Bipartite Entanglement

Bipartite entanglement is nonseparability across a specified split into two subsystems, $A$ and $B$. For pure states, the Schmidt decomposition solves the structural problem. For mixed states, separability is a convex problem and no equally simple universal diagnostic exists.

That difference determines the chapter’s organization:

$$\begin{aligned} \text{pure state} &\longrightarrow \text{Schmidt data}, \\ \text{mixed state} &\longrightarrow \text{separability criteria}. \end{aligned}$$

The first question should therefore be whether the joint state is pure or mixed. The second should be whether the task is to detect, quantify, or operationally use entanglement. Those are different tasks and may require different tools.

Helpful background. State classification distinguishes separability from product form; partial trace gives local states; singular-value decomposition constructs Schmidt coefficients. The formulas below first assume finite-dimensional factors and a fixed physical split.

Choose a bipartite calculation

To learn how to…	Read
express a pure state in Schmidt form	Schmidt Decomposition
classify pure-state factorization	Schmidt Rank
quantify pure-state bipartite entanglement	Entanglement Entropy
compare spectral moments	Rényi Entropies
quantify total correlation	Mutual Information
use the special two-qubit mixed-state formula	Concurrence for Two Qubits
test partial-transpose positivity	Negativity and the PPT Criterion
detect entanglement through observables	Entanglement Witnesses
interpret entanglement as a resource	LOCC Preview

Pure States: Schmidt Data Solve the Problem

Every normalized pure state of a finite-dimensional bipartite system admits a Schmidt decomposition

$$\lvert\psi\rangle_{AB} = \sum_{r=1}^{R} \sqrt{\lambda_r}\, \lvert u_r\rangle_A \lvert v_r\rangle_B,$$

where

$$\lambda_r>0, \qquad \sum_{r=1}^{R}\lambda_r=1,$$

and the two Schmidt families are orthonormal. The positive integer $R$ is the Schmidt rank.

The reduced states are diagonal in the Schmidt bases:

$$\begin{aligned} \rho_A &= \sum_{r=1}^{R}\lambda_r \lvert u_r\rangle\langle u_r\rvert, \\ \rho_B &= \sum_{r=1}^{R}\lambda_r \lvert v_r\rangle\langle v_r\rvert. \end{aligned}$$

Thus the nonzero spectra of $\rho_A$ and $\rho_B$ agree. The Schmidt coefficients determine pure-state bipartite properties that are invariant under local unitaries. One nonzero coefficient means a product state; two or more mean entanglement.

Pure-State Quantifiers

Entanglement entropy

For a bipartite pure state, the entanglement entropy is

$$\begin{aligned} E(\psi) &= S(\rho_A) = S(\rho_B) \\ &= -\sum_r\lambda_r\log\lambda_r. \end{aligned}$$

It is zero for product states and reaches $\log d$ when the Schmidt spectrum is uniform over $d=\min(d_A,d_B)$ nonzero coefficients. The logarithm base sets the unit: base two gives bits and base $e$ gives nats.

Rényi entropies

For $\alpha>0$ with $\alpha\neq1$,

$$\begin{aligned} S_\alpha(\rho_A) &= \frac{1}{1-\alpha} \log\operatorname{Tr}(\rho_A^\alpha) \\ &= \frac{1}{1-\alpha} \log\sum_r\lambda_r^\alpha. \end{aligned}$$

Different orders weight the Schmidt spectrum differently. In particular,

$$S_2(\rho_A) = -\log\operatorname{Tr}(\rho_A^2).$$

The limit $\alpha\to1$ gives the von Neumann entropy under the usual continuity conditions. Rényi entropies are useful in many-body numerics, experiments, and replica constructions, but different orders are not interchangeable summaries.

A One-Parameter Pure-State Example

Consider

$$\begin{aligned} \lvert\psi(\theta)\rangle &= \cos\theta\,\lvert00\rangle + \sin\theta\,\lvert11\rangle, \\ 0&\leq\theta\leq\frac{\pi}{4}. \end{aligned}$$

This is already in Schmidt form, with

$$\lambda_1=\cos^2\theta, \qquad \lambda_2=\sin^2\theta.$$

Its main diagnostics are

$$\begin{aligned} R&= \begin{cases} 1,&\theta=0,\\ 2,&0<\theta\leq\pi/4, \end{cases} \\ E(\theta) &= -\cos^2\theta\log(\cos^2\theta) - \sin^2\theta\log(\sin^2\theta), \\ C(\theta)&=\sin(2\theta), \end{aligned}$$

where $C$ is the pure two-qubit concurrence. At $\theta=0$ the state is product. At $\theta=\pi/4$ it is a Bell state with a uniform Schmidt spectrum and maximal two-qubit entanglement.

This family is useful because rank, entropy, and concurrence agree on the endpoints while encoding different kinds of information between them.

Mixed States: Separate Detection from Quantification

For mixed states, local mixedness is inconclusive. A reduced state may be mixed because of entanglement, classical correlation, preparation uncertainty, or combinations of these.

No elementary analogue of Schmidt rank classifies arbitrary mixed states. Use a claim whose strength and scope match the available criterion.

