Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 15E (Quantum Information and Computation)

Orthogonal projection gives the unique componentsChooseThen $a,b\ge 0$, $a^2+b^2=1$, and the required decomposition follows. If one projection vanishes, its coefficient is zero and the corresponding normalized vector may be chosen arbitrarily in that subspace; the nonzero projected components remain unique.

Conjugating a reflection by a unitary reflects in the transported vector. Since $A|0^n⟩=|\Omega ⟩$,and every vector orthogonal to $|\Omega ⟩$ is fixed. Hencein particular both sides act identically on $|\Omega ⟩$.

Using $|{\psi }_b⟩$ as the horizontal axis and $|{\psi }_g⟩$ as the vertical axis,Thus $\sin \theta =\sqrt{p}$ and

The operator $R_g$ reverses the $|{\psi }_g⟩$ component and fixes $|{\psi }_b⟩$, so it is reflection in the horizontal $|{\psi }_b⟩$ axis. The operator $R_{|\Omega ⟩}$ reverses the component parallel to $|\Omega ⟩$ and fixes its perpendicular line; multiplication by $-1$ reverses those roles. Hence $-R_{|\Omega ⟩}$ is reflection in the line spanned by $|\Omega ⟩$.

Both operators map linear combinations of $|{\psi }_g⟩$ and $|{\psi }_b⟩$ to linear combinations of the same two vectors: $R_g$ merely changes one coefficient, whilehas $|\Omega ⟩\in P$. Therefore each leaves $P$ invariant.

The product of reflections in two lines meeting at angle $\theta$ is a rotation through $2\theta$. Here $Q=-R_{|\Omega ⟩}{R}_g$ rotates toward the good axis by $2\theta$. Since $|\Omega ⟩$ initially makes angle $\theta$ with the bad axis,A single use replaces the angle $\theta$ by $3\theta$.

The amplitude amplification success probability isChoose $k$ to be the nearest integer to $\pi /(4\theta )-1/2$. Then $(2k+1)\theta$ differs from $\pi /2$ by at most $\theta$, so $P_k\ge {\cos }^2\theta =1-p$. Since $\theta =\arcsin \sqrt{p}\sim \sqrt{p}$, this usesiterations and makes the good amplitude arbitrarily close to one when $p$ is small.