Entanglement Entropy

Entanglement entropy is the von Neumann entropy of a subsystem of a pure bipartite state. The first density-operator introduction to the formula is Entropy Overview. The broader reduced-state quantity is Subsystem Entropy; this page specializes it to pure-state bipartite entanglement. If the joint state on $AB$ is pure,

$$\rho_{AB} = \lvert\Psi\rangle\langle\Psi\rvert,$$

then the entanglement entropy across the $A\vert B$ split is

$$S_A = S(\rho_A) = -\operatorname{Tr}(\rho_A\log\rho_A), \qquad \rho_A=\operatorname{Tr}_B\rho_{AB}.$$

For pure bipartite states, this quantity measures how entangled $A$ is with $B$. For mixed joint states, the entropy of a subsystem is not by itself an entanglement measure.

Definition for Pure Bipartite States

Let

$$\lvert\Psi\rangle_{AB} \in \mathcal H_A\otimes\mathcal H_B$$

be a normalized pure state. Define

$$\rho_A=\operatorname{Tr}_B \bigl( \lvert\Psi\rangle\langle\Psi\rvert \bigr), \qquad \rho_B=\operatorname{Tr}_A \bigl( \lvert\Psi\rangle\langle\Psi\rvert \bigr).$$

The entanglement entropy is

$$E(\Psi) = S(\rho_A) = S(\rho_B).$$

The equality $S(\rho_A)=S(\rho_B)$ holds because the nonzero spectra of $\rho_A$ and $\rho_B$ are the same for a pure bipartite state.

Von Neumann Entropy

For a density operator $\rho$ with eigenvalues $\lambda_k$, the von Neumann entropy is

$$S(\rho) = -\operatorname{Tr}(\rho\log\rho) = -\sum_k \lambda_k\log\lambda_k.$$

The convention is

$$0\log0=0,$$

understood as a limit.

The base of the logarithm fixes the units:

• $\log_2$ gives entropy in bits;
• $\ln$ gives entropy in nats.

This page states numerical examples in bits unless otherwise noted.

Equality of Subsystem Entropies

Use the Schmidt decomposition

$$\lvert\Psi\rangle = \sum_r s_r \lvert u_r\rangle_A\lvert v_r\rangle_B.$$

The reduced density operators are

$$\rho_A = \sum_r s_r^2 \lvert u_r\rangle\langle u_r\rvert_A, \qquad \rho_B = \sum_r s_r^2 \lvert v_r\rangle\langle v_r\rvert_B.$$

Thus $\rho_A$ and $\rho_B$ have the same nonzero eigenvalues $s_r^2$. Entropy depends only on eigenvalues, so

$$S(\rho_A)=S(\rho_B).$$

This equality is special to pure joint states. If $\rho_{AB}$ is mixed, the subsystem entropies need not be equal.

Schmidt-Coefficient Formula

Let

$$p_r=s_r^2.$$

Then

$$E(\Psi) = -\sum_r p_r\log p_r.$$

In base $2$,

$$E_2(\Psi) = -\sum_r p_r\log_2 p_r \quad \text{bits}.$$

The formula shows that pure-state entanglement entropy is the Shannon entropy of the squared Schmidt coefficients.

Product State Entropy

For a product pure state,

$$\lvert\Psi\rangle = \lvert\alpha\rangle_A\otimes\lvert\beta\rangle_B,$$

the Schmidt coefficients are just

$$s_1=1.$$

Thus

$$E(\Psi) = -(1)\log(1) = 0.$$

Product pure states have zero entanglement entropy.

Bell State Entropy

For the Bell state

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr),$$

the Schmidt probabilities are

$$p_1=p_2=\frac12.$$

Therefore

$$\begin{aligned} E_2(\Phi^+) &= -\frac12\log_2\frac12 -\frac12\log_2\frac12\\ &= 1. \end{aligned}$$

A Bell state has one bit of entanglement entropy. It is common to say it contains one ebit of bipartite entanglement.

Partially Entangled Two-Qubit State

For

$$\lvert\Psi_\theta\rangle = \cos\theta\,\lvert00\rangle + \sin\theta\,\lvert11\rangle, \qquad 0\le\theta\le\frac{\pi}{2},$$

the Schmidt probabilities are

$$p_1=\cos^2\theta, \qquad p_2=\sin^2\theta.$$

The entanglement entropy in bits is

$$E_2(\Psi_\theta) = -\cos^2\theta\,\log_2(\cos^2\theta) -\sin^2\theta\,\log_2(\sin^2\theta).$$

It is zero at $\theta=0$ and $\theta=\pi/2$, and it reaches one bit at $\theta=\pi/4$.

