Discussion
Consider two particles with angular momenta \(j_{1} = 2\hbar\) and \(j_{2} = \hbar/2\). If the expression
\[ \begin{aligned} |j = 5/2, m = 3/2\rangle = \{ &c_{1}|j_{1} = 2, m_{1} = 1\rangle|j_{2} = 1/2, m_{2} = 1/2\rangle + \\ &c_{2}|j_{1} = 2, m_{1} = 2\rangle|j_{2} = 1/2, m_{2} = -1/2\rangle \end{aligned} \]
gives an eigenstate of the total angular momentum of the two particles, using standard notation. Which of the following is true?
(Hint: \(\hat{J}_{\pm}|j, m\rangle = \sqrt{j(j+1) - m(m \pm 1)}\,|j, m \pm 1\rangle\))
Answer: A
The coefficients are \(c_{1} = 2/\sqrt{5}\) and \(c_{2} = 1/\sqrt{5}\).
Start from the stretched state \(|5/2, 5/2\rangle = |2, 2\rangle|1/2, 1/2\rangle\) and apply \(\hat{J}_{-} = \hat{J}_{1-} + \hat{J}_{2-}\) to both sides using the hint:
\[ \begin{aligned} \hat{J}_{-}|\tfrac{5}{2}, \tfrac{5}{2}\rangle &= \sqrt{\tfrac{35}{4} - \tfrac{15}{4}}\,|\tfrac{5}{2}, \tfrac{3}{2}\rangle = \sqrt{5}\,|\tfrac{5}{2}, \tfrac{3}{2}\rangle \\ \hat{J}_{1-}|2, 2\rangle &= \sqrt{6 - 2}\,|2, 1\rangle = 2|2, 1\rangle \\ \hat{J}_{2-}|\tfrac{1}{2}, \tfrac{1}{2}\rangle &= 1 \cdot |\tfrac{1}{2}, -\tfrac{1}{2}\rangle \end{aligned} \]
\[ \begin{aligned} \sqrt{5}\,|\tfrac{5}{2}, \tfrac{3}{2}\rangle = \ &2|2,1\rangle|\tfrac{1}{2},\tfrac{1}{2}\rangle \\ &+ |2,2\rangle|\tfrac{1}{2},-\tfrac{1}{2}\rangle \end{aligned} \]
This gives \(c_{1} = 2/\sqrt{5}\) and \(c_{2} = 1/\sqrt{5}\). The swapped pair belongs to the orthogonal state \(|3/2, 3/2\rangle\) (up to sign), equal weights do not follow from the ladder factors, and \(c_{1} = 0\) is not an eigenstate of \(J^{2}\).
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