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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.

Each of the $M$ vortices contributes one bulk Majorana zero mode. A finite fermionic system must have an even total number of Majorana zero modes. The boundary Chiral Majorana edge mode has no zero momentum in the antiperiodic sector but has one $k=0$ Majorana mode in the periodic sector. Thereforeor compactly $c(x+L)=(-1{)}^{M+1}c(x)$. This boundary condition of a chiral Majorana edge mode supplies the extra boundary zero mode exactly when the vortex count is odd.