Inner and outer codes and resource-overhead data

Let $H_L,H_P,H_O$ be nonzero finite-dimensional complex inner-product spaces, each with a specified ordered orthonormal basis. Write $\mathcal{L}(H)$ for the complex-linear endomorphisms of $H$, and use the inner product linear in its first argument. For vectors $u,v$, write $\mathrm{\mid }u⟩⟨v\mathrm{\mid }$ for the operator $x\mapsto ⟨x,v⟩u$; ket basis labels refer to the specified bases. An inner code consists of a linear isometry $V:H_L\to H_P$; regard $H_P$ as a one-factor physical register unless a finer ordered finite nonempty register is specified. Fix quantum channels $\mathcal{N}:\mathcal{L}(H_P)\to \mathcal{L}(H_O)$ and $\mathcal{R}:\mathcal{L}(H_O)\to \mathcal{L}(H_L)$, describing one noise cycle and its decoder, and a positive cycle duration $\tau$. A quantum channel is a complex-linear, completely positive, trace-preserving map. The decoded inner channel is $\mathcal{E}=\mathcal{R}\circ \mathcal{N}\circ {\mathrm{Ad}}_V$, where ${\mathrm{Ad}}_V(Y)=VY{V}^{\ast }$.

An outer code is a linear isometry $W:K_L\to B$, where $K_L$ is a nonzero finite-dimensional complex inner-product space and $B={⨂}_{b=1}^n{K}_b$ is an ordered register with $n\ge 1$ and nonzero finite-dimensional complex inner-product factors, all with specified ordered orthonormal bases. For every outer factor implemented by an inner block, specify its inner-code data and a unitary identification $J_b:H_{L,b}\to K_b$. The effective channel on that factor is ${\mathcal{E}}_b={\mathrm{Ad}}_{J_b}\circ {\mathcal{E}}_b^{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}\circ {\mathrm{Ad}}_{J_b^{\ast }}$. Thus the decoded channel of an inner block is the one-cycle noise interface supplied to its outer physical factor.

For one cycle on the full register, specify a joint channel $\mathrm{\Phi }:\mathcal{L}(B)\to \mathcal{L}(B)$. Write ${\mathrm{Tr}}_{\mathrm{\ne }b}$ for the partial trace over all factors except $K_b$, using their specified order and wire permutations. A local marginal channel ${\mathrm{\Phi }}_b:\mathcal{L}(K_b)\to \mathcal{L}(K_b)$ exists in the input-independent sense used here only if ${\mathrm{Tr}}_{\mathrm{\ne }b}\mathrm{\Phi }(Y)={\mathrm{\Phi }}_b({\mathrm{Tr}}_{\mathrm{\ne }b}Y)$ for every $Y\in \mathcal{L}(B)$. When local marginals exist, their consistency with the inner-block interface is an additional requirement ${\mathrm{\Phi }}_b={\mathcal{E}}_b$ on implemented factors. Cross-block independence for this joint channel means that channels ${\mathrm{\Phi }}_b$ exist with $\mathrm{\Phi }={⨂}_b{\mathrm{\Phi }}_b$; failure of this factorization is called cross-block correlation here. An arbitrary joint channel need not have input-independent local marginals. This definition concerns the joint channel for the declared cycle or protocol; later cycles also require the memory data below.

For a finite horizon of $N\ge 1$ cycles, a sequential noise process is specified by nonzero finite-dimensional complex inner-product memory spaces $M_0,\dots ,M_N$ with ordered orthonormal bases, a unit vector $m_0\in M_0$, and channels ${\mathcal{T}}_t:\mathcal{L}(M_{t-1}\otimes B)\to \mathcal{L}(M_t\otimes B)$ for $1\le t\le N$. Initialize memory by the isometry $v\mapsto m_0\otimes v$, apply the channels in time order, and trace out the final memory. Memory can include a shared environment. The process also specifies the response to interventions: for any nonzero finite-dimensional complex inner-product ancillary space $A$ with a specified ordered orthonormal basis, carry its identity channel through each ${\mathcal{T}}_t$, allow an arbitrary channel on $B\otimes A$ between successive cycles, and trace out memory at any prefix of the sequence. Temporal independence means that there exist fixed channels ${\mathrm{\Lambda }}_1,\dots ,{\mathrm{\Lambda }}_N$ on $B$ whose memory-free sequence gives the same reduced output map at every prefix for every $A$ and every such intervention sequence. Failure of this equality is called temporal correlation here. A one-block, one-cycle channel does not determine either the joint channel or this intervention response.

