In the defining representation,
$$
H=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad
E_+=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad
E_-=\begin{pmatrix}0&0\\1&0\end{pmatrix}.
$$
Multiplying the three factors in the <disentangling identity> and comparing with $e^{i\theta\sigma_1/2}$ gives
$$
\alpha=\gamma=\tan\frac\theta2,
\qquad
\exp\!\left(\frac{i\beta H}{2}\right)
=\begin{pmatrix}\cos(\theta/2)&0\\0&\sec(\theta/2)\end{pmatrix}.
$$
This factorization is the coordinate patch where $\cos(\theta/2)\ne0$.

Solution

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

Since $-\theta T_1=i\theta\sigma_1/2$, the same <disentangling identity> applies in every <Lie algebra representation>. The rightmost $E_+$ kills $|\Lambda\rangle$, the leftmost $E_-$ kills $\langle\Lambda|$, and the middle factor acts by its highest weight. Therefore
$$
\langle\Lambda|e^{-\theta d(T_1)}|\Lambda\rangle
=\left(\cos\frac\theta2\right)^\Lambda.
$$

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

A disentangling identity factors the exponential of a sum of noncommuting Lie-algebra generators into an ordered product of exponentials. For $\mathfrak{sl}_2$, a common order is $e^{aE_-}e^{bH}e^{cE_+}$.


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