35 Rem 12: Noncommuting Operators Cannot Be Deformed
This chapter formalizes the observation that a Pauli operator \(P\) which does not commute with the logical operator \(L\) cannot have a deformed version. The obstruction is topological: if \(P\) anticommutes with \(L = \prod _v X_v\), then \(|\mathcal{S}_Z \cap V_G|\) is odd, but any path boundary \(\partial \gamma \) always has even cardinality, so no edge-path \(\gamma \) with \(\partial \gamma = \mathcal{S}_Z \cap V_G\) can exist.
The support of a binary vector \(f : V' = V \to \mathbb {Z}_2\) is the set of vertices where \(f\) is nonzero:
For any \(x \in \mathbb {Z}_2\), we have \(x \neq 0\) if and only if \(x = 1\).
We prove both directions. For the forward direction, assume \(x \neq 0\). We case-split on \(x \in \mathbb {Z}_2\): if \(x = 0\) we reach a contradiction with the assumption, so \(x = 1\) and this holds by reflexivity. For the reverse direction, assume \(x = 1\); rewriting, \(1 \neq 0\) holds by computation.
The boundary support of an edge-path \(\gamma \) is the set of vertices where the boundary is nonzero:
For any edge-path \(\gamma \), the boundary support equals the set of vertices where the boundary value is \(1\):
By extensionality, it suffices to show membership equivalence for an arbitrary vertex \(v\). Simplifying the definition of boundary support, both directions follow from the equivalence \(x \neq 0 \Leftrightarrow x = 1\) in \(\mathbb {Z}_2\): the forward direction applies this equivalence, and the reverse direction rewrites with \(\partial \gamma (v) = 1\) and verifies \(1 \neq 0\) by computation.
For any edge-path \(\gamma \), the sum of boundary values over all vertices is zero:
This follows directly from boundary_sum_zero.
For any edge-path \(\gamma \), the boundary support has even cardinality:
This captures the paper’s statement that “a path boundary always has even cardinality.”
We first establish that the sum of boundary values is zero: \(\sum _{v \in V} \partial \gamma (v) = 0\) (from boundary_values_sum_zero). We then show that this sum equals \(|\mathrm{boundarySupport}(G,\gamma )|\) in \(\mathbb {Z}_2\). To see this, we split the sum over \(V\) into the sum over vertices where \(\partial \gamma (v) \neq 0\) and vertices where \(\partial \gamma (v) = 0\). The latter sum is zero (each summand is zero). The filter for nonzero values is exactly the boundary support. On the boundary support, each value equals \(1\) (by the \(\mathbb {Z}_2\) characterization), so the sum equals \(|\mathrm{boundarySupport}(G,\gamma )| \cdot 1\). Combining, we get \(|\mathrm{boundarySupport}(G,\gamma )| = 0\) in \(\mathbb {Z}_2\). Extracting the \(\mathbb {Z}_2\) value via ZMod.val yields the result \(|\mathrm{boundarySupport}(G,\gamma )| \bmod 2 = 0\).
For any edge-path \(\gamma \), \(|\mathrm{boundarySupport}(G, \gamma )|\) is even.
Rewriting the definition of Even using Nat.even_iff, this reduces to \(|\mathrm{boundarySupport}(G,\gamma )| \bmod 2 = 0\), which is exactly boundary_cardinality_even.
If \(\gamma \) is a valid deforming path for a \(Z\)-support set \(S \subseteq V\), then
By extensionality, it suffices to show membership equivalence for arbitrary \(v\). Simplifying the boundary support definition and using the characterization of valid deforming paths (which gives \(\partial \gamma (v) = \chi _S(v)\) for all \(v\)), we rewrite the boundary value. We then show equivalence with membership in \(S\) using the \(Z\)-support vector: if \(v \notin S\), then \(\chi _S(v) = 0\) and \(0 \neq 0\) is false; if \(v \in S\), then \(\chi _S(v) = 1\) and \(1 \neq 0\) holds by computation.
If \(|S| \bmod 2 = 1\) (i.e., \(S\) has odd cardinality), then \(S\) is not the boundary support of any edge-path:
Suppose for contradiction there exists \(\gamma \) with \(\mathrm{boundarySupport}(G,\gamma ) = S\). By boundary_cardinality_even, \(|\mathrm{boundarySupport}(G,\gamma )| \bmod 2 = 0\). Rewriting with the equality gives \(|S| \bmod 2 = 0\), contradicting \(|S| \bmod 2 = 1\) by integer arithmetic.
A Pauli operator \(P\) (represented by its \(Z\)-support on \(V\)) anticommutes with \(L\) if its \(Z\)-support has odd cardinality:
A Pauli operator \(P\) (represented by its \(Z\)-support on \(V\)) commutes with \(L\) if its \(Z\)-support has even cardinality:
For any \(Z\)-support set, either \(P\) commutes with \(L\) or \(P\) anticommutes with \(L\):
Unfolding the definitions, both conditions concern \(|\mathcal{S}_Z| \bmod 2\) being \(0\) or \(1\). The result follows by integer arithmetic (a natural number modulo \(2\) is either \(0\) or \(1\)).
