Toric Code Anyons, Logical Operators, and Protection

The horizontal and vertical edges are indexed by two disjoint copies of ${\mathbb{Z}}_L^2$, so there are $2L^2$ edges; the vertex and face sets each have $L^2$ indices. At $q_{i,j}$ the four incident edges are $h_{i-1,j},h_{i,j},v_{i,j-1},v_{i,j}$. Their endpoints are distinct because $L\ge 3$.

The faces adjacent to $h_{i,j}$ are $f_{i,j-1}$ and $f_{i,j}$, and the faces adjacent to $v_{i,j}$ are $f_{i-1,j}$ and $f_{i,j}$. These pairs are distinct for $L\ge 3$; each displayed face boundary has four distinct edges. Thus the stated incidences define a square cellulation without repeated incidences.

Repeated horizontal and vertical steps modulo $L$ connect any pair of vertices. The same coordinate moves connect any pair of face indices through shared edges. Therefore both graphs defined above are connected.

For each row $j$, sum the cycle equations over $i\in \mathbb{Z}$. The horizontal terms cancel in pairs, giving $S_j=S_{j-1}$, where $S_j:={\sum }_i{V}_{i,j}$. Finite support gives $S_j=0$ for all sufficiently large $\mathrm{\mid }j\mathrm{\mid }$, so $S_j=0$ for every $j$. Define a face coefficient $F_{i,j}:={\sum }_{k>i}{V}_{k,j}$ in ${\mathbb{F}}_2$. For fixed $j$, this coefficient is zero for $i$ sufficiently large by finite support and for $i$ sufficiently negative because then it equals $S_j=0$. It is also zero for all but finitely many rows. Hence $F$ is a finitely supported face chain.

The vertical edge $v_{i,j}$ belongs to the boundaries of $f_{i-1,j}$ and $f_{i,j}$, so its coefficient in $\mathrm{\partial }F$ is $F_{i-1,j}+F_{i,j}=V_{i,j}$. Thus $\mathrm{\partial }F$ and the given chain agree on every vertical edge.

The horizontal edge $h_{i,j}$ belongs to the boundaries of $f_{i,j-1}$ and $f_{i,j}$. Its coefficient in $\mathrm{\partial }F$ is $F_{i,j-1}+F_{i,j}={\sum }_{k>i}(V_{k,j-1}+V_{k,j})$. The cycle equation at $(k,j)$ replaces each summand by $H_{k,j}+H_{k-1,j}$; summing over $k>i$ telescopes to $H_{i,j}$ because the sequence is finitely supported. Thus $\mathrm{\partial }F$ and the given chain agree on every horizontal edge, proving $\mathrm{\partial }F=(H,V)$. Translating the dual square lattice by $(1\mathrm{/}2,1\mathrm{/}2)$ identifies it with the same integer grid, so the identical construction proves the dual assertion.

By [F1], each displayed check acts on four distinct edge factors. By [F2] and [F3], its factors are defined endomorphisms of the ordered physical register, so both $A_{i,j}$ and $B_{i,j}$ are well-defined tensor-product operators.

Each factor in either product is Hermitian and squares to the identity by [F2]. The factors act on distinct tensor coordinates, so [F4] makes their two compositions agree on every elementary tensor; those tensors span the register by [F5]. Therefore each full product is Hermitian and squares to the identity. These are the star and plaquette checks used below.

For any $\tilde{p}\in {\mathbb{R}}^2$, choose an open square $U$ about $\tilde{p}$ of side length less than $L$. Its translates by distinct vectors in $L{\mathbb{Z}}^2$ are disjoint, and the quotient map restricts to a homeomorphism from each translate onto the same open neighborhood of $\pi (\tilde{p})$. These neighborhoods cover the torus. For a path, pull back this cover to the compact interval, take a finite subcover and a Lebesgue number, and partition the interval into pieces of smaller length. Each piece maps into one such neighborhood; once its initial lift is fixed, the inverse on the unique sheet through that lift defines the lift on the piece. These pieces agree at endpoints, and uniqueness follows from uniqueness of the local inverse. For a homotopy square, use a finite grid of rectangles whose images each lie in one evenly covered neighborhood. Starting from a lifted corner, the local inverses extend across adjacent rectangles; on a common edge two extensions agree because they are lifts agreeing at one endpoint. Thus paths and homotopies lift after their initial data are fixed.

