Parallel QMA amplification remains sound for entangled witnesses

Put $\tau =(c+s)\mathrm{/}2=s+\gamma \mathrm{/}2$. By [F6], the midpoint test accepts at $\lceil r\tau \rceil$. Its CNOT fanout prepares a separate register $\mathrm{\mid }x⟩$ for every verifier copy from the classical basis input, and the outputs are evaluated using the same predicate ${\mathrm{Acc}}_n(x,\cdot )$, so each block realizes the acceptance effect $E_x$ of [F1]. On a YES input, choose a witness $\rho$ with acceptance $p\ge c$ and give the product witness ${\rho }^{\otimes r}$. Each copy acts on a distinct input, witness and ancilla block, so its accept bit is an independent Bernoulli$(p)$ variable by the product rule in [F2] and the finite POVM law. If the midpoint test rejects, then $S<\lceil r\tau \rceil$, hence $S<r\tau$ and $S-rp\le -r\gamma \mathrm{/}2$. The lower-tail inequality in [F3] bounds this event by $\exp (-r{\gamma }^2\mathrm{/}2)$.

Fix a NO input and any joint witness density $\sigma$ on the $r$ blocks. By the exact basis-state fanout in [F6], every verifier copy starts with its own fixed $\mathrm{\mid }x⟩$ input and its own clean ancillas; hence its acceptance effect on its witness block is the same $E_x$ from [F1]. On one block, [F1] and [F4] give $⟨E_{x}v,v⟩\le s$ for every unit witness vector $v$; as $E_x$ is an effect, this means $0\le E_x\le sI$. Let $A_j=I\otimes \cdots \otimes E_x\otimes \cdots \otimes I$ on the full $r$-block space. An eigenbasis of $E_x$ from [F9], tensored with the other block bases, shows $0\le A_j\le sI$. Distinct $A_j$ act on distinct tensor factors, so they commute by [F2].

Define the full-space effects $B_{j,1}=A_j$ and $B_{j,0}=I-A_j$. The joint outcome $b\in \{0,1{\}}^r$ has effect ${\prod }_j{B}_{j,b_j}$ on this same space: equivalently it is ${⨂}_j{C}_{b_j}$ for the one-block effects $C_1=E_x$, $C_0=I-E_x$. This product/tensor identity follows on elementary tensors by [F2]. The finite POVM trace rule and normalization [F7]–[F8] therefore give, for $t>0$, ${\mathbb{E}}_{\sigma }{e}^{tS}=\mathrm{Tr}\text{\hspace{0.17em}⁣}\left(\sigma {\sum }_b{e}^{t\mathrm{\mid }b\mathrm{\mid }}{\prod }_j{B}_{j,b_j}\right)=\mathrm{Tr}\text{\hspace{0.17em}⁣}\left(\sigma {\prod }_{j=1}^r\right(I+(e^t-1)A_j\left)\right)$. Each factor is positive and at most $(1-s+s{e}^t)I$ by step 1.2. In the product of the single-block eigenbases from [F9], their product is diagonal with entries bounded by $(1-s+s{e}^t{)}^r$, so the product is at most that scalar times identity. Taking its trace against the density $\sigma$ gives ${\mathbb{E}}_{\sigma }{e}^{tS}\le (1-s+s{e}^t{)}^r$, without any tensor product of full-space effects.

Apply the single-Bernoulli moment bound in [F3] with parameter $s$: $e^{-ts}(1-s+s{e}^t)\le e^{t^2\mathrm{/}8}$. Thus ${\mathbb{E}}_{\sigma }{e}^{t(S-rs)}\le e^{r{t}^2\mathrm{/}8}$. Since $S\ge \lceil r\tau \rceil$ implies $S-rs\ge r\gamma \mathrm{/}2$, the finite-sum Markov bound gives ${\mathrm{Pr}}_{\sigma }(S\ge \lceil r\tau \rceil )\le \exp (-tr\gamma \mathrm{/}2+r{t}^2\mathrm{/}8)\mathrm{.}$ Taking $t=2\gamma$ makes this at most $e^{-r{\gamma }^2\mathrm{/}2}$. No independence assumption is used for $\sigma$.

The choice $r\ge 2{\gamma }^{-2}\ln (1\mathrm{/}\delta )$ gives $e^{-r{\gamma }^2\mathrm{/}2}\le \delta$ on both promise sides by [F5]. The threshold is the exact integer rule in [F6], and repeating a polynomial-size full-list verifier $r$ times gives polynomial gate count, witness width, and generator time whenever $r$ is polynomially bounded. For the least admissible integer, $r\le 2{\gamma }^{-2}\ln (1\mathrm{/}\delta )+1$, which is polynomial under the resource hypotheses of the Statement.