
\section{Representations for weight states}\label{sec:repweight}
This section discusses border rank and MPS representations for the weight states. For the convenience of the reader, we recall the definition of the (unnormalized) weight states from the main text:
\begin{defi}[Weight states]
Given $a,L,\mathfrak{d}\in\mathds{N}_0$ the \emph{weight state} of weight $a$ on $L$ sites of local dimension $\mathfrak{d}+1$ is:
\begin{align*}
     \ket{\chi_{a,\mathfrak{d},L}}=\sum\limits_{\substack{i_1+i_2+\ldots+i_L=a \\0\leq i_1,\ldots,i_L\leq \mathfrak{d}}}  \ket{i_1,i_2,\ldots,i_L} \in \left( \mathds{C}^{\mathfrak{d}+1}\right)^{\otimes L} .
\end{align*}
Write $\ket{\chi_{a,L}}=\ket{\chi_{a,a,L}}$.
\end{defi}

A state $\ket{\psi}\in\left( \mathds{C}^{d}\right)^{\otimes L}$ admits a representation of tensor rank $r$~\cite{Lan:TensorBook} if there exist product vectors $\ket{u_1},\ldots,\ket{u_r}\in \left( \mathds{C}^{d}\right)^{\otimes L}$ such that
\begin{align}\label{equ:rankdecomposition}
\ket{\psi}=\ket{u_1} + \cdots + \ket{u_r}.
\end{align}

A state $\ket{\psi}\in\left( \mathds{C}^{d}\right)^{\otimes L}$ admits a representation of border rank $r$~\cite{Bini_1980,christandl2018tensor} if there exist product vectors $\ket{u_1(\epsilon)},\ldots,\ket{u_r(\epsilon)}\in \left( \mathds{C}^{d}\right)^{\otimes L}$, depending polynomially on $\epsilon$, and a nonnegative integer $a$ such that
\begin{align}\label{equ:borderrankdecomposition}
\ket{\psi}= \lim _{\epsilon\to0}\frac{1}{\epsilon^a} \left( \ket{u_1(\epsilon)} + \cdots + \ket{u_r(\epsilon)}\right).
\end{align}

The minimal integer $r$ such that $\ket{\psi}$ has a representation of tensor rank (resp. border rank) is called tensor rank (resp. border rank) of $\ket{\psi}$, denoted $\mathrm{R}( \ket{\psi})$ (resp. $\underline{\mathrm{R}}(\ket{\psi})$).

Note that $\ket{\chi_{a,\mathfrak{d},L}}$ can be obtained from $\ket{\chi_{a,L}}$ applying local projection maps $P_{\mathfrak{d}}:\mathds{C}^{a+1}\to \mathds{C}^{\mathfrak{d}+1}$ defined by 
\begin{align*}
    P_{\mathfrak{d}}={\textstyle \sum}_{k=0}^{\mathfrak{d}}\ketbra{k}{k}.
\end{align*}
In particular, a border rank $r$ representation of $\ket{\chi_{a,L}}$ immediately provides a border rank $r$ representation of $\ket{\chi_{a,\mathfrak{d},L}}$. Thus, we focus on efficient representations of $\ket{\chi_{a,L}}$.

\begin{prop}[Border rank of weight states]\label{thm:degwweight}
Let $\ket{\chi_{a,L}}\in\left( \mathds{C}^{a+1}\right)^{\otimes L}$ be a weight state. For $\epsilon \in \mathds{C}$, let $\ket{\phi(\epsilon)}\in\mathds{C}^{a+1}$ be the state
\begin{align*}
    \ket{\phi_a(\epsilon)}=\sum\limits_{i=0}^a\epsilon^i\ket{i}.
\end{align*}
Then
\begin{equation}\label{eqn: border rank decomp weight}
    \ket{\chi_{a,L}}= \frac{1}{a!} \lim _{\epsilon \to 0}\frac{1}{\epsilon^a} \left[ \sum_{j=0}^a (-1)^{a-j}{\binom{a}{j}}\ket{\phi(j\epsilon)}^{\otimes L}\right].
\end{equation}
In particular, $\underline{\mathrm{R}}(\ket{\chi_{a,L}}) = a+1$.
\end{prop}
\begin{proof}
The lower bound is immediate via a standard flattening argument, see e.g.~\cite[Section 7.2]{BalBerChrGes:PartiallySymRkW} or~\cite{efremenko94barriers}.
We verify \eqref{eqn: border rank decomp weight} via a direct calculation:
\begin{align*}
\sum_{j=0}^a (-1)^{a-j}{\binom{a}{j}}\ket{\phi(j\epsilon)}^{\otimes L} &= \sum_{j=0}^a (-1)^{a-j} \binom{a}{j} \left[ {\textstyle \sum}_{i=0}^a (j \epsilon)^i \ket{i} \right]^{\otimes L}  \\
&=\sum_{j=0}^a (-1)^{a-j} \binom{a}{j} \left[ {\textstyle \sum}_{i_1 , \dots , i_L } j^{i_1 + \cdots +i_L} \epsilon^{i_1 + \cdots +i_L} \ket{i_1 , \dots , i_L} \right]  \\
&=\sum_{j=0}^a (-1)^{a-j} \binom{a}{j} \sum_{\alpha \geq 0} j^\alpha \epsilon^\alpha \left[ {\textstyle \sum}_{i_1 + \cdots + i_L =\alpha} \ket{i_1 , \dots , i_L} \right]  \\
&= \sum_{\alpha \geq 0} \epsilon^\alpha  \ket{ \chi_{\alpha,a,L}} \left[ {\textstyle \sum}_{j=0}^a  (-1)^{a-j} \binom{a}{j} j^\alpha \right].
\end{align*}
To conclude the proof we show that the coefficients $c_{\alpha,a} = {\textstyle \sum}_{j=0}^a  (-1)^{a-j} \binom{a}{j} j^\alpha$ satisfy $c_{\alpha,a} = 0$ if $\alpha < a$ and $c_{a,a} = a!$. To see this, we use a double induction on $a$ and $\alpha$.

If $(\alpha,a) = (0,0)$ then clearly $c_{\alpha,a}=1$, and if $\alpha = 0$ and $a > 0$, then $c_{\alpha,a} =  {\textstyle \sum}_{j=0}^a  (-1)^{a-j} \binom{a}{j} = (1-1)^a = 0$ by Newton's binomial formula.