Fix $p\in D$ and a lift $\tilde{p}$. For each $q\in D$, lift a path from $p$ to $q$ starting at $\tilde{p}$. Any two such paths are homotopic relative to their endpoints inside $D$, since the disk is contractible. By the homotopy lifting construction in step 1.1, the lifted endpoint varies continuously through the discrete fiber over $q$, so it is constant during the homotopy. Thus the endpoint is independent of the chosen path. These endpoints define a section $s:D\to {\mathbb{R}}^2$ of $\pi$ over $D$; local inverse charts from step 1.1 make the section continuous and one-to-one.

The image $\tilde{D}$ is compact because $D$ is compact and the section is continuous, hence it is bounded. The square and dual square tilings are locally finite, so only finitely many of their edges meet $\tilde{D}$. The lift preserves local edge incidences by step 1.1. Therefore every finite primal or dual cycle supported in $D$ lifts to a finite edge chain with even incidence at each lifted vertex.

Define $S:{\mathbb{F}}_2^V\to {\mathbb{F}}_2^E$ by sending a coordinate vector at a vertex to its star-support vector and extending linearly. The coefficient of an edge joining vertices $q,q^{\mathrm{\prime }}$ in $S(a)$ is $a_q+a_{q^{\mathrm{\prime }}}$. Thus $\ker S$ consists exactly of assignments constant along every edge. By [F1] the vertex graph is connected, so $\ker S$ is the one-dimensional space of constant assignments. Since $\dim {\mathbb{F}}_2^V=L^2$, [F2] gives $\dim \mathrm{im}S=L^2-1$. The all-vertex sum is zero because each edge has two endpoints, so this is the only star relation.

Define $T:{\mathbb{F}}_2^F\to {\mathbb{F}}_2^E$ by sending a face coordinate vector to that face's boundary and extending linearly. An edge separating faces $f,f^{\mathrm{\prime }}$ has coefficient $b_f+b_{f^{\mathrm{\prime }}}$ in $T(b)$. Thus $\ker T$ consists exactly of assignments constant along adjacent faces. By [F1] the face-adjacency graph is connected, so $\ker T$ is the one-dimensional space of constant assignments. Since $\dim {\mathbb{F}}_2^F=L^2$, [F2] gives $\dim \mathrm{im}T=L^2-1$. The all-face sum is zero because each edge borders two faces, so this is the only face-boundary relation. The dual face-adjacency graph has the same square-grid incidence, proving its incidence rank is also $L^2-1$.

A Hermitian involution has only eigenvalues $+1$ and $-1$. Therefore each check eigenvalue on the given nonzero simultaneous eigenvector is one of these two signs, so the two sets in the Definition are well-defined finite subsets of vertices and faces.

Let $\varphi \mathrm{\ne }0$ be a common $+1$ check eigenvector, and let $E$ be a Pauli error. For any check $g$, write $Eg=(-1{)}^{s}gE$ with $s\in {\mathbb{F}}_2$. Then $g(E\varphi )=(-1{)}^{s}E(g\varphi )=(-1{)}^{s}E\varphi$. Since $E$ is invertible, $E\varphi \mathrm{\ne }0$; its negative-check sets therefore consist exactly of the checks with commutation bit $s=1$. On an independent generating list these are the syndrome bits of [F2]. Thus the eigenvalue labels defined here are compatible with that error-syndrome convention, with stars and plaquettes kept separate.

For a fixed vertex $q_{i,j}$, the only faces sharing its incident edges are $f_{i,j},f_{i-1,j},f_{i,j-1},f_{i-1,j-1}$. Their boundaries overlap the star support respectively in $\{h_{i,j},v_{i,j}\}$, $\{h_{i-1,j},v_{i,j}\}$, $\{h_{i,j},v_{i,j-1}\}$, and $\{h_{i-1,j},v_{i,j-1}\}$; every other face has empty overlap. Hence every star-plaquette overlap has even size. By [F2], each star commutes with each plaquette. Two stars commute because all their nonidentity factors are $X$; two plaquettes commute because all their nonidentity factors are $Z$.