A Pauli operator does not commute with \(L\) if and only if it anticommutes with \(L\):
Unfolding the definitions, this reduces to \(|\mathcal{S}_Z| \bmod 2 \neq 0 \Leftrightarrow |\mathcal{S}_Z| \bmod 2 = 1\), which holds by integer arithmetic.
A Pauli operator does not anticommute with \(L\) if and only if it commutes with \(L\):
Unfolding the definitions, this reduces to \(|\mathcal{S}_Z| \bmod 2 \neq 1 \Leftrightarrow |\mathcal{S}_Z| \bmod 2 = 0\), which holds by integer arithmetic.
If a Pauli operator \(P\) anticommutes with \(L\) (i.e., \(|\mathcal{S}_Z \cap V_G|\) is odd), then no valid deforming path exists:
This formalizes the remark’s central claim: there is no deformed version of a Pauli operator that does not commute with the logical \(L\).
Suppose for contradiction there exists \(\gamma \) with \(\mathrm{IsValidDeformingPath}(G, \mathcal{S}_Z, \gamma )\). By valid_path_boundary_support_eq, we have \(\mathrm{boundarySupport}(G,\gamma ) = \mathcal{S}_Z\). By boundary_cardinality_even, \(|\mathrm{boundarySupport}(G,\gamma )| \bmod 2 = 0\). Rewriting, \(|\mathcal{S}_Z| \bmod 2 = 0\). But anticommutation gives \(|\mathcal{S}_Z| \bmod 2 = 1\), a contradiction by integer arithmetic.
Restatement of the above using the existing theorem no_valid_path_if_odd from Def 4:
This follows directly from no_valid_path_if_odd.
The contrapositive: existence of a valid deforming path implies \(P\) commutes with \(L\):
Unfolding the definition of commutesWithL, we need \(|\mathcal{S}_Z| \bmod 2 = 0\). This follows directly from zSupport_even_of_valid_path_exists applied to \(G\), \(\mathcal{S}_Z\), and \(\gamma \).
If \(P\) anticommutes with \(L\), then for every edge-path \(\gamma \), \(\gamma \) is not a valid deforming path for \(\mathcal{S}_Z\). This captures the paper’s claim: “there is no way to multiply such a \(P\) with stabilizers \(Z_e\) and \(s_j\) to make it commute with all the Gauss’s law operators \(A_v\).”
Let \(\gamma \) be an arbitrary edge-path. From no_deforming_path_for_anticommuting_operator, we know there is no valid deforming path at all. If \(\gamma \) were valid, it would witness the existence of such a path, yielding a contradiction.
The Gauss law operators \(A_v\) form an obstruction: if \(P\) anticommutes with \(L\), then for any edge-path \(\gamma \), either there exists a vertex \(v\) at which the deformed Gauss law symplectic condition fails (i.e., the symplectic inner product is not even), or \(\gamma \) is not a valid deforming path:
We take the right disjunct. By stabilizer_product_cannot_fix_anticommutation, \(\gamma \) is not a valid deforming path for \(\mathcal{S}_Z\).
Complete formalization of Remark 12: There is no deformed version of a Pauli operator \(P\) that does not commute with the logical \(L\). Specifically:
If \(P\) anticommutes with \(L\), then \(|\mathcal{S}_Z \cap V_G|\) is odd.
A path boundary \(\partial \gamma \) always has even cardinality.
Therefore no edge-path \(\gamma \) with \(\partial \gamma = \mathcal{S}_Z \cap V_G\) exists.
Hence \(P\) has no well-defined deformed version.
This follows directly from no_deforming_path_for_anticommuting_operator.
The logical structure of the argument consists of four parts, all holding simultaneously:
\(\mathrm{anticommutesWithL}(\mathcal{S}_Z) \implies |\mathcal{S}_Z| \bmod 2 = 1\).
\(\forall \gamma ,\; |\mathrm{boundarySupport}(G,\gamma )| \bmod 2 = 0\).
\(\forall \gamma ,\; \mathrm{IsValidDeformingPath}(G,\mathcal{S}_Z,\gamma ) \implies \mathrm{boundarySupport}(G,\gamma ) = \mathcal{S}_Z\).
\(\mathrm{anticommutesWithL}(\mathcal{S}_Z) \implies \nexists \gamma ,\; \mathrm{IsValidDeformingPath}(G,\mathcal{S}_Z,\gamma )\).
We verify each part. Part 1: the anticommutation hypothesis is exactly the statement that \(|\mathcal{S}_Z| \bmod 2 = 1\), so this holds by the identity function. Part 2: this is exactly boundary_cardinality_even. Part 3: this is exactly valid_path_boundary_support_eq. Part 4: this is exactly no_deforming_path_for_anticommuting_operator.
If there exists a valid deforming path for \(\mathcal{S}_Z\), then \(P\) must commute with \(L\):
From the hypothesis, we obtain \(\gamma \) and \(h_\gamma \) witnessing the valid deforming path. We then apply deformation_requires_commutation to \(G\), \(\mathcal{S}_Z\), \(\gamma \), and \(h_\gamma \).
The existence of a valid deforming path implies commutation:
Assume the hypothesis. This follows directly from valid_path_implies_commutes_with_L.
The forward direction, which is the key result of the remark: if \(P\) anticommutes with \(L\), then no valid deforming path exists.
This follows directly from no_deforming_path_for_anticommuting_operator.