 Solution

Because $H_n|0^n⟩=|{\psi }_0⟩$,which is the reflection operator in the line spanned by $|{\psi }_0⟩$. Meanwhile $I_G$ changes the sign of every good basis state and fixes every bad basis state; onit is the reflection in the $|{\psi }_B⟩$ axis. Both reflections preserve $\mathcal{P}$, so their product $Q$ does too.

WriteIn Euclidean geometry, the composition of reflections in two lines through the origin has a rotation matrix through twice the oriented angle between the lines. Hence $Q$ rotates the good-bad plane by $2\theta$ from $|{\psi }_B⟩$ toward $|{\psi }_G⟩$. In particular, the Grover rotation angle formula is