Schmidt truncation, transfer operators and parent Hamiltonians

For a nonzero vector, reshape its coefficient array as a matrix with row index $i_1$ and column index $(i_2,\dots ,i_N)$. By [F1] this gives an exact finite sum of products of a first-site coefficient and a residual coefficient array. Reshape the residual array at site $2$ with row index $(a_1,i_2)$ and remaining sites as column index, apply [F1] again, and store the resulting left coefficient array as the entries of $A_2^{i_2}$ with virtual indices $(a_1,a_2)$. Repeating this finite operation through site $N-1$ leaves a final array defining $A_N^{i_N}$; every SVD factorization is exact, so the original coefficient of each ordered product-basis vector is the displayed matrix product. For $N=1$ take the coefficients as scalar matrices. For the zero vector take all bond dimensions equal to $1$ and set the first site matrices to zero. For the nonzero construction choose each internal $D_j$ as the corresponding positive Schmidt rank; in both constructions use $D_0=D_N=1$ with $\mathrm{\ell }=r=(1)$. Thus all finite vectors admit a representation with suitably chosen bond dimensions.

For any integer $q\ge 1$ and positive semidefinite $Y\in M_q(\mathbb{C})\otimes M_D(\mathbb{C})$, the amplification is $({\mathrm{i}\mathrm{d}}_{M_q}\otimes \mathcal{E})(Y)={\sum }_{i=1}^d(I_q\otimes A^i)Y(I_q\otimes A^{i\ast })$. Each summand is positive semidefinite, so every amplification is positive and $\mathcal{E}$ is completely positive. The displayed definition also fixes its exact input and output spaces and makes no trace-preserving or unitality assertion.

The map ${\mathrm{\Gamma }}_L$ is linear and its image is a finite-dimensional subspace of $({\mathbb{C}}^d{)}^{\otimes L}$. By [F2], choose an orthonormal basis $u_1,\dots ,u_s$ of that image and define $P_{L}v:={\sum }_{a=1}^s⟨v,u_a⟩u_a$. Orthonormality gives $P_L^{\ast }=P_L$ and $P_L^2=P_L$, and $v-P_{L}v$ is orthogonal to every $u_a$. Thus $h_L=I-P_L$ is the orthogonal projection onto the complement, with $⟨h_{L}v,v⟩=\mathrm{\parallel }{h}_{L}v{\mathrm{\parallel }}^2\ge 0$ and kernel $\mathrm{im}{\mathrm{\Gamma }}_L$.

Fix a finite open MPS made from the same tensor and a block of $L$ consecutive sites. For each fixed physical-basis configuration outside that block, contraction of the outside tensors leaves a row $\alpha$ on the left and a column $\beta$ on the right. With $X=\beta \alpha$, direct coordinates give $\mathrm{tr}(X{A}^{i_1}\cdots A^{i_L})={\sum }_{u,v}{\beta }_u{\alpha }_v(A^{i_1}\cdots A^{i_L}{)}_{vu}=\alpha A^{i_1}\cdots A^{i_L}\beta$, so that block slice is ${\mathrm{\Gamma }}_L(X)$. The outside basis configurations are orthonormal, so the full vector is a sum of outside basis vectors tensored with block vectors in $\mathrm{im}{\mathrm{\Gamma }}_L$; the corresponding $h_L$ therefore annihilates it. Each term in $H_{N,L}$ is positive semidefinite by step 1.3, and their sum annihilates the MPS. If that vector is nonzero, its normalized vector has quadratic form $0$, while every unit vector has nonnegative quadratic form; hence $E_0(H_{N,L})=0$, and it is a zero-energy ground-state vector by the Definition. The zero vector is annihilated but is excluded from ground-state vectors by definition. For example, if all tensor matrices vanish, then ${\mathrm{\Gamma }}_L=0$, $h_L=I$ and $H_{N,L}=(N-L+1)I$, whose ground energy is positive. No uniqueness or positive-gap conclusion for a nonzero MPS follows from this argument.

