Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 302 / 3 / e

Table of contents
- Solution e

Solution

Because

d (T_{1})

commutes with itself and is anti-Hermitian,

⟨ φ ∣ θ ⟩ = ⟨ Λ∣ e^{φ d (T_{1})} e^{- θ d (T_{1})} ∣Λ ⟩ = ⟨ Λ∣ e^{- (θ - φ) d (T_{1})} ∣Λ ⟩ .

Applying the preceding highest-weight matrix coefficient yields

⟨ φ ∣ θ ⟩ = [cos \frac{θ - φ}{2}]^{Λ} .

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