---
title: "Why half of a pure quantum state can look random"
description: "Calculate a reduced state and its entropy, then compare entanglement, classical mixtures and thermal states with identical local statistics."
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language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/holography/entanglement-and-heat/"
last_modified: "2026-10-28"
---

[Holography and quantum gravity](https://sandboxphysics.com/stories/holography/)

L1 · Chapter 2

# Why half of a pure quantum state can look random

Sandbox Physics · 2026-10-28

A completely specified joint quantum state can give random outcomes to someone who sees only half of it. A reduced density matrix records those local statistics. For a pure bipartite state, its entropy measures entanglement. This definition needs neither gravity nor a temperature.

## Before you begin

Begin with probabilities that add to one. A complex amplitude carries a phase; a density matrix records measurement statistics. Both are explained with a two-qubit example. This chapter can be studied independently before connecting it to the geometry.

- Calculate local measurement statistics and entropy from a joint pure state.
- Explain why local randomness alone establishes neither entanglement nor temperature.

## Random separately, correlated together

Give two observers one qubit each, with basis states $\lvert0\rangle$ and $\lvert1\rangle$. Prepare $\lvert\Psi\rangle=(\lvert00\rangle+\lvert11\rangle)/\sqrt2$. A measurement in this basis gives either observer zero or one with equal probability. Their results always agree when compared. The relative phase also matters for correlations measured in other bases.

The full list of joint amplitudes specifies a pure state. To predict experiments performed by just one observer, sum over every outcome of the other observer. This partial trace retains everything needed for local measurements and removes correlations accessible only through joint measurements.

$$
\rho_{AB}=|\Psi\rangle\langle\Psi|,\qquad \rho_A=\operatorname{Tr}_B\rho_{AB}=\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix}
$$

Diagonal entries are probabilities in this basis; off-diagonal entries contain coherence. Orthogonality of the other qubit’s basis states removes the off-diagonal terms in the reduced state.

## Quantify the missing information

If the density matrix has eigenvalues $p_i$, its von Neumann entropy is $-\sum_i p_i\log p_i$. These eigenvalues act as probabilities in a basis that diagonalizes the matrix. Natural logarithms give entropy in nats. Base-two logarithms give bits.

The two local eigenvalues are $1/2$, so local entropy is $\log2$ nats, or one bit. The joint pure state has one nonzero eigenvalue equal to one and therefore zero entropy. Local entropies do not simply add to the joint entropy when correlations are present.

Worked example: prepare $\sqrt{3/4}\lvert00\rangle+\sqrt{1/4}\lvert11\rangle$. The partial trace leaves probabilities three quarters and one quarter, giving about $0.562$ nats. Keep the state normalized as the second probability falls to zero. The local outcome becomes certain and the entanglement entropy approaches zero.

$$
S(\rho)=-\operatorname{Tr}(\rho\log\rho),\qquad S_A=-\frac34\log\frac34-\frac14\log\frac14\simeq0.562
$$

A zero eigenvalue contributes zero by continuity. For a pure bipartite state, the two local entropies are equal and measure entanglement.

Source: Xi Dong · Shinsei Ryu and Tadashi Takayanagi

## The same local matrix can come from different joint states

Instead, use a classical random number to send either $\lvert00\rangle$ or $\lvert11\rangle$, each half the time. Each observer gets the same local matrix as before. The mixture has no coherence between its two terms. Measurements in that single basis cannot distinguish it from the entangled state; correlations in another basis can.

The mixture has joint entropy $\log2$, whereas the entangled pure state has joint entropy zero. For a mixed joint state, $S_A$ can include classical uncertainty as well as entanglement. It is then insufficient as a measure of entanglement by itself.

$$
\rho_{\rm mix}=\frac12|00\rangle\langle00|+\frac12|11\rangle\langle11|
$$

This is a probabilistic mixture of product states. It has the same local statistics as the Bell state but different joint coherence.

