Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 (Topological Quantum Matter)

Use the standard toric code conventionAn open $Z$ string anticommutes with $A_v$ at its endpoints and creates the electric particles $e$ there; an open dual-lattice $X$ string creates magnetic particles $m$ at its endpoint plaquettes. Joining two strings cancels every shared Pauli because $X_j^2=Z_j^2=1$. Bringing two identical endpoints together therefore annihilates them:

Taking $e$ around $m$ makes its closed $Z$ string cross the $X$ string ending on $m$ once. At that crossing $ZX=-XZ$, so the state acquires $-1$. The same argument with the roles reversed givesIdentical $Z$ strings commute with one another, as do identical $X$ strings. Their exchanges can be deformed without a crossing between anticommuting operators, so $e^{i{\theta }_{ee}}=e^{i{\theta }_{mm}}=1$. Thus $e$ and $m$ are bosons and are mutual semions, as summarized by the surface-code anyon model.

The composite is $f=e\times m$. Exchanging two composites exchanges the two $e$ constituents and the two $m$ constituents and produces one mutual winding between unlike constituents. The identical-particle phases are both $+1$, while the mutual contribution is $-1$, henceThe topological spin obeys $e^{2\pi i{s}_f}=e^{i{\theta }_{ff}}$, soThis is precisely the two-dimensional spin-statistics relation for a fermion, so the result is consistent.

In the charge-flux composite model, particle $1$ acquires the Aharonov-Bohm effect phase $q_1{\Phi }_2/\hslash$ when it circles the flux of particle $2$. The reciprocal Aharonov-Casher effect contributes $q_2{\Phi }_1/\hslash$. Thus the full braid isThe assumption for $a=(q_0,0)$ and $b=(0,{\Phi }_0)$ saysFor any allowed particle $(n{q}_0,m{\Phi }_0)$, the full braid with $a^2=(2q_0,0)$ is $e^{2im{q}_0{\Phi }_0/\hslash }=1$; similarly every particle braids trivially with $b^2=(0,2{\Phi }_0)$. Under the stated operational identification,For $c=a\times b=(q_0,{\Phi }_0)$, exchanging two identical composites is half their full braid and givesTherefore $a$ and $b$ reproduce the bosonic $e$ and $m$ particles, in either order, and $c$ reproduces their fermionic fusion product $f$.

Forthe vectors $q_e=(1,0{)}^T$ and $q_m=(0,1{)}^T$ haveThe K-matrix formula therefore gives bosonic self-exchange for $e,m$ and mutual full-braiding phaseFusion adds vectors. Becauseboth double fusions lie in $K{\mathbb{Z}}^2$ and braid trivially with every vector; hence they represent the vacuum in the anyon lattice of an Abelian Chern--Simons theory. Finally $q_f=q_e+q_m$ has $q_f^T{K}^{-1}{q}_f=1$, so its exchange phase is $e^{i\pi }=-1$. These are exactly the fusion and braiding data of the surface-code anyon model.

Every element of the $n$-qubit Pauli group squares to either $1$ or $-1$. If a stabilizer generator $S_j$ had $S_j^2=-1$, closure of the group would put $-1$ in $S$, contrary to the definition. Thus $S_j^2=1$. Since Pauli operators are unitary,and $S_j^{-1}=S_j$ proves $S_j=S_j^{†}$.

If two stabilizers $g,h$ anticommute and a nonzero vector $|\psi ⟩$ belonged to the codespace, thena contradiction. Hence a nonzero stabilizer code requires $S$ to be Abelian. Its centralizer of a stabilizer group is

Measure each generator $S_j$. If a Pauli error $E$ has occurred, its outcome is $(-1{)}^{s_j}$, whereThe bit vector $s=(s_1,\dots ,s_m)$ is the error syndrome. Choose one representative $E_s$ with this syndrome. Another Pauli $E$ has the same syndrome exactly when $E_s^{†}E$ commutes with every $S_j$, namely when $E_s^{†}E\in C(S)$. ThereforeThis stabilizer-syndrome coset is all the syndrome reveals: multiplication by a stabilizer changes nothing on the code, while multiplication by an element of $C(S)\setminus S$ can change the logical state without changing any syndrome bit.

Fix the reference bit-flip chainand assign to each lattice edge $e=v{v}^{\prime }$ the bond signwhere $n_e(E)=1$ if $E$ contains $X_e$ and is zero otherwise. A product of star stabilizers is specified by Ising spins ${\sigma }_v$: choose the star at $v$ when ${\sigma }_v=-1$. The resulting error has edge occupationFor $N$ edges, its independent bit-flip probability isbecause $e^{\beta J}=\sqrt{(1-p)/p}$. Summing over products of stars therefore gives the Surface-code decoding as a random-bond Ising model identitywhere $r$ is the number of Ising configurations representing the same stabilizer product, usually $r=2$ on a closed connected lattice because a global spin flip changes no bond. The class-independent prefactor cancels when the two logical classes $q=0,1$ are compared.

Choose both circular ends of the cylinder to be electric, or rough, boundaries. In the convention of part 1, retain plaquette generatorswith their boundary truncations, and use star generatorsonly at vertices not lying on an electric boundary. Omitting the endpoint star checks allows a $Z$ string to terminate at either boundary, which is precisely anyon condensation at a boundary for $e$.

Relative homology now has one nontrivial class represented by a primal $Z$ path $\gamma$ joining the two boundaries. The dual absolute homology has one class represented by an $X$ loop $\tilde{\gamma }$ around the cylinder. Their strings intersect once and anticommute, so they form one logical pair $\bar{Z},\bar{X}$. Every other closed or boundary-ending string is a product of stabilizers, so the code has exactly one logical qubit.

In the Random-bond Ising model mapping, there is no Ising spin for an omitted boundary star. An edge $e$ joining an interior vertex $v$ to an electric boundary is toggled by only $A_v$, so its bond factor ${\eta }_e{\sigma }_v{\sigma }_{v^{\prime }}$ is replaced by the boundary-field factor ${\eta }_e{\sigma }_v$. Equivalently, attach exterior spins fixed to $+1$ and retain the same bond formula. Edges wholly on a boundary contribute fixed constants. The bulk Hamiltonian is therefore supplemented bywith the signs still determined by $E_s{\bar{X}}^q$. This is the required boundary modification.

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.