= Solution
Use the standard <toric code> convention
$$
A_v=\prod_{j\ni v}X_j,
\qquad
B_p=\prod_{j\in\partial p}Z_j.
$$
An 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:
$$
e\times e=1,
\qquad
m\times m=1.
$$
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 gives
$$
e^{2i\theta_{em}}=e^{2i\theta_{me}}=-1.
$$
Identical $Z$ 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>.
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