K_{s} = (tan α / tan β 0 01), K_{f} = (1 - tan^{2} α / tan^{2} β 0 00) .

Since

0 < tan α / tan β < 1

, these are valid and satisfy

K_{s}^{†} K_{s} + K_{f}^{†} K_{f} = I

. Alice sends Bob the success or failure outcome. The successful unnormalized branch is

(K_{s} \otimes I) ∣ ψ ⟩ = \frac{sin α cos β}{sin β} ∣00 ⟩ + sin α ∣11 ⟩ = \frac{sin α}{sin β} ∣ ϕ ⟩ .

Therefore the normalized success state is exactly the target and

p_{succ} = \frac{sin ^{2} α}{sin ^{2} β} = \frac{s}{t} = p_{m a x} .

Bob needs no quantum correction in this known Schmidt basis. The failure branch is proportional to

∣00 ⟩

1 - s / t

. The successful branch saturates the previous bound:

p_{succ} E_{2} (ϕ) = E_{2} (ψ)

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