Manuscript En04-A@1 · En04-A.tex · theorem · bab77211

Theorem: First-order residual response witness

A genuine Clifford criterion and a responding deformation › A general sufficient response condition · lines 1231–1278 · proof included or linked

En04-A@1#thm:linearresponse, En04-A@1#eq:linearresponse

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Let $V_{0} : C \to R B$ have constant complement, let $K = K^{†}$ be bounded, and define

V_{λ} = e^{- i λ K} V_{0} .

Fix code states $ρ_{0}, ρ_{1}$ , put $X = ρ_{0} - ρ_{1}$ , and choose a bounded Hermitian residual observable $O$ with $∥ O ∥ \leq 1$ . Suppose

m = Tr [O (- i Tr_{R} [K, V_{0} X V_{0}^{†}])] \neq = 0.

Then

δ_{full} (V_{λ}) \geq max {0, \frac{∣ m ∣ ∣ λ ∣}{4} - ∥ K ∥^{2} λ^{2}} .

In particular, the complement is nonconstant for

0 < ∣ λ ∣ < \frac{∣ m ∣}{4 ∥ K ∥ ^{2}} .

Proof. The derivative of the conjugated trace-class operator is $- i [K, V_{0} X V_{0}^{†}]$ . The double commutator has trace norm at most $4 ∥ K ∥^{2} ∥ X ∥_{1}$ . Taylor's integral remainder is therefore bounded by $2 λ^{2} ∥ K ∥^{2} ∥ X ∥_{1} \leq 4 λ^{2} ∥ K ∥^{2}$ .

The baseline residual image of $X$ is zero by constancy. Hence

∣ Tr (O C_{λ} (X)) ∣ \geq ∣ m ∣ ∣ λ ∣ - 4 λ^{2} ∥ K ∥^{2} .

Apply (eq:observablewitness). The assumption $m \neq = 0$ entails $∥ K ∥ > 0$ .

Displayed formulas in this passage

En04-A@1#eq:linearresponse · line 1249
$δ_{full} (V_{λ}) \geq max {0, \frac{∣ m ∣ ∣ λ ∣}{4} - ∥ K ∥^{2} λ^{2}} .$
unlabeled · line 1235
$V_{λ} = e^{- i λ K} V_{0} .$
unlabeled · line 1242
$m = Tr [O (- i Tr_{R} [K, V_{0} X V_{0}^{†}])] \neq = 0.$
unlabeled · line 1257
$0 < ∣ λ ∣ < \frac{∣ m ∣}{4 ∥ K ∥ ^{2}} .$
unlabeled · line 1272
$∣ Tr (O C_{λ} (X)) ∣ \geq ∣ m ∣ ∣ λ ∣ - 4 λ^{2} ∥ K ∥^{2} .$

Nearby in this section

section: A general sufficient response condition