Manuscript Em03-E@1 · Em03-E-v1.tex · lemma · c4b07e6a

Lemma: Pauli comparison

E4: explicit decoder and optimal referenced recovery › A Pauli-channel distance identity · lines 841–861 · proof included or linked

Em03-E@1#lem:Pauli

Back to manuscript

For qubit Pauli channels

Φ = P \sum a_{P} Ad_{P}, Ψ = P \sum b_{P} Ad_{P},

d_{⋄} (Φ, Ψ) = \frac{1}{2} P \sum ∣ a_{P} - b_{P} ∣.

In particular, $d_{⋄} (Φ, id) = 1 - a_{I}$ .

Proof. The triangle inequality and $∥ Ad_{P} ∥_{⋄} = 1$ give the upper bound. On a maximally entangled input, the four Pauli outputs are orthogonal Bell projectors. Their trace-distance difference is exactly the stated classical distance, proving equality.

Displayed formulas in this passage

unlabeled · line 844
$Φ = P \sum a_{P} Ad_{P}, Ψ = P \sum b_{P} Ad_{P},$
unlabeled · line 848
$d_{⋄} (Φ, Ψ) = \frac{1}{2} P \sum ∣ a_{P} - b_{P} ∣.$

Nearby in this section

section: A Pauli-channel distance identity
theorem: Theorem: Exact optimal recovery
section: A Pauli-channel distance identity