LemmaStatement: AI-generatedProof: AI-generated

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Telescoping error budget for repeated finite channels

Statement

Let $d \geq 1$ and let $E, F : M_{d} (C) \to M_{d} (C)$ be quantum channels: complex-linear, completely positive, trace-preserving maps. For a linear map $T$ between finite matrix spaces define $∥ T ∥_{1 \to 1} = \sup_{X \neq 0} ∥ T (X) ∥_{1} / ∥ X ∥_{1}$ , where $∥ X ∥_{1} = Tr \sqrt{X^{*} X}$ . For every integer $N \geq 0$ , $∥ E^{N} - F^{N} ∥_{1 \to 1} \leq N ∥ E - F ∥_{1 \to 1}$ , with zeroth powers equal to the identity map. For a fixed reference dimension $r \geq 1$ , use the reference-first ordered product basis of $C^{r} \otimes C^{d}$ , identifying its operator space with $M_{r d} (C)$ . For $X \in M_{r d} (C)$ , define its system blocks $X_{i j} \in M_{d} (C)$ by $(X_{i j})_{a b} = X_{(i, a), (j, b)}$ for $0 \leq i, j < r$ and $0 \leq a, b < d$ . For any complex-linear $T : M_{d} (C) \to M_{d} (C)$ , define its reference extension ${\hat{T}}_{r} : M_{r d} (C) \to M_{r d} (C)$ by ${\hat{T}}_{r} (X)_{(i, a), (j, b)} = T (X_{i j})_{a b}$ . Thus the extension leaves reference matrix-unit indices unchanged and applies $T$ to each system block. The same error budget holds for these extensions: $∥ {\hat{E}}_{r}^{N} - {\hat{F}}_{r}^{N} ∥_{1 \to 1} \leq N ∥ {\hat{E}}_{r} - {\hat{F}}_{r} ∥_{1 \to 1}$ , with zeroth powers again the identity.

For $q \in [0, 1]$ , define the qubit dephasing channel $D_{q}$ by $D_{q} (\begin{matrix} a & b \\ c & d \end{matrix}) = \begin{matrix} a & q b \\ q c & d \end{matrix}$ . Then $D_{q}^{N} = D_{q^{N}}$ and $1 - q^{N} \leq N (1 - q)$ . For a normalized pure input $∣ ψ ⟩ = α ∣ 0 ⟩ + β ∣ 1 ⟩$ , defining its output fidelity as $⟨ ψ ∣ D_{q}^{N} (∣ ψ ⟩ ⟨ ψ ∣) ∣ ψ ⟩$ , the corresponding infidelity is $2 ∣ α ∣^{2} ∣ β ∣^{2} (1 - q^{N})$ and is at most $N$ times its one-cycle value. A fidelity value for one input alone does not imply a corresponding induced-norm error bound.

Facts & Assumptions

Given: The channels, reference space, integer $N$ , parameter $q$ , and normalized qubit vector in the Statement.

Goal: Establish the norm budget, the dephasing identities and bound, and the limitation of a single-input fidelity.

[F1]

A finite quantum channel is a complex-linear completely positive trace-preserving map, with trace preservation on every input matrix. Finite-dimensional quantum channel

[F2]

Every finite quantum channel contracts the trace norm on every complex input matrix. Schwarz and trace-norm channel contraction

[F3]

Finite ordered tensor registers have an ordered orthonormal product basis with the product inner product, so a reference and system operator space may be represented in that basis as one finite matrix space. Ordered tensor registers, their inner product, and wire permutations

[F4]

On every finite matrix space, $∥ X ∥_{1} = Tr \sqrt{X^{*} X}$ is a norm; in particular it is positive for $X \neq 0$ , homogeneous, and satisfies the triangle inequality. Finite matrix trace-norm ideal inequalities

[F5]

The qubit register is $Q_{2} = C^{2}$ with ordered computational basis $∣ 0 ⟩ = e_{0} = (1, 0)^{T}$ , $∣ 1 ⟩ = e_{1} = (0, 1)^{T}$ , the basis used in the matrix formula for $D_{q}$ . Kets are coordinate columns, bras their conjugate-transpose rows, and $∣ x ⟩ ⟨ y ∣$ has coordinate matrix $x y^{*}$ . Finite registers and their ordered computational bases Dirac Notation Conversion

[F6]

For a map $T$ on the system operator space, its amplification $T^{(r)}$ is defined on the system-first space $C^{d} \otimes C^{r}$ by applying $T$ to each system block indexed by reference indices. Complete positivity means that $T^{(r)}$ sends nonnegative operators to nonnegative operators for every $r \geq 1$ . Completely positive maps

[F7]

A map between finite-dimensional complex operator spaces given by a finite sum $T (X) = \sum_{a} K_{a} X K_{a}^{*}$ of linear operators $K_{a}$ is completely positive. A finite Kraus sum is completely positive

[F8]

An operator is nonnegative when it is self-adjoint and its quadratic form is nonnegative on every vector. Non-negative and positive operators

Proof

1.1givenF2F4algebra

For $N \geq 1$ , cancellation of consecutive terms in the finite sum gives $E^{N} - F^{N} = \sum_{j = 0}^{N - 1} E^{N - 1 - j} (E - F) F^{j}$ . By [F2], every power of either channel contracts the trace norm; the zeroth power is the identity. By [F4], the trace-norm denominator is positive on nonzero inputs. Applying its triangle inequality to the displayed sum and then taking the supremum over those inputs gives the stated norm bound. For $N = 0$ both sides are zero.

