OP5 HHL returns the QLSP state within the requested error

First take the exact sine preparation, exact simulations and exact flag rotation. Before reversing phase estimation, different decoded labels are orthogonal. For a fixed eigenvalue $\lambda$, the well-flag projection keeps only the $f_{\kappa }$ coordinate, so the squared error is ${\sum }_y{p}_y(\lambda )\mathrm{\mid }{f}_{\kappa }({\tilde{\lambda }}_y)-f_{\kappa }(\lambda ){\mathrm{\mid }}^2\le {\sum }_y{p}_y(\lambda )\mathrm{\parallel }m({\tilde{\lambda }}_y)-m(\lambda ){\mathrm{\parallel }}_2^2\mathrm{.}$ By [F2] and [F3], this is at most $(10\kappa \mathrm{/}{t}_0{)}^{2}300\mathrm{\mid }{f}_{\kappa }(\lambda ){\mathrm{\mid }}^2$. Summing with the eigenbasis weights shows that the exact-kernel well vector differs from $v$ by at most $(10\sqrt{300}\kappa \mathrm{/}{t}_0)\mathrm{\parallel }v\mathrm{\parallel }$. Reversing the phase transform is unitary and preserves this norm bound.

Replacing the exact preparation, simulations and flag map by their implementations changes the attempt on its clean initial input by at most the sum of their norm errors. In the full-space mode this is ordinary unitary telescoping; in the clean-workspace mode [F7] applies because all ideal other gates leave simulator work untouched and ideal simulations return it to zero. The actual intermediate work need not be clean. In either mode the sum consists of two sine-table preparation uses, the total forward/reverse simulation budget, and one flag map, at most $4ϵ\mathrm{/}(4096\kappa )$. The success projection is contractive and fixes $v$: in rectangular mode this follows from [F8], and otherwise it is just the well flag. Applying it to both the exact-kernel comparison and the implementation error therefore gives $\frac{\mathrm{\parallel }u-v\mathrm{\parallel }}{\mathrm{\parallel }v\mathrm{\parallel }}\le \frac{10\sqrt{300}\kappa }{t_0}+\frac{4ϵ\mathrm{/}(4096\kappa )}{1\mathrm{/}(2\kappa )}<(\frac{180}{2048}+\frac{1}{512})ϵ<0.09ϵ=:q,$ using $t_0=2048\kappa \mathrm{/}ϵ$ and $\sqrt{300}<18$. Hence $\mathrm{\parallel }u\mathrm{\parallel }\ge (1-q)\mathrm{\parallel }v\mathrm{\parallel }>1\mathrm{/}(4\kappa )$ and the implemented well probability exceeds $1\mathrm{/}(16{\kappa }^2)$.

The exact filter on supported eigenvalues is $v=(2\kappa {)}^{-1}{H}^{+}\mathrm{\mid }b⟩$, so $v\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }$ is the target. The normalization inequality in [L1] applies because step 2.1 gives relative error below $0.09ϵ$: directly, $\mathrm{\parallel }u\mathrm{/}\mathrm{\parallel }u\mathrm{\parallel }-v\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }\le 2q\mathrm{/}(1-q)<ϵ$. By [F4], every successful amplified branch is the same ray as $u\mathrm{/}\mathrm{\parallel }u\mathrm{\parallel }$; choose the representative $u\mathrm{/}\mathrm{\parallel }u\mathrm{\parallel }$ to obtain the stated distance. The preparation work factors as $\mathrm{\mid }{g}_B⟩$ and is untouched by the HHL circuit; the ideal uncomputation leaves the other algorithmic registers zero and the measured flag in $\mathrm{\mid }\mathrm{w}\mathrm{e}\mathrm{l}\mathrm{l}⟩$. Tensoring these unit vectors does not change any norm above. For a rectangular map with $H=H_A\mathrm{/}\mathrm{\Lambda }$ and $\mathrm{\Lambda }>0$, Hermitian dilation for a rectangular linear system gives $H_A^{+}(b,0)=(0,A^{+}b)$; the singular values of the scaled dilation are ${\sigma }_j\mathrm{/}\mathrm{\Lambda }$, so $H^{+}(b,0)=\mathrm{\Lambda }(0,A^{+}b)$. This positive scalar cancels on normalization, leaving the ideal QLSP ray in the tagged domain summand. The actual successful vector $u$ also lies in that summand because ${\mathrm{\Pi }}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}$ includes its domain projection. Tracing out work gives the corresponding output density state.