The <Dirac field> Lagrangian is
$$
\mathcal L_D=i\bar\psi\gamma^\mu\partial_\mu\psi-m\bar\psi\psi.
$$
Choose the <Majorana spinor>
$$
\psi_M=\binom{u_L}{i\sigma^2u_L^*}.
$$
Substitution shows that its two chiral kinetic terms agree after integration by parts and that
$$
-m\bar\psi_M\psi_M
=i m\{u_L^T\sigma^2u_L-u_L^\dagger\sigma^2u_L^*\}.
$$
Thus $\mathcal L_D[\psi_M]$ equals twice the displayed one-Weyl-field Lagrangian, up to a total derivative.

The lower chiral component of the <Dirac equation> is
$$
i\bar\sigma^\mu\partial_\mu u_L-mu_R=0.
$$
Putting $u_R=i\sigma^2u_L^*$ gives exactly $i\bar\sigma^\mu\partial_\mu u_L-i m\sigma^2u_L^*=0$; the upper component is its <complex conjugate>.

Solution

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

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A Majorana spinor equals its charge conjugate. In the Weyl representation it can be written $\psi_M=(u_L,i\sigma^2u_L^*)^T$ from one left-handed Weyl spinor.


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