The finite coefficient sum defining ${\mathcal{E}}_O(X)$ is complex-linear in both $O$ and $X$. For $O\ge 0$, write its spectral decomposition as $O={\sum }_{\alpha }\mathrm{\mid }{v}_{\alpha }⟩⟨v_{\alpha }\mathrm{\mid }$, absorbing the square roots of the nonnegative eigenvalues into $v_{\alpha }$, and put $v_{\alpha ,I}=⟨v_{\alpha },e_I⟩$. Then $O_{I,J}={\sum }_{\alpha }{v}_{\alpha ,I}\overline{v_{\alpha ,J}}$, so with $K_{\alpha }:={\sum }_J\overline{v_{\alpha ,J}}{A}^J$ the insertion map satisfies ${\mathcal{E}}_O(X)={\sum }_{\alpha }{K}_{\alpha }X{K}_{\alpha }^{\ast }\ge 0$ whenever $X\ge 0$. Diagonalizing $\mathrm{\ell }$ by [F7] and expanding $\mathrm{tr}(\mathrm{\ell }{\mathcal{E}}_O(\rho ))$ as a weighted sum of nonnegative diagonal expectations shows ${\omega }_k(O)\ge 0$. For $O=I_{d^k}$, $O_{I,J}=1$ if $I=J$ and $0$ otherwise, so the insertion sums over all length-$k$ words and equals ${\mathcal{E}}^k$. The fixed-point equations and $\mathrm{tr}(\mathrm{\ell }\rho )=1$ give ${\omega }_k(I)=\mathrm{tr}(\mathrm{\ell }{\mathcal{E}}^k(\rho ))=1$. Thus each contiguous-block functional is positive and normalized.

Tensor-factor order in the definition gives ${\mathcal{E}}_{O\otimes P}={\mathcal{E}}_O\circ {\mathcal{E}}_P$ for consecutive blocks: expanding both sides gives the same ordered product $A^{J_O}{A}^{J_P}X(A^{I_P}{)}^{\ast }(A^{I_O}{)}^{\ast }$. Inserting an identity site on the right composes with $\mathcal{E}$ on the input, which leaves $\rho$ fixed; inserting one on the left composes with $\mathcal{E}$ on the output, which leaves the trace pairing with $\mathrm{\ell }$ fixed. Repeating these identities shows that the value is independent of the chosen containing interval and that restrictions to nested finite regions agree. A positive observable remains positive when identity factors are inserted, so the local functionals on arbitrary finite regions are positive and normalized as well. The empty-region convention is consistent because every local functional assigns one to the identity.

By [F2]–[F4] each local positive normalized functional has a unique trace-one density ${\rho }_{\mathrm{\Lambda }}$. For nested regions, step 2.1 gives ${\mathrm{tr}}_{{\mathrm{\Lambda }}^{\mathrm{\prime }}}((O\otimes I){\rho }_{{\mathrm{\Lambda }}^{\mathrm{\prime }}})={\mathrm{tr}}_{\mathrm{\Lambda }}(O{\rho }_{\mathrm{\Lambda }})$ for every $O\in {\mathcal{A}}_{\mathrm{\Lambda }}$. Applying the defining partial-trace pairing [F5] shows ${\mathrm{tr}}_{\mathrm{\Lambda }}(O{\mathrm{Tr}}_{{\mathrm{\Lambda }}^{\mathrm{\prime }}\setminus \mathrm{\Lambda }}({\rho }_{{\mathrm{\Lambda }}^{\mathrm{\prime }}}))={\mathrm{tr}}_{\mathrm{\Lambda }}(O{\rho }_{\mathrm{\Lambda }})$ for all $O$. Taking $O$ to be each matrix unit makes every matrix entry of the two matrices equal, so the density family is consistent.

Apply [F6] to the consistent finite-region density family from step 3.1. Its hypotheses hold by [A1], [F2], [F3], and the positive trace-one property just proved. It yields a unique infinite algebraic state on the specified quasi-local algebra, with exactly these local restrictions and no chosen global Hilbert-space representation. This establishes that the transfer contractions in the Definition are well defined.