1.2givenF1F7algebra

With $Z = diag (1, - 1)$ , put $K_{0} = \sqrt{(1 + q) / 2} I_{2}$ and $K_{1} = \sqrt{(1 - q) / 2} Z$ . For $q \in [0, 1]$ these are linear operators on $C^{2}$ , and direct matrix multiplication gives $D_{q} (X) = K_{0} X K_{0}^{*} + K_{1} X K_{1}^{*}$ for every $X \in M_{2} (C)$ . The finite-sum theorem [F7] therefore makes $D_{q}$ completely positive. The same matrix formula shows complex linearity and preserves the two diagonal entries, hence the trace of every input matrix. Thus $D_{q}$ is a channel by [F1]. Its off-diagonal entries gain a factor $q$ per application, giving $D_{q}^{N} = D_{q^{N}}$ , including $N = 0$ . The geometric-sum identity gives $1 - q^{N} = (1 - q) \sum_{j = 0}^{N - 1} q^{j} \leq N (1 - q)$ , with an empty sum when $N = 0$ .

1.3givenF1F3algebra

The block formula in the Statement assigns every entry of ${\hat{T}}_{r} (X)$ uniquely and is complex-linear because $T$ is. It is therefore a well-defined linear map on $M_{r d} (C)$ using the product basis of [F3]. For another complex-linear $S : M_{d} (C) \to M_{d} (C)$ , applying the rule twice gives ${\hat{S}}_{r} {\hat{T}}_{r} = {\hat{S \circ T}}_{r}$ , and the extension of the identity is the identity. Hence ${\hat{T}}_{r}^{N} = {\hat{T^{N}}}_{r}$ for every $N \geq 0$ . If $T$ preserves trace, then $Tr ({\hat{T}}_{r} (X)) = \sum_{i = 0}^{r - 1} Tr (T (X_{i i})) = \sum_{i = 0}^{r - 1} Tr (X_{i i}) = Tr X$ . In particular the two channel extensions preserve trace by [F1].

2.1givenF1F3F6F8step 1.1step 1.3algebra

To compare factor orders with [F6], fix $s \geq 1$ and define the coordinate permutation $P$ from $(C^{r} \otimes C^{d}) \otimes C^{s}$ to $C^{d} \otimes C^{r s}$ by $P ((e_{i} \otimes e_{a}) \otimes e_{k}) = e_{a} \otimes e_{i s + k}$ . Here $0 \leq i < r$ , $0 \leq a < d$ and $0 \leq k < s$ . The product bases and their inner products in [F3] make this a bijection of orthonormal coordinate bases, so $P^{*} P = P P^{*} = I$ . Coordinate multiplication and [F8] show that conjugation by $P$ or $P^{*}$ preserves nonnegativity: self-adjointness is preserved and $v^{*} P Y P^{*} v = (P^{*} v)^{*} Y (P^{*} v) \geq 0$ when $Y$ is nonnegative. For every $Y$ , the block rules in the Statement and [F6] give $({\hat{T}}_{r})^{(s)} (Y) = P^{*} T^{(r s)} (P Y P^{*}) P$ : the system block indexed by $(i, k), (j, ℓ)$ on either side is obtained by applying $T$ to the matrix with entries $Y_{((i, a), k), ((j, b), ℓ)}$ as $a, b$ vary. For $T = E$ or $F$ , [F6] makes $T^{(r s)}$ positive, so this identity proves positivity of every amplification of ${\hat{T}}_{r}$ . With $s = 1$ it also explicitly identifies the reference-first extension with the supplied system-first amplification under the factor swap. Thus both extensions are completely positive. By step 1.3 and [F1] they are channels, and applying step 1.1 to them proves the stated reference-extended inequality.

3.1givenF4F5step 1.2algebra∎

Using the coordinate and outer-product conventions in [F5], put $p = ∣ α ∣^{2}$ , so $∣ β ∣^{2} = 1 - p$ . Direct multiplication gives $⟨ ψ ∣ D_{q}^{N} (∣ ψ ⟩ ⟨ ψ ∣) ∣ ψ ⟩ = p^{2} + (1 - p)^{2} + 2 p (1 - p) q^{N}$ . Its infidelity is therefore $2 p (1 - p) (1 - q^{N})$ , and step 1.2 bounds it by $N$ times the same expression at $N = 1$ . For the limitation, the identity channel and $D_{0}$ both return $∣ 0 ⟩ ⟨ 0 ∣$ exactly, giving fidelity one on that input, but their difference sends $∣ 0 ⟩ ⟨ 1 ∣$ to itself and hence, by the positive trace norm of this nonzero matrix in [F4], has induced norm at least one. Thus that single fidelity cannot imply a zero induced-norm error bound.

Depends on

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