Suppose $\alpha >0$; then the $0$-th term of the summation is $0$. We obtain
\begin{align*}
 c_{\alpha,a} & = \sum_{j=1}^a (-1)^{a-j} \binom{a}{j} j^\alpha  \\ 
 &=\sum_{j=1}^a (-1)^{a-j} \frac{a!}{j!(a-j)!} j^\alpha  \\
 &=a \sum_{j=1}^a (-1)^{(a-1)-(j-1)} \frac{(a-1)!}{(j-1)!(a-j)!} j^{\alpha-1}   \\
 &=a \sum_{j=0}^{a-1} (-1)^{(a-1)-j} \frac{(a-1)!}{j!((a-1)-j)!} (j+1)^{\alpha-1}   \\
 &=a \sum_{j=0}^{a-1} (-1)^{(a-1)-j} \frac{(a-1)!}{j!((a-1)-j)!} \left[ \sum_{\beta=0}^{\alpha-1} \binom{\alpha-1}{\beta} j^\beta  \right] =\\
 &=a \sum_{\beta=0}^{\alpha-1} \binom{\alpha-1}{\beta} \sum_{j=0}^{a-1} (-1)^{(a-1)-j} \binom{a-1}{j} j^\beta = a \sum_{\beta=0}^{\alpha-1} \binom{\alpha-1}{\beta}  c_{\beta, a-1}.
\end{align*}
By the inductive hypothesis, $c_{\beta,a-1} =0$ when $\beta < \alpha-1$ and $c_{a-1,a-1} = (a-1)!$. Since $\beta \leq \alpha-1$, we deduce that the summation above vanishes whenever $\alpha < a$. When $\alpha = a$, the summation reduces to $c_{\alpha, a} = a c_{\alpha-1,a-1} = a(a-1)! = a!$. This concludes the proof. 
\end{proof}

A consequence of Proposition \ref{thm:degwweight} is that $\underline{\mathrm{R}}(\ket{\chi_{a,\mathfrak{d},L}}) \leq a+1$, and in particular it is bounded from above independently of $L$. Moreover, all product states in the border rank expression of Proposition \ref{thm:degwweight} are symmetric. Regarding tensor rank, it is a classical fact \cite{Sylv:PrinciplesCalculusForms} that $\mathrm{R}( \ket{\chi_{1,L}}) = L+1$, which implies that $\mathrm{R}(\ket{\chi_{a,L}}) \geq L+1$ grows (at least) linearly in the system size $L$; on the other hand, since the border rank expression for $\ket{\chi_{a,L}}$ has degree $L-a$ in $\epsilon$, a standard interpolation argument (see, e.g., \cite[Thm. 8]{Christandl2017_TR} and \cite[Prop. 6.2]{ChrGesJen:BorderRankNonMult}) shows that $\mathrm{R}(\ket{\chi_{a,L}}) \leq (L-a+1) (a+1)$.

The following result provides a matrix product state representation with open boundary conditions for the weight states. As before, we only deal with the case where $\mathfrak{d}=a$. This representation gives rise to the MPS contraction strategy described in the section \emph{Performing  computations  in  the  ansatz  class} of the main text.

\begin{lemma}\label{lemma: MPS for chi}
Fix $a,L\in\mathds{N}$ and let $P_L$ be the graph consisting of an open chain with $L$ nodes indexed by $\{1 , \dots , L\}$. For $k = 1 , \dots , L$, define a local map $\mathcal{A}^{k}:\mathds{C}^{a+1}\otimes \mathds{C}^{a+1}\to \mathds{C}^{a+1}$ with $\mathcal{A}^{k}=\sum\limits_{j=0}^{a}\ket{j}\otimes A^{(k)}_{j}$ where 
 \begin{align*}
 A^{(1)}_j &= \ket{j}\otimes \bra{j}, \\
 A^{(k)}_{j}&=\sum \limits_{i=0}^{a-j}\ket{i}\otimes \bra{i+j} \text{ for }  2\leq k\leq L-1 ,\\
 A^{(L)} _j &=\sum_{i=0}^{a} \ket{a-j}\otimes \bra{i}.
 \end{align*}
Let $\ket{\phi} = ({\textstyle \bigotimes}_{k =1}^L \mathcal{A}_k ) \ket{\Omega^{(P_L)}_{a+1}}$ be the resulting tensor network state. Then $\ket{\phi} = \ket{\chi_{a,L}}$.
\end{lemma}
\begin{proof}
Write $A_j$ for $A_j^{(k)}$ when $k = 2 , \dots , L-1$ and define $A_j = 0$ for $j > a$. It is easy to see that  $A_jA_{j'}=A_{j+j'}$ for all $j,j'$. Therefore, if $i_2+i_3+\ldots+i_{L-1}>a$, we obtain
\begin{align*}
\langle i_1\ldots i_L| \phi\rangle=0.
\end{align*}
Therefore, suppose $i_2+i_3+\ldots+i_{L-1} = b \leq a$, so that $A_{i_2}A_{i_3}\ldots A_{i_{L-1}}=A_b$. Then
\begin{align*}
\langle i_1\ldots i_L| \phi\rangle=1
\end{align*}
if $i_1+i_{L}= a-b$ and $0$ otherwise. This concludes the proof. 
\end{proof}

Regarding matrix product states on the ring $C_L$, Lemma \ref{lemma: MPS for chi} implies that $\bTNS^{C_L}_{D,a,\mathfrak{d},d} \subseteq \TNS^{C_L}_{(a+1)D,d}$, relying simply on the fact that $\ket{\Omega^{(C_L)}_{D(a+1)}} =\ket{ \Omega^{(C_L)}_D} \otimes \ket{\Omega^{(P_L)}_{a+1}} \otimes \ket{\Omega^{(e)}_{a+1}}$, where $e$ is the edge which makes the chain $P_L$ into the ring $C_L$.


\section{Separations in efficiency of representations}\label{sec:efficiencyseparations}
As mentioned in the main text, for matrix product states representations on the cycle $C_L$, the maximal separation between the border bond dimension and bond dimension is at most quadratic. For the sake of completeness, we now provide a proof of this folklore result.

\begin{prop}[Separation between border bond and bond dimension]\label{prop: bond quadratic in border bond}
Let $\ket{\psi}\in\left( \mathds{C}^d\right)^{\otimes L}$ be a state with $\bbond^{C_L} (\ket{\psi}) = D$. Then $\bond^{C_L}(\ket{\psi}) \leq D^2$. In particular
\[
 \cTNS^{C_L}_{D} \subseteq \TNS^{C_L}_{D^2}.
\]

\end{prop}
\begin{proof}
Let $P_L$ be the chain obtained from $C_L$ by removing a single edge $e_L$. Then $\TNS^{P_L}_{D^2}$ is a closed set \cite{Landsberg_geometry} and clearly $\TNS^{P_L}_{D^2} \subseteq \TNS_{D^2}^{C_L}$. We show that $\cTNS^{C_L}_D \subseteq \TNS^{P_L}_{D^2}$ and this will conclude the proof. 

Note that $\TNS^{P_L}_{D^2}$ can be interpreted as having two copies of maximally entangled stated of bond dimension $D$ on the chain $P_L$. By performing an entanglement swapping protocol, we can then generate a maximally entangled state between sites $1$ and $L$ with one of the maximally entangled states, which shows that we can generate any TNS on the cycle by starting with two copies on the chain.


More formally, let $ T : \mathbb{C}^D\otimes \mathbb{C}^D \to \mathbb{C}$ be the map defined by $T= \sum_1^D \bra{i} \otimes \bra{i}$. Define a family of linear maps $(\mathcal{A}_k : k =1 , \dots , L)$ by $\mathcal{A}^1 = \mathcal{A}^k = \text{id}_{\mathbb{C}^D}$ and $\mathcal{A}^k = T$ for $k =2 , \dots , L-1$. Then $(\bigotimes_k \mathcal{A}^k) \ket{\Omega^{(P_L)} _D }= \ket{\Omega^{(e_L)}_D}$. 