Since $\mathrm{\parallel }\psi \mathrm{\parallel }=1$ and every cap $D_k\ge 1$, each current remainder has a positive singular value and the algorithm retains at least one; hence all subsequent remainders are nonzero and all retained bond dimensions are positive. The loop makes one SVD for each $k=1,\dots ,N-1$, exactly $N-1$ in total. Put ${\mathcal{H}}_{\le k}:=H_1\otimes \cdots \otimes H_k$ and ${\mathcal{R}}_k:=H_{k+1}\otimes \cdots \otimes H_N$, with ${\mathcal{H}}_{\le 0}=\mathbb{C}$. Let $U_k:S_k\to S_{k-1}\otimes H_k$ have the retained left singular columns, so $U_k^{\ast }{U}_k=I_{S_k}$ and ${\psi }^{(k)}=(U_k^{\ast }\otimes I_{{\mathcal{R}}_k}){\psi }^{(k-1)}$. Define $W_0=I_{\mathbb{C}}$ and $W_k=(W_{k-1}\otimes I_{H_k})U_k:S_k\to {\mathcal{H}}_{\le k}$. Each $W_k$ is an isometry by induction. Under the canonical $\mathbb{C}\otimes \mathcal{H}\cong \mathcal{H}$ identification, set $x_0=\psi$, $P_0=I$, and, for $k\ge 1$, $x_k=(W_k\otimes I_{{\mathcal{R}}_k}){\psi }^{(k)}$ and $P_k=(W_k{W}_k^{\ast })\otimes I_{{\mathcal{R}}_k}$ on the full chain. The range of $W_k$ lies in $\mathrm{ran}{W}_{k-1}\otimes H_k$, so $\mathrm{ran}{P}_k\subseteq \mathrm{ran}{P}_{k-1}$ and $P_k{P}_{k-1}=P_{k-1}{P}_k=P_k$. The remainder recurrence gives $x_k=P_k{x}_{k-1}=P_k\psi$ inductively. Expanding $W_{N-1}$ in the stored site tensors shows that $x_{N-1}$ is exactly their contraction with the final remainder, hence is the returned unnormalized $\varphi$ in algorithm instruction 3. For $N=1$, $x_0=\varphi =\psi$.

Across cut $k$, the coefficient matrix of an exact MPS factors as $L_k{R}_k$ by summing over its $D_k$-dimensional virtual index. Its matrix rank is therefore at most $D_k$. By [F3]–[F4], the Schmidt rank is the number of positive singular values, so it is at most $D_k$. Writing the Schmidt expansion as $\psi ={\sum }_{j=1}^q{s}_j{u}_j\otimes v_j$ gives $q\le D_k$, $p_j:=s_j^2>0$, and ${\sum }_j{p}_j=1$. Expanding the partial trace in [F5] gives ${\rho }_{1\cdots k}={\sum }_j{p}_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$, so these $p_j$ are exactly its positive eigenvalues.

Put ${\epsilon }_k:=x_{k-1}-x_k=(P_{k-1}-P_k)\psi$ in the full chain. This vector lies in $\mathrm{ran}{P}_{k-1}$ and is orthogonal to $\mathrm{ran}{P}_k$. For $j>k$, nesting gives ${\epsilon }_j\in \mathrm{ran}{P}_{j-1}\subseteq \mathrm{ran}{P}_k$, so $⟨{\epsilon }_k,{\epsilon }_j⟩=0$. Thus the embedded discarded components from different cuts are pairwise orthogonal.

At cut $k$, ${\epsilon }_k=(W_{k-1}\otimes I_{H_k\otimes {\mathcal{R}}_k})({\psi }^{(k-1)}-(U_k\otimes I_{{\mathcal{R}}_k}){\psi }^{(k)})$. The map outside the parentheses is an isometry by step 1.1, and the vector in parentheses is the discarded Schmidt part of the current compressed remainder. Hence [F3] and [F1] give $\mathrm{\parallel }{\epsilon }_k{\mathrm{\parallel }}^2={\delta }_k$. Since $x_0=\psi$ and $x_{N-1}=\varphi$, telescoping gives $\psi -\varphi ={\sum }_{k=1}^{N-1}{\epsilon }_k$; orthogonality from step 2.1 yields $\mathrm{\parallel }\psi -\varphi {\mathrm{\parallel }}^2={\sum }_k\mathrm{\parallel }{\epsilon }_k{\mathrm{\parallel }}^2={\sum }_k{\delta }_k$. For $N=1$ both sides are zero, so the equality also covers the empty sweep.

