Two-gate padded history state

At the algebraic macro-step level, the padding count is $K=\lceil 2\cdot 2\mathrm{/}1\rceil =4$, so $N=6$. The prefix states are $V_0\mathrm{\mid }0⟩=\mathrm{\mid }0⟩$, $V_1\mathrm{\mid }0⟩=X\mathrm{\mid }0⟩=\mathrm{\mid }1⟩$, and $V_2\mathrm{\mid }0⟩=ZX\mathrm{\mid }0⟩=Z\mathrm{\mid }1⟩=-\mathrm{\mid }1⟩$. Since $U_3,\dots ,U_6$ are identities, all later prefix states remain $-\mathrm{\mid }1⟩$. Substitution in the history-state formula of [F1] gives the displayed normalized vector.

For $\mathrm{\ell }=1$, the endpoint patterns are $a_1=00,b_1=10$ on clock bits $(1,2)$; for $\mathrm{\ell }=2,3,4,5$, the internal patterns are $a_{\mathrm{\ell }}=100,b_{\mathrm{\ell }}=110$ on bits $(\mathrm{\ell }-1,\mathrm{\ell },\mathrm{\ell }+1)$; for $\mathrm{\ell }=6$, the endpoint patterns are $a_6=10,b_6=11$ on bits $(5,6)$. On every adjacent pair $(c_{\mathrm{\ell }-1},c_{\mathrm{\ell }})$, the data amplitudes in step 1.1 satisfy ${\varphi }_{\mathrm{\ell }}=U_{\mathrm{\ell }}{\varphi }_{\mathrm{\ell }-1}$: the first two transitions apply $X$ and $Z$, and the last four apply $I$.

By [F1], each $h_{\mathrm{\ell }}$ restricted to its pattern pair has quadratic form $\frac{1}{2}\mathrm{\parallel }{\varphi }_{\mathrm{\ell }}-U_{\mathrm{\ell }}{\varphi }_{\mathrm{\ell }-1}{\mathrm{\parallel }}^2$. Step 2.1 makes this zero for each $\mathrm{\ell }=1,\dots ,6$; all other legal clock amplitudes are outside that term's two-pattern support. Thus every propagation term annihilates $\mathrm{\mid }\eta ⟩$, and the sum $H_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}$ does as well.