Probability difference between state ψ\psi appears at b 0b_0 and state ϕ\phi appears at b 0b_0
Any two states can be expressed in term of base states:
[ψ ϕ] =[b 0 b 1][B 11 B 12 B 21 B 22] ⟨b 0|ψ⟩ =B 11 ⟨b 0|ϕ⟩ =B 12\begin{aligned}\begin{bmatrix}\psi&\phi\end{bmatrix}&= \begin{bmatrix}b_0&b_1\end{bmatrix} \begin{bmatrix}B_{11}&B_{12}\\B_{21}&B_{22}\end{bmatrix}\\\langle b_0|\psi \rangle &= B_{11}\\\langle b_0|\phi \rangle &= B_12\end{aligned}
Therefore,
Pr(0|ψ) =|⟨b 0|ψ⟩| 2 Pr(0|ϕ) =|⟨b 0|ϕ⟩| 2 |Pr(0|ψ)−Pr(0|ϕ)| =||⟨b 0|ψ⟩| 2−|⟨b 0|ϕ⟩| 2|=|(|⟨b 0|ψ⟩|+|⟨b 0|ϕ⟩|)(|⟨b 0|ψ⟩|−|⟨b 0|ϕ⟩|)| =(|⟨b 0|ψ⟩|+|⟨b 0|ϕ⟩|)⋅|(|⟨b 0|ψ⟩|−|⟨b 0|ϕ⟩|)| ≤2||⟨b 0|ψ⟩|−|⟨b 0|ϕ⟩|| =2||⟨b 0|ψ⟩|−|⟨b 0|−ϕ⟩|| ≤2|⟨b 0|ψ⟩+⟨b 0|−ϕ⟩| =2|⟨b 0|ψ−ϕ⟩| ≤2|b 0|⋅|ψ−ϕ|\begin{aligned}\Pr(0 |\psi) &= |\langle b_0| \psi \rangle|^2\\\Pr(0 |\phi) &= |\langle b_0 |\phi \rangle|^2\\|\Pr(0| \psi) -\Pr(0 |\phi)| &= ||\langle b_0 |\psi \rangle|^2 - |\langle b_0 |\phi \rangle|^2|= |\left(|\langle b_0 |\psi \rangle| + |\langle b_0 |\phi \rangle|\right)\left(|\langle b_0 |\psi \rangle| - |\langle b_0 |\phi \rangle|\right)|\\&=\left(|\langle b_0 |\psi \rangle| + |\langle b_0 |\phi \rangle|\right) \cdot |\left(|\langle b_0 |\psi \rangle| - |\langle b_0 |\phi \rangle|\right)|\\&\leq 2 ||\langle b_0 |\psi\rangle| - |\langle b_0 |\phi \rangle||\\&= 2 ||\langle b_0 |\psi \rangle| - |\langle b_0 |-\phi \rangle||\\&\leq 2 |\langle b_0 |\psi \rangle + \langle b_0 |-\phi \rangle|\\&= 2 |\langle b_0 |\psi - \phi \rangle|\\&\leq 2 |b_0| \cdot |\psi - \phi|\end{aligned}
12Pr(0|ψ)+12Pr(1|ϕ) =12Pr(0|ψ)+12(1−Pr(0|ϕ)) =12+12(Pr(0|ψ)−Pr(0|ϕ)) ≤12+12|Pr(0|ψ)−Pr(0|ϕ)| ≤12+|b 0|⋅|ψ−ϕ|\begin{aligned}\frac{1}{2} \Pr(0| \psi) + \frac{1}{2} \Pr(1| \phi) &= \frac{1}{2} \Pr(0| \psi) + \frac{1}{2} \left( 1 - \Pr(0| \phi)\right)\\&= \frac{1}{2} + \frac{1}{2} \left( \Pr(0| \psi) -\Pr(0| \phi) \right)\\&\leq \frac{1}{2} + \frac{1}{2} |\Pr(0| \psi) - \Pr(0| \phi)|\\&\leq \frac{1}{2} + |b_0| \cdot |\psi - \phi|\end{aligned}
