For the displayed boost,
$$
S[B(\beta)]
=\exp\!\left(-\frac\beta2\gamma^0\gamma^1\right)
=\begin{pmatrix}
e^{\beta\sigma^1/2}&0\\
0&e^{-\beta\sigma^1/2}
\end{pmatrix}.
$$
At rest,
$$
p_1\mathbin\cdot\sigma=p_1\mathbin\cdot\overline\sigma=mI_2,
\qquad
u(p_1)=\sqrt m\begin{pmatrix}\xi\\\xi\end{pmatrix}.
$$
For $p_2=(m\cosh\beta,m\sinh\beta,0,0)$, the <Pauli matrix> identity $e^{\beta\sigma^1}=\cosh\beta I_2+\sinh\beta\sigma^1$ gives
$$
p_2\mathbin\cdot\sigma=me^{\beta\sigma^1},
\qquad
p_2\mathbin\cdot\overline\sigma=me^{-\beta\sigma^1}.
$$
Taking the positive matrix square roots,
$$
u(p_2)=\sqrt m
\begin{pmatrix}e^{\beta\sigma^1/2}\xi\\e^{-\beta\sigma^1/2}\xi\end{pmatrix}
=S[B(\beta)]u(p_1),
$$
as required.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 301 / 3 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-301/3/ii/solution