A nucleus of spin is coupled to the spin of an unpaired electron,

PHYSICS · Subjective · Fermi University Physics · Quantum Mechanics · Symmetry, angular momentum and spin · Symmetry, angular momentum and spin

The Structure of Quantum Mechanics, Ch Adding angular momenta, tensor operators and Wigner-Eckart

Problem

A nucleus of spin j1=32j_1=\tfrac32 is coupled to the spin of an unpaired electron, j2=12j_2=\tfrac12. Treat this purely as the addition of two angular momenta, with the uncoupled basis ∣m1⟩∣m2⟩|m_1\rangle|m_2\rangle (m1=32,12,−12,−32m_1=\tfrac32,\tfrac12,-\tfrac12,-\tfrac32; m2=12,−12m_2=\tfrac12,-\tfrac12) and the coupled basis ∣J,M⟩|J,M\rangle. Recall the ladder eigenvalue J^−∣j,m⟩=ℏj(j+1)−m(m−1) ∣j,m−1⟩\hat J_-|j,m\rangle=\hbar\sqrt{j(j+1)-m(m-1)}\,|j,m-1\rangle, and that J^−=J^1−+J^2−\hat J_-=\hat J_{1-}+\hat J_{2-} on the product space. (a) By tabulating M=m1+m2M=m_1+m_2 for all eight product states and peeling multiplets off from the top, derive the decomposition j1⊗j2=⨁JJj_1\otimes j_2=\bigoplus_J J for this pair. Check that ∑J(2J+1)\sum_J(2J+1) equals the dimension of the product space. (b) Starting from the unique top state ∣J=2,M=2⟩=∣m1=32⟩∣m2=12⟩|J=2,M=2\rangle=|m_1=\tfrac32\rangle|m_2=\tfrac12\rangle, apply J^−\hat J_- once to write ∣2,1⟩|2,1\rangle as a combination of the two product states with M=1M=1. Then use orthogonality inside that two-dimensional M=1M=1 subspace, fixing the sign by the Condon-Shortley convention that the coefficient of ∣m1=32⟩∣m2=−12⟩|m_1=\tfrac32\rangle|m_2=-\tfrac12\rangle is positive, to construct ∣J=1,M=1⟩|J=1,M=1\rangle.

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