Preserving the logical marginal forces a gauge-only channel

Fix unit vectors $g\in H_G$ and $\mathrm{\ell }\in H_L$, and write $P_{\mathrm{\ell }}x=⟨x,\mathrm{\ell }⟩\mathrm{\ell }$. Extend $\mathrm{\ell }$ to an orthonormal basis $(e_0,e_1,\dots ,e_{q-1})$ of $H_L$ with $e_0=\mathrm{\ell }$, using [F3]. For each $j$, expand $K_j(g\otimes \mathrm{\ell })={\sum }_{k<q}{y}_{jk}\otimes e_k$. Apply the marginal hypothesis to the rank-one density operator $P_{g\otimes \mathrm{\ell }}$. By [F4] the output is ${\sum }_j{P}_{K_j(g\otimes \mathrm{\ell })}$. In each diagonal entry $k,k$, the partial-trace formula [F1] sums the squared coordinates of $y_{jk}$ over an orthonormal basis of $H_G$, so that entry is ${\sum }_j\mathrm{\parallel }{y}_{jk}{\mathrm{\parallel }}^2$. The required marginal is $P_{\mathrm{\ell }}$, whose $k$th diagonal entry is zero for every $k>0$. Since each squared norm is nonnegative, $y_{jk}=0$ for all $j$ and $k>0$. Hence $K_j(g\otimes \mathrm{\ell })={\eta }_{j,g,\mathrm{\ell }}\otimes \mathrm{\ell }$ for some ${\eta }_{j,g,\mathrm{\ell }}\in H_G$.

Fix a unit $g$. If unit vectors ${\mathrm{\ell }}_1,{\mathrm{\ell }}_2$ are linearly independent, then ${\mathrm{\ell }}_1+{\mathrm{\ell }}_2\mathrm{\ne }0$; normalize their sum and apply step 1.1 to ${\mathrm{\ell }}_1,{\mathrm{\ell }}_2$ and $({\mathrm{\ell }}_1+{\mathrm{\ell }}_2)\mathrm{/}\mathrm{\parallel }{\mathrm{\ell }}_1+{\mathrm{\ell }}_2\mathrm{\parallel }$. Linearity of $K_j$ gives ${\eta }_{j,g,{\mathrm{\ell }}_1}\otimes {\mathrm{\ell }}_1+{\eta }_{j,g,{\mathrm{\ell }}_2}\otimes {\mathrm{\ell }}_2={\eta }_{j,g,{\mathrm{\ell }}_{+}}\otimes ({\mathrm{\ell }}_1+{\mathrm{\ell }}_2)$. Extend ${\mathrm{\ell }}_1,{\mathrm{\ell }}_2$ to a basis and apply its two corresponding coordinate functionals to the second tensor factor; this gives ${\eta }_{j,g,{\mathrm{\ell }}_1}={\eta }_{j,g,{\mathrm{\ell }}_{+}}={\eta }_{j,g,{\mathrm{\ell }}_2}$. If ${\mathrm{\ell }}_2=c{\mathrm{\ell }}_1$ for a unit scalar $c$, linearity and $c\mathrm{\ne }0$ give the same equality directly. Thus ${\eta }_{j,g,\mathrm{\ell }}$ is independent of the unit vector $\mathrm{\ell }$.

Fix a unit ${\mathrm{\ell }}_0\in H_L$. For unit $g$, step 1.1 shows $K_j(g\otimes {\mathrm{\ell }}_0)$ lies in $H_G\otimes \mathrm{span}\{{\mathrm{\ell }}_0\}$. Scaling gives the same for every $g$. Define $A_{j}g$ by $K_j(g\otimes {\mathrm{\ell }}_0)=A_{j}g\otimes {\mathrm{\ell }}_0$; it is unique because $u\mapsto u\otimes {\mathrm{\ell }}_0$ is injective. The linearity of $K_j$ makes $A_j$ linear. By step 2.1, $K_j(g\otimes \mathrm{\ell })=A_{j}g\otimes \mathrm{\ell }$ for all simple tensors. Since simple tensors span by [F8], $K_j=A_j\otimes I_{H_L}$.

Trace preservation [F5] and step 3.1 give $({\sum }_j{A}_j^{\ast }{A}_j)\otimes I_{H_L}=I_{H_G}\otimes I_{H_L}$, using the tensor-adjoint rule [F7]. Apply both sides to $g\otimes {\mathrm{\ell }}_0$ for arbitrary $g$ and a fixed unit ${\mathrm{\ell }}_0$. Injectivity of $u\mapsto u\otimes {\mathrm{\ell }}_0$ yields ${\sum }_j{A}_j^{\ast }{A}_j=I_{H_G}$.

