---
title: "Why a chain of spins outgrows any computer, and why its ground state does not"
description: "Each added spin doubles the amplitudes a quantum state needs. Ground states still fit in kilobytes, because they carry little entanglement across any cut."
canonical_url: "https://sandboxphysics.com/stories/tensor-networks/many-spins-many-amplitudes/"
language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/tensor-networks/many-spins-many-amplitudes/"
last_modified: "2026-10-29"
---

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

L1 · Chapter 1

# Why a chain of spins outgrows any computer, and why its ground state does not

A chain of spin one-half particles needs two to the power of the chain length amplitudes, one for every pattern of up and down. That list doubles with each spin and passes the memory of any computer within a few dozen spins. Ground states of local chains escape this count because the two halves of the chain share little entanglement. Most other states do not.

Sandbox Physics · 2026-10-29

In this chapter

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

Before you start

Basic quantum mechanics of a single spin: the states up and down, superposition, and the Pauli matrices. Powers of two and logarithms.

## Every added spin doubles the list

Take a row of spins, each of which can point up or down along a chosen axis. A quantum state of the row assigns one complex amplitude to every pattern of ups and downs, and the squared magnitude of that amplitude is the probability of finding that pattern. Two spins have four patterns, three spins have eight, and each further spin doubles the count.

Stored in double precision, a complex number takes sixteen bytes. Twenty spins then need about seventeen million bytes, which any laptop holds. Thirty-two spins need about 69 billion bytes, and forty spins need about eighteen trillion. No machine stores a general state of a hundred spins.

$$
|\psi\rangle=\sum_{s_1,\dots,s_L\in\{\uparrow,\downarrow\}}\psi_{s_1s_2\cdots s_L}\,|s_1s_2\cdots s_L\rangle,\qquad \#\text{amplitudes}=2^{L}
$$

## The two chains this course uses

The first chain is the transverse-field Ising chain. Neighbouring spins want to align along one axis, and a field along a perpendicular axis tries to tip every spin. The coupling sets the energy unit. With the field below the coupling the chain orders; above it, the spins follow the field and there is an energy gap to the first excitation; at equal values the chain is critical and the gap closes.

The second chain is the spin one-half Heisenberg antiferromagnet, in which neighbouring spins prefer to point opposite each other in every direction. Both chains here have open ends, so the first and last spins have only one neighbour each. These two models were chosen because exact answers exist for them: free fermions for the Ising chain and the Bethe ansatz for the Heisenberg chain.

$$
H_{\rm Ising}=-J\sum_{j=1}^{L-1}\sigma^x_j\sigma^x_{j+1}-h\sum_{j=1}^{L}\sigma^z_j,\qquad H_{\rm Heis}=J\sum_{j=1}^{L-1}\mathbf S_j\cdot\mathbf S_{j+1}
$$

Worked example

## Count the memory for a 32-spin ground state

The Hamiltonians above have only real entries when written in the up–down basis, so their ground states can be stored with real amplitudes of eight bytes each. A full list for 32 spins still takes about 34 billion bytes.

The DMRG Lab instead stores a matrix product state, the format built in chapter three. Its default run keeps at most sixteen states across each link between neighbouring spins, and fewer near the ends. Count the entries of the matrices along the finished chain: they total 11,048 real numbers, about 88 kilobytes. With that storage the energy of the critical Ising chain matches the exact free-fermion value to within a few parts in ten billion of the coupling.

The saving is not a trick of the format. The same format applied to a random state of the same length would need almost the full list. The next section says why.

$$
2^{32}\times8\ \text{bytes}\approx3.4\times10^{10}\ \text{bytes},\qquad N_{\rm MPS}=\sum_{j=1}^{32}2\,\chi_{j-1}\chi_j=11\,048
$$

## What makes ground states special

Cut the chain between two neighbouring spins. Chapter two defines a number, the entanglement entropy, that measures how strongly the two sides depend on each other. It is zero when the state is a product of a left state and a right state. For a twelve-spin chain cut in the middle it can be at most six times the natural logarithm of two, about 4.16.

The ground states in the compression Lab sit far below that ceiling. The critical Ising chain carries 0.40 across its middle cut, and the gapped chain at a field of 2.5 carries only 0.06. A random twelve-spin state carries 3.66. Schollwöck traces this to the area law: in one dimension, the entanglement of a gapped ground state stays bounded as the chain grows, while a random state has extensive entanglement, growing in proportion to the size of the block.

Verstraete and Cirac turned this into a statement about errors. They bound how well a matrix product state of a given size can approximate a state, using the entanglement of its blocks, and conclude that ground states of one-dimensional chains are represented faithfully. The compression Lab shows the contrast directly. The critical ground state keeps 296 of its 4,096 numbers with a squared error of about six in a million. The random state keeps 2,728 and still misses about half of its squared norm.

$$
S_{\rm max}=\ln 2^{L/2}=\tfrac{L}{2}\ln2\approx4.16\quad(L=12)
$$

An experiment to think through

## Predict, change, explain

1. Open the compression Lab on the critical chain. Predict how many numbers you must store before the squared error falls below one in a million.
2. Switch to the random state and raise the bond dimension as far as it goes.
3. Explain why the random state still has a large error while the critical ground state needed only a few hundred numbers.

[Cut the bonds of a spin chain and watch the error](https://sandboxphysics.com/labs/entanglement-compression/) [Sweep DMRG down to the ground-state energy](https://sandboxphysics.com/labs/dmrg-ground-state-sweeps/)

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. How many bytes does a general complex state of twenty spins need in double precision? And forty spins?

Read an explained answer

Two to the twentieth power is 1,048,576 amplitudes, which at sixteen bytes each is about 16.8 million bytes. For forty spins the count is about 1.1 trillion amplitudes, or about eighteen trillion bytes.

$$
2^{20}\times16=16\,777\,216,\qquad 2^{40}\times16\approx1.8\times10^{13}
$$

### 2. A product state gives each spin its own state, independent of the others. How many numbers does it need for twelve spins?

Read an explained answer

Two per spin, so 24. That is exactly what the compression Lab reports when the bond dimension is one, the setting that can only describe product states.

### 3. The random state carries 3.66 of entanglement across its middle cut, out of a possible 4.16. Why does this make it hard to compress?

Read an explained answer

High entanglement means the weight across the cut is spread over many terms of nearly equal size. Keeping a few of them throws away most of the state, so the error stays large until almost every term is kept.

## 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.

[Next Measure entanglement with one matrix decomposition](https://sandboxphysics.com/stories/tensor-networks/schmidt-decomposition/)