Amplitude-amplification invariant plane

Put $c=\sqrt{1-a}=\cos \theta$ and $s=\sqrt{a}=\sin \theta$. The two components ${\mathrm{\Pi }}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}\mathrm{\mid }\psi ⟩$ and $(I-{\mathrm{\Pi }}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}})\mathrm{\mid }\psi ⟩$ are orthogonal because ${\mathrm{\Pi }}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}$ is an orthogonal projection. Their nonzero normalized vectors therefore form an orthonormal pair, and $\mathrm{\mid }\psi ⟩=c\mathrm{\mid }B⟩+s\mathrm{\mid }G⟩$. The angle $\theta$ is unique in $(0,\pi \mathrm{/}2)$: sine is continuous there, has endpoint values $0,1$, and is strictly increasing because its derivative is the positive cosine before the first positive cosine zero.

In the ordered basis $(\mathrm{\mid }B⟩,\mathrm{\mid }G⟩)$, ${\mathrm{\Pi }}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}=\mathrm{diag}(0,1)$, so $S_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}=\mathrm{diag}(1,-1)$. Also $A{S}_0{A}^{†}=I-2\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$, since $A\mathrm{\mid }0⟩=\mathrm{\mid }\psi ⟩$. The matrix of this latter reflection is [given, F1, F2, F3, F4, step 1.1, algebra] $\left(\begin{array}{cc}1-2c^2& -2cs\\ -2cs& 1-2s^2\end{array}\right)=\left(\begin{array}{cc}-\cos (2\theta )& -\sin (2\theta )\\ -\sin (2\theta )& \cos (2\theta )\end{array}\right)\mathrm{.}$ The product $Q=-(A{S}_0{A}^{†})S_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}$ is therefore the displayed rotation matrix. This proves invariance and the claimed restriction.

If $a=0$, then ${\mathrm{\Pi }}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}\mathrm{\mid }\psi ⟩=0$, so $S_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}\mathrm{\mid }\psi ⟩=\mathrm{\mid }\psi ⟩$, while $A{S}_0{A}^{†}\mathrm{\mid }\psi ⟩=-\mathrm{\mid }\psi ⟩$; hence $Q\mathrm{\mid }\psi ⟩=\mathrm{\mid }\psi ⟩$. If $a=1$, then $S_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}\mathrm{\mid }\psi ⟩=-\mathrm{\mid }\psi ⟩$ and $A{S}_0{A}^{†}\mathrm{\mid }\psi ⟩=-\mathrm{\mid }\psi ⟩$, hence $Q\mathrm{\mid }\psi ⟩=-\mathrm{\mid }\psi ⟩$. No zero component is normalized.