For every $x>1$, apply [F8] to $\log$ on $[1,x]$; [F6]–[F7] give $\log x=(x-1)\mathrm{/}c\le x-1$ for some $c\in (1,x)$. For $0<x<1$, apply [F8] on $[x,1]$ to get $-\log x=(1-x)\mathrm{/}c$ for some $c\in (x,1)$, hence $\log x\le x-1$; equality holds at $x=1$. Apply this inequality to $x=(q{p}_j{)}^{-1}$, multiply by $p_j>0$, and sum: ${\sum }_j{p}_j\log ((q{p}_j{)}^{-1})\le {\sum }_j(1\mathrm{/}q-p_j)=0$. The logarithm product law [F6] then gives $\log q-S({\rho }_{1\cdots k})={\sum }_j{p}_j\log (q{p}_j)\ge 0$. Since $q\le D_k$ and $\log$ is increasing, $S({\rho }_{1\cdots k})\le \log q\le \log D_k$. The zero-eigenvalue convention in [F5] includes rank-deficient reductions.

Define the linear functional $\phi (X):=\mathrm{tr}(\mathrm{\ell }X)$ and the rank-one map $P(X):=\rho \phi (X)$. Since $\mathrm{tr}\rho =1$, $\rho \mathrm{\ne }0$; the pairing normalization $\phi (\rho )=1$ gives $P^2=P$. The fixed-point equations give $\mathcal{E}P=P$ and $P\mathcal{E}=P$, since $\phi (\mathcal{E}(X))=\mathrm{tr}({\mathcal{E}}^{\ast }(\mathrm{\ell })X)=\phi (X)$. Hence $M_D(\mathbb{C})=\mathbb{C}\rho \oplus \ker \phi$ is an $\mathcal{E}$-invariant direct sum, $\mathcal{E}$ is the identity on $\mathbb{C}\rho$, and $P$ is zero on $\ker \phi$. If the restriction of $\mathcal{E}$ to $\ker \phi$ had eigenvalue $1$, it would have a nonzero eigenvector $X$ there. Since $\phi (X)=0$ and $\phi (\rho )=1$, $X$ and $\rho$ would be linearly independent eigenvectors of $\mathcal{E}$ for $1$, so $\dim E_1(\mathcal{E})\ge 2$. But [F7]–[F8] identify the Statement's simple-root hypothesis with algebraic multiplicity one, and [F9] bounds this geometric multiplicity by the algebraic multiplicity. This contradiction shows that the restriction has no eigenvalue $1$. Therefore ${\mathcal{E}}^g-P$ is zero on $\mathbb{C}\rho$ and equals $(\mathcal{E}{\mathrm{\mid }}_{\ker \phi }{)}^g$ on the invariant complement.

