Manuscript Psy03-A@1 · Psy03-A.tex · theorem · b30eb6e9

Theorem: Exact node-record repair pilot

Exact coherence-sensitive pilots › Coherence converted to a node population · lines 1761–1810 · proof included or linked

Psy03-A@1#thm:nodepilot, Psy03-A@1#eq:nodecontrast

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For $0 \leq γ < 2Ω$ , write

ω_{d} = Ω^{2} - γ^{2} /4 .

The coherence-sensitive node contrast is

C_{P_{0}} (T) = \frac{Ω}{2 ω _{d}} e^{- γ T /2} sin (ω_{d} T) .

It is positive for $0 < T < π / ω_{d}$ .

For the equal-prior inputs $(I \pm Y) /2$ , node readout with labels assigned to $P_{0}, P_{1}$ has success

Q_{node} (T) = \frac{1}{2} + \frac{Ω}{2 ω _{d}} e^{- γ T /2} sin (ω_{d} T) .

Removing the coupling, $Ω = 0$ , gives success $1/2$ . Restoring a coupling in the stated positive interval therefore gives a strict task improvement.

There is no global monotonicity in coupling strength or sensing time: the sine term can change sign.

Proof. For the initial $+ Y$ Bloch vector, its $y, z$ components satisfy

y ˙ = - γ y - Ω z, z ˙ = Ω y, y (0) = 1, z (0) = 0.

Thus

z ¨ + γ z ˙ + Ω^{2} z = 0, z ˙ (0) = Ω,

whose underdamped solution is $z (T) = (Ω/ ω_{d}) e^{- γ T /2} sin (ω_{d} T)$ . The reference $I /2$ remains fixed. The node contrast is $z (T) /2$ . The opposite input gives the opposite $z$ , proving the binary success formula. At zero coupling the initial populations stay equal.

Displayed formulas in this passage

Psy03-A@1#eq:nodecontrast · line 1768
$C_{P_{0}} (T) = \frac{Ω}{2 ω _{d}} e^{- γ T /2} sin (ω_{d} T) .$
unlabeled · line 1764
$ω_{d} = Ω^{2} - γ^{2} /4 .$
unlabeled · line 1778
$Q_{node} (T) = \frac{1}{2} + \frac{Ω}{2 ω _{d}} e^{- γ T /2} sin (ω_{d} T) .$
unlabeled · line 1794
$y ˙ = - γ y - Ω z, z ˙ = Ω y, y (0) = 1, z (0) = 0.$
unlabeled · line 1800
$z ¨ + γ z ˙ + Ω^{2} z = 0, z ˙ (0) = Ω,$

Theorem: Exact node-record repair pilot

Displayed formulas in this passage

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