greatitself.
Continue
research35 min studyLesson 799 in Physics

Ryu–Takayanagi entropy in semiclassical models

Track the surface, homology and saddle assumptions behind classical and quantum entropy formulas.

Before you begin
On this page

The question for today

What you’ll understand

  • Specify the static RT surface and its units.
  • Compute an AdS3 interval entropy difference.
  • Distinguish classical area minimization from quantum generalized-entropy extremization.

In appropriate semiclassical holographic models, boundary entanglement can be computed using bulk geometry. The static Ryu–Takayanagi prescription relates a boundary region's entropy to a minimal surface with specified anchoring and homology. Quantum and time-dependent extensions require more structure. These formulas are conditional results within a duality and gravitational saddle framework, not direct measurements of hidden spatial surfaces in arbitrary quantum systems. A finite-cutoff entropy also needs a regulator and a consistent convention for additive terms.

Choose the surface problem before computing an area

Take a time-reflection-symmetric static state with an Einstein bulk in d plus one dimensions and a spatial boundary region A. On the corresponding bulk time slice, search for a codimension-two surface anchored on the region's boundary and homologous to A. The homology condition constrains which topological candidates are allowed; the locally shortest-looking surface is not sufficient if it violates that requirement. Compare the areas of the admissible minimal candidates at the same cutoff.

In natural units the leading entropy is area divided by four Newton constant. The surface area has length to the d minus one units, exactly canceling the effective Newton units. In a time-dependent Lorentzian setting, a covariant extremal-surface prescription and its selection rules replace simple minimization on an arbitrary slice. A surface extremizing a functional is not automatically the dominant allowed saddle; anchoring, homology and the state remain part of the calculation.

S(A)=Area⁡(γA)4Gd+1at leading static Einstein order,∂γA=∂A,γA∼AS(A)=\frac{\operatorname{Area}(\gamma_A)}{4G_{d+1}}\quad\text{at leading static Einstein order},\qquad \partial\gamma_A=\partial A,\quad \gamma_A\sim A
Entropy is dimensionless with kB = 1 and the natural gravitational constant convention. The symbol γA ∼ A denotes the homology requirement, not approximate equality of shapes.↔ Scroll wide equations horizontally.

Benchmark a vacuum interval in AdS3

For a single interval in the vacuum on an infinite two-dimensional boundary, the AdS3 static surface is a geodesic. Its regulated length is logarithmic in interval size divided by a common short-distance cutoff. The Brown–Henneaux central-charge relation converts its coefficient into the familiar boundary logarithm. An additive constant depends on regulator and normalization choices, so an entropy difference at fixed cutoff is often a cleaner comparison than an isolated absolute number.

Doubling the interval changes the logarithm by a fixed amount independent of its original length, provided the vacuum infinite-line assumptions and scale hierarchy remain valid. Finite circles, finite temperature, multiple intervals and other states give different expressions or competing surface saddles. An ultraviolet divergence in the entropy is expected for sharp spatial partitions; it does not mean the quantum state is unnormalized. Universal coefficients and finite differences require their own carefully specified comparison.

Length⁡(γA)≃2Lln⁡(ℓ/ϵ),c=3L2G3,S(A)=c3ln⁡(ℓ/ϵ)+Cscheme\operatorname{Length}(\gamma_A)\simeq2L\ln(\ell/\epsilon),\qquad c=\frac{3L}{2G_3},\qquad S(A)=\frac c3\ln(\ell/\epsilon)+C_{\rm scheme}
The boundary has 1 + 1 dimensions and the bulk has 2 + 1. ℓ and ε are lengths with ℓ ≫ ε; Cscheme denotes a fixed additive convention. G3 has length units.↔ Scroll wide equations horizontally.

Add quantum terms without discarding saddle selection

Beyond leading area order, bulk quantum entanglement contributes to generalized entropy, together with consistent renormalization of area and local terms. A quantum extremal surface makes this generalized entropy stationary. Comparing the admissible extremized candidates then selects the relevant saddle under the prescription's assumptions. Minimizing the classical area alone can pick a different candidate when the bulk entropy contribution matters, especially near a competition between saddles.

