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

For each complex fermion define two Majorana fermion operatorsThey obey $c_r^{†}=c_r$, $\{c_r,c_s\}=2{\delta }_{rs}$, and $a_j=(c_{2j-1}+i{c}_{2j})/2$. Substituting these inverse relations expands every hopping and pairing monomial as a bilinear in the $c_r$. Diagonal terms $c_r^2=1$ contribute only a constant.

For $r\ne s$, $c_r{c}_s$ is anti-Hermitian. Hermiticity of the original quadratic fermion Hamiltonian therefore makes its coefficient purely imaginary, so it can be written $i{A}_{rs}$ with $A_{rs}$ real. Sincethe symmetric part again changes only the constant. Absorbing conventional factors into $A$ gives

Position-space particle-hole symmetry complex-conjugates plane waves, so the Fourier transform sends $k$ to $-k$. Fourier transforming the stated relation therefore givesIf $H_{\mathrm{BdG}}(k)u_k={\epsilon }_k{u}_k$, complex conjugation and multiplication by ${\Sigma }_1$ yieldThus the Particle-hole symmetry of a Bogoliubov--de Gennes Hamiltonian imposesor, band by band after a suitable relabelling, ${\epsilon }_{-k}=-{\epsilon }_k$.

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.

The two new boundaries surround opposite sides of the same bulk. Their induced orientations are opposite, so their chiral edge modes counterpropagate and $\mathrm{sgn}{v}_1=-\mathrm{sgn}{v}_2$.

For $v_1=v$ and $v_2=-v$, define the massIt approaches $+t$ on the left and $-t$ on the right. The zero-energy first-order equations have one normalizable real solution, whose envelope can be chosen asafter choosing the constant Majorana spinor with the appropriate relative sign. The corresponding operatoris self-adjoint and commutes with the Hamiltonian, so it is a Majorana zero mode at a mass domain wall.

Outside the core it decays onIf the phase varies approximately linearly across the core, then $m(x)\simeq -\pi tx/R_c$ near zero and the central envelope is Gaussian with widthThus the spatial extent is of order $\max (\xi ,{\ell }_{\mathrm{core}})$, up to constants depending on the detailed vortex profile. When $|v_1|\ne |v_2|$, continuously changing their magnitudes to equality never changes their opposite signs or closes the asymptotic mass gap. The mass domain wall therefore retains its odd, particle-hole-protected zero mode; only its two component amplitudes and localization lengths change.

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.