 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/33b/a/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Put

J = j_{1} + j_{2}

. The only state with

m = J

∣ J, J ⟩ = ∣ j_{1}, j_{1} ⟩_{1} ∣ j_{2}, j_{2} ⟩_{2} .

Applying

J_{-} = J_{1 -} + J_{2 -}

and normalizing gives

∣ J, J - 1 ⟩ = \frac{j _{1}}{J}, ∣ j_{1}, j_{1} - 1 ⟩_{1} ∣ j_{2}, j_{2} ⟩_{2} + \frac{j _{2}}{J}, ∣ j_{1}, j_{1} ⟩_{1} ∣ j_{2}, j_{2} - 1 ⟩_{2} .

The orthonormal vector in the same two-dimensional

m = J - 1

subspace is

∣ J - 1, J - 1 ⟩ = \frac{j _{2}}{J}, ∣ j_{1}, j_{1} - 1 ⟩_{1} ∣ j_{2}, j_{2} ⟩_{2} - \frac{j _{1}}{J}, ∣ j_{1}, j_{1} ⟩_{1} ∣ j_{2}, j_{2} - 1 ⟩_{2} .

It is annihilated by

J_{+}

, so it is the highest-weight state of the

j = J - 1

multiplet. These are all states with

m \geq J - 1

. The last state exists when both subsystem spins are nonzero.

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