A resource-overhead record fixes a target logical task, an error criterion and target tolerance, an architecture and noise process, and a counting convention. It records the physical degrees of freedom, operation counts, and elapsed time used to perform that task under those choices; any ratio called an overhead must also state its reference task and a positive reference resource count. An inner-code comparison fixes the encoding, noise, decoder, cycle duration, and the joint and sequential noise data along with the task and counting convention.

For a one-cycle error comparison, write $\mathsf{C}(H_L)$ for the set of channels on $\mathcal{L}(H_L)$. Explicitly supply a reference channel $\mathrm{\Psi }\in \mathsf{C}(H_L)$ and a formula for an error functional $ϵ:\mathsf{C}(H_L)\times \mathsf{C}(H_L)\to [0,1]$ with $ϵ(\mathrm{\Theta },\mathrm{\Theta })=0$ for every $\mathrm{\Theta }\in \mathsf{C}(H_L)$. The value compared is $ϵ(\mathcal{E},\mathrm{\Psi })$, using the same functional and reference for both channels on the same logical space, or after a declared unitary identification. If this value is called one-cycle infidelity, the record must specify its fidelity convention and the resulting formula for $ϵ$; no particular fidelity is implicit. Lower one-cycle infidelity means a smaller such value. It is one datum in a resource comparison and by itself specifies no quantitative overhead law.

Given: The spaces, bases, isometries, channels, memory data, error functional and resource conventions in the Definition.

Goal: Verify typing and show explicitly why one-cycle channels do not determine spatial or temporal correlations or an overhead law.

1.1 The hypotheses of [F1] now apply to each inner block, so its decoded map is a channel. Unitary identifications and their inverses are isometries, hence their conjugation maps are channels by [F1]. Their composition gives the displayed channel on $K_b$. The outer encoding satisfies [F3] because its logical space is nonzero and its physical register is finite and nonempty with the structure in [F4]. [given, F1, F2, F3, F4]

1.2 The local-marginal equality is a stated compatibility condition, not a conclusion for arbitrary joint channels. By [F8], tensor products of linear maps act factorwise on elementary tensors. For finite spaces $H,K$, [F7] identifies $\mathcal{L}(H)\otimes \mathcal{L}(K)$ with $\mathcal{L}(H\otimes K)$ by sending tensors of matrix units to the corresponding tensor operators: these are bases on both sides, so this map is a linear isomorphism. Under this identification, $\mathrm{\Phi }\otimes \mathrm{\Psi }$ is the unique linear map satisfying $(\mathrm{\Phi }\otimes \mathrm{\Psi })(A\otimes C)=\mathrm{\Phi }(A)\otimes \mathrm{\Psi }(C)$; iterate this construction for more factors, including identity maps on ancillas. All process operations are channels: tensor products of channels with Kraus families $(L_a)$ and $(Q_c)$ have Kraus family $(L_a\otimes Q_c)$, whose normalization follows from ${\sum }_{a,c}(L_a^{\ast }{L}_a)\otimes (Q_c^{\ast }{Q}_c)=I\otimes I$. This Kraus map agrees with the tensor channel on elementary operator tensors by [F8], hence everywhere by [F7]. The adjoint and multiplication rules used here follow by evaluating on product vectors with [F4] and [F8]. Tracing out $M$ has Kraus maps $C_s(m\otimes v)=⟨m,e_s⟩v$ in its supplied basis; [F5] gives the trace formula and ${\sum }_s{C}_s^{\ast }{C}_s=I$. Memory initialization is conjugation by an isometry. By [F6] these maps, their ancillary extensions and the declared interventions are channels; composition is a channel by [F1]. The resulting reduced maps are therefore well-typed at every prefix. [given, F1, F4, F5, F6, F7, F8]

