Entanglement entropy and area laws
Calculate entropy from Schmidt probabilities and distinguish finite quantum-information measures from regulator-dependent field-theory entanglement.
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The question for today
What you’ll understand
- Compute a reduced-state entropy using explicitly normalized Schmidt weights.
- Explain the short-distance origin and dimensional scaling of a leading area-law divergence.
- Identify the extra assumptions behind interval entropy, gauge-theory subsystem choices, and finite mutual information.
A complete pure state can look mixed when only one region is observed. Quantum fields add infinitely many short-distance degrees of freedom near the boundary. We first calculate a two-qubit example, then distinguish its exact statements from continuum formulas requiring a regulator, state, region, and specified operator algebra.
Trace out a subsystem, then diagonalize
For a bipartite pure state, the Schmidt decomposition pairs orthogonal states in the two subsystems with nonnegative probabilities summing to one. Tracing out one subsystem leaves those probabilities as eigenvalues of the other subsystem's density matrix. The von Neumann entropy is their Shannon entropy. A relative phase between two Schmidt terms does not change these eigenvalues, whereas changing their magnitudes does. This is why amplitudes must be squared before they enter an entropy.
In the two-term example, p equal to zero or one gives a product state and zero entropy. Equal weights maximize the entropy at the natural logarithm of two. These statements concern a pure global state with a specified tensor-product split. If the global state is mixed, a subsystem's entropy can include ordinary thermal or classical uncertainty, so a nonzero entropy alone no longer certifies entanglement. Distinguishing those contributions requires additional information or an appropriate entanglement measure.
The continuum boundary contains ultraviolet correlations
In a local field-theory vacuum, degrees of freedom on opposite sides of a region's boundary correlate at arbitrarily short distances. A spatial cutoff epsilon makes the entropy finite. In more than one spatial dimension, a common leading behavior is proportional to boundary area divided by epsilon raised to the boundary dimension. Its coefficient generally depends on the regulator and the microscopic field content. Calling this an area law does not make its leading divergent coefficient a universal observable.
A useful special result occurs for a single interval of length ell in the vacuum of a unitary conformal field theory on an infinite one-dimensional line. The entropy has logarithmic growth with coefficient c/3, where c is the central charge, plus a nonuniversal constant. Finite temperature, finite system size, boundaries, excited states, or multiple intervals change the formula. The logarithm is not obtained by setting the boundary area to zero in a higher-dimensional power law; it is a distinct dimensional case.
Choose quantities and subsystems carefully
For separated regions, mutual information combines entropies as . Local boundary divergences can cancel, giving a finite quantity under appropriate separation and state assumptions. Adjacent regions can reintroduce short-distance divergences. Relative entropy is another powerful finite comparison in suitable settings, but its mathematical domain and reference state matter. Subtracting two enormous regulated numbers numerically is not by itself a proof that a continuum limit exists.
Gauge constraints add a further subtlety: the physical Hilbert space need not factor into a simple product of independent spatial-region Hilbert spaces. One must specify the observable algebra or an extended-space prescription and account for boundary or center degrees of freedom. Entropy can depend on that choice. Universal terms, mutual information, and carefully defined relative entropy often provide more robust comparisons. Extending such tools through interacting continuum and gravitational systems remains an active research direction, rather than a reason to omit the subsystem definition.
Work it through
Entropy of unequal Schmidt weights
Use the pure two-qubit state with Schmidt probabilities p = 1/4 and 1 − p = 3/4. Compute the reduced entropy and compare it with the equal-weight state.
Tracing out subsystem B gives a diagonal reduced density matrix with eigenvalues 0.25 and 0.75. Their sum is one, so no extra normalization is needed.
Compute −0.25 log(0.25), approximately 0.3465736, and −0.75 log(0.75), approximately 0.2157616. Add them to obtain 0.5623351 nats.
The equal-weight entropy is log 2, approximately 0.6931472 nats. The unequal state is entangled but has about 0.1308120 fewer nats of reduced entropy.
The entropy depends on the Schmidt probabilities, not a relative phase. This finite calculation contains no spatial cutoff and should not be mistaken for the divergent entropy of a continuum region.
The living notebook
Explore the model
Interactive lab · probability
Inspect the two weights in a reduced-state spectrum
Set trial count n to one in the exact binomial model and vary success probability p. Its two bars then display p and 1−p, matching the two numerical Schmidt weights. The lab itself is a classical probability calculation: it does not construct a bipartite quantum state, trace a field region, or establish entanglement from a histogram.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
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 forThe same probability entropy enters this pure bipartite example, but the quantum state and subsystem assumptions are what give the weights an entanglement interpretation.
Pause & reason
Check your understanding
Your turn
Try it yourself
In the stated one-dimensional conformal vacuum with c = 1, double the interval length while holding the cutoff fixed. What entropy difference is predicted? What assumption would a finite-temperature experiment violate?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Subtract the two logarithms so the nonuniversal additive constant cancels.
Compare with the solution
The difference is (1/3) log 2, approximately 0.2310491 nats. Both the cutoff and the additive constant cancel in this difference within the stated formula. A finite-temperature state violates the vacuum assumption and introduces a thermal length scale, so the infinite-line vacuum logarithm cannot be used unchanged at arbitrary interval lengths.
Keep the idea
Take it with you
- Schmidt probabilities determine the subsystem entropy of a bipartite pure state.
- Continuum entanglement generally needs ultraviolet regulation, and leading area coefficients need not be universal.
- Gauge constraints and finite-comparison constructions make the subsystem definition part of the physics.
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