Nonnegative monotone ground vector of the history path

For the constructed path, the vectors ${\eta }_{\mathrm{\ell }}$ are orthonormal because the prefixes $W_{\mathrm{\ell }}$ are unitary and the clock states are orthonormal. The clock penalty vanishes on them. The input penalty annihilates ${\eta }_0$ because its data input is $\mathrm{\mid }{0}^n⟩$, and annihilates ${\eta }_{\mathrm{\ell }}$ for $\mathrm{\ell }>0$ because the first clock bit is $1$. Also $H_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{k}0}{\eta }_{\mathrm{\ell }}=\mathrm{\ell }{\eta }_{\mathrm{\ell }}$. Each propagation term $h_{\mathrm{\ell }}$ sends ${\eta }_{\mathrm{\ell }-1}$ to $({\eta }_{\mathrm{\ell }-1}-{\eta }_{\mathrm{\ell }})\mathrm{/}2$ and ${\eta }_{\mathrm{\ell }}$ to $({\eta }_{\mathrm{\ell }}-{\eta }_{\mathrm{\ell }-1})\mathrm{/}2$, and annihilates the other history basis vectors, since $W_{\mathrm{\ell }}=U_{\mathrm{\ell }}{W}_{\mathrm{\ell }-1}$. Thus ${\mathcal{S}}_0$ is invariant and the restricted matrix is $a\mathrm{diag}(0,1,\dots ,N)+t{L}_N$. For every $N\ge 1$, the quadratic form of this abstract matrix at $x$ is $a{\sum }_{i=0}^{N}i\mathrm{\mid }{x}_i{\mathrm{\mid }}^2+t{\sum }_{i=0}^{N-1}\mathrm{\mid }{x}_i-x_{i+1}{\mathrm{\mid }}^2$, so both summands are positive semidefinite.

If $s=0$, then $H_0(0)=\mathrm{diag}(0,1,\dots ,N)$ and its unique normalized nonnegative ground vector is $e_0$. If $s=1$, then $H_0(1)=L_N\mathrm{/}2$; its quadratic form is one half the sum of squared adjacent differences, so its kernel consists exactly of constant vectors, giving the stated uniform ground vector.

Suppose $0<s<1$, so $a,t>0$. The matrix is real symmetric with neighboring off-diagonal entries $-t$. By [F1], a complex eigenvector for the smallest eigenvalue exists; its real or imaginary part is a nonzero real eigenvector because the matrix and eigenvalue are real. Normalize that real eigenvector. Expanding any vector in the eigenbasis from [F1] shows its Rayleigh quotient is a weighted average of the eigenvalues, so this real eigenvector minimizes the quotient. Replacing its coordinates by their absolute values does not increase the Rayleigh quotient, since each edge contribution is $t(x_i-x_{i+1}{)}^2$ and $(\mathrm{\mid }{x}_i\mathrm{\mid }-\mathrm{\mid }{x}_{i+1}\mathrm{\mid }{)}^2\le (x_i-x_{i+1}{)}^2$; therefore a nonnegative ground vector $v$ exists. Its eigenvalue equation at any zero coordinate forces every adjacent coordinate to be zero, and connectedness of the path would then force $v=0$. Hence $v_i>0$ for every $i$.

The endpoint equation at vertex $0$ determines $v_1$ uniquely from $v_0$, and each interior equation determines $v_{i+1}$ from $v_i,v_{i-1}$. If $v_0=0$, the endpoint equation gives $v_1=0$ and recurrence gives $v=0$, impossible. Thus every eigenvector at a fixed eigenvalue is determined by its first coordinate, so the ground eigenspace is one-dimensional.

Write $E$ for the ground eigenvalue and $d_i=v_i-v_{i+1}$ for $0\le i<N$. The endpoint and interior eigenvalue equations give ${\displaystyle d_0=\frac{E}{t}{v}_0,d_i-d_{i-1}=\frac{E-ai}{t}{v}_i(1\le i<N),d_{N-1}=\frac{aN-E}{t}{v}_N\mathrm{.}}$ The ground energy satisfies $E\ge 0$ because both terms in $H_0(s)$ are positive semidefinite. Testing the normalized constant vector gives $E\le aN\mathrm{/}2$, because its Laplacian energy is zero and the average of $0,1,\dots ,N$ is $N\mathrm{/}2$. Therefore $d_0\ge 0$ and $d_{N-1}\ge 0$.

Since $a>0$ and each $v_i>0$, the increments $d_i-d_{i-1}$ in step 3.2 are nonnegative while $ai\le E$ and nonpositive after $ai\ge E$. Thus the sequence $(d_i)$ first does not decrease and then does not increase; every one of its entries is at least the smaller of its two endpoint entries. Step 3.2 makes both endpoints nonnegative, so every $d_i\ge 0$. This proves $v_i\ge v_{i+1}$ and completes the remaining case.