Maximum Value

If the smaller subsystem dimension is

$$d=\min(\dim\mathcal H_A,\dim\mathcal H_B),$$

then

$$0\le E_2(\Psi)\le \log_2 d.$$

The upper bound occurs when the nonzero Schmidt probabilities are all equal:

$$p_r=\frac1d, \qquad r=1,\ldots,d.$$

Then

$$E_2(\Psi)=\log_2 d.$$

For two qubits, this maximum is $1$ bit. For two qutrits, it is $\log_2 3$ bits.

What It Measures and What It Does Not

For a pure bipartite state, entanglement entropy measures the amount of entanglement across the chosen split. It is invariant under local unitaries:

$$\lvert\Psi\rangle \mapsto (U_A\otimes U_B)\lvert\Psi\rangle.$$

It is also additive for independent pure pairs:

$$E(\Psi_1\otimes\Psi_2) = E(\Psi_1)+E(\Psi_2).$$

However, $S(\rho_A)$ is not a general mixed-state entanglement measure. If $\rho_{AB}$ is mixed, subsystem entropy can include ordinary classical uncertainty, thermal entropy, environmental noise, and correlations that are not entanglement. Mixed-state entanglement requires different tools.

Many-Body and QFT Preview

In many-body physics, one often studies the entropy of a spatial region or a collection of sites in a pure many-body state. The same formula appears:

$$S_A=-\operatorname{Tr}(\rho_A\log\rho_A).$$

The interpretation becomes richer. Entanglement entropy can diagnose area laws, critical scaling, topological order, tensor-network efficiency, and thermalization behavior.

In QFT, entanglement entropy is subtle because local regions have infinitely many degrees of freedom. The entropy often depends on a short-distance regulator and can diverge. The finite-dimensional formula here is the conceptual starting point, not the complete field-theoretic story.

For spatial partitions, exact many-body benchmarks, area and volume laws, critical scaling, and tensor-network consequences, see Entanglement Entropy in Many-Body Systems. For continuum-region, modular-Hamiltonian, relative-entropy, and holography cautions, see Entanglement in QFT Preview.

Common Mistakes

• Forgetting to specify the subsystem split.
• Using $S(\rho_A)$ as an entanglement measure when the global state is mixed.
• Leaving the logarithm base ambiguous in numerical answers.
• Thinking zero local expectation values imply maximal entanglement.
• Confusing Schmidt coefficients $s_r$ with probabilities $p_r=s_r^2$.
• Treating entanglement entropy as a directly measured observable rather than a nonlinear function of the state.

Cross-Links

• Schmidt Decomposition
• Entanglement Measures
• Entropy Overview
• Schmidt Rank
• Renyi Entropies
• Mutual Information
• Concurrence for Two Qubits
• Negativity and PPT Criterion
• LOCC Preview
• Local Unitary Equivalence
• Reduced Density Operators
• Partial Trace
• Subsystem Entropy
• Marginals and Correlations
• Entanglement Entropy in Many-Body Systems
• Entanglement in Many-Body Physics
• Entanglement in QFT Preview
• Entangled States
• Bell States
• Product States
• Pure Versus Mixed States
• Schmidt Decomposition as Linear Algebra
• Core Formulas Index

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865-942, 2009.
• J. Eisert, M. Cramer, and M. B. Plenio, “Area Laws for the Entanglement Entropy,” Reviews of Modern Physics 82, 277-306, 2010.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.

Exercises

1. Compute the entanglement entropy of a product pure state.

Solution

A product pure state has one Schmidt probability, $p_1=1$. Therefore

$$E = -1\log 1 = 0.$$

2. Compute the entanglement entropy in bits of

$$\lvert\Psi\rangle = \frac{1}{\sqrt3} \bigl( \lvert00\rangle+\lvert11\rangle+\lvert22\rangle \bigr).$$

Solution

The Schmidt probabilities are

$$p_1=p_2=p_3=\frac13.$$

Thus

$$E_2 = -3\left(\frac13\log_2\frac13\right) = \log_2 3.$$

3. For

$$\lvert\Psi_\theta\rangle = \cos\theta\,\lvert00\rangle + \sin\theta\,\lvert11\rangle,$$

show that the entropy vanishes at $\theta=0$.

Solution

At $\theta=0$, the state is $\lvert00\rangle$. The Schmidt probabilities are $p_1=1$ and $p_2=0$. Therefore

$$E = -1\log 1-0\log0 = 0.$$

4. Why is $S(\rho_A)$ not automatically an entanglement measure when $\rho_{AB}$ is mixed?

Solution

If $\rho_{AB}$ is mixed, $\rho_A$ may be mixed because of classical uncertainty, thermal noise, environmental correlations, entanglement, or a combination of these. The entropy of $\rho_A$ does not distinguish these origins. For pure $\rho_{AB}$, a mixed reduced state has only one possible source: entanglement with the other subsystem.