣2+14∣⟨{ΔL^x,ΔL^y}⟩∣2(\Delta L_x)^2(\Delta L_y)^2\ge\tfrac14\left|\langle[\hat L_x,\hat L_y]\rangle\right|^2+\tfrac14\left|\langle\{\Delta\hat L_x,\Delta\hat L_y\}\rangle\right|^2 for this state, and show that the anticommutator term vanishes while the commutator term alone saturates the inequality.

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