Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 / 3 / c / Solution

The group velocity is $v_g={\hslash }^{-1}{\partial }_k{\epsilon }_k$. Since it is positive everywhere on the branch, every wave packet travels in the same direction: this is a Chiral Majorana edge mode.

Near $k=0$, particle-hole symmetry gives ${\epsilon }_k=\hslash vk+O(k^3)$. Let $c_k$ annihilate the positive-$k$ part of this branch. Particle-hole symmetry identifies $c_k^{†}=c_{-k}$. Hencesatisfies $c(x{)}^{†}=c(x)$ and is a Majorana field. Fourier transforming the linear dispersion givesThe continuum analogue of the real antisymmetric matrix $A$ is the real anti-self-adjoint differential kernelup to the normalization convention for the Majorana anticommutator.