Since $\ket{\Omega^{(C_L)}_{D}} = \ket{\Omega^{(P_L)}_{D}} \otimes \ket{\Omega^{(e_L)}_D}$, one immediately has $\ket{\Omega^{(C_L)}_{D} }\in \TNS_{D^2}^{P_L}$ by applying the family $(\mathcal{A}_k : k =1 , \dots , L)$ on one factor of $\ket{\Omega^{(P_L)}_{D^2}} = \ket{{\Omega^{(P_L)}_{D}}}^{\otimes 2}$. Therefore, 
\[
 \TNS_D^{C_L} \subseteq \TNS_{D^2}^{P_L}.
\]
Passing to the closures, we obtain $ \cTNS_D^{C_L} \subseteq \cTNS_{D^2}^{P_L}$, and since $\cTNS_{D^2}^{P_L} = \TNS_{D^2}^{P_L}$ is closed we conclude.
\end{proof}

The proof above is purely topological and not constructive: it only ensures the existence of a matrix product states representation of bond dimension $D^2$ under the hypothesis of the existence of a border bond dimension $D$ representation. 

The same argument can be generalized to arbitrary graphs as follows. Given a graph $G$, let $T$ be a spanning tree of $G$, i.e., a tree on all vertices of $G$. In particular, if $e$ is an edge of $G$ which is not an edge of $T$, then $T \cup \{e \}$ contains a cycle; the entanglement swapping argument of Proposition \ref{prop: bond quadratic in border bond} shows that $\Omega^{(e)}_D \in \TNS_D^{T}$. In this way, one can obtain the maximally entangled states $\Omega_D^{(e)}$ corresponding to all edges of $G$ which are not edges of $T$.  However, without any assumptions on $G$, one might need a full $\Omega_D^{(T)}$ to generate each edge.

Since $T$ does not contain any cycle, the set $\TNS_D^T$ is a closed set and we conclude
\[
\cTNS^{G}_D \subseteq \TNS^T_{D^\eta} \subseteq \TNS^G_{D^\eta} 
\]
where $\eta$ is the number of edges of $G$ which are not edges of $T$. As any tree with $|V|$ vertices has $|E|=|V|-1$ edges, we conclude that  for a graph $G=(V,E)$
\begin{align*}
\cTNS^{G}_D \subseteq \TNS^G_{D^{|E|-|V|+2}}. 
\end{align*}

\subsection{Examples of system-size dependent separations in efficiency of representation}\label{sec:analyticalexamples}
In this section, we discuss an example of a state $\ket{\psi_L} \in (\mathbb{C}^3)^{\otimes L}$ for odd $L$ which admits a border bond dimension $2$ translation invariant matrix product state representation with approximation degree $a=1$, yet having translation invariant bond dimension growing with $L$. More precisely, 
\begin{align*}
\bbond^{\text{TI-}C_L}(\ket{\psi_L}) &= 2 \text{ (with $a = \mathfrak{d} = 1$)} \\
\bond^{\text{TI-}C_L}(\ket{\psi_L}) & = \Omega \left( \textfrac{L^{1/3}}{\log(L)} \right) .
\end{align*}
Moreover, the state $\ket{\psi_L}$ arises as ground state of a translation invariant local (nearest neighbor) Hamiltonian on the cycle. In particular, we obtain $\ket{\psi_L} \in \bTNS^{\text{TI-}C_L}_{2,1,1,3}$ and $\ket{\psi_L} \notin \TNS^{\text{TI-}C_L}_{D,3}$ if $D$ grows slower than $\Omega( \frac{L^{1/3}}{\log(L)})$.

In the non-translation invariant case, there are no known examples of separations that are larger than a constant factor. 

The separation in the translation invariant setting, and in particular the lower bound on $\bond^{\text{TI-}C_L}(\ket{\psi})$ relies on the analogous result for the $W$-state, proved with the same argument as \cite[Corollary 1]{mapsgarcia} via the results of \cite{sanz2010quantum,Rahaman_2020,Michalek_2019_Wielandt}.

\begin{lemma}\label{lem:repwstate}
Let $C_L$ the cycle on $L$ nodes. Then 
\begin{align*}
 (a+1)L\geq\bond^{\text{TI-}C_L}(\ket{\chi_{a,L}}) &= \Omega  \left(\frac{L^{1/3}}{\log(L)}\right) ,\\
\bbond^{\text{TI-}C_L} (\ket{\chi_{a,L}}) &\leq a+1.
\end{align*}
\end{lemma}
\begin{proof}
The upper bound $\bbond^{\text{TI-}C_L} (\ket{\chi_{a,L}})\leq a+1$ is obtained directly from the border rank expression of Proposition \ref{thm:degwweight}.  This is because the superposition of $a+1$ symmetric product states can be written as a TI-MPS of bond dimension $a+1$.

The upper bound $\bond^{\text{TI-}C_L}(\ket{\chi_{a,L}}) \leq (a+1)L$ is obtained symmetrizing the non-translation invariant representation obtained in Lemma \ref{lemma: MPS for chi}.

As for the lower bound $\bond^{\text{TI-}C_L}(\ket{\chi_{a,L}}) = \Omega  \left(\frac{L^{1/3}}{\log(L)}\right)$, first note that 
the (local and translation invariant) projection map $P^{\otimes L}$ where $P = \ketbra{0}{0} + \ketbra{1}{a}$ sends $\ket{\chi_{a,L}}$ to the $W$-state $\ket{\chi_{1,L}}$. In particular, lower bounds on $\bond^{\text{TI-}C_L}(\ket{\chi_{1,L}})$ are lower bounds for $\ket{\chi_{a,L}}$ as well.

In~\cite[Corollary 1]{mapsgarcia}, the authors related $\bond^{\text{TI-}C_L}(\ket{\chi_{1,L}})$ to quantum Wielandt inequalities~\cite{sanz2010quantum,Rahaman_2020,Michalek_2019_Wielandt}. Although the version of the inequality stated in~\cite[Conjecture 2]{mapsgarcia}, on which \cite[Corollary 1]{mapsgarcia} relies on, is only conjectured, in~\cite{Michalek_2019_Wielandt} the authors show that~\cite[Conjecture 2]{mapsgarcia} holds with $f(D)=\mathcal{O}(D^2\log(D))$.

Reproducing the proof of~\cite[Corollary 1]{mapsgarcia} provides 
\begin{align*}
    (D)^3\log(D)^3=\Omega(L),
\end{align*}
whenever $D \geq \bond^{\text{TI-}C_L}(\ket{\chi_{1,L}})$. This implies $ \bond^{\text{TI-}C_L}(\ket{\chi_{1,L}}) =\Omega(e^{\mathcal{W}(L^{1/3})})$, where $\mathcal{W}$ is the Lambert $\mathcal{W}$ (see, e.g., \cite{hoorfar2008inequalities}). Indeed, one can show \cite[(1.1)]{hoorfar2008inequalities} that
\begin{align*}
\mathcal{W}(x)\geq \log(x)-\log \log(x)
\end{align*}
for all $x\geq e$, which implies that $e^{\mathcal{W}(L^{1/3})}=\Omega\left(\frac{L^{1/3}}{\log(L)}\right) $.
\end{proof}

From Lemma \ref{lem:repwstate}, already $\ket{\chi_{a,L}}$ provides an example of a state for which we have a separation. 