Tool	Output	Correct scope
explicit separable decomposition	proves separability	sufficient in every finite bipartite dimension
negative partial transpose	proves entanglement	sufficient in every finite bipartite dimension
positive partial transpose	proves separability only in low dimensions	necessary in general; sufficient for two-by-two and two-by-three systems
negative witness expectation	proves entanglement	detects states separated by that witness
concurrence	detects and quantifies in the specified two-qubit setting	not a general high-dimensional measure
mutual information	quantifies total correlation	includes both classical and quantum correlation

The absence of a positive detection result does not usually prove separability. For example, a PPT state in higher dimension may still be entangled.

Mutual Information Measures Total Correlation

For a finite-dimensional bipartite state,

$$I(A{:}B) = S(\rho_A)+S(\rho_B)-S(\rho_{AB}).$$

It is nonnegative and vanishes exactly for product states in finite dimensions. For a pure bipartite state,

$$I(A{:}B) = 2S(\rho_A),$$

so it is twice the entanglement entropy. For a mixed state, it includes classical and quantum correlations and is not an entanglement measure.

The distinction is visible in two states with identical marginals. Using base-two logarithms,

Joint state	$S(A)$	$S(B)$	$S(AB)$	$I(A{:}B)$
diagonal mixture of matched bits	$1$	$1$	$1$	$1$
Bell state	$1$	$1$	$0$	$2$

The diagonal mixture is separable; the Bell state is entangled. Mutual information correctly distinguishes their total correlation but does not, in general, isolate the entangled part.

Local Operations and Classical Communication

LOCC protocols allow local quantum operations on $A$ and $B$, together with classical messages that can influence later local operations. They cannot create entanglement from a separable input.

For a separable state

$$\rho_{AB} = \sum_k p_k\, \rho_A^{(k)}\otimes\rho_B^{(k)},$$

each local branch maps product operators to product operators, and classical mixing preserves convex combinations. Consequently, every LOCC output obtained from a separable input remains separable.

For entanglement measures that are LOCC monotones, entanglement cannot increase on average over outcomes. A selected successful branch can have more entanglement than the input, at the cost of its success probability. This motivates the resource viewpoint: entanglement is something joint preparation may supply and LOCC cannot freely manufacture. Exact state-conversion criteria, asymptotic rates, catalytic effects, and protocol design are developed in Quantum Information; LOCC Preview introduces the allowed operations.

Exercises

Read Schmidt data directly

For

$$\lvert\psi\rangle = \sqrt{\frac12}\lvert00\rangle + \sqrt{\frac13}\lvert11\rangle + \sqrt{\frac16}\lvert22\rangle,$$

find the Schmidt rank, the reduced-state spectrum, and the entanglement entropy.

Solution

The state is already in Schmidt form. All three coefficients are nonzero, so $R=3$. Both reduced states have spectrum

$$\left\{ \frac12,\frac13,\frac16 \right\}.$$

Therefore

$$E(\psi) = -\frac12\log\frac12 -\frac13\log\frac13 -\frac16\log\frac16.$$

The logarithm base determines whether the answer is reported in bits or nats.

Partial transpose of a Bell state

Find the eigenvalues of the partial transpose of

$$\rho_{\Phi^+} = \lvert\Phi^+\rangle\langle\Phi^+\rvert, \qquad \lvert\Phi^+\rangle = \frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2}.$$

Use them to compute the negativity $\mathcal N=(\|\rho^{T_B}\|_1-1)/2$, the sum of the magnitudes of the negative eigenvalues. Negativity and PPT explains why a negative eigenvalue establishes entanglement.

Solution

In the computational basis,

$$\rho_{\Phi^+}^{T_B} = \frac12 \begin{pmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{pmatrix}.$$

Its eigenvalues are

$$\left\{ \frac12,\frac12,\frac12,-\frac12 \right\}.$$

The negative eigenvalue proves entanglement. The trace norm is $2$, so

$$\mathcal N = \frac{2-1}{2} = \frac12.$$

Local-unitary invariance of Schmidt data

Let

$$\lvert\psi'\rangle = (U_A\otimes U_B)\lvert\psi\rangle.$$

Show that $\lvert\psi\rangle$ and $\lvert\psi'\rangle$ have the same Schmidt coefficients.

Solution

If

$$\lvert\psi\rangle = \sum_r\sqrt{\lambda_r}\, \lvert u_r\rangle_A\lvert v_r\rangle_B,$$

then

$$\lvert\psi'\rangle = \sum_r\sqrt{\lambda_r}\, (U_A\lvert u_r\rangle_A) (U_B\lvert v_r\rangle_B).$$

Unitary maps preserve inner products, so the transformed local families remain orthonormal. This is a Schmidt decomposition with the same $\{\lambda_r\}$. Consequently Schmidt rank, entanglement entropy, and every function only of the Schmidt spectrum are local-unitary invariants.

References

• M. Horodecki, P. Horodecki, and R. Horodecki, “Separability of Mixed States: Necessary and Sufficient Conditions,” Physics Letters A 223, 1–8, 1996.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865–942, 2009.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.
• A. Peres, “Separability Criterion for Density Matrices,” Physical Review Letters 77, 1413–1415, 1996.
• E. Schmidt, “Zur Theorie der linearen und nichtlinearen Integralgleichungen,” Mathematische Annalen 63, 433–476, 1907.
• G. Vidal and R. F. Werner, “Computable Measure of Entanglement,” Physical Review A 65, 032314, 2002.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.
• W. K. Wootters, “Entanglement of Formation of an Arbitrary State of Two Qubits,” Physical Review Letters 80, 2245–2248, 1998.