By [F3], every edge has exactly two endpoints, so in the product of all stars its $X$ factor occurs twice. Since $X^2=I$, the total product is $I$. Every edge borders exactly two faces, so the same argument with $Z^2=I$ gives ${\prod }_{i,j}{B}_{i,j}=I$. The first total product is also ${\prod }_{i,j}{A}_{i,j}=I$.

The primal vertex-edge graph has the explicit spanning tree $T=\{h_{i,0}:0\le i<L-1\}\cup \{v_{i,j}:i\in {\mathbb{Z}}_L,0\le j<L-1\}$. The horizontal edges form a path through row zero, and each other row is attached to that path by one vertical column path; hence $T$ is connected and has no cycle. For each edge $e\mathrm{\notin }T$, add $e$ to the unique tree path between its endpoints. This gives a cycle ${\gamma }_e$ containing $e$ and no other edge outside $T$. These cycles are independent because the coefficient of each non-tree edge in a linear combination is its own coefficient. They span $Z_1$: subtract from any cycle the ${\gamma }_e$ corresponding to its non-tree edges; the remainder is an even-incidence chain supported on the tree, and a nonempty finite forest has a leaf of odd degree, so the remainder is zero. Thus $\dim Z_1=\mathrm{\mid }E\mathrm{\mid }-(\mathrm{\mid }V\mathrm{\mid }-1)=L^2+1$.

Every face boundary has even incidence at each vertex, so $B_1\subseteq Z_1$. By [F2], $\dim B_1=L^2-1$. For a cycle $c$, define $w_h^k(c)={\sum }_{j}c(h_{k,j})$ and $w_v^{\mathrm{\ell }}(c)={\sum }_{i}c(v_{i,\mathrm{\ell }})$, with all sums in ${\mathbb{F}}_2$. Summing the even-incidence equations over column $k$ shows $w_h^k=w_h^{k-1}$; summing over row $\mathrm{\ell }$ shows $w_v^{\mathrm{\ell }}=w_v^{\mathrm{\ell }-1}$. Thus the two parities $w_h,w_v$ are independent of the chosen cuts. Each face boundary crosses every such cut zero or two times, so both parities vanish on $B_1$.

The horizontal fundamental loop ${\gamma }_h={\sum }_i{h}_{i,0}$ has $(w_h,w_v)=(1,0)$, and the vertical fundamental loop ${\gamma }_v={\sum }_j{v}_{0,j}$ has $(w_h,w_v)=(0,1)$. Therefore no nonzero combination of these loops lies in $B_1$. A basis of $B_1$ together with ${\gamma }_h,{\gamma }_v$ is an independent family of $(L^2-1)+2=L^2+1$ cycles, which is the dimension of $Z_1$ from step 1.1. It is therefore a basis of $Z_1$, proving that the winding map identifies $Z_1\mathrm{/}{B}_1$ with ${\mathbb{F}}_2^2$. The dual square cellulation has the same coordinate graph and incidence counts, so the identical argument proves its two winding coordinates.

Each coefficient $-J_e$ and $-J_m$ is real, and each summand is self-adjoint by [F1]. Their finite sum is therefore self-adjoint.

The terms commute pairwise by [F2], so $H_L$ is a finite sum of commuting local check operators. Its locality follows from the four-edge supports in [F1].

A product of retained checks with identity binary support must have zero X support and zero Z support separately. By [F3], this forces the selected star subset and plaquette subset both to be empty, so the retained checks are independent. By [F1] and [F2] they are commuting Hermitian involutions. For a retained check $g_a$ and sign $s_a$, put $p_{a,s_a}=(I+(-1{)}^{s_a}{g}_a)\mathrm{/}2$. Direct expansion using $g_a^{\ast }=g_a$ and $g_a^2=I$ gives $p_{a,s_a}^{\ast }=p_{a,s_a}$ and $p_{a,s_a}^2=p_{a,s_a}$. The checks commute, so their factors commute; hence $P_s$ is a self-adjoint idempotent, thus an orthogonal projection.

