Full-space history-path gap

Let $\mathcal{S}$ be the subspace of legal unary clocks. The prefix conjugation from History state and padded output preserves each clock basis state and makes $H_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}$ equal to $L_N\mathrm{/}2$ on the clock. Since the input penalty acts only at $c_0$ and the prefix at $c_0$ is the identity, the compression of $H_1(s):=H(s)-J{H}_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{k}}$ to $\mathcal{S}$ decomposes, in the computational basis $x\in \{0,1{\}}^n$, as ${\displaystyle H_x(s)=H_0(s)+\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }{c}_0⟩⟨c_0\mathrm{\mid },}$ where $H_0(s)$ is the zero-input clock block and $\mathrm{\mid }x\mathrm{\mid }$ is Hamming weight.

The $x=0$ block has simple ground energy $E_0$ and gap $\mathrm{\Delta }\ge 1\mathrm{/}(48N^2)$ by [F1]–[F2]. Write $p=\mathrm{\mid }{v}_0{\mathrm{\mid }}^2$. Since $0\le v_i\le v_0$ and ${\sum }_i\mathrm{\mid }{v}_i{\mathrm{\mid }}^2=1$, $p\ge 1\mathrm{/}(N+1)>0$. Also $\mathrm{\Delta }\le 2$: the compression of $H_0(s)$ to $S=\mathrm{span}\{e_0,e_1\}$ is positive and has trace at most $2$ (for $N=1$ its trace is $a+2t=1$, and for $N\ge 2$ it is $a+3t=1+s\mathrm{/}2\le 3\mathrm{/}2$). If $u_0$ is a normalized ground vector, the two-dimensional space $S$ contains a nonzero vector $z\perp u_0$. Its Rayleigh quotient is at least $E_1$ by expansion in an orthonormal eigenbasis, while the quotient is at most the largest eigenvalue of the compression to $S$, hence at most its trace $2$. Thus $E_1\le 2$ and, since $E_0\ge 0$, $\mathrm{\Delta }=E_1-E_0\le 2$.

For any $x\mathrm{\ne }0$, put $r=\mathrm{\mid }x\mathrm{\mid }\ge 1$ and $R=\mathrm{\mid }{c}_0⟩⟨c_0\mathrm{\mid }$. The spectral gap of the zero-input block gives $H_0-E_{0}I\ge \mathrm{\Delta }(I-\mathrm{\mid }v⟩⟨v\mathrm{\mid })$, and $rR\ge R$. Thus ${\displaystyle H_x-E_{0}I\ge B:=\mathrm{\Delta }(I-\mathrm{\mid }v⟩⟨v\mathrm{\mid })+R\mathrm{.}}$ If $p<1$, set $w=(c_0-\sqrt{p}v)\mathrm{/}\sqrt{1-p}$. On the two-dimensional span of $v$ and $c_0$, the matrix of $B$ in the orthonormal basis $v,w$ has trace $1+\mathrm{\Delta }$ and determinant $p\mathrm{\Delta }$. Its smaller eigenvalue is at least its determinant divided by its trace, hence at least $p\mathrm{\Delta }\mathrm{/}(1+\mathrm{\Delta })$. On the orthogonal complement, $B=\mathrm{\Delta }I$, so this bound applies to the whole space. If $p=1$, then $c_0=v$ and $B$ has eigenvalues $1$ on $v$ and $\mathrm{\Delta }$ on $v^{\perp }$, so $B\ge \min \{1,\mathrm{\Delta }\}I\ge \mathrm{\Delta }\mathrm{/}(1+\mathrm{\Delta })I$. Thus in either case $B\ge p\mathrm{\Delta }\mathrm{/}(1+\mathrm{\Delta })I$. By $\mathrm{\Delta }\le 2$, every nonzero-input block has ground energy at least $E_0+\mathrm{\Delta }p\mathrm{/}3\ge E_0+1\mathrm{/}[144N^2(N+1)]$.

The legal compression therefore has unique ground in the $x=0$ block and gap at least ${\displaystyle \min \{\frac{1}{48N^2},\frac{1}{144N^2(N+1)}\}=\frac{1}{144N^2(N+1)}\ge {\mathrm{\Delta }}_0\mathrm{.}}$ The last inequality is equivalent to $3(N+1{)}^2\ge 2N^2$, which holds for every $N\ge 1$.

On the full register, $H_2=H_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{k}}$ is positive, vanishes exactly on $\mathcal{S}$, and is at least $I$ on ${\mathcal{S}}^{\perp }$: every illegal binary clock string contains at least one adjacent $01$ pair. To bound $H_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}$, fix an assignment of all clock bits outside the local support of a term $h_{\mathrm{\ell }}$. On the two local pattern states $a_{\mathrm{\ell }},b_{\mathrm{\ell }}$, the block is $\frac{1}{2}\left(\begin{array}{cc}I& -U_{\mathrm{\ell }}^{\ast }\\ -U_{\mathrm{\ell }}& I\end{array}\right)$; it has eigenvalues $0$ and $1$ because $U_{\mathrm{\ell }}$ is unitary, and $h_{\mathrm{\ell }}$ is zero on the other local patterns. Thus $0\le h_{\mathrm{\ell }}\le I$, $\mathrm{\parallel }{h}_{\mathrm{\ell }}\mathrm{\parallel }\le 1$, and $0\le H_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}\le NI$. Also $\mathrm{\parallel }{H}_{\mathrm{i}\mathrm{n}\mathrm{p}\mathrm{u}\mathrm{t}}\mathrm{\parallel }\le n$ and $\mathrm{\parallel }{H}_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{k}0}\mathrm{\parallel }\le N$, by their sums of $n$ and $N$ projections. Hence $H_1(s)=H_{\mathrm{i}\mathrm{n}\mathrm{p}\mathrm{u}\mathrm{t}}+(1-s)H_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{k}0}+s{H}_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}$ is self-adjoint and $\mathrm{\parallel }{H}_1(s)\mathrm{\parallel }\le n+(1-s)N+sN=n+N\le K_0$. The legal dimension is $2^n(N+1)\ge 2$, so [F4] applies to its first two eigenvalues with $K=K_0$.

For this $J$, $J>2K_0$ and ${\displaystyle {ϵ}_J=\frac{K_0^2}{J-2K_0}<\frac{{\mathrm{\Delta }}_0}{4}\mathrm{.}}$ The perturbation bounds in [F4] give ${\lambda }_1(H(s))\le {\mu }_1$ and ${\lambda }_2(H(s))\ge {\mu }_2-{ϵ}_J$, where ${\mu }_2-{\mu }_1\ge {\mathrm{\Delta }}_0$ by step 4.1. Thus the full gap is at least ${\mathrm{\Delta }}_0-{ϵ}_J>3{\mathrm{\Delta }}_0\mathrm{/}4$ for every $s$. Its positivity makes the ground state unique, and ${\mathrm{\Delta }}_0=\mathrm{\Theta }(N^{-3})$ gives the stated asymptotic gap.