Phase-matching condition for amplitude-amplification rotations

In the ordered basis $(\mathrm{\mid }G⟩,\mathrm{\mid }B⟩)$, write $u=e^{i\varphi }$ and $v=e^{i\phi }$. Since $A\mathrm{\mid }0⟩=\mathrm{\mid }\psi ⟩$, [given, F1, F2, algebra] $A{S}_0(\varphi )A^{†}=I+(u-1)\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }\mathrm{.}$ Multiplying on the right by $S_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}(\phi )=\mathrm{diag}(v,1)$ and including the scalar minus gives $Q=-\left(\begin{array}{cc}v((1-a)+au)& (u-1)\sqrt{a(1-a)}\\ v(u-1)\sqrt{a(1-a)}& a+(1-a)u\end{array}\right)\mathrm{.}$ This calculation also shows directly that the good/bad plane is invariant.

Equality of the diagonal entries is equivalent to [given, F3, F4, step 1.1, algebra] $v=\frac{a+(1-a)u}{(1-a)+au}\mathrm{.}$ For $\varphi \mathrm{\ne }\pi$, $\cos (\varphi \mathrm{/}2)\mathrm{\ne }0$, and after multiplying numerator and denominator by $e^{-i\varphi \mathrm{/}2}$, the quotient is $\frac{x+iy}{x-iy},x=\cos (\varphi \mathrm{/}2),y=(1-2a)\sin (\varphi \mathrm{/}2)\mathrm{.}$ By [F3], this equals $e^{i\phi }$ exactly when $\tan (\phi \mathrm{/}2)=y\mathrm{/}x=(1-2a)\tan (\varphi \mathrm{/}2)$; cross-multiplication of $(1+it)\mathrm{/}(1-it)$ shows the equivalence in both directions. On $[0,2\pi )\setminus \{\pi \}$, the half-angle tangent takes each real value exactly once, so the branch is unique.

If $\varphi =0$, the quotient in step 2.1 is $1$, hence $\phi =0$. If $\varphi =\pi$, the diagonal equality becomes $2a-1=-e^{i\phi }(2a-1)$. For $a\mathrm{\ne }1\mathrm{/}2$, this is equivalent to $e^{i\phi }=-1$, hence $\phi =\pi$; for $a=1\mathrm{/}2$, both diagonal entries are zero for every phase. At $a=0$ or $1$, one normalized component in [F1] is absent, so a two-dimensional diagonal comparison is not defined.