If $s\mathrm{\ne }t$, choose an index $a$ with $s_a\mathrm{\ne }{t}_a$. Then $p_{a,s_a}{p}_{a,t_a}=(I-g_a^2)\mathrm{/}4=0$, so $P_s{P}_t=0$. Summing over all sign strings and distributing the finite product gives ${\sum }_s{P}_s={\prod }_a(p_{a,0}+p_{a,1})=I$. Thus the ranges are pairwise orthogonal and their direct sum is the whole Hilbert space; zero projections are allowed.

By [F2], the omitted star is the product of all retained stars, so on $\mathrm{ran}{P}_s$ its eigenvalue is the product of their signs; the product of the eigenvalues over all stars is therefore $+1$. A finite list of signs in $\{+1,-1\}$ has product $+1$ exactly when it contains an even number of minus signs. The same argument for the omitted plaquette gives even parity for the negative plaquette checks. By [F4], these are the e and m defect parities. The projectors therefore decompose the register into the stated full-check syndrome sectors.

In the mod-two support of a primal trail, each internal visit to a vertex contributes one incoming and one outgoing edge, hence even incidence. After pairing arrivals and departures at every visit, the distinct endpoints have odd incidence and every other vertex has even incidence. By [F3], a $Z$ string anticommutes with a star exactly when the number of shared edges is odd. Thus it anticommutes with exactly $A_{q_0}$ and $A_{q_1}$. It commutes with every plaquette because both use only $Z$ factors.

In the dual graph, the same incidence count gives odd overlap exactly at the two endpoint faces. A dual edge crosses the primal edge carrying the same qubit, so [F3] makes the $X$ string anticommute with exactly the two endpoint plaquettes. It commutes with all stars because both use only $X$ factors. By [F2], these signs are precisely the stated e and m endpoint syndromes.

For a closed edge-simple primal trail, every vertex has even degree in its mod-two edge support: arrivals and departures pair at every visit, including the common initial and final vertex. Thus every star overlap is even, and every plaquette overlap is even because both strings use only $Z$. For a closed dual trail, the same incidence count gives even overlap with every plaquette, and its X string commutes with all stars. Hence both closed strings have zero syndrome.

Omit the fixed star $A_{0,0}$ and fixed plaquette $B_{0,0}$. By [F2], the remaining $L^2-1$ stars are independent, and the remaining $L^2-1$ plaquettes are independent; [F1] expresses each omitted check as the product of the retained checks of its type. Their combined binary supports are independent because an X-support and a Z-support occupy separate coordinates by [F4]. Thus these checks generate all checks with $r=2L^2-2$ independent generators.

Let $S$ be their generated subgroup. By [F1] it is abelian and generated by Hermitian Paulis. If a product of selected generators had zero binary label, its X-support and Z-support would each be zero by [F4]. The two independent families in [F2] then force both selected subsets to be empty, so that product is $I$. In particular, no product is $-I$, and $-I\mathrm{\notin }S$. Hence $S$ satisfies the stabilizer hypotheses in [F3].

There are $n=2L^2$ edge qubits by [F6]. The dimension theorem [F5] therefore gives $\dim \mathcal{C}(S)=2^{2L^2-(2L^2-2)}=4=2^2$; [F3] and step 2.1 ensure that its hypotheses hold, and step 1.1 supplies the independent rank. The code is nonzero and encodes exactly two qubits.

The supports of ${\bar{Z}}_1$ and ${\bar{Z}}_2$ close because their successive horizontal and vertical edges meet at vertices modulo $L$, by [F7]. The supports of ${\bar{X}}_1$ and ${\bar{X}}_2$ close on the dual graph because the crossed edges join successive adjacent faces modulo $L$, also by [F7]. By [F1] and [F2], a primal Z cycle commutes with every star and plaquette, and a dual X cycle commutes with every plaquette and star. Their winding pairs are nonzero by [F3], so none is a boundary. If a pure Z loop were in the stabilizer, its X support would be zero; by [F4] the selected star factors would either be empty or the full star family, and [F8] makes that full product $I$. The loop would therefore be a product of plaquettes, contrary to its nonzero winding. Interchanging X and Z gives the same conclusion for the dual loops. Hence all four lie in $N(S)\setminus Z_{n}S$ and define nontrivial logical classes by [F6].

The code is nonzero by [F5], so these normalizer classes act on its four-dimensional code space. The four formulas fix the representatives whose pairings are calculated in the subsequent logical-algebra result.

