A finite J/CZ circuit is simulated branch by branch by its adaptive pattern

By [F8], the graph state's finite product of edge gates may be reordered without changing it. For round $s$, the vertical edges from $M_s$ act on column $s-1$ before that column is measured, and the horizontal edges from column $s-1$ to $s$ form one one-bit transport on each wire. The tensor factors of these $n$ transports are disjoint; later vertical edges act on later columns and do not change an earlier measurement projector. The ordered pattern in [F4] therefore realizes the round-wise sequence of matching and horizontal transports.

Suppose the logical state entering round $s$ is represented with frame $P(x^{s-1},z^{s-1})$. The matching relation [F7] moves that frame past $C{Z}_{M_s}$, contributes ${\mu }_s=(-1{)}^{q_{M_s}(x^{s-1})}$, and changes its $Z$ vector to $z^{s,\ast }=z^{s-1}\oplus A_s{x}^{s-1}$. Applying [F6] on wire $j$ with angle ${\alpha }_{sj}$ and outcome $b_j^s$ gives the new frame bits $x_j^s=b_j^s\oplus z_j^{s,\ast }$ and $z_j^s=x_j^{s-1}$ and contributes $2^{-1\mathrm{/}2}\gamma (x_j^{s-1},z_j^{s,\ast },{\alpha }_{sj})$. Tensoring the $n$ one-wire identities gives round factor $2^{-n\mathrm{/}2}{\mu }_s{\prod }_j\gamma (x_j^{s-1},z_j^{s,\ast },{\alpha }_{sj})$ and applies the round unitary after the prior logical unitary.

Initially $x^0=z^0=0$ and the branch map is the identity. Induction on $s$ using step 1.2 gives, for the first $s$ outcome blocks, the scalar $2^{-ns\mathrm{/}2}$, the recursively specified Pauli frame, and the chronological product of the first $s$ normalized round unitaries. At $s=R$, [F2] turns that logical product into ${\sigma }_C{U}_C$, so the accumulated scalar equals the explicit ${\lambda }_{\mathbf{b}}$ of the encoding. Therefore $K_{\mathbf{b}}=2^{-nR\mathrm{/}2}{\lambda }_{\mathbf{b}}{P}_{\mathbf{b}}{U}_C$.

Both $P_{\mathbf{b}}$ and $U_C$ are unitary and $\mathrm{\mid }{\lambda }_{\mathbf{b}}\mathrm{\mid }=1$. Thus a normalized pure input has branch probability $2^{-nR}$ by [F9]. For a density operator $\rho$, the branch output is $2^{-nR}{P}_{\mathbf{b}}{U}_C\rho U_C^{†}{P}_{\mathbf{b}}^{†}$; unitary conjugation is a similarity, so [F10] gives its trace as $2^{-nR}\mathrm{tr}(\rho )=2^{-nR}$. The transcript has $nR$ binary outcomes, hence $2^{nR}$ leaves and total probability one.

Applying $P_{\mathbf{b}}^{†}$ to each branch leaves $2^{-nR\mathrm{/}2}{\lambda }_{\mathbf{b}}{U}_C$. On pure inputs this differs from $U_C\mathrm{\mid }\psi ⟩$ only by a nonzero scalar, so it has the same output ray. On density inputs, summing the corrected branch maps gives $2^{-nR}{\sum }_{\mathbf{b}}{U}_C\rho U_C^{†}=U_C\rho U_C^{†}$, because there are $2^{nR}$ transcripts. If $R=0$, there is one empty transcript and the schedule contains only empty layers; if $n=0$, there is one empty transcript on the one-dimensional register. The formula gives the identity in both boundary cases.