Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 342 / 1

At a boundary parallel to lattice links, define each star as the product over links of the retained lattice incident on the vertex,A boundary vertex has three rather than four retained incident links. For a plaquette adjacent to the boundary, retainincluding its boundary link. A boundary link belongs to only one bulk plaquette, so applying $X$ on that link flips one $B_p$ rather than two. A magnetic string made from $X$ operators can therefore terminate at the boundary and its endpoint can be created or removed by a local boundary operator. This is precisely an $m$-condensing, or magnetic, boundary. By contrast, the same local operation does not permit an isolated electric endpoint.

With identical $m$-condensing boundaries, the cylinder supports one logical qubit and hence hasOne logical operator is an electric $Z$ string around the circumference; its conjugate is a magnetic $X$ string joining the two boundaries. The perturbation $h{\sum }_j{X}_j$ can generate the latter only after a virtual magnetic anyon traverses the length of the cylinder. Degenerate perturbation theory therefore gives the ground-state splitting of a surface-code cylinderwhere the factor $C$ counts translated shortest paths and nonuniversal order-one factors have been suppressed.

If both boundaries instead condense $e$, the unperturbed degeneracy remains two. The logical operator made solely from $X$ is now a magnetic loop winding around the circumference, so the same perturbation first acts nontrivially at order $C$:Thus exchanging the condensed anyon exchanges the geometrical length controlling this perturbative splitting.

The block matrix isIf both $q\in \mathcal{L}$ and $q^{\prime }$ condense, choose their boundary excursions so that the two string operators cross once. Their commutator is their mutual full-braiding phase:Both operators act as the identity on every ground state, so consistency requiresWriting $q^{\prime }=(q_{\mathrm{top}}^{\prime },q_{\mathrm{bot}}^{\prime })$, this saysSet $l_{\mathrm{top}}=M^{-1}{q}_{\mathrm{bot}}^{\prime }$ and choose $l_{\mathrm{bot}}=0$. Then $Kl=(0,q_{\mathrm{bot}}^{\prime }{)}^T$, andTherefore every additionally condensable anyon has the formThe $Kl$ term is a local particle in the anyon lattice of an Abelian Chern--Simons theory, so the condensate is maximal modulo local excitations, as required for a Lagrangian subgroup of Abelian anyons.

Since $\det M=-3$, the condensed top-sector anyons modulo local particles formwhich has order $|\det M|=3$. The same lower bound follows directly from Wilson-operator algebra. TakeBecausetheir crossing operators obeyActing repeatedly with one operator on an eigenstate of the other produces three states with distinct eigenvalues; they are linearly independent and have the same energy. HenceFor these identical maximal boundaries the bound is saturated, although only the lower bound was requested.