On a joint-check eigenspace with $N_e$ negative star signs and $N_m$ negative plaquette signs, each star sum is $L^2-2N_e$ and each plaquette sum is $L^2-2N_m$. By [F1] the Hamiltonian acts there by the scalar $-L^2(J_e+J_m)+2J_e{N}_e+2J_m{N}_m$.

The all-positive sector is the code space by definition, is nonzero by [F3], and has energy $E_g=-L^2(J_e+J_m)$. Every other nonzero sector has at least one defect species present; by [F2] its defect count is even, hence at least two. Step 1.1 then places its energy at least $4\min (J_e,J_m)$ above $E_g$.

If $J_e\le J_m$, choose a one-edge primal path and apply its Z string to a nonzero code vector. By [F4] the result is a nonzero joint eigenvector with exactly two e defects, so its energy is $E_g+4J_e$. If $J_m<J_e$, use a one-edge dual path and its X string instead; it yields a nonzero joint eigenvector with exactly two m defects and energy $E_g+4J_m$. These states attain the lower bound from step 2.1, proving ${\mathrm{\Delta }}_L=4\min (J_e,J_m)$.

The basis matrix units $\mathrm{\mid }x⟩⟨y\mathrm{\mid }$ span all operators on the finite register by [F7]. Each is the tensor product of one-qubit matrix units by [F8]. The four identities in [F6] therefore express every global matrix unit as a linear combination of tensor Pauli operators. Hence tensor Paulis span the full operator space. If $O$ is supported on $S$, apply the same expansion to $O_S$ in [F5] and tensor with identity outside $S$; every resulting Pauli term is supported on a subset of $S$.

Consider one tensor Pauli $E$ from this expansion. If it anticommutes with a check $g$, then the commutation fact [F2] and the common-$+1$ code definition [F3] make the projector relation [F4] applicable. It gives $\mathrm{\Pi }E\mathrm{\Pi }=\mathrm{\Pi }gE\mathrm{\Pi }=-\mathrm{\Pi }Eg\mathrm{\Pi }=-\mathrm{\Pi }E\mathrm{\Pi }$, so $\mathrm{\Pi }E\mathrm{\Pi }=0$. If $E$ commutes with all checks, write it as a phase times $X^x{Z}^z$. By [F1] and [F9], $z$ has even incidence at every primal vertex and $x$ has even incidence at every dual vertex. They are therefore primal and dual cycles. Their supports lie among the qubits in $S$; by [F12] both cycles lift to finite square-grid cycles. By [F11], the lifted Z cycle is a sum of square face boundaries and the lifted X cycle is a sum of dual face boundaries. Projecting these finite face chains to the torus gives products of plaquette checks and star checks with exactly the original support labels. Indeed, the quotient is a chain map because the boundary of each unit face maps to the stated four-edge torus boundary by [F13]; repeated image faces cancel in pairs over ${\mathbb{F}}_2$. Thus $E$ differs by a scalar phase from a product of checks, by [F10], and $\mathrm{\Pi }E\mathrm{\Pi }$ is a scalar multiple of $\mathrm{\Pi }$ by [F3] and [F4].

Every term in the Pauli expansion of $O$ from step 1.1 has either zero compression or scalar compression by step 2.1. Summing these finitely many terms gives $\mathrm{\Pi }O\mathrm{\Pi }=c(O)\mathrm{\Pi }$. The code projector is nonzero by [F14], so the scalar can be chosen as the sum of the scalar coefficients of those terms.

The X and Z supports meet at exactly one qubit, so their binary symplectic pairing is $1\in {\mathbb{F}}_2$. By [F2], $Z(P)X(Q)=(-1{)}^{1}X(Q)Z(P)=-X(Q)Z(P)$.

The sign is obtained from the finite Pauli commutator alone. No trajectory, Hamiltonian evolution, or dynamical exchange is part of the hypotheses, so the result asserts the mutual string algebra and no braiding experiment.

