Manuscript Em05-B@1 · Em05-B-v1.tex · theorem · 6c050f5b

Theorem: Balanced decoder and improved recovery sandwich

Improved intermediate recovery control › A balanced decoder with an exact Pauli error · lines 1301–1383 · proof included or linked

Em05-B@1#thm:balanced, Em05-B@1#eq:balancedPauliI, Em05-B@1#eq:balancedUpper, Em05-B@1#eq:improvedSandwich, Em05-B@1#eq:endpointOptima

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The balanced decoder is CPTP. Its decoded channel is unital and Pauli diagonal, with Bloch multipliers

c_{η} = \frac{η}{1 + t ^{2}}, c_{η}, s_{η} = \frac{1 - t ^{2}}{1 + t ^{2}} .

Its Pauli probabilities are

p_{I} p_{Z} p_{X} = p_{Y} = \frac{1}{2 ( 1 + t ^{2} )} + \frac{η}{2 1 + t ^{2}}, = \frac{1}{2 ( 1 + t ^{2} )} - \frac{η}{2 1 + t ^{2}}, = \frac{t ^{2}}{2 ( 1 + t ^{2} )} .

Its exact error is

U (η) = 1 - \frac{1}{2 ( 1 + t ^{2} )} - \frac{η}{2 1 + t ^{2}} .

Thus

B (η) \leq ε_{rec} (N_{η}) \leq min {t, \frac{3}{4}, U (η)} .

The endpoints are exact:

ε_{rec} (N_{0}) = \frac{3}{4}, ε_{rec} (N_{1}) = 0.

Proof. A complete Kraus family for the balanced decoder is

G_{0} W^{†}, G_{1} W^{†}, ∣1 ⟩ ⟨ 01∣, ∣1 ⟩ ⟨ 10∣,

whose effects sum to $I_{R}$ .

For a general value $x$ , its decoded excitation probability on input (eq:input) is

x + p (1 - x) (1 - t^{2}),

and its coherence is $η 1 - x z$ . At $x = t^{2} / (1 + t^{2})$ , the populations become $x + p s_{η}$ , with $x = (1 - s_{η}) /2$ . The channel is therefore unital with the stated Bloch multipliers. Expanding a unital Pauli-diagonal qubit channel gives (eq:balancedPauliI) and its companions.

The probabilities are nonnegative. In particular,

η 1 + t^{2} = (1 - t) 1 + t^{2} \leq (1 - t) (1 + t) = 1 - t^{2} \leq 1,

which proves $p_{Z} \geq 0$ . Lemma lem:witnesses gives its exact error $1 - p_{I} = U (η)$ . The lower bound is Theorem thm:optimalBell; the other upper bounds were proved above.

At $η = 0$ , both $B$ and $U$ equal $3/4$ . At $η = 1$ , both equal zero.

Displayed formulas in this passage

Em05-B@1#eq:balancedPauliI · line 1313
$p_{I} p_{Z} p_{X} = p_{Y} = \frac{1}{2 ( 1 + t ^{2} )} + \frac{η}{2 1 + t ^{2}}, = \frac{1}{2 ( 1 + t ^{2} )} - \frac{η}{2 1 + t ^{2}}, = \frac{t ^{2}}{2 ( 1 + t ^{2} )} .$
Em05-B@1#eq:balancedUpper · line 1324
$U (η) = 1 - \frac{1}{2 ( 1 + t ^{2} )} - \frac{η}{2 1 + t ^{2}} .$
Em05-B@1#eq:improvedSandwich · line 1332
$B (η) \leq ε_{rec} (N_{η}) \leq min {t, \frac{3}{4}, U (η)} .$
Em05-B@1#eq:endpointOptima · line 1341
$ε_{rec} (N_{0}) = \frac{3}{4}, ε_{rec} (N_{1}) = 0.$
unlabeled · line 1305
$c_{η} = \frac{η}{1 + t ^{2}}, c_{η}, s_{η} = \frac{1 - t ^{2}}{1 + t ^{2}} .$
unlabeled · line 1351
$G_{0} W^{†}, G_{1} W^{†}, ∣1 ⟩ ⟨ 01∣, ∣1 ⟩ ⟨ 10∣,$
unlabeled · line 1359
$x + p (1 - x) (1 - t^{2}),$
unlabeled · line 1370
$η 1 + t^{2} = (1 - t) 1 + t^{2} \leq (1 - t) (1 + t) = 1 - t^{2} \leq 1,$

Theorem: Balanced decoder and improved recovery sandwich

Displayed formulas in this passage

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