However, in the main text we presented a local translationally invariant Hamiltonian whose unique translationally invariant ground state presents this behaviour as well. Let us recall its definition and discuss some of its properties. The weight states themselves give separations in the required bond dimension and will be used as stepping stones to prove separations for these examples. But the next examples require a bond dimension $D>1$ to be represented in the ansatz class introduced  class, unlike the weight states.

Let $P :\mathbb{C}^3 \otimes \mathbb{C}^3 \to \mathbb{C}^3 \otimes \mathbb{C}^3$ be the projection defined by
\begin{align}\label{equ:definitonprojection}
P=\ketbra{01}{01}+\ketbra{10}{10}+\ketbra{02}{02}+\ketbra{21}{21};
\end{align}
denote by $P_{i,i+1}$ the linear map $P$ acting on the two factors of $(\mathbb{C}^3)^{\otimes L}$ corresponding to sites $i$ and $i+1$ of the ring $C_L$. 
Define $H_{i,i+1} = \text{id} - P_{i,i+1} + \ketbra{2}{2}_i$, where the third summand acts on the $i$-th site. Let $H = \sum_i H_i$: this is a diagonal, translationally invariant Hamiltonian. 

We determine its translationally invariant ground state in the case where $L$ is odd.
\begin{prop}
Let $L$ be odd. Then $H$ has a unique translationally invariant ground state given by
\begin{align}\label{equ:TIgroundstate}
\ket{\psi_L}=\frac{1}{\sqrt{L}}\sum\limits_{i=0}^{L-1}S^i\ket{2101010\ldots 10},
\end{align}
where $S$ is the cyclic permutation of the tensor factors, i.e., the shift operator.
\end{prop}
\begin{proof}
Since $H$ is diagonal, every product state in the standard basis is and eigenstate for $H$. Moreover, using the fact that the $P_{i,i+1}$ commute, one obtains that the spectrum of $H$ is contained in the set
\begin{align*}
\left\{\ell+\frac{m}{2L}|\ell,m \in\mathds{N}_{0}\right\}.
\end{align*}
Given a product state in the standard basis, we compute its energy,i.e. the corresponding eigenvalue of $H$. 

Note that every state involving two adjacent $\ket{0}$'s or two adjacent $\ket{1}$'s has energy at least $1$, since $1$ is the eigenvalue of the corresponding $H_i$. Moreover, since $L$ is odd, if the state does not involve any $\ket{2}$, there are necessarily adjacent $\ket{0}$'s or $\ket{1}$'s. Every state involving at least two $\ket{2}$'s has energy at least $2 \cdot 1/(2L)$, arising from the $\ketbra{2}{2}$ summands in $H$. There are only two basis state, up to cyclic permutation, involving exactly one $\ket{2}$ and having no adjacent $\ket{0}$'s nor $\ket{1}$'s: they are $\ket{20101 \ldots 01}$ and $\ket {21010 \ldots 10}$. The first has energy at least $1$ arising from $H_{12}$. The state $\ket {21010 \ldots 10}$ has energy $1/(2L)$. 

We conclude that $\{ S^k\ket{2101010\ldots 10} : k = 0 , \dots , L-1\}$ span the ground state space of $H$. The only translation invariant state in this span is $\ket{\psi_L}$.
\end{proof}

The following result gives upper and lower bounds for the translation invariant bond and border bond dimension of $\ket{\psi_L}$:
\begin{prop}
Let $\ket{\psi_L} \in (\mathbb{C}^3)^{\otimes L}$ be the state of \eqref{equ:TIgroundstate}. Then
 \begin{align*}
\bbond^{\text{TI-}C_L}(\ket{\psi_L}) &= 2;\\
  \bond^{\text{TI-}C_L}(\ket{\psi_L}) &= \Omega\left( \frac{L^{\frac{1}{3}}}{\log(L)}\right).
 \end{align*}
In fact, $\ket{\psi_L} \in \bTNS^{\text{TI-}C_L}_{2,1,1,3}$ and, for $L$ large enough, $\ket{\psi_L} \notin \TNS^{\text{TI-}C_L}_{D,3}$ if $D$ does not depend on $L$.
\end{prop}
\begin{proof}
Let $\mathcal{Q}: \mathbb{C}^3 \to \mathbb{C}^2$ be the local map defined by 
\begin{align*}
\mathcal{Q}=\ketbra{0}{1}+\ketbra{0}{0}+\ketbra{1}{2}.
\end{align*}
Then $\mathcal{Q}^{\otimes L} \ket{\psi_L}=\ket{\chi_{1,L}}$ is the $W$-state. In particular, the lower bound for $\ket{\chi_{1,L}}$ obtained in Lemma~\ref{lem:repwstate} holds for $\bond^{\text{TI-}C_L}(\ket{\psi_L})$ as well.

Now, define $\mathcal{A}(\epsilon): \mathds{C}^{2}\otimes \mathds{C}^{ 2} \to\mathds{C}^3$ by
\begin{align*}
\mathcal{A}(\epsilon)=\ketbra{0}{01}+\ketbra{1}{10}+\epsilon \ketbra{2}{00}.
\end{align*}
One can readily check that 
\begin{align*}
\lim_{\epsilon \to 0} \epsilon^{-1} \left( \mathcal{A}(\epsilon)^{\otimes L}\right) \ket{\Omega_2^{(C_L)}}=\ket{\psi_L},
\end{align*}
giving a representation of $\ket{\psi_L}$ of border bond dimension $2$ and approximation degree $1$. In particular, $\ket{\psi_L} \in \bTNS^{C_L}_{2,1,1,3}$ and concludes the proof.
\end{proof}
One can define explicitly a local map $\mathcal{B} : (\mathds{C}^{2}\otimes \mathds{C}^{ 2} )  \otimes \mathbb{C}^2 \to \mathbb{C}^3$ by 
\[
(\ketbra{0}{01}+\ketbra{1}{10}) \otimes \bra{0}+\ketbra{2}{00} \otimes \bra{1}
\]
and observe $\ket{\psi_L} = \mathcal{B}^{\otimes L}\ket{ \Omega^{(C_L)}_2} \otimes \ket{\chi_{1,L}}$.

This shows that there are unbounded separations in the bond dimension required to represent states that arise as ground states of local Hamiltonians in the new ansatz class for translationally invariant matrix product state when compared with standard matrix product states. 

\subsection{An explicit example of separation between bond and border bond dimension}

For small system size and small local dimension, several explicit examples of separation between bond and border bond dimension are known. In \cite{Str:RelativeBilComplMatMult}, Strassen proved that the GHZ state of level three on three parties, $\ket{\mathrm{ghz}_3} =3^{-\frac{1}{2}}\left( \ket{000} + \ket{111} + \ket{222}\right)$, satisfies $\bbond^{C_3} (\ket{\mathrm{ghz}}) = 2$, realized by a degeneration with approximation degree $a=2$; in fact, one can achieve the same result with a degeneration of approximation degree $a=1$ and in addition in \cite{christandl2018tensor} it was proved that $\bond^{C_3} (\ket{\mathrm{ghz}}) = 3$, providing a separation. 

