Clean-ancilla postselection has normalized branch weights

From the product inner product [F2] and unit norm [F5], $\mathrm{\parallel }Jx{\mathrm{\parallel }}^2=⟨\mathrm{\mid }{0}_A⟩\otimes x,\mathrm{\mid }{0}_A⟩\otimes x⟩=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ for every $x$, so $J$ is an isometry by its definition in [F3]. Therefore $J^{†}J=I_S$ by [F3]. It follows that ${\mathrm{\Pi }}_0=J{J}^{†}$ is self-adjoint and idempotent, ${\mathrm{\Pi }}_0\varphi =Jq$, and $\varphi ={\mathrm{\Pi }}_0\varphi +(I-{\mathrm{\Pi }}_0)\varphi$.

The two summands are orthogonal because ${\mathrm{\Pi }}_0(I-{\mathrm{\Pi }}_0)=0$ and ${\mathrm{\Pi }}_0^{†}={\mathrm{\Pi }}_0$. Expanding their squared norm by [F4] gives $1=\mathrm{\parallel }\varphi {\mathrm{\parallel }}^2=\mathrm{\parallel }{\mathrm{\Pi }}_0\varphi {\mathrm{\parallel }}^2+\mathrm{\parallel }(I-{\mathrm{\Pi }}_0)\varphi {\mathrm{\parallel }}^2=\mathrm{\parallel }q{\mathrm{\parallel }}^2+p_{\perp }=p_0+p_{\perp }$. Both summands are nonnegative, so each lies in $[0,1]$.

If $p_0>0$, then $\mathrm{\parallel }q\mathrm{/}\sqrt{p_0}{\mathrm{\parallel }}^2=p_0\mathrm{/}{p}_0=1$. If $p_0=0$, definiteness of the inner-product norm gives $q=0$, so normalization is undefined and the definition correctly assigns no conditional vector.