research31 min studyLesson 70 in Physics

Entanglement and tensor-network states

Use Schmidt weights to quantify bipartite entanglement and understand when matrix-product states compress quantum matter.

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What you’ll understand

  • Compute a reduced-state entropy from Schmidt probabilities.
  • Relate bond dimension to retained information across a cut.
  • Identify the regimes where a compact matrix-product ansatz becomes costly.

A chain of L spin-half sites has exponentially many basis amplitudes, but many physically important states occupy a much more structured part of that space. Tensor networks exploit structure in correlations across partitions. Their usefulness is governed by entanglement and the accuracy required, rather than by the number of particles alone. We will start with an exact two-site decomposition, turn its singular values into a truncation rule, and then explain why one-dimensional ground states are often more accessible than generic long-time dynamics.

A cut turns amplitudes into singular values

Divide a pure state into subsystems A and B and arrange its amplitudes into a matrix. Singular-value decomposition writes the state as a sum of paired orthonormal states, with nonnegative coefficients λα\lambda_\alpha. Their squares pαp_\alpha sum to one. Tracing out B leaves A with eigenvalues pαp_\alpha. A global pure state can therefore give a mixed local state: the uncertainty comes from correlations with the omitted subsystem, not from assuming an unknown classical preparation of the whole system.

The von Neumann entropy of that reduced state measures bipartite entanglement for a globally pure state. A product state has one nonzero Schmidt weight and zero entropy. A maximally entangled pair of qubits has two equal weights and entropy log⁡2\log 2 using natural logarithms. If the full state is already mixed, reduced-state entropy also contains classical and thermal uncertainty; it is no longer by itself a unique measure of entanglement between the parts. Always state the purity assumption before interpreting the number.

∣ψ⟩=∑αλα∣αA⟩∣αB⟩,pα=λα2,SA=−∑αpαlog⁡pα|\psi\rangle=\sum_\alpha\lambda_\alpha|\alpha_A\rangle|\alpha_B\rangle,\qquad p_\alpha=\lambda_\alpha^2,\qquad S_A=-\sum_\alpha p_\alpha\log p_\alpha
The Schmidt coefficients are nonnegative and normalized by Σp = 1. Entropy is measured in nats here; dividing by log 2 converts it to bits.↔ Scroll wide equations horizontally.

Bond dimension is a controlled compression choice

Repeated decompositions along a one-dimensional chain produce a matrix-product state, or MPS. For an open-chain MPS, a virtual bond index links neighboring tensors and carries correlations across that cut. If its dimension is χ\chi, the Schmidt rank across that bond is at most χ\chi and its entropy is at most log⁡χ\log\chi. For finite open chains every state has an exact MPS representation if sufficiently large bond dimensions are permitted. The computational advantage comes only when a much smaller representation achieves the desired accuracy.

For one cut, retain the largest Schmidt weights and discard the rest. The discarded weight is the sum of the omitted probabilities. After normalizing the truncated state, its squared overlap with the original at that cut is one minus that weight. This is a precise local compression statement. A long sequence of truncations, a time-evolution algorithm, or an approximate variational optimizer needs additional error accounting; one small discarded weight is not a universal bound on every later observable or on a simulation's total error.

SA≤log⁡χ,εdisc=∑α>χpα,∣⟨ψ∣ψχ⟩∣2=1−εdiscS_A\leq\log\chi,\qquad \varepsilon_{\mathrm{disc}}=\sum_{\alpha>\chi}p_\alpha,\qquad |\langle\psi|\psi_\chi\rangle|^2=1-\varepsilon_{\mathrm{disc}}
The entropy bound applies to a state of Schmidt rank at most χ. The overlap formula describes one optimal truncation across a specified cut with the retained state normalized, not arbitrary repeated truncations.↔ Scroll wide equations horizontally.

Why geometry and time change the cost

For one-dimensional chains with bounded finite-dimensional local spaces and bounded finite-range interactions, a unique ground state separated by a uniformly positive gap obeys an entanglement area law: entropy across a cut stays bounded as chain length grows. This motivates compact MPS approximations, though accuracy is governed by the full Schmidt tail rather than entropy alone. Critical chains can show logarithmically growing entanglement, and approaching small gaps can demand larger bonds. Density-matrix renormalization group optimizes an MPS variationally; it is a numerical method with convergence questions, not an exact solution merely because an area law is expected.

Generic highly excited states and long-time unitary dynamics can develop much more entanglement. A global quench often produces approximately linear growth before finite-size saturation, forcing the bond dimension needed for fixed accuracy to increase rapidly. Representing a two-dimensional system as a one-dimensional ordering also creates wider cuts and growing cost. Tensor-network variants exploit other geometries, but their contractions and optimization introduce further challenges. Report size, boundary conditions, retained bonds, discarded weights, and observable convergence rather than advertise compression without an error target.

Compress one entangled pair

For |ψ⟩ = √0.8|00⟩ + √0.2|11⟩, calculate the entropy of the first qubit and the fidelity after keeping only the largest Schmidt component.

  1. The displayed basis is already a Schmidt basis. The reduced state is diagonal with probabilities p1=0.8p_1=0.8 and p2=0.2p_2=0.2.

  2. Compute S=−0.8log⁡0.8−0.2log⁡0.2≈0.5004S=-0.8\log0.8-0.2\log0.2\approx0.5004 nats, or about 0.72190.7219 bits. This is below the qubit maximum log⁡2≈0.6931\log2\approx0.6931 nats.

  3. Keeping one component discards weight εdisc=0.2\varepsilon_{\mathrm{disc}}=0.2. The normalized retained state is ∣00⟩|00\rangle, whose squared overlap with the original is 0.8=1−εdisc0.8=1-\varepsilon_{\mathrm{disc}}.

The result

Bond dimension two represents the pair exactly across this cut. Bond dimension one produces a product state and a measurable loss of overlap; it cannot preserve the entanglement.

Explore the model

Interactive lab

Interactive lab · probability

Read two Schmidt weights as probabilities

Set the binomial trial count to one and choose success probability p = 0.8. The two exact bars are then 0.2 and 0.8, matching the two Schmidt probabilities up to order in the worked example. The lab is a binomial probability model, not an entangled-state simulator, and it does not calculate entropy, tensor contractions, or truncation error.

Read two Schmidt weights as probabilitiesThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forSchmidt probabilities form a normalized spectrum. Quantum entanglement additionally requires the global pure-state structure, which these probability bars alone do not establish.

Check your understanding

What does a small discarded Schmidt weight certify for one normalized optimal truncation across one cut?

Try it yourself

Replace the pair by (|00⟩+|11⟩)/√2. Calculate its entropy and the best squared overlap of a normalized one-component Schmidt truncation.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Both Schmidt probabilities are one half. Retaining either component gives the same answer.

Compare with the solution

The entropy is −2 × (1/2) log(1/2) = log 2 nats, or one bit. A one-component truncation discards weight 1/2 and has squared overlap 1/2 with the original. Bond dimension two is necessary for an exact representation across this cut, while bond dimension one can only represent a product state.

Take it with you

  • Schmidt probabilities quantify bipartite entanglement of a pure state.
  • Bond dimension limits rank, while discarded weights diagnose a particular compression.
  • Area laws, dimension, dynamics, and required accuracy determine when tensor networks remain efficient.

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