A degenerate gapped Ising ground mixture does not cluster

Each interaction term is an orthogonal projection of norm one. The interaction is self-adjoint, has support size two, range one and per-site sum $\kappa =2$. It satisfies the finite-range hypotheses in [F2], so its thermodynamic dynamics ${\alpha }_t$ exists. This conclusion requires no choice of a finite-volume ground state.

Let $\mathcal{F}$ be the set of finite subsets of $\mathbb{Z}$. It is countable under [A1], by enumerating finite tuples and taking their ranges. Complete the finitely supported complex functions on $\{+1,-1\}\times \mathcal{F}$ for their linear-first sum inner product, obtaining $\mathcal{H}={\mathrm{\ell }}^2(\{+1,-1\}\times \mathcal{F})$ by [F3]. Denote its orthonormal basis by $\mathrm{\mid }\sigma ,F⟩$. Define $\pi (Z_j)\mathrm{\mid }\sigma ,F⟩=\sigma (-1{)}^{{\mathbf{1}}_F(j)}\mathrm{\mid }\sigma ,F⟩,\pi (X_j)\mathrm{\mid }\sigma ,F⟩=\mathrm{\mid }\sigma ,F\mathrm{△}\{j\}⟩\mathrm{.}$ Here $X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, and $\mathrm{△}$ denotes symmetric difference. These operators are self-adjoint unitaries, anticommute at the same site and commute at distinct sites. Thus they give the usual local Pauli matrix representations, including $Y_j=i{X}_j{Z}_j$. On a finite region $\mathrm{\Lambda }$, decompose $\mathcal{H}$ into the subspaces with fixed $\sigma$ and fixed finite $F\setminus \mathrm{\Lambda }$; each has the $2^{\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}$ basis vectors indexed by $F\cap \mathrm{\Lambda }$ and is a standard matrix representation of ${\mathcal{A}}_{\mathrm{\Lambda }}$. Therefore $\mathrm{\parallel }\pi (A)\mathrm{\parallel }=\mathrm{\parallel }A\mathrm{\parallel }$ locally. These compatible representations extend by norm continuity and [F3] to a unital contractive $\ast$-representation of $\mathcal{A}$.

For $F\in \mathcal{F}$, put $w(F)=\mathrm{\mid }\{j\in \mathbb{Z}:{\mathbf{1}}_F(j)\mathrm{\ne }{\mathbf{1}}_F(j+1)\}\mathrm{\mid }$. This number is finite, is zero exactly when $F=\mathrm{\varnothing }$, and is at least two otherwise: a nonempty finite set has both a left and a right boundary. More precisely it is twice the number of finite interval components of $F$. Define the diagonal operator $K\mathrm{\mid }\sigma ,F⟩=w(F)\mathrm{\mid }\sigma ,F⟩,D(K)=\{\xi :{\sum }_{\sigma ,F}w(F{)}^2\mathrm{\mid }{\xi }_{\sigma ,F}{\mathrm{\mid }}^2<\mathrm{\infty }\}\mathrm{.}$ Finite-support vectors make the domain dense and the quadratic form is nonnegative. If $\xi$ lies in the adjoint domain with $K^{\ast }\xi =\zeta$, testing on each basis vector gives ${\zeta }_{\sigma ,F}=w(F){\xi }_{\sigma ,F}$, so $\xi \in D(K)$. Conversely that square-summability condition makes the adjoint identity hold by sum Cauchy–Schwarz. Thus $K=K^{\ast }$ on exactly this domain.

