Discussion
The spin-orbit interaction in a hydrogen-like atom is given by the Hamiltonian
\(H' = -k\vec{L}\cdot\vec{S}\)
where \(k\) is a real constant. The splitting between levels \({}^{2}p_{3/2}\) and \({}^{2}p_{1/2}\) due to this interaction is
Answer: B
The splitting is \(\dfrac{3}{2}k\hbar^{2}\).
With J = L + S, the operator L·S = (J2 - L2 - S2)/2, so for a p electron (l = 1, s = 1/2):
\[ \begin{aligned} \langle\vec{L}\cdot\vec{S}\rangle_{j=3/2} &= \tfrac{\hbar^{2}}{2}(\tfrac{15}{4} - 2 - \tfrac{3}{4}) = \tfrac{1}{2}\hbar^{2} \\ \langle\vec{L}\cdot\vec{S}\rangle_{j=1/2} &= \tfrac{\hbar^{2}}{2}(\tfrac{3}{4} - 2 - \tfrac{3}{4}) = -\hbar^{2} \end{aligned} \]
The two levels differ in L·S by 3ℏ2/2, so with H′ = -kL·S the splitting is (3/2)kℏ2 in magnitude (with k > 0 the j = 1/2 level lies higher).
1/2 kℏ2 is the shift of the j = 3/2 level alone. 3/4 and 2 come from slips in j(j + 1), l(l + 1) or s(s + 1).
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