Subsystem codes with spatially local generators

We study subsystem codes whose gauge group has local generators in two-dimensional (2D) geometry. It is shown that there exists a family of such codes defined on lattices of size $L\times L$ with the number of logical qubits $k$ and the minimum distance $d$ both proportional to $L$. The gauge group of these codes involves only two-qubit generators of type $\mathit{XX}$ and $\mathit{ZZ}$ coupling nearest-neighbor qubits (and some auxiliary one-qubit generators). Our proof is not constructive as it relies on a certain version of the Gilbert-Varshamov bound for classical codes. Along the way, we introduce and study properties of generalized Bacon-Shor codes that might be of independent interest. Secondly, we prove that any 2D subsystem $[n,k,d]$ code with spatially local generators obeys upper bounds $\mathit{kd}=O(n)$ and $d^2=O(n)$. The analogous upper bound proved recently for 2D stabilizer codes is ${\mathit{kd}}^2=O(n)$. Our results thus demonstrate that subsystem codes can be more powerful than stabilizer codes under the spatial locality constraint.

Figure 1

The cells $c$ and $c^{\prime }$ represent a pair of consecutive qubits in some row of $A$ that are not nearest neighbors. The cells $c_1,\dots ,c_p$ are empty.

Figure 2

The conjugation by $U$ maps generators of $\mathrm{G̃}$ to generators of ${\mathcal{G}}^{\prime }$ since ${\mathit{UX}}_a{U}^{†}=X_q{X}_a$ and ${\mathit{UZ}}_a{U}^{†}=Z_a$.

Figure 3

Partition of the lattice $\Lambda =\mathit{ABC}$. If the block $A$ supports no dressed logical operators, $l(A)=0$, and the perimeter of $A$ is small compared with the distance $d$ (such that $\left|B\right|+|\partial \overline{A}|<d$), one can show that the larger region $\mathit{AB}$ also supports no dressed logical operators, $l(\mathit{AB})=0$. This argument can be applied inductively until $A$ has linear size comparable with the distance $d$.

Figure 4

The partition of the lattice $\Lambda =\mathit{AB}$ used to prove the bound $\mathit{kd}=O(n)$. The region $A$ consists of square blocks $A_1,\dots ,A_m$ of size $R\times R$ with $R=\Omega (d)$ such that $l(A_i)=0$. Adjacent blocks are separated from each other by distance $r=O(1)$. This implies $l_{\mathrm{bare}}(A)=0$ and thus $l(B)=2k$. This is possible only if $\left|B\right|⩾k$, which yields $\mathit{kd}=O(n)$.

Figure 5

The partition of the lattice $\Lambda =\mathit{ABC}$ used to prove the bound ${\mathit{kd}}^2=O(n)$. The regions $A$,$B$ consist of blocks $A_1,\dots ,A_m$ and $B_1,\dots ,B_m$, respectively, of size $R\times R$ with $R=\Omega (d)$ such that $l(A_i)=0$ and $l(B_i)=0$. The region $C$ consists of disks of radius $max\{r,r_s\}$ so that adjacent blocks in $A$ and adjacent blocks in $B$ are separated from each other by distance $max\{r,r_s\}$. This implies $l_{\mathrm{bare}}(A)=0$ and thus $l(\mathit{BC})=2k$. If $k$ violates the upper bound, one would have $l(B)⩾l(\mathit{BC})-2\left|C\right|>0$. Assuming that $\mathcal{S}$ has spatially local generators, this is possible only if $l(B_i)>0$ for some block $B_i$, which is a contradiction.