In \cite{christandl2018tensor}, a separation was shown for the state $\ket{\lambda} = \sum_{\sigma \in \mathfrak{S}_3} (-1)^\sigma \ket{\sigma(0)\sigma(1)\sigma(2) } + |222\rangle$ as well; here $\mathfrak{S}_3$ denotes the symmetric group on three elements and $(-1)^\sigma$ is the sign of a permutation $\sigma$; in particular $\sum_{\sigma \in \mathfrak{S}_3} (-1)^\sigma \ket{\sigma(0)\sigma(1)\sigma(2) }$ is the unique (up to scaling) alternating tensor in $\mathbb{C}^3 \otimes \mathbb{C}^3 \otimes \mathbb{C}^3 $. It was proved that $\bbond^{C_3} (\ket{\lambda}) =2$ and $\bond^{C_3} = 3$. 

For these examples, comparing the numerical results in the standard tensor network ansatz class with the ones in the new ansatz class, we observe improvements both in the distance and in the convergence. However, in this case, the class $\bTNS^{C_3} _{2,1,1,3}$ is dense in $\mathbb{C}^3 \otimes \mathbb{C}^3 \otimes \mathbb{C}^3$, so the improvements occur for a trivial reason, as the optimization procedure can just ``walk straight'' to the global minimum. 

In order to see an example where the optimization procedure is not trivial, we consider the state 
\begin{equation}\label{eqn: tensor degen 3x3 mamu}
\begin{aligned}
 \ket{T} &=\sqrt{17}^{-1}( \ket{005}+\ket{016}+\ket{040}+\ket{126}+\ket{160}+\ket{227}+\ket{251}+\ket{262}+\ket{338}\\
 &+\ket{373}+\ket{384}+\ket{430}+\ket{501}+\ket{632}+\ket{703}+\ket{714}+\ket{824}) \in \mathbb{C}^9 \otimes \mathbb{C}^9 \otimes \mathbb{C}^9,
 \end{aligned}
\end{equation}
introduced in the \emph{Numerical examples} section of the main text. Lemma \ref{lemma: bond and border bond of T} below shows $\ket{T} \notin \TNS^{C_3}_{3,9}$ and $\ket{T} \in \bTNS^{C_3}_{3,1,9}$. Moreover, the set $\bar{\bTNS}^{C_3}_{3,1,9}$ is a proper subvariety of $\mathbb{C}^9 \otimes \mathbb{C}^9 \otimes \mathbb{C}^9$, therefore the new ansatz class is non-trivial. 

\begin{lemma}\label{lemma: bond and border bond of T}
Let $\ket{T}$ be the state defined in \eqref{eqn: tensor degen 3x3 mamu}. Then 
\begin{align*}
\bond^{C_3}{(\ket{T})} &\geq 4 , \\
\bbond^{C_3}{(\ket{T})}& = 3 \text{ and the degeneration has approximation degree $a = 1$}.
\end{align*}
In particular, $\ket{T} \notin \TNS^{C_3}_{3,9}$ and $\ket{T} \in \bTNS^{C_3}_{3,1,9}$.
\end{lemma}
\begin{proof}
Assume by contradiction $\bond^{C_3}(\ket{T}) \leq 3$ and observe that $\ket{\Omega^{(C_3)}_3} \in \mathbb{C}^9 \otimes \mathbb{C}^9 \otimes \mathbb{C}^9$, where each $\mathbb{C}^9$ has a local structure $\mathbb{C}^3 \otimes \mathbb{C}^3$. For $i=1,2,3$, let $\mathcal{A}_i : (\mathbb{C}^3 \otimes \mathbb{C}^3) \to \mathbb{C}^9$ be the linear maps such that $(\bigotimes \mathcal{A}^i) (\ket{\Omega^{(C_3)}_3}) = \ket{T}$. A direct calculation shows that there is no proper subspace $U \subsetneq \mathbb{C}^9$ such that $\ket{T} \in U \otimes \mathbb{C}^9 \otimes \mathbb{C}^9$ or $\ket{T} \in \mathbb{C}^9 \otimes U \otimes \mathbb{C}^9$ or $\ket{T} \in \mathbb{C}^9 \otimes \mathbb{C}^9 \otimes U$. Thus, the maps $\mathcal{A}^i$ are invertible, and $\ket{T}$ is a point of the orbit of $\ket{\Omega^{(C_3)}_3}$ under the action of $GL_9 \times GL_9 \times GL_9$. As a consequence, the stabilizer of $\ket{T}$ and the one of $\ket{\Omega^{(C_3)}_3}$ are conjugate and in particular they have the same dimension. But this is not the case: a direct calculation shows the stabilizer of $\ket{\Omega^{(C_3)}_3}$ is $26$-dimensional whether the stabilizer of $\ket{T}$ is $28$-dimensional. This shows the lower bound $\bond^{C_3} (\ket{T}) \geq 4$. 

We provide a degeneration of $\ket{\Omega^{(C_3)}_3}$ with approximation degree $a=1$ which realizes $\ket{T}$. In other words, we determine three linear maps $\mathcal{A}^{i}(\epsilon) : \mathbb{C}^3 \otimes \mathbb{C}^3 \to \mathbb{C}^9$ such that $\lim _{\epsilon \to 0} ( \frac{1}{\epsilon} \bigotimes _i \mathcal{A}^i(\epsilon)) \ket{\Omega^{(C_3)}_3} = \ket{T}$. Write $\mathcal{A}^i(\epsilon) = \mathcal{A}_0^{i} +  \epsilon \mathcal{A}_1^i$ and express $\mathcal{A}_0^i$ and $\mathcal{A}_1^i$ as matrices with entries in $\mathbb{C}^9$:
\begin{align*}
 \mathcal{A}_0^1 = \mathcal{A}_0^2 = \left(\begin{array}{ccc}
                          \ket{0} & \ket{1} & \\
                          & \ket{2} & \\
                          & & \ket{3}
                         \end{array} \right)
                         & \quad 
 \mathcal{A}_0^3 = \left(\begin{array}{ccc}
                         & \ket{1}   & \ket{3} \\
                          & & \ket{4} \\
                          \ket{0} & \ket{2} & 
                         \end{array} \right) \\
 \mathcal{A}_1^1 = \mathcal{A}_1^2 = \left(\begin{array}{ccc}
                         & & \ket{4} \\
                         \ket{5} & & \ket{6} \\
                          \ket{7} & \ket{8} & 
                         \end{array} \right)
&          \quad               
\mathcal{A}_1^3 = \left(\begin{array}{ccc}
                          \ket{5} &  & \\
                          \ket{6} & \ket{7} & \\
                          & & \ket{8}
                         \end{array} \right)
                         & \quad 
 \end{align*}
A direct calculation shows that $\lim _{\epsilon \to 0} ( \frac{1}{\epsilon} \bigotimes _i \mathcal{A}^i(\epsilon)) \ket{\Omega^{(C_3)}_3} = \ket{T}$.
\end{proof}


\section{Approximate representations}\label{sec:approximaterep}
As we already showed, there exist states that arise as ground states of local Hamiltonians and admit a more efficient \emph{exact} representation in $\textrm{bTNS}$. However, by definition, if a state lies on the boundary of the TNS variety, it can be approximated arbitrarily well by ``usual" TNS, albeit with local maps whose entries diverge as we increase the precision. 