Use $\mathrm{\parallel }X{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}:={\sup }_{\mathrm{\parallel }v\mathrm{\parallel }=1}\mathrm{\parallel }Xv\mathrm{\parallel }$ on $M_D(\mathbb{C})$ and the induced norm $\mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }:={\sup }_{\mathrm{\parallel }X{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1}\mathrm{\parallel }\mathrm{\Phi }(X){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ on its linear maps. If $\ker \phi =\{0\}$, then $\mathcal{E}=P=I$ and ${\mathcal{E}}^g-P=0$ for every $g\ge 0$; for every allowed $r$, take $K_r=0$. Otherwise the restriction is a finite-dimensional complex endomorphism of positive dimension $m$. Its characteristic polynomial is monic of degree $m$ by [F7]. By [F4] it has a root $\mu$. Write this characteristic polynomial as $p(x)={\sum }_{k=0}^m{c}_k{x}^k$. Since $p(\mu )=0$, the identities $x^k-{\mu }^k=(x-\mu ){\sum }_{t=0}^{k-1}{x}^{k-1-t}{\mu }^t$ give $p(x)=(x-\mu )h(x)$ with $\mathrm{deg}h=m-1$. Induction on degree therefore shows that this characteristic polynomial splits into linear factors over $\mathbb{C}$. Now [F5] gives a similarity to finitely many Jordan blocks $J_s(\lambda )=\lambda I+N$ with $N^s=0$. Every such $\lambda$ belongs to ${\sigma }_{\mathbb{C}}(\mathcal{E})\setminus \{1\}$, since an eigenvector of the restriction is also an eigenvector of $\mathcal{E}$ and step 1.1 excludes $1$ from the restricted spectrum; hence $\mathrm{\mid }\lambda \mathrm{\mid }\le q$. If $q=0$, all Jordan powers of the restriction vanish for sufficiently large $g$, so ${\mathcal{E}}^g-P=0$ for those $g$. Since every allowed $r>q=0$ is positive, the finitely many remaining quantities $\mathrm{\parallel }{\mathcal{E}}^g-P\mathrm{\parallel }{r}^{-g}$ have a finite maximum; take this maximum as $K_r$ to obtain the required bound for all $g$. If $q>0$, fix $r>q$. For a block with $\lambda \mathrm{\ne }0$ and $g\ge s$, the binomial expansion gives $J_s(\lambda {)}^g={\sum }_{j=0}^{s-1}\left(\genfrac{}{}{0px}{}{g}{j}\right){\lambda }^{g-j}{N}^j$. For every fixed $j$, $g^j(\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}r{)}^g$ is bounded because the ratio of consecutive terms tends to $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}r<1$; hence each matrix entry is bounded by a constant times $r^g$. Blocks with $\lambda =0$ vanish for $g\ge s$, and all finitely many smaller $g$ are absorbed into the constant. In the maximum-coordinate norm of the Jordan basis, the norm of the block-diagonal power is bounded by a constant times $r^g$, since there are finitely many blocks and each has finitely many entries; the zero action of ${\mathcal{E}}^g-P$ on $\mathbb{C}\rho$ adds only a one-dimensional zero block. The fixed change-of-basis map and its inverse have finite induced norms, so submultiplicativity transfers this bound to the stated superoperator norm: $\mathrm{\parallel }{\mathcal{E}}^g-P\mathrm{\parallel }\le K_r{r}^g$ for a finite $K_r$ and every $g\ge 0$.

By [F1]–[F2], the transfer contraction for the two blocks with a gap of $g$ identities is $\mathrm{tr}(\mathrm{\ell }{\mathcal{E}}_{O_A}{\mathcal{E}}^g{\mathcal{E}}_{O_B}(\rho ))$. The product of the separate expectations equals $\mathrm{tr}(\mathrm{\ell }{\mathcal{E}}_{O_A}P{\mathcal{E}}_{O_B}(\rho ))$, since $P(X)=\rho \mathrm{tr}(\mathrm{\ell }X)$. Subtracting gives $C_g=\mathrm{tr}(\mathrm{\ell }{\mathcal{E}}_{O_A}({\mathcal{E}}^g-P){\mathcal{E}}_{O_B}(\rho ))$.

Diagonalize $\mathrm{\ell }\ge 0$ using [F6] as $\mathrm{\ell }={\sum }_j{t}_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$ with $t_j>0$. For every matrix $X$, $\mathrm{\mid }\mathrm{tr}(\mathrm{\ell }X)\mathrm{\mid }=\mathrm{\mid }{\sum }_j{t}_j⟨X{u}_j,u_j⟩\mathrm{\mid }\le \mathrm{tr}(\mathrm{\ell })\mathrm{\parallel }X{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. The block insertion maps are finite sums of matrix multiplications, so their induced norms in step 2.1 are finite; submultiplicativity gives $\mathrm{\mid }{C}_g\mathrm{\mid }\le \mathrm{tr}(\mathrm{\ell })\mathrm{\parallel }{\mathcal{E}}_{O_A}\mathrm{\parallel }{K}_r\mathrm{\parallel }{\mathcal{E}}_{O_B}(\rho ){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}{r}^g$. All factors before $r^g$ are finite and independent of $g$, so their product is an admissible $C$. This proves the claimed exponential clustering bound, including Jordan blocks and the case $q=0$.