Qubitized simulation correctness

If $t=0$, [F1] returns the identity circuit with no signal query, and $e^{-itH}=I$; hence its clean-input isometry error is zero. If $t\mathrm{\ne }0$, continue with the certified OP4 output.

On an eigenvector of $H$ with eigenvalue $\lambda$, $H\mathrm{/}\alpha$ has eigenvalue $x=\lambda \mathrm{/}\alpha \in [-1,1]$, by [F8] and [F9]. Put $Q=E-iO$ and $B=1+2\rho$. The OP4 certificate gives $\mathrm{\parallel }Q-f_{\tau }{\mathrm{\parallel }}_{[-1,1]}\le \rho$ by [F1]–[F2], and $\mathrm{\mid }{f}_{\tau }(x)\mathrm{\mid }=1$. Since $M=Q\mathrm{/}B$, $\mathrm{\mid }M(x)-f_{\tau }(x)\mathrm{\mid }\le \mathrm{\mid }Q(x)-f_{\tau }(x)\mathrm{\mid }\mathrm{/}B+(B-1)\mathrm{\mid }{f}_{\tau }(x)\mathrm{\mid }\mathrm{/}B\le 3\rho \mathrm{/}(1+2\rho )<3\rho \mathrm{.}$ In the orthonormal eigenbasis, this error is diagonal; for any vector with coordinates $c_j$, its squared output norm is the sum of $\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$ times the squared scalar errors. Thus the uniform scalar bound gives $\mathrm{\parallel }M(H\mathrm{/}\alpha )-e^{-itH}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<3\rho$. For each real QSVT extraction, [F3] bounds the exact phase-list polynomial mismatch plus the implemented phase-slot errors by ${\delta }_{\mathrm{q}\mathrm{s}\mathrm{p}}=\rho \mathrm{/}64$ uniformly on $[-1,1]$. For the bounded mode, apply [F12] to the exact $\mathrm{\Phi }$ and $-\mathrm{\Phi }$ phase sequences on the supplied block and on the target block; each branch changes by at most $4d\sqrt{{ϵ}_{\mathrm{B}\mathrm{E}}}$, so their average has the same block-perturbation bound. The coefficient-qubit average therefore gives a complete block $M_{\mathrm{p}\mathrm{h}}\mathrm{/}2$ with $\mathrm{\parallel }{M}_{\mathrm{p}\mathrm{h}}-M(H\mathrm{/}\alpha ){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2{\delta }_{\mathrm{q}\mathrm{s}\mathrm{p}}+8d\sqrt{{ϵ}_{\mathrm{B}\mathrm{E}}}\mathrm{.}$ For even $p_E$, $p_E(\sigma )=p_E(-\sigma )$; for odd $p_O$, the singular vector of a negative eigenvalue has left vector $u=-v$, so $p_O(\sigma )\mathrm{\mid }u⟩⟨v\mathrm{\mid }=-p_O(\sigma )\mathrm{\mid }v⟩⟨v\mathrm{\mid }=p_O(-\sigma )\mathrm{\mid }v⟩⟨v\mathrm{\mid }$, and $p_O(0)=0$. Thus these transforms are the ordinary polynomial blocks on $H\mathrm{/}\alpha$. The certified error is consequently $\mathrm{\parallel }{M}_{\mathrm{p}\mathrm{h}}-e^{-itH}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<3\rho +2{\delta }_{\mathrm{q}\mathrm{s}\mathrm{p}}+8d\sqrt{{ϵ}_{\mathrm{B}\mathrm{E}}}\le {\eta }_{\mathrm{O}\mathrm{A}\mathrm{A}}\le {\left(\frac{3ϵ}{128}\right)}^2\mathrm{.}$

Since $0<ϵ\le 1$, ${\eta }_{\mathrm{O}\mathrm{A}\mathrm{A}}\le 9\mathrm{/}16384<1\mathrm{/}100$. Apply [F10] to the unitary $W$ whose clean block is $M_{\mathrm{p}\mathrm{h}}\mathrm{/}2$, with target unitary $V=e^{-itH}$. The phase-corrected OAA circuit therefore has clean-input isometry error at most $10\sqrt{{\eta }_{\mathrm{O}\mathrm{A}\mathrm{A}}}\le 30ϵ\mathrm{/}128=15ϵ\mathrm{/}64$, including its retained-work leakage.

The per-occurrence certificates in [F1] cover every remaining unitary primitive in the fully unrolled post-amplification circuit, counting repeated, inverse and controlled calls separately. If there are $N>0$ such occurrences, each contributes at most $ϵ\mathrm{/}(64N)$ in full operator norm. The telescoping product identity, [F11], and unitary norm one bound the difference between the actual and phase-certified OAA unitaries by ${\sum }_{j=1}^Nϵ\mathrm{/}(64N)=ϵ\mathrm{/}64$; for $N=0$ the difference is zero. Thus the actual clean-input isometry error is at most $15ϵ\mathrm{/}64+ϵ\mathrm{/}64=ϵ\mathrm{/}4<ϵ$.

If $t=0$, [F1] returns the exact identity and the conclusion holds with zero error. If $t\mathrm{\ne }0$, step 3.1 gives the clean-input isometry bound. Apply [F7] in either case; it includes arbitrary reference systems and discarding the work registers, so the trace distance is at most $ϵ$.