Set $U_t\mathrm{\mid }\sigma ,F⟩=e^{itw(F)}\mathrm{\mid }\sigma ,F⟩$. It is a unitary group. For each $\xi$, dominated convergence with majorant $4\mathrm{\mid }{\xi }_{\sigma ,F}{\mathrm{\mid }}^2$ proves strong continuity. For $\xi \in D(K)$, $\mathrm{\mid }{e}^{itw}-1\mathrm{\mid }\mathrm{/}\mathrm{\mid }t\mathrm{\mid }\le w$ and [F4] show its norm derivative at zero is $iK\xi$. Conversely, if that norm derivative exists, its coordinates must be $iw(F){\xi }_{\sigma ,F}$, whose square sum is finite, so the exact generator domain is $D(K)$. The kernel has basis $\mathrm{\mid }+,\mathrm{\varnothing }⟩,\mathrm{\mid }-,\mathrm{\varnothing }⟩$. Every other diagonal energy is at least two and energy two occurs for a singleton $F$. For $0<\lambda <2$, the diagonal inverse $(w(F)-\lambda {)}^{-1}$ is bounded by $\max (1\mathrm{/}\lambda ,1\mathrm{/}(2-\lambda ))$ and maps into $D(K)$ since $w\mathrm{/}(w-\lambda )$ is bounded. The same diagonal argument excludes negative real and nonreal spectrum. Consequently $\sigma (K)\subseteq \{0\}\cup [2,\mathrm{\infty })$, with excitation gap exactly two.

The representation is covariant for the actual dynamics in step 1.1. Every interaction term is diagonal in the Pauli $Z$ basis, and these terms commute. Finite-volume conjugation of a local observable therefore depends only on the finitely many bonds meeting its support. In particular the evolution of $Z_j$ is constant. On $X_j\mathrm{\mid }\sigma ,F⟩$ the phase of $U_t\pi (X_j)U_t^{\ast }$ is $e^{it(w(F\mathrm{△}\{j\})-w(F))}$; the energy difference involves only the two bonds adjacent to $j$. It is exactly the phase from every finite-volume Hamiltonian containing these bonds. The local Pauli matrices generate every finite matrix algebra, so the covariance identity holds for every local $A$, and the norm completion gives $\pi ({\alpha }_t(A))=U_t\pi (A)U_t^{\ast }(A\in \mathcal{A})\mathrm{.}$ This verifies the relation to the thermodynamic dynamics rather than choosing an unrelated diagonal operator.

Let $\mathrm{\Omega }=(\mathrm{\mid }+,\mathrm{\varnothing }⟩+\mathrm{\mid }-,\mathrm{\varnothing }⟩)\mathrm{/}\sqrt{2}$. It is a unit zero-energy vector fixed by $U_t$, so its state $\omega (A)=⟨\pi (A)\mathrm{\Omega },\mathrm{\Omega }⟩$ is a positive-energy ground state. Its finite restrictions have density ${\rho }_{\mathrm{\Lambda }}=\frac{1}{2}\mathrm{\mid }+\cdots +⟩⟨+\cdots +\mathrm{\mid }+\frac{1}{2}\mathrm{\mid }-\cdots -⟩⟨-\cdots -\mathrm{\mid },$ with the empty-region density one; these are the consistent equal mixture of the two product phases. This representation is cyclic. Given finite $F$ and $\sigma \in \{+1,-1\}$, choose $j\mathrm{\notin }F$ and let $P_{\sigma ,j}=(I+\sigma Z_j)\mathrm{/}2$. Then $\sqrt{2}\pi \left({\prod }_{i\in F}{X}_i{P}_{\sigma ,j}\right)\mathrm{\Omega }=\mathrm{\mid }\sigma ,F⟩\mathrm{.}$ Thus every displayed basis vector is generated by a local observable, proving cyclicity and identifying the gap in step 3.1 as a gap of the state's own ground representation.

In the two product vectors $Z_i$ has respective expectations $+1$ and $-1$, whereas $Z_i{Z}_j$ has expectation one in each for $i\mathrm{\ne }j$. Hence the equal mixture has $\omega (Z_j)=0$ and $\omega (Z_i{Z}_j)=1$. For every integer $r\ge 1$ its connected $Z_0,Z_r$ correlation is therefore one. No finite $C_0$ and positive $\mu$ can satisfy $1\le C_0{e}^{-\mu r}$ for all $r$. The two product vector states are distinct positive-energy ground states, so uniqueness fails exactly where needed. This gives a gapped degenerate counterexample without altering any assertion about transport of an entire gapped ground band.