Most rigorous results available in the literature only ascertain that tensor network states of a polynomial bond dimension \emph{approximate} ground states well~\cite{Verstraete_2006}. Thus, it is of high importance to investigate the interplay between the speed at which the local maps diverge and how well we can approximate the target state by the restrictions arising from degenerations. This issue that was also raised in~\cite{Hackbusch_2019} recently. In other words, it is important to understand how small one should pick $\epsilon$ so that $\ket{\phi(\epsilon)}\simeq \ket{\phi(0)}$ for the task at hand. First, recall that to estimate most quantities of physical interest up to an additive error $\delta>0$,  such as the energy or two-point correlation functions of the ground state, it usually suffices to estimate states to a precision of $\mathcal{O}(\delta L^{-\alpha})$ in the Euclidean norm, where again $L$ is the system size and $\alpha\in\mathds{N}_0$. This is because these quantities are expectation values of observables with operator norm  $\mathcal{O}(L^{\alpha})$.

With that in mind, we are ready to discuss how to pick $\epsilon$ for a degeneration to ensure the desired precision. To gain some intuition about how the local maps behave in the regime of interest, we consider the case of the $W$-state, before considering the general setting.

Consider the family $\ket{\psi(\epsilon)}$ of TI-MPS on a cycle of size $L$ with bond dimension $2$ and matrices given by:
\begin{align}\label{equ:MPSdefining}
    A_0(\epsilon)= \left(\begin{matrix}1&0\\0& e^{\frac{\pi i}{L}}\end{matrix}\right), \quad A_1(\epsilon)=\left(\begin{matrix}\epsilon&0\\0&0\end{matrix}\right).
\end{align}
One can readily check that
\begin{align*}
    \lim\limits_{\epsilon\to 0}\epsilon^{-1}\ket{\psi(\epsilon)}=\ket{W},
\end{align*}
where $\ket{W}$ is the unnormalized $W$-state:
\begin{align*}
\ket{W}=\ket{100\ldots 0}+\ket{010\ldots 0}+\ldots+\ket{000\ldots 1}.
\end{align*}
Moreover, the transfer matrix $\mathcal{E}(\epsilon)$ of this family of MPS is given by:
\begin{align*}
   \mathcal{E}(\epsilon)= \left(\begin{matrix}1+\epsilon^2&0&0&0\\0& e^{\frac{\pi i}{L}}&0&0\\0&0& e^{\frac{\pi i}{L}}&0\\0&0&0&1\end{matrix}\right).
\end{align*}
Thus, we have that:
\begin{align*}
    \|\psi(\epsilon)\|^2=\text{tr}{\mathcal{E}(\epsilon)^L}=(1+\epsilon^2)^L-1.
\end{align*}
It is also straightforward to compute the overlap of $\ket{\psi(\epsilon)}$ with the $W$-state, as:
\begin{align*}
    \scalar{W}{\psi(\epsilon)}=L\text{tr}{A_0^{L-1}A_1}=L\epsilon.
\end{align*}
Combining these expressions and normalizing all involved states we have:
\begin{align}\label{equ:overlapWdeg}
    \left\|\frac{\ket{\psi(\epsilon)}}{\|\ket{\psi(\epsilon)}\|}-\frac{\ket{W}}{\sqrt{L}}\right\|^2=2\lb1-\frac{L\epsilon}{\left( L\left( \left( 1+\epsilon^2\right)^L-1\right)\rb^{\frac{1}{2}}}\right)=2\frac{\left( \left( 1+\epsilon^2\right)^L-1\right)^{\frac{1}{2}}-\sqrt{L}\epsilon}{\left( \left( 1+\epsilon^2\right)^L-1\right)^{\frac{1}{2}}}
\end{align}
Performing a Taylor expansion, we see that
\begin{align*}
2\frac{\left( \left( 1+\epsilon^2\right)^L-1\right)^{\frac{1}{2}}-\sqrt{L}\epsilon}{\left( \left( 1+\epsilon^2\right)^L-1\right)^{\frac{1}{2}}}=\mathcal{O}(L\epsilon^2).
\end{align*}
Thus, in order to ensure that
\begin{align*}
 \left\|\frac{\ket{\psi(\epsilon)}}{\|\ket{\psi(\epsilon)}\|}-\frac{\ket{W}}{\sqrt{L}}\right\|^2\leq \frac{\delta^2}{L^{2\alpha}}
\end{align*}
for some $\alpha\in \mathds{N}_0$, we need to pick $\epsilon=\mathcal{O}(\delta L^{-\frac{2\alpha+1}{2}})$.


Inserting this bound in the definitions of the matrices in Eq. \eqref{equ:MPSdefining} and normalizing the underlying MPS, we see that in order to obtain an approximation up to an error $\delta L^{-\alpha}$ to the W-state in Euclidean norm we need to pick the matrices with entries scaling like:
\begin{align*}
        A_0= \left(\frac{e^{\log(L)\frac{2\alpha+1}{2}}}{\delta}\right)^{\frac{1}{L}}\left(\begin{matrix}1&0\\0& e^{\frac{\pi i}{L}}\end{matrix}\right), \quad A_1=\left(\frac{e^{-\log(L)\frac{2\alpha+1}{2}}}{\delta}\right)\left(\begin{matrix}1&0\\0&0\end{matrix}\right).
\end{align*}
Note that $A_0$ has entries of constant order, while $A_1$ has entries of inverse polynomial order for $\alpha$ fixed. Moreover, the norm of the underlying MPS will be of order $\delta L^{-\alpha}$, which implies that trying to approximate the $W$-state with this family of MPS leads to ill-conditioned MPS. Setting the vanishing entries to $0$ leads to a state that is \emph{orthogonal} to the state of interest.
We expect that this behavior is universal, i.e., that \emph{any} approximation of states on the boundary of the set of TNS of a given bond dimension up to inverse polynomial in system size  precision leads to tensors with entries that are polynomially small and ill-conditioned. 
Unfortunately, we still lack the tools to prove such statements for any approximation of a degeneration and can only prove it for some given degeneration, as we did for the $W$ state:
\begin{lemma}[Approximate representation from degeneration]\label{lem:approximate_rep}
Let $\mathcal{A}^k(\epsilon):\left( \mathds{C}^{D}\right)^{\otimes k_v}\to \mathds{C}^d$ be a family of local maps whose entries depend polynomially on $\epsilon$ defining a degeneration to $\ket{\phi_0}$ with $\|\phi_0\|=1$. Denote the expansion of 
\begin{align*}
\ket{\psi(\epsilon)}=\left(\bigotimes_{v\in V}\mathcal{A}^v(\epsilon)\right)\left(\bigotimes\limits_{y\in E} \ket{\Omega^y}   \right)
\end{align*}
as a polynomial in terms of $\epsilon$ by:
\begin{align}\label{equ:expansionhigherorder}
\ket{\psi(\epsilon)}=\epsilon^{a}\ket{\phi_0}+\sum\limits_{l=1}^{e}\epsilon^{a+l}\ket{\phi_l}
\end{align}
for some $a\in\mathds{N}_0$ and set
\begin{align*}
\tau(\epsilon)=\sum\limits_{l=1}^{e}\epsilon^{l}\|\ket{\phi_l}\|,
\end{align*}
Then, for $\tau(\epsilon)\leq\frac{1}{2}$ we have:
\begin{align}\label{equ:firstapprox}
\left\|\frac{\ket{\psi(\epsilon)}}{\|\ket{\psi(\epsilon)}\|}-\ket{\phi_0}\right\|\leq 4\tau(\epsilon)
\end{align}
\end{lemma}
\begin{proof}

