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$\newcommand{\complex}{\mathbb{C}}\newcommand{\ket}[1]{|#1\rangle}$ Let $\ket{\psi}\in(\complex^d)^{\otimes n}$ be a pure quantum state. It is well-known that $\ket{\psi}$ is a matrix product state with bond dimension $r$ if and only if $\ket{\psi}$ has Schmidt rank $r$ with respect to the bipartite cuts $(1 | 2,\dots, n), (1,2|3,\dots,n),\dots,(1,\dots,n-1|n)$. It therefore seems natural to expect that if the sum of all but the first $r$ singular values of $\ket{\psi}$ with respect to each of these bipartite cuts is at most $\epsilon$, then $\ket{\psi}$ is $f(\epsilon)$-close to a matrix product state with bond dimension $r$, where $f$ is some (non-trivial) function that might also depend on $n$ and $r$. I am looking for references that compute such a function $f$.

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The answer is found in F. Verstraete and J.I. Cirac, Matrix product states represent ground states faithfully, Lemma 1 on pg. 5 in the arxiv version.

It states that the total error in approximating a state $\lvert\psi\rangle$ by one with Schmidt rank $r$ in every cut, $\lvert\psi_D\rangle$, is $$ \| \lvert\psi\rangle - \lvert \psi_D \rangle \|^2 \le (N-1) \delta\ , $$ where $\delta$ is the sum of the square of the singular values.

(The last point -- sum of squares -- is not what you ask in your question, but it is the right quantity; if you really need the sum of the singular values themselves, you can bound them relative to the sum of squares, and this is likely the best you can do.)

(It is worth noting that this bound scales better in $N$ than what one would get from the triangle inequality of the norm; the reason is that the errors from different truncations are orthogonal, see also the proof discussed below.)

I might try to post an alternative proof to the one in the paper here in the next days if I find time. A step-by-step sketch can be found in https://schuch.univie.ac.at/ss21-qmb/, exercise sheet 2, problem 4.

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