Manuscript En07-B@1 · En07-B.tex · theorem · 45dff4a1

Theorem: Exact complete-complement criterion

Complete preservation and exact outgoing recovery · lines 385–498 · proof included or linked

En07-B@1#thm:exact, En07-B@1#eq:exact-decoder

For $V : C \to R \otimes B$ , the following conditions are equivalent:

A deterministic CPTP decoder satisfies $D \circ N = id_{C}$ .
$E = T_{σ}$ for one density operator $σ$ on $B$ .
$E (E_{ij}) = δ_{ij} σ$ for all $i, j$ .
$J_{n} (E) = I_{Q} / d \otimes σ$ .
For every $O \in B (B)$ , $V^{†} (I_{R} \otimes O) V$ is a scalar multiple of $I_{C}$ .

When these conditions hold, there are an auxiliary space $K$ , a state $σ_{K}$ , and an isometry $A : C \otimes K \to R$ such that

N (X) = A (X \otimes σ_{K}) A^{†} .

An exact deterministic decoder on all of $R$ is

D (Y) = Tr_{K} (A^{†} Y A) + Tr ((I_{R} - A A^{†}) Y) τ_{C} .

If $r = rank σ$ , then $dim R \geq d r$ .

Proof. Matrix units span $B (C)$ , proving the equivalence of (ii) and (iii). Expanding

P_{Φ_{d}} = \frac{1}{d} i, j \sum E_{ij} \otimes E_{ij}

proves the equivalence with (iv).

If (ii) holds, then

Tr (O E (X)) = Tr (X) Tr (O σ) .

Trace duality gives (v). Conversely, under (v), evaluating the compressed observable at $τ_{C}$ determines its scalar as $Tr (O E (τ_{C}))$ . Since observables separate operators,

E (X) = Tr (X) E (τ_{C}) .

Thus (v) implies (ii).

Assume (ii). Diagonalize

σ = a = 1 \sum^{r} p_{a} ∣ b_{a} ⟩ ⟨ b_{a} ∣, p_{a} > 0, a \sum p_{a} = 1.

The $B$ marginal of each $V ∣ i ⟩$ is supported on $supp σ$ , so write

V ∣ i ⟩ = a = 1 \sum^{r} p_{a} ∣ u_{ia} ⟩_{R} ∣ b_{a} ⟩_{B} .

The coefficient of $∣ b_{a} ⟩ ⟨ b_{b} ∣$ in $E (E_{ij}) = δ_{ij} σ$ gives

⟨ u_{j b}, u_{ia} ⟩ = δ_{ij} δ_{ab} .

There are therefore $d r$ orthonormal vectors in $R$ .

Let $K$ have basis ${∣ a ⟩}_{a = 1}^{r}$ , define

A (∣ i ⟩ ∣ a ⟩) = ∣ u_{ia} ⟩, σ_{K} = a \sum p_{a} ∣ a ⟩ ⟨ a ∣ .

Then $A$ is an isometry, and tracing $B$ gives the asserted normal form for $N$ .

Both terms of (eq:exact-decoder) are completely positive. Their traces add to $Tr Y$ . Substitution of the normal form shows $DN (X) = X$ for every $X$ .

Conversely, suppose (i). Choose a Kraus family ${D_{α}}$ for $D$ and its isometric dilation

L : R ⟶ C \otimes G, L ∣ r ⟩ = α \sum D_{α} ∣ r ⟩ \otimes ∣ α ⟩ .

The composite isometry $Z = (L \otimes I_{B}) V$ dilates the identity on $C$ . The decoded marginal of $Z ∣ i ⟩$ is the pure state $∣ i ⟩ ⟨ i ∣$ , so

Z ∣ i ⟩ = ∣ i ⟩ \otimes ∣ ξ_{i} ⟩_{GB} .

Exact recovery of $E_{ij}$ then gives $⟨ ξ_{j}, ξ_{i} ⟩ = 1$ . All normalized vectors $ξ_{i}$ are one vector $ξ$ . Tracing $C \otimes G$ yields

E (E_{ij}) = δ_{ij} Tr_{G} ∣ ξ ⟩ ⟨ ξ ∣,

which is (iii).

All identities are operator identities; tensoring with $id_{Q}$ proves reference preservation.

Displayed formulas in this passage

En07-B@1#eq:exact-decoder · line 405
$D (Y) = Tr_{K} (A^{†} Y A) + Tr ((I_{R} - A A^{†}) Y) τ_{C} .$
unlabeled · line 401
$N (X) = A (X \otimes σ_{K}) A^{†} .$
unlabeled · line 419
$P_{Φ_{d}} = \frac{1}{d} i, j \sum E_{ij} \otimes E_{ij}$
unlabeled · line 425
$Tr (O E (X)) = Tr (X) Tr (O σ) .$
unlabeled · line 433
$E (X) = Tr (X) E (τ_{C}) .$
unlabeled · line 439
$σ = a = 1 \sum^{r} p_{a} ∣ b_{a} ⟩ ⟨ b_{a} ∣, p_{a} > 0, a \sum p_{a} = 1.$
unlabeled · line 445
$V ∣ i ⟩ = a = 1 \sum^{r} p_{a} ∣ u_{ia} ⟩_{R} ∣ b_{a} ⟩_{B} .$
unlabeled · line 453
$⟨ u_{j b}, u_{ia} ⟩ = δ_{ij} δ_{ab} .$
unlabeled · line 461
$A (∣ i ⟩ ∣ a ⟩) = ∣ u_{ia} ⟩, σ_{K} = a \sum p_{a} ∣ a ⟩ ⟨ a ∣ .$
unlabeled · line 475
$L : R ⟶ C \otimes G, L ∣ r ⟩ = α \sum D_{α} ∣ r ⟩ \otimes ∣ α ⟩ .$
unlabeled · line 482
$Z ∣ i ⟩ = ∣ i ⟩ \otimes ∣ ξ_{i} ⟩_{GB} .$
unlabeled · line 489
$E (E_{ij}) = δ_{ij} Tr_{G} ∣ ξ ⟩ ⟨ ξ ∣,$

Theorem: Exact complete-complement criterion

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