2.1 For a spatial example let $K_1=K_2={\mathbb{C}}^2$, let $S\mathrm{\mid }0⟩=\mathrm{\mid }1⟩$ and $S\mathrm{\mid }1⟩=\mathrm{\mid }0⟩$, and set $D(Y)=(Y+SY{S}^{\ast })\mathrm{/}2$. Define ${\mathrm{\Phi }}_{\mathrm{i}\mathrm{n}\mathrm{d}}=D\otimes D$ and ${\mathrm{\Phi }}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}}(Y)=(Y+(S\otimes S)Y(S\otimes S{)}^{\ast })\mathrm{/}2$. The Kraus families $I\mathrm{/}\sqrt{2},S\mathrm{/}\sqrt{2}$ for $D$, and $I\otimes I\mathrm{/}\sqrt{2},S\otimes S\mathrm{/}\sqrt{2}$ for ${\mathrm{\Phi }}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}}$, are normalized, so these are channels by [F6] and step 1.2. On $A\otimes C$, tracing out the second factor gives $D(A)\mathrm{tr}C$ for both joint channels, because conjugation by $S$ permutes the diagonal entries and preserves trace; the first-factor calculation is identical. By [F5] and [F7], both have local channel $D$ on all input operators, including entangled inputs. Yet on $\mathrm{\mid }00⟩⟨00\mathrm{\mid }$, the independent output assigns diagonal weight $1\mathrm{/}4$ to each of $00,01,10,11$, whereas the correlated output assigns weight $1\mathrm{/}2$ to each of $00,11$ and zero to the others. Thus the channels are distinct with the same local marginals. [F4, F5, F6, F7, step 1.2, construct, algebra]

3.1 For two temporal uses of $B={\mathbb{C}}^2$, initialize a memory qubit in $m_0=(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$ and in each cycle apply the unitary $U=\mathrm{\mid }0⟩⟨0\mathrm{\mid }\otimes I+\mathrm{\mid }1⟩⟨1\mathrm{\mid }\otimes S$ on memory and system. It is unitary since its two orthogonal memory sectors carry the unitaries $I,S$. Tracing the memory after one use gives $D$ by [F5]. For an initial system operator of trace one, replacing the system input with a fresh input before the second use also gives $D$: the memory's diagonal weights remain $1\mathrm{/}2,1\mathrm{/}2$, and tracing memory removes its off-diagonal blocks. With no intervention, two uses give the identity channel since $U^2=I$. Compare the memory-free sequence $D,D$: it has the same fresh-input one-use channels, but its two-use channel is $D\circ D=D$, which differs from the identity on $\mathrm{\mid }0⟩⟨0\mathrm{\mid }$. Replacement by a unit vector $\psi$ is the channel $Y\mapsto \mathrm{tr}(Y)\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$, with normalized Kraus maps $\mathrm{\mid }\psi ⟩⟨j\mathrm{\mid }$. Varying $\psi$ determines a linear map on all operators, since the projectors onto basis vectors and their real and imaginary two-vector superpositions span the matrix units. The shared-memory process cannot be temporal-independent as defined: its first prefix forces ${\mathrm{\Lambda }}_1=D$, and fresh-input replacement interventions force ${\mathrm{\Lambda }}_2=D$, contradicting the no-intervention output. This proves temporal nonuniqueness within the declared process type. [F1, F4, F5, F6, F7, step 1.2, step 2.1, construct, algebra]

4.1 The error functional and reference are separately supplied data with the domain of the decoded channel from step 1.1; a smaller value is meaningful only under that fixed choice. The resource record likewise separately supplies task, tolerance, architecture, process and counts, and requires a positive denominator for any ratio. None of these definitions supplies a numerical relation from one error value to resource counts. Steps 2.1 and 3.1 explicitly show why marginal one-cycle data cannot replace the joint and sequential process specifications. No physical realization or empirical performance follows from these mathematical constructions. [given, F1, F2, F3, step 1.1, step 2.1, step 3.1] ∎