By the triangle inequality we have:
\begin{align}\label{equ:unnormalized2}
\left\|\frac{\ket{\psi(\epsilon)}}{\|\psi(\epsilon)\|}-\ket{\phi_0}\right\|\leq\left\|\frac{\ket{\psi(\epsilon)}}{\epsilon^a}-\ket{\phi_0}\right\|+\left\|\frac{\ket{\psi(\epsilon)}}{\|\ket{\psi(\epsilon)}\|}-\frac{\ket{\psi(\epsilon)}}{\epsilon^a}\right\|
\end{align}
Let us analyze the first term in the expression above. Expanding the polynomials we see that:
\begin{align}\label{equ:firsttriangle}
\left\|\frac{\ket{\psi(\epsilon)}}{\epsilon^a}-\ket{\phi_0}\right\|=\left\|\sum\limits_{l=1}^{e}\epsilon^{l}\ket{\phi_l}\right\|\leq \tau(\epsilon)
\end{align}
by a triangle inequality.
Let us now analyze the second term. We have:
\begin{align}\label{equ:normalized}
&\left\|\frac{\ket{\psi(\epsilon)}}{\|\ket{\psi(\epsilon)}\|}-\frac{\ket{\psi(\epsilon)}}{\epsilon^a}\right\|=\left|\frac{1}{\|\ket{\psi(\epsilon)}\|}-\frac{1}{\epsilon^a}\right|\|\ket{\psi(\epsilon)}\|=\left|\frac{1}{\epsilon^{-a}\|\ket{\psi(\epsilon)}\|}-1\right|\epsilon^{-a}\|\ket{\psi(\epsilon)}\|
\end{align}
Again by a triangle inequality and the assumption that $\|\phi_0\|=1$, it follows from eq.~\eqref{equ:expansionhigherorder} that:
\begin{align*}
\|\epsilon^{-a}\psi(\epsilon)\|\leq 1+\tau(\epsilon).
\end{align*}

Moreover, by a reverse triangle inequality
\begin{align*}
\left|\epsilon^{-a}\|\ket{\psi(\epsilon)}\|-\|\ket{\phi_0}\|\right|=\left|\epsilon^{-a}\|\ket{\psi(\epsilon)}\|-1\right|\leq\|\epsilon^{-a}\ket{\psi(\epsilon)}-\ket{\phi_0}\|\leq \tau(\epsilon)
\end{align*}
and we assumed that $\|\phi_0\|=1$. 
This yields
\begin{align*}
\epsilon^{-a}\|\psi(\epsilon)\|\geq 1-\tau(\epsilon)
\end{align*}
and we obtain
\begin{align*}
\left|\frac{1}{\epsilon^{-a}\|\ket{\psi(\epsilon)}\|}-1\right|\leq\frac{\tau(\epsilon)}{1-\tau(\epsilon)}.
\end{align*}
Together with Eq.~\eqref{equ:normalized} we conclude that
\begin{align}\label{equ:boundonnormalized}
\left\|\frac{\ket{\psi(\epsilon)}}{\|\ket{\psi(\epsilon)}\|}-\frac{\ket{\psi(\epsilon)}}{\epsilon^a}\right\|\leq \frac{\tau(\epsilon)(1+\tau(\epsilon))}{1-\tau(\epsilon)}\leq 3\tau(\epsilon),
\end{align}
as we assumed that $\tau(\epsilon)\leq 1/2$.
Inserting the bounds of Equation~\eqref{equ:boundonnormalized} and~\eqref{equ:firsttriangle} into~\eqref{equ:unnormalized2} we obtain the claim in~\eqref{equ:firstapprox}.

\end{proof}
Thus, we see that in order to obtain an approximation up to an error $\mathcal{O}(\delta L^{-\alpha})$, it suffices to pick $\epsilon_0=\frac{\delta}{L^{\alpha}M}$, where $M=\max \|\phi_l\|$. Indeed, for this choice of parameters, we have $\tau(\epsilon_0)=\mathcal{O}(\delta L^{-\alpha})$. Moreover, we have:
\begin{align*}
\|\ket{\psi(\epsilon_0)}\|=\mathcal{O}\left(\frac{\delta^a}{L^{a\alpha}M^a}\right).
\end{align*}
and we expect that the condition number of the underlying maps scales in a similar way. This implies, for instance, that in order to obtain an estimate of the energy of the normalized state $\ket{\psi(\epsilon_0)}/\|\psi(\epsilon_0)\|$, it will be necessary to perform the multiplication $\scalar{\psi(\epsilon_0)}{H\psi(\epsilon_0)}\times\|\psi(\epsilon_0)\|^{-2}$, which entails multiplying two numbers of order $L^{-2a}\delta^{2a}$ and $L^{2a}\delta^{-2a}$, respectively.
For $a$ constant this requires a polynomial precision in the computation of expectation values and in case $a$ is at least linear in system size, this requires exponential precision.

We conclude that approximating states arising from degenerations by restrictions requires a great amount of precision when computing expectation values and will lead to numerical instabilities as the system size or degree of the degenerations increases. On the other hand, our methods remain stable.

This is even more pronounced in the case of tensor networks that are hard to contract exactly. In those cases, it is not feasible to contract the restrictions to the precision we just established are necessary. However, as highlighted in Section~\ref{sec:stableinterpolation}, approximate contractions yield good approximations in bTNS.

