Introduction to Quantum Mechanics · Problem 4.65

Quantum Mechanics in Three Dimensions — David J. Griffiths

Physics — Quantum Mechanics · addition_angular_momentum

Problem

If you combine three spin-1/2 particles, you can get a total spin of 3/2 or 1/2 (and the latter can be achieved in two distinct ways). Construct the quadruplet and the two doublets, using the notation of Equations 4.175 and 4.176: {∣32 32⟩=??∣32 12⟩=??∣32 −12⟩=??∣32 −32⟩=??s=32 (quadruplet)\begin{cases} |\frac{3}{2}\,\frac{3}{2}\rangle = ?? \\ |\frac{3}{2}\,\frac{1}{2}\rangle = ?? \\ |\frac{3}{2}\,-\frac{1}{2}\rangle = ?? \\ |\frac{3}{2}\,-\frac{3}{2}\rangle = ?? \end{cases} \quad s = \frac{3}{2} \text{ (quadruplet)} {∣12 12⟩1=??∣12 −12⟩1=??s=12 (doublet 1)\begin{cases} |\frac{1}{2}\,\frac{1}{2}\rangle_1 = ?? \\ |\frac{1}{2}\,-\frac{1}{2}\rangle_1 = ?? \end{cases} \quad s = \frac{1}{2} \text{ (doublet 1)} {∣12 12⟩2=??∣12 −12⟩2=??s=12 (doublet 2)\begin{cases} |\frac{1}{2}\,\frac{1}{2}\rangle_2 = ?? \\ |\frac{1}{2}\,-\frac{1}{2}\rangle_2 = ?? \end{cases} \quad s = \frac{1}{2} \text{ (doublet 2)} Hint: The first one is easy: ∣32 32⟩=∣ ⁣↑↑↑⟩|\frac{3}{2}\,\frac{3}{2}\rangle = |\!\uparrow\uparrow\uparrow\rangle; apply the lowering operator to get the other states in the quadruplet. For the doublets you might start with the first two in the singlet state, and tack on the third: ∣12 12⟩1=12(∣ ⁣↑↓⟩−∣ ⁣↓↑⟩)∣ ⁣↑⟩.|\frac{1}{2}\,\frac{1}{2}\rangle_1 = \frac{1}{\sqrt{2}}(|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle)|\!\uparrow\rangle. Take it from there—make sure ∣12 12⟩2|\frac{1}{2}\,\frac{1}{2}\rangle_2 is orthogonal to ∣12 12⟩1|\frac{1}{2}\,\frac{1}{2}\rangle_1 and to ∣32 12⟩|\frac{3}{2}\,\frac{1}{2}\rangle. Note: the two doublets are not uniquely determined—any linear combination of them would still carry spin 1/2. The point is to construct two independent doublets.

Related theory

Browse the problem library