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

In the Weyl representation of the gamma matrices,A Lorentz transformation acts as ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ with $S(\Lambda )=\exp [-\frac{i}{4}{\omega }_{\mu \nu }{\sigma }^{\mu \nu }]$. In this basis a boost is block diagonal and gives $u_L\mapsto e^{\mathbf{\chi }\cdot \mathbf{\sigma }/2}{u}_L$ and $u_R\mapsto e^{-\mathbf{\chi }\cdot \mathbf{\sigma }/2}{u}_R$.

For boosts, the two exponentials cancel in $u_L^{†}{u}_R$; for rotations, the unitary rotation and its inverse cancel. Hence this bilinear is a Lorentz scalar. The Pauli matrices obey the identityshows that $i{\sigma }^2{u}_L^{\ast }$ acquires the right-handed boost matrix and the usual rotation matrix, so it is right-handed.

The identities $S_L^{†}{\bar{\sigma }}^{\mu }{S}_L={\Lambda }^{\mu }{}_{\nu }{\bar{\sigma }}^{\nu }$ and $S_L^T{\sigma }^2{S}_L={\sigma }^2$ show respectively that the kinetic term and both mass bilinears are Lorentz invariant. Infinitesimally, the first identity makes $u_L^{†}{\bar{\sigma }}^{\mu }{u}_L$ transform by the vector law stated in the question, while the second makes $u_L^T{\sigma }^2{u}_L$ a scalar.

Varying the anticommuting components of $u_L^{†}$ givesThe factor two from varying the antisymmetric quadratic form cancels the $1/2$ in the mass term. This is a Majorana mass term: under $u_L\mapsto e^{iq\alpha }{u}_L$ it carries charge $2q$, so a nonzero mass is compatible with an unbroken electric charge only for $q=0$. A charged massive fermion instead needs an independent Weyl field of opposite chirality to form a Dirac mass.

The Dirac field Lagrangian isChoose the Majorana spinorSubstitution shows that its two chiral kinetic terms agree after integration by parts and thatThus ${\mathcal{L}}_D[{\psi }_M]$ equals twice the displayed one-Weyl-field Lagrangian, up to a total derivative.

The lower chiral component of the Dirac equation isPutting $u_R=i{\sigma }^2{u}_L^{\ast }$ gives exactly $i{\bar{\sigma }}^{\mu }{\partial }_{\mu }{u}_L-im{\sigma }^2{u}_L^{\ast }=0$; the upper component is its complex conjugate.