The displayed $R(\varphi)$ rotates the coordinate components in the $xy$ plane by angle $-\varphi$. Since
$$
\gamma^1\gamma^2
=-i\begin{pmatrix}\sigma^3&0\\0&\sigma^3\end{pmatrix},
$$
its spin representative is
$$
S[R(\varphi)]
=\exp\!\left(-\frac\varphi2\gamma^1\gamma^2\right)
=\begin{pmatrix}
e^{i\varphi\sigma^3/2}&0\\
0&e^{i\varphi\sigma^3/2}
\end{pmatrix}.
$$
At a full turn,
$$
S[R(2\pi)]=-I_4.
$$
Thus a $2\pi$ spatial rotation changes the sign of a spin-one-half state, while a $4\pi$ rotation returns it to itself. This realizes the fact that the <Spin group> is a double cover of the proper orthochronous Lorentz group; observable spinor bilinears are unchanged by the sign.

Solution

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

 Solution

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