## A thermal state also needs an energy operator

Temperature controls the probabilities assigned to different energies in equilibrium. Calling a state thermal therefore requires a Hamiltonian, the operator that defines energy, and a specified equilibrium setting. At very high temperature, a two-level system gives almost equal probabilities and can have the same matrix form as the reduced state above.

For energies zero and $E$, let $\beta=1/(k_{\rm B}T)$. The excited-to-ground probability ratio is $e^{-\beta E}$. At $\beta E=\log3$, the ground probability is three quarters and the excited probability one quarter, matching the reduced-state example. The matrix alone does not identify how the randomness was prepared.

In holography, some boundary thermal states correspond to bulk black holes. A subregion of the boundary vacuum has an entropy too. These involve different states and geometric conditions. We next introduce boundary correlation functions before relating an entropy to a geometric length.

$$
\rho_\beta=\frac{e^{-\beta H}}{Z},\qquad Z=\operatorname{Tr}e^{-\beta H}
$$

This Gibbs state is defined relative to a Hamiltonian. For the two-level example, the specified energy gap and equilibrium population ratio together fix temperature.

Source: João Penedones

## Try the reasoning

### [Compare one boundary node with a joint encoding](https://sandboxphysics.com/labs/boundary-quantum-code/)

1. Keep every boundary node and change the logical state direction. Observe the reconstructed state.
2. Erase one node. Compare the information available to the erased subsystem with what the remaining subsystem can recover.
3. Retain fewer nodes, predict when recovery will fail, and explain why reading one local probability is a weaker test than recovering a quantum state.

## Exercises and explanations

1. What entropy does one half of a product pure state have?

   Read the explanation

   The partial trace leaves that half in a pure state with one nonzero eigenvalue equal to one. Its entropy is zero regardless of how complicated the other factor is.

2. Change the relative phase of the two joint amplitudes by $e^{i\varphi}$. Do the local eigenvalues change? Must all joint measurements stay the same?

   Read the explanation

   The partial trace removes the cross terms, so local eigenvalues and entropy stay the same. Joint measurements in suitable bases still detect the phase. An arbitrary quantum state also requires its off-diagonal coherences.

[Previous: Why light reaches the AdS boundary in finite coordinate time](https://sandboxphysics.com/stories/holography/clocks-and-curvature/) [Next: Reading boundary correlations from a bulk field](https://sandboxphysics.com/stories/holography/fields-at-the-boundary/)

## Sources and recalculated scope

The course derives its worked examples and draws its own schematics. The numerical references recalculate specified analytic geometry and a finite five-qubit encoding. The source locations below distinguish those calculations from theoretical reading topics.

1. [João Penedones, TASI lectures on AdS/CFT, arXiv:1608.04948v1 (2016)](https://arxiv.org/html/1608.04948v1)

   Sections 3.1–3.3: global and Poincaré coordinates, fixed-background fields, source boundary conditions and dynamical gravity; section 4.1: conditions for a CFT with a weakly coupled bulk description. The course derives its radial optical paths and scalar near-boundary exponents from these metrics and field equations.

2. [Shinsei Ryu and Tadashi Takayanagi, Holographic Derivation of Entanglement Entropy from AdS/CFT, hep-th/0603001v2; PRL 96, 181602 (2006)](https://arxiv.org/html/hep-th/0603001v2)

   Equation (5), static area prescription; equations (6)–(10) and Figure 1, vacuum AdS3 interval geometry and the circle result. We independently integrate the Poincaré geodesic and compare regulator limits. Our equal-height finite-cutoff anchors are a separately defined geometric diagnostic. Multiple-interval minima use Headrick, not equation (12) of this early paper.

3. [Xi Dong, Daniel Harlow and Aron C. Wall, Bulk Reconstruction in the Entanglement Wedge in AdS/CFT, arXiv:1601.05416v2; PRL 117, 021601 (2016)](https://arxiv.org/html/1601.05416v2)

   Sections I–II and Figure 1 define causal and entanglement wedges and a perturbative code subspace. Sections III–IV state the relative-entropy assumptions behind reconstruction. We explain these assumptions; the finite code checks are a separate exact model.