Manuscript Psy03-B@1 · Psy03-B-v1.tex · theorem · 8ca3caf9

Theorem: Coherence-sensitive structural-memory contrast

An exact one-edge structural-memory pilot › Common future challenge and task performance · lines 2197–2247 · proof included or linked

Psy03-B@1#thm:pilot, Psy03-B@1#eq:C, Psy03-B@1#eq:pilotQ

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The unscarred-sector contrast is

C (T) = Tr [P_{0} Λ_{0} (T) (ρ_{+ y} - I /2)] = \frac{Ω}{2 ω _{d}} e^{- γ T /2} sin (ω_{d} T) .

The scarred-sector contrast is zero.

After $t$ damage updates, the unconditional coherent-versus- dephased node contrast is $w_{t} C (T)$ . For equal-prior phase-label inputs $(I \pm Y) /2$ , with node labels assigned to $P_{0}, P_{1}$ , success is

Q_{t} (T) = \frac{1}{2} + w_{t} C (T) .

At $γ = 0$ and $T = π / (2Ω)$ ,

C (T) = \frac{1}{2}, Q_{t} (T) = \frac{1}{2} + \frac{( 1 - h ) ^{t}}{2} .

Proof. In sector zero, the $y, z$ Bloch coordinates obey

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,

and

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

The node contrast is $z (T) /2$ . Sector one retains equal node populations, and $I /2$ is fixed in both sectors. Average over the structural law. The opposite phase input gives opposite $z$ , proving the success formula.

Displayed formulas in this passage

Psy03-B@1#eq:C · line 2200
$C (T) = Tr [P_{0} Λ_{0} (T) (ρ_{+ y} - I /2)] = \frac{Ω}{2 ω _{d}} e^{- γ T /2} sin (ω_{d} T) .$
Psy03-B@1#eq:pilotQ · line 2214
$Q_{t} (T) = \frac{1}{2} + w_{t} C (T) .$
unlabeled · line 2219
$C (T) = \frac{1}{2}, Q_{t} (T) = \frac{1}{2} + \frac{( 1 - h ) ^{t}}{2} .$
unlabeled · line 2227
$y ˙ = - γ y - Ω z, z ˙ = Ω y, y (0) = 1, z (0) = 0.$
unlabeled · line 2232
$z ¨ + γ z ˙ + Ω^{2} z = 0, z ˙ (0) = Ω,$
unlabeled · line 2237
$z (T) = \frac{Ω}{ω _{d}} e^{- γ T /2} sin (ω_{d} T) .$

Theorem: Coherence-sensitive structural-memory contrast

Displayed formulas in this passage

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