Define ${\mathcal{T}}_G(X)={\sum }_j{A}_{j}X{A}_j^{\ast }$. Its finite Kraus form makes it completely positive by [F9], and the normalization from step 4.1 makes it trace-preserving by [F5]. Thus it is a channel. On a simple operator tensor $X_G\otimes X_L$, step 3.1 gives $\mathrm{\Phi }(X_G\otimes X_L)={\sum }_j(A_j{X}_G{A}_j^{\ast })\otimes X_L={\mathcal{T}}_G(X_G)\otimes X_L$. By [F11], tensor products of operator matrix units form a basis of the full operator space, so linearity gives $\mathrm{\Phi }={\mathcal{T}}_G\otimes {\mathrm{id}}_{\mathcal{B}(H_L)}$.

For a finite reference space $H_R$, choose orthonormal bases of $H_G,H_L,H_R$ by [F3]; [F8] gives the product basis of $H_L\otimes H_R$. Use the canonical reassociation in [F11] and the operator matrix-unit basis there to write any operator $\rho$ on $(H_G\otimes H_L)\otimes H_R$ as $\rho ={\sum }_{s,t}\mathrm{\mid }{g}_s⟩⟨g_t\mathrm{\mid }\otimes {\rho }_{st}$, with ${\rho }_{st}\in \mathcal{B}(H_L\otimes H_R)$. Write $A_j{g}_s={\sum }_r{a}_{rs}^{(j)}{g}_r$. By the first-factor partial-trace formula [F1], the coefficient of ${\rho }_{st}$ after applying $\mathrm{\Phi }\otimes {\mathrm{id}}_{\mathcal{B}(H_R)}$ and tracing out $G$ is ${\sum }_{j,r}{a}_{rs}^{(j)}\overline{a_{rt}^{(j)}}={\sum }_j(A_j^{\ast }{A}_j{)}_{ts}={\delta }_{ts}$. Therefore the output marginal is ${\sum }_s{\rho }_{ss}=({\mathrm{Tr}}_G\otimes {\mathrm{id}}_{\mathcal{B}(H_R)})(\rho )$. The calculation applies to every operator, hence to every density operator and every finite reference.

For the gauge-reset assertion, choose an orthonormal basis $(g_s)$ of $H_G$ and define $C_s:H_G\otimes H_L\to H_L$ by $C_s(g\otimes \mathrm{\ell })=⟨g,g_s⟩\mathrm{\ell }$. Its adjoint is $C_s^{\ast }\mathrm{\ell }=g_s\otimes \mathrm{\ell }$, so ${\sum }_s{C}_s^{\ast }{C}_s(g\otimes \mathrm{\ell })={\sum }_s⟨g,g_s⟩g_s\otimes \mathrm{\ell }=g\otimes \mathrm{\ell }$; simple tensors span, hence ${\sum }_s{C}_s^{\ast }{C}_s=I_{H_G\otimes H_L}$. On the product basis, the matrix entries of ${\sum }_s{C}_{s}Y{C}_s^{\ast }$ are exactly the first-factor contraction entries in [F1], hence ${\mathrm{Tr}}_G(Y)={\sum }_s{C}_{s}Y{C}_s^{\ast }$. Let $J_{g_0}:H_L\to H_P$ be $J_{g_0}\mathrm{\ell }=J(g_0\otimes \mathrm{\ell })$. It is an isometry by [F6]. Choose finite Kraus operators $D_r$ for $\mathcal{D}$ using [F10]. Since $J_{g_0}Y{J}_{g_0}^{\ast }=J(P_{g_0}\otimes Y)J^{\ast }$, the gauge-reset map equals $X\mapsto {\sum }_{s,r}{J}_{g_0}{C}_s{D}_{r}X{D}_r^{\ast }{C}_s^{\ast }{J}_{g_0}^{\ast }$, so it has Kraus operators $J_{g_0}{C}_s{D}_r$, and ${\sum }_{s,r}(J_{g_0}{C}_s{D}_r{)}^{\ast }(J_{g_0}{C}_s{D}_r)={\sum }_r{D}_r^{\ast }({\sum }_s{C}_s^{\ast }{J}_{g_0}^{\ast }{J}_{g_0}{C}_s)D_r={\sum }_r{D}_r^{\ast }{D}_r=I_{H_O}$. Thus the map is completely positive by [F9] and trace-preserving by [F5], so it is a channel. Its range is contained in $\mathrm{im}J$ by construction.