Because $d(T_1)$ commutes with itself and is anti-Hermitian,
$$
\langle\varphi|\theta\rangle
=\langle\Lambda|e^{\varphi d(T_1)}e^{-\theta d(T_1)}|\Lambda\rangle
=\langle\Lambda|e^{-(\theta-\varphi)d(T_1)}|\Lambda\rangle.
$$
Applying the preceding highest-weight matrix coefficient yields
$$
\boxed{\langle\varphi|\theta\rangle
=\left[\cos\frac{\theta-\varphi}{2}\right]^\Lambda}.
$$


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-302/3/e/solution