Finite sweep termination and computable norm error

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.