---
title: "How much does cutting the bonds cost?"
description: "Dropped Schmidt weights set a lower and an upper bound on the error of a compressed state. Local readings stay far more accurate than the whole state."
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language: "en"
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  zh-CN: "https://sandboxphysics.com/zh/stories/tensor-networks/truncation-and-local-errors/"
last_modified: "2026-10-29"
---

[Tensor networks · A course with experiments](https://sandboxphysics.com/stories/tensor-networks/)

L2 · Chapter 4

# How much does cutting the bonds cost?

At a single cut, keeping the largest Schmidt weights is the best you can do, and the squared error equals the weight you dropped. Over a whole chain the squared error lies between the largest dropped weight at any one cut and twice the sum over all cuts. A reading that lives on one side of a cut feels the truncation only at second order; a correlation that spans the cut feels it at first order.

Sandbox Physics · 2026-10-29

In this chapter

[All eight chapters](https://sandboxphysics.com/stories/tensor-networks/)

Before you start

Chapters two and three: Schmidt weights, and a matrix product state as a chain of truncated decompositions. Expectation values and the norm of a state.

## One cut: keep the largest weights

Suppose you may keep only a fixed number of Schmidt terms at one cut. Which approximation comes closest to the original state? Schollwöck states the answer in Eq. 19 of his review: keep the terms with the largest singular values and set the rest to zero. The squared distance to the original state then equals the sum of the dropped Schmidt weights, called the tail.

No other choice does better at that cut. Any state with that many terms or fewer across the cut is at least the tail away from the original. The tail is therefore a floor, set by the state itself, that no algorithm can beat for that bond dimension.

$$
\varepsilon_c(\chi)=\sum_{\alpha>\chi}\lambda^{(c)}_\alpha,\qquad \min_{\operatorname{rank}_c\phi\le\chi}\big\lVert\psi-\phi\big\rVert^2=\varepsilon_c(\chi)
$$

## Many cuts: a lower and an upper bound

A matrix product state truncates every cut. The best single cut already gives a lower bound: the squared error cannot be smaller than the largest tail along the chain. For an upper bound, Verstraete and Cirac proved in Lemma 1 of their 2006 paper that the squared error is at most twice the sum of the tails over all cuts. Schollwöck repeats the bound as Eq. 58.

Take the critical twelve-spin chain of the compression Lab. At bond dimension four the largest tail is 1.58 millionths, the measured squared error is 5.66 millionths and the sum of tails is 6.16 millionths, so twice the sum is 12.3 millionths. The measured error sits inside the bounds, and closer to the upper one, because every cut adds its own tail.

$$
\max_c\,\varepsilon_c(\chi)\ \le\ \lVert\psi-\tilde\psi\rVert^2\ \le\ 2\sum_{c=1}^{L-1}\varepsilon_c(\chi)
$$

Worked example

## Why local readings stay accurate

Truncate a single cut and split the state into the kept part and the dropped part. The two parts use different Schmidt terms, so their left states are orthogonal and so are their right states. Take an operator that acts only on spins to the left of the cut. In the cross term between the kept and dropped parts, the right states meet only each other, and their overlap is zero. The cross term vanishes, and the reading changes only by terms of the size of the dropped weight.

A correlation between the two ends of the chain acts on both sides of every cut. Its cross term survives, and its error can be as large as the distance between the states, which is the square root of the squared error. The Lab shows the difference. At bond dimension four the squared error is 5.66 millionths, so the distance is about 0.0024. The energy per site is off by 2.2 millionths and the largest single-site magnetization by 2.1 millionths, both of the size of the squared error. The end-to-end correlation is off by 0.0015, of the size of the distance. The energy has one more protection: the exact state is an eigenstate, so the first-order change of its energy cancels for any small change of the state.

$$
\psi=\tilde\psi+\delta:\quad\langle\psi|O|\psi\rangle-\langle\tilde\psi|O|\tilde\psi\rangle=2\,\mathrm{Re}\,\langle\tilde\psi|O|\delta\rangle+\langle\delta|O|\delta\rangle,\qquad \langle\tilde\psi|O_{\rm L}|\delta\rangle=0
$$

## The tail decides what to keep

How fast the Schmidt weights fall decides how much you must keep. For the gapped chain at a field of 2.5, bond dimension two already gives a squared error of about ten millionths. The critical chain needs bond dimension four for a similar error, and the random state has no short tail at all: it still misses more than half its squared norm at bond dimension sixteen.

Choose the bond dimension for the measurement you need. Local energies and magnetizations converge at the rate of the squared error. Long-range correlations converge at its square root and need a larger bond dimension for the same accuracy.

An experiment to think through

## Predict, change, explain

1. In the compression Lab on the critical chain, predict which readout improves fastest as you raise the bond dimension from two to eight: the energy per site or the end-to-end correlation.
2. Step the bond dimension through two, four and eight and note both readouts and the squared error.
3. Explain the different rates: why does the energy err only at second order, while the end-to-end correlation, which spans every cut, errs at first order?

[Cut the bonds of a spin chain and watch the error](https://sandboxphysics.com/labs/entanglement-compression/)

The Labs simulate open spin-1/2 chains with energies in units of the coupling and time in units of the reduced Planck constant divided by the coupling. Each Lab states its model and limits in its notes and can export the data behind its readouts.

## Try explaining it yourself

### 1. A chain has non-zero tails at only three cuts: one, two and three millionths. Between which values must the squared error of the truncated state lie?

Read an explained answer

At least the largest tail, three millionths, and at most twice the sum, twelve millionths.

$$
3\times10^{-6}\le\lVert\psi-\tilde\psi\rVert^2\le2\,(1+2+3)\times10^{-6}=1.2\times10^{-5}
$$

### 2. In the cat-state preset at bond dimension one, the lower bound is about one half. Why can no product state come closer to this ground state?

Read an explained answer

Its middle cut has two Schmidt weights of nearly one half each. A product state keeps one term at that cut, so it drops at least the other weight, about one half.

### 3. At bond dimension eight, the end-to-end correlation error of the critical chain is about ten millionths, while the squared error is about nine ten-billionths. Is this consistent with the cross-term argument?

Read an explained answer

Yes. The distance is the square root of the squared error, about 30 millionths. A correlation spanning every cut can be off by up to about that much, and ten millionths lies below it. Local readings are off by about the squared error itself.

## Sources to read alongside

1. [U. Schollwöck, The density-matrix renormalization group in the age of matrix product states, Annals of Physics 326, 96 (2011)](https://arxiv.org/abs/1008.3477)

   The standard review. Section 3 uses the area law to explain why DMRG works in one dimension; Section 4.1 builds matrix product states from successive singular value decompositions and bounds the truncation error in Eq. 58; Section 6 sets out the ground-state search as a variational sweep; Section 7 covers time evolution and its limits.

2. [F. Verstraete and J. I. Cirac, Matrix product states represent ground states faithfully, Physical Review B 73, 094423 (2006)](https://arxiv.org/abs/cond-mat/0505140)

   Lemma 1 bounds the squared distance between a state and its bond-dimension-D truncation by twice the sum of the discarded Schmidt weights over all cuts; the paper then argues why gapped ground states need only modest D.

[Previous Turn a wave function into a chain of small matrices](https://sandboxphysics.com/stories/tensor-networks/matrix-product-states/) [Next Find a ground state by sweeping, and decide whether to believe it](https://sandboxphysics.com/stories/tensor-networks/dmrg-variational-sweeps/)