Replica derivations and quantum extremal-surface arguments require assumptions about semiclassical expansion, allowed saddles and analytic continuation. Applying them to evaporating black holes or information recovery adds state, bath and reconstruction questions; a small classroom area calculation does not settle those questions universally. The optimization lab illustrates convergence of one quadratic iteration. It contains no bulk manifold, homology, geometric area or bulk density matrix, and cannot find an RT or quantum extremal surface.

Sgen[γ]=Area⁡(γ)4Gren+Sbulk,ren[Σγ]+⋯ ,δγSgen=0S_{\rm gen}[\gamma]=\frac{\operatorname{Area}(\gamma)}{4G_{\rm ren}}+S_{\rm bulk,ren}[\Sigma_\gamma]+\cdots,\qquad \delta_\gamma S_{\rm gen}=0
The terms use one consistent renormalization scheme; the ellipsis allows local or higher-order contributions appropriate to the model. One compares admissible extremized saddles rather than minimizing only the first term.↔ Scroll wide equations horizontally.

Work it through

A fixed-cutoff interval comparison

Use the leading AdS3 formula with L/G3 = 20 and set the additive convention to zero. For ℓ/ε = 100 find c and S. At the same cutoff, what entropy change follows if the interval is doubled while remaining in the vacuum infinite-line regime?

  1. The central charge is c=3(20)/2=30c=3(20)/2=30, so the logarithm coefficient is c/3=10c/3=10.

  2. At the declared cutoff ratio, the leading entropy is 10ln⁡100≈46.051710\ln100\approx46.0517 in natural entropy units. Its absolute value includes the stated regulator and additive convention.

  3. Doubling the interval adds 10ln⁡2≈6.9314710\ln2\approx6.93147. The common cutoff and additive term cancel from this difference. It remains a leading semiclassical model calculation, with quantum and finite-cutoff corrections requiring separate control.

The result

The central charge is 30, the stipulated leading entropy is about 46.0517 and the fixed-cutoff doubling change is 10 ln 2 ≈ 6.93147.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

An iteration analogy, not a surface search

Reset and set BOTH Step size η = 0.50 and Initial position x₀ = 2; predict convergence toward zero. Reset and set η = 2.20 and x₀ = 2; predict alternating growth and eventual Outside plot reporting. This is gradient descent on x²/2, not RT area minimization, a homology test or quantum generalized-entropy extremization.

An iteration analogy, not a surface searchGradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Current readings: Position x0: 2; Objective x²/2: 2; Iteration behavior: Stable step. All parameters are adjustable with the labeled controls below.−4−202402.174.356.528.7Parameter xf(x) = x²/2n = 0
ObjectiveDiscrete iterate
Position x0
2
Objective x²/2
2
Iteration behavior
Stable step
n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. 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 forA convergent numerical procedure still needs the correct physical functional and admissible domain.

Pause & reason

Check your understanding

What must be included when quantum extremal-surface candidates compete beyond leading classical area order?

Your turn

Try it yourself

Two already extremized, admissible candidates in the same state and common scheme have area-entropy terms 80 and 90, and bulk-entropy terms 12 and 1 respectively. Neglect other terms. Compare their generalized entropies and select the smaller candidate under this stated prescription.

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

Give me a hint

Add the two contributions for each candidate before comparing; both candidates have already met the same boundary and homology requirements.

Compare with the solution

The first generalized entropy is 80+12=9280+12=92, while the second is 90+1=9190+1=91. The second is preferred within the stated candidate set even though its classical area term is larger. This arithmetic does not locate either surface, prove their extremality or establish that no further saddle exists; those properties were supplied as assumptions. Comparing candidates with different subtraction conventions would not be meaningful because their generalized entropies would not share a common reference.

Keep the idea

Take it with you

  • RT specifies an anchored homologous surface in a particular semiclassical regime.
  • Entropy differences require a shared cutoff and state assumptions.
  • Quantum saddle comparisons use generalized entropy rather than classical area alone.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.