However, note that in principle some of the low order terms of $\tau(\epsilon)$ may vanish as well, which would lead to better convergence estimates than the ones discussed above. Let us illustrate this phenomenon in the case of the weight states:

\subsection{Approximate representations for the weight states}

Let us analyze more closely the trade-off between the rank of border rank representations of weight states and the speed at which they converge to the state. As noted in~\cite{Hackbusch_2019}, the representation given in~\eqref{equ:MPSdefining} for the W-state is suboptimal w.r.t. to the convergence speed. Indeed, one can check that:
\begin{align}\label{equ:betterrepW}
\frac{1}{2\epsilon}\left(\lb \ket{0}+\epsilon \ket{1}\right)^{\otimes L}-\left( \ket{0}-\epsilon \ket{1}\right)^{\otimes L}\right)=\ket{W}+\mathcal{O}(\epsilon^2),
\end{align}
while the degeneration given in~\eqref{equ:MPSdefining} only converges up to $\mathcal{O}(\epsilon)$.
This improved representation is closely connected with the fact that we can also obtain the weight states as:
\begin{align}\label{equ:representationderivative}
\ket{\chi_{a,L}}=a!\frac{d^a}{d \epsilon^a}\ket{\Gamma_a(\epsilon)}\big |_{\epsilon=0},
\end{align}
where
\begin{align*}
\ket{\Gamma_a(\epsilon)}=(\ket{0}+\epsilon \ket{1}+\ldots+\epsilon^{a}\ket{a})^{\otimes L}.
\end{align*}
For instance, the degeneration given in Eq.~\eqref{equ:betterrepW} arises from the symmetric difference formula for the derivative of a function:
\begin{align*}
\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}=f'(x)+\mathcal{O}(\epsilon^2).
\end{align*}
Evaluating this formula for $\ket{\Gamma_a(\epsilon)}$ we obtain this improved degeneration. More generally, so-called central finite difference formulas~\cite[Section 10.10]{Quarteroni_2006} allow us to estimate the $a$-th derivative of a function up to an error of order $\mathcal{O}(\epsilon^{2(k+1)})$ by evaluating the function at $2\lfloor \frac{a+1}{2}\rfloor+2k$ points. This immediately yields:
\begin{lemma}\label{lem:betterdeg}
For $a,L\in \mathds{N}$ and $k\in\mathds{N}_0$ the weight states $\ket{\chi_{a,L}}$ admit a representation of symmetric border rank  $2\lfloor \frac{a+1}{2}\rfloor+2k$, $\ket{\chi_{a,L,k}(\epsilon)} $,  s.t.
\begin{align*}
\|\epsilon^{-a}\ket{\chi_{a,L,k}(\epsilon)} -\ket{\chi_{a,L}}\|=\mathcal{O}(\epsilon^{2(1+k)}).
\end{align*}
\end{lemma}
\begin{proof}
This immediately follows from the central finite difference formulas together with the representation of the weight states given in~\eqref{equ:representationderivative}.
\end{proof}
Moreover, it is possible to explicitly compute central difference formulas for a desired order of the derivative and accuracy by solving a system of linear equations~\cite{Quarteroni_2006}.
One corollary of Lemma~\ref{lem:betterdeg} is that, even though the weight states do not admit exact system-size independent TI representations as matrix product states, the accuracy with which we can approximate them increases exponentially with the bond dimension.

In principle these representations can be used to obtain more efficient approximate contractions schemes for states in bTNS. Indeed, by tensoring with a state with $\epsilon^{-a}\ket{\chi_{a,L,k}(\epsilon)}$ for $k$ large enough instead of $\ket{\chi_{a,L}}$ for small enough $\epsilon$ will lead to approximate contraction schemes that do not depend on $L$ and do not blow up the bond dimension of the underlying states. But it remains unclear at this stage how numerically stable these methods are.


\section{Stable Interpolation}\label{sec:stableinterpolation}
In this section, we discuss the stability result related to the border rank contraction strategy presented in the main text to compute the expectation values of an observable for a state in the new ansatz class. Given a state $\ket{\phi} \in \bTNS^{G}_{D,a,\mathfrak{d},d}$, the border rank contraction strategy considers the state $\ket{\phi(\epsilon)} = \ket{\Omega^{(G)}_D} \otimes \ket{\chi_{a,\mathfrak{d},L}(\epsilon)}$ where $\ket{\chi_{a,\mathfrak{d},L}(\epsilon)}$ is a border rank expression for $\ket{\chi_{a,\mathfrak{d},L}}$ (e.g., the one arising from Proposition \ref{thm:degwweight}. Given an observable $O : (\mathbb{C}^d)^{\otimes L} \to (\mathbb{C}^d)^{\otimes L}$, define 
\begin{align}\label{equ:expectationvalueeps}
p(\epsilon) = \epsilon^{-2a}\scalar{\phi(\bar{\epsilon})}{O\phi(\epsilon)};
\end{align}
then $p(0) = \scalar{\phi}{O\phi}$ is the desired expectation value. 

Rather than evaluating $p(0)$, in the border rank strategy we recover the value of $p(0)$ via Lagrange interpolation, by evaluating $p(\epsilon)$ at $2(L\mathfrak{d}-a) +1$ points. Expanding the polynomial, we observed that the evaluation of $p(\epsilon)$ can be performed in terms of $(a+1)^2$ overlaps of states in $\TNS^{G}_{D,d}$.

In the main text, we only considered exact contraction schemes to compute such overlaps. However, beyond the case of 1D systems, exact contraction of tensor network states is known to be hard~\cite{PhysRevLett.98.140506,Haferkamp_2020} and, in practice, approximate contraction schemes are used~\cite{Orus_practical}. Thus, it is of paramount importance for the applicability of the border rank contraction scheme that the interpolation step is robust: in other words, even if the values of $p(\epsilon)$ at the $2(L\mathfrak{d} -a)$ interpolation points are only exact up to an error $\delta$, one should still be able to recover the value at $0$ up to an error $\mathcal{O}(\delta)$. 

Choosing the interpolation points to be roots of unity achieves this result. 

\begin{lemma}[Stable polynomial interpolation]
Let $p:\mathds{C}\to\mathds{C}$ be a polynomial of degree $k$ and $z_j=e^{\frac{2\pi ij}{k}}$ for $0\leq j \leq k-1$ be $k$-th roots of unity. Let $c_0 , \dots , c_{k-1} \in\mathds{C}$ be $k$ points such that there exists $\delta > 0$ with  
\begin{align}\label{equ:localapprox}
    \left|p(z_j)-c_j\right|\leq \delta.
\end{align}
Then:
\begin{align*}
    \left|p(0) -\frac{1}{k}{\textstyle \sum}_0^{k-1} c_j\right|\leq\delta.
\end{align*}
\end{lemma}
\begin{proof}
Via Lagrange interpolation 
\begin{align*}
    p(0)=\frac{1}{k}{\textstyle \sum}_0^{k-1} p(z_j).
\end{align*}
Therefore 
\begin{align*}
 \left|p(0) - \frac{1}{k}{\textstyle \sum}_0^{k-1} c_j\right| = \frac{1}{k}\left|{\textstyle \sum}_0^{k-1} ( p(z_j) - c_j ) \right| \leq \frac{1}{k}{\textstyle \sum}_0^{k-1}\bigl|p(z_j) - c_j\bigr| \leq \delta
\end{align*}
by a triangle inequality.
\end{proof}

This result shows that, choosing the interpolation points to be the $2aL$-th roots of unity, if the approximate contraction scheme only provides the values of $z_j^{-2a}p(z_j)$ up to an error $\delta$.
Note that $z_j^{-2a}$ is a complex number of modulus $1$, so  $c_j$ satisfying $|p(z_j)-c_j|\leq\delta$ also satisfies $|z_j^{-2a}p(z_j)-c_jz_j^{-2a}|$.
The resulting approximate value of $p(0)$ is up to an error $\delta$ as well.


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