For any unit code vector $\psi$, each check has eigenvalue $+1$ by [F2], so [F1] gives $⟨\psi ,H_L\psi ⟩=-L^2(J_e+J_m)=E_{\mathrm{c}\mathrm{o}\mathrm{d}\mathrm{e}}$. For any Pauli prefix $E$ and any check $g$, [F5] gives $gE\psi =\pm Eg\psi =\pm E\psi$. Thus $E\psi$ has a joint check syndrome, and each term of $H_L$ has a fixed expectation on it. The sign pattern depends on $E$ and the checks, not on the choice of code vector $\psi$. Therefore $\epsilon (E)$ is well-defined and independent of that choice.

Every $Q$ in [F3] is a finite tensor product of one-edge Pauli factors $I,X,Y,Z$ from [F4], up to scalar phase. Applying its nonidentity factors one edge at a time gives at least one elementary path, so the set of paths in the Definition is nonempty.

By [F1] and [F2], the energy excess of any prefix is $2J_{e}a+2J_{m}b$, where $a,b\in \{0,\dots ,L^2\}$ count its violated stars and plaquettes. This is a finite set of possible real values. Hence the set of peak excesses over the nonempty path family is a nonempty subset of a finite set, and its minimum exists.

Let $E=i^a{X}^x{Z}^z$ be a Pauli in the normalizer. Commuting with every star forces the Z-support $z$ to have even incidence at every primal vertex; commuting with every plaquette forces the X-support $x$ to have even incidence at every dual vertex, by [F1] and [F4]. Thus $z$ is a primal cycle and $x$ is a dual cycle.

If both cycles were boundaries, their supports would be products of plaquette and star checks respectively, so $E$ would be a scalar-phase stabilizer. Hence a nontrivial logical Pauli has at least one nonboundary cycle by [F2] and [F3]. That cycle has a nonzero winding parity. For a horizontal winding, each of the $L$ disjoint vertical cuts is crossed an odd number of times; for a vertical winding, each of the $L$ disjoint horizontal cuts is crossed an odd number of times. In either case its support contains at least one distinct edge from each cut, hence at least $L$ edges. The Pauli support contains the support of each nonzero X or Z chain, even when both bits occur on one qubit, so [F7] gives $\mathrm{wt}(E)\ge L$.

The operator ${\bar{Z}}_1$ from [F5] has exactly $L$ distinct edge factors and is logically nontrivial. Thus the set defining [F6] is nonempty and contains a weight-$L$ element. The code encodes two qubits by [F8], so [F6] applies; combining this upper bound with step 2.1 gives $d(S)=L$.

By [F1], ${\bar{Z}}_1$ and ${\bar{X}}_1$ share only $h_{0,0}$, while ${\bar{Z}}_2$ and ${\bar{X}}_2$ share only $v_{0,0}$. [F6] gives anticommutation for these equal-index pairs. The cross pairs $({\bar{Z}}_1,{\bar{X}}_2)$ and $({\bar{Z}}_2,{\bar{X}}_1)$ have disjoint physical-edge supports, so [F2] gives commutation for unequal indices. Two Z strings commute with one another and two X strings commute with one another because their factors have the same Pauli type.

Suppose a product ${\bar{Z}}_1^a{\bar{Z}}_2^b{\bar{X}}_1^c{\bar{X}}_2^d$, with $a,b,c,d\in {\mathbb{F}}_2$, represented the identity class in [F3]. It would then commute with every normalizer representative. Commuting with ${\bar{X}}_1,{\bar{X}}_2,{\bar{Z}}_1,{\bar{Z}}_2$ in turn and using step 1.1 forces $a=b=c=d=0$. Hence the four classes are independent.

By [F4] the code has $k=2$, and [F5] gives logical quotient dimension $2k=4$. The four independent classes from step 2.1 therefore form a basis of that quotient.

Suppose $J_e\le J_m$. Apply $Z_{h_{0,0}},Z_{h_{1,0}},\dots ,Z_{h_{L-1,0}}$ in that order. After $k$ factors with $1\le k<L$, the partial string is a primal trail from $q_{0,0}$ to $q_{k,0}$; these endpoints are distinct, so [F3] gives exactly two e defects and no m defects. By [F5], changing one star eigenvalue from $+1$ to $-1$ changes its energy contribution from $-J_e$ to $+J_e$, an increase of $2J_e$; the excess is therefore $4J_e$. After the last factor the path is closed and has zero syndrome by [F3]. This is an elementary path in [F1] whose peak excess is at most $4J_e$.