In the <Weyl representation of the gamma matrices>,
$$
\gamma^\mu=\begin{pmatrix}0&\sigma^\mu\\\bar\sigma^\mu&0\end{pmatrix},
\qquad
\gamma^5=\begin{pmatrix}-I&0\\0&I\end{pmatrix},
\qquad
\psi=\binom{u_L}{u_R}.
$$
A <Lorentz transformation> acts as $\psi'(x')=S(\Lambda)\psi(x)$ with $S(\Lambda)=\exp[-\tfrac i4\omega_{\mu\nu}\sigma^{\mu\nu}]$. In this basis a boost is block diagonal and gives $u_L\mapsto e^{\boldsymbol\chi\cdot\boldsymbol\sigma/2}u_L$ and $u_R\mapsto e^{-\boldsymbol\chi\cdot\boldsymbol\sigma/2}u_R$.

For boosts, the two exponentials cancel in $u_L^\dagger u_R$; for rotations, the unitary rotation and its inverse cancel. Hence this bilinear is a <Lorentz scalar>. The <Pauli matrices> obey the identity
$$
i\sigma^2(\sigma^i)^*=-\sigma^i i\sigma^2
$$
shows that $i\sigma^2u_L^*$ acquires the right-handed boost matrix and the usual rotation matrix, so it is right-handed.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 301 / 4 / a / Solution

{c}

In the Weyl representation,
$$
\gamma^\mu=\begin{pmatrix}0&\sigma^\mu\\\bar\sigma^\mu&0\end{pmatrix},
\qquad
\gamma^5=\begin{pmatrix}-I&0\\0&I\end{pmatrix}.
$$
A Dirac spinor therefore splits into left- and right-handed two-component spinors.


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 Incoming links: Weyl representation of the gamma matrices