Discussion
A system of three non-identical spin \(\frac{1}{2}\) particles has the Hamiltonian \(H = \frac{A}{\hbar^{2}}(\vec{S}_{1} + \vec{S}_{2}) \cdot \vec{S}_{3}\), where \(\vec{S}_{1}, \vec{S}_{2}\) and \(\vec{S}_{3}\) are the spin operators of particles labelled 1, 2 and 3 respectively and \(A\) is a constant with appropriate dimensions. The set of possible energy eigenvalues of the system is
Answer: A
The eigenvalues are \(0, \frac{A}{2}, -A\).
Let \(\vec{S}_{12} = \vec{S}_{1} + \vec{S}_{2}\), with \(s_{12} = 0\) or 1, and total spin \(\vec{S} = \vec{S}_{12} + \vec{S}_{3}\). Then
\[ \begin{aligned} \vec{S}_{12}\cdot\vec{S}_{3} = \frac{\hbar^{2}}{2}[&s(s+1) \\ &- s_{12}(s_{12}+1) - \tfrac{3}{4}] \end{aligned} \]
- \(s_{12} = 0\), \(s = \frac{1}{2}\): \(E = 0\).
- \(s_{12} = 1\), \(s = \frac{3}{2}\): \(E = \frac{A}{2}(\frac{15}{4} - 2 - \frac{3}{4}) = \frac{A}{2}\).
- \(s_{12} = 1\), \(s = \frac{1}{2}\): \(E = \frac{A}{2}(\frac{3}{4} - 2 - \frac{3}{4}) = -A\).
The sets containing \(-\frac{A}{2}\) or \(\pm\frac{3A}{2}\) come from using the wrong \(s(s+1)\) values or dropping the factor \(\frac{1}{2}\).
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