Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 / 1

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.