Manuscript X-J-ST@1 · X-J-ST.tex · corollary · 8f0d1a36

Corollary: A gap of 3/16

Section C: complete destructive instruments and the positive gap › An exact finite calibration instance · lines 1308–1336 · proof included or linked

X-J-ST@1#cor:benchmark

Take

ν = 1, r = \frac{1}{4}, d = \frac{σ}{2} lo g 3,

and evaluate $C_{2}$ on $ϕ_{(0, d)}$ . The signal state and laws are

∣ e_{σ} (d)⟩ = \frac{1}{2} ∣ 0 ⟩ + \frac{3}{2} ∣ 1 ⟩,

(P_{d} (F), P_{d} (+), P_{d} (-)) = (\frac{1}{4}, \frac{9}{16}, \frac{3}{16}), (P_{0} (F), P_{0} (+), P_{0} (-)) = (\frac{1}{4}, \frac{3}{8}, \frac{3}{8}) .

Consequently $Δ (d) = 3/16$ .

Proof. The coefficient is $d$ by Theorem thm:localmatching. Since $tanh ((lo g 3) /2) = 1/2$ , substitution proves every displayed equality.

Displayed formulas in this passage

unlabeled · line 1311
$ν = 1, r = \frac{1}{4}, d = \frac{σ}{2} lo g 3,$
unlabeled · line 1317
$∣ e_{σ} (d)⟩ = \frac{1}{2} ∣ 0 ⟩ + \frac{3}{2} ∣ 1 ⟩,$
unlabeled · line 1321
$(P_{d} (F), P_{d} (+), P_{d} (-)) = (\frac{1}{4}, \frac{9}{16}, \frac{3}{16}), (P_{0} (F), P_{0} (+), P_{0} (-)) = (\frac{1}{4}, \frac{3}{8}, \frac{3}{8}) .$