Concurrence for Two Qubits

Concurrence is a special entanglement measure for two qubits. It is valuable because it gives a closed-form answer for all two-qubit mixed states, a rare feature in mixed-state entanglement theory.

For pure two-qubit states, concurrence is a compact way to encode the Schmidt coefficients. For mixed two-qubit states, Wootters’ formula solves a convex-roof problem that is usually difficult in larger systems.

This page treats concurrence as a concrete diagnostic for the two-qubit case. Full operational entanglement measures, distillation, and resource-conversion theory belong to quantum information.

Why Two Qubits Are Special

For bipartite pure states, the Schmidt decomposition gives a direct entanglement classification. A two-qubit pure state has at most two Schmidt coefficients:

$$\lvert\psi\rangle = s_0\lvert0_A0_B\rangle + s_1\lvert1_A1_B\rangle, \qquad s_0,s_1\ge 0, \qquad s_0^2+s_1^2=1.$$

It is product when one coefficient vanishes and maximally entangled when

$$s_0=s_1=\frac{1}{\sqrt2}.$$

For mixed states, separability is harder because one must ask whether a density operator can be written as a convex mixture of product states. Two qubits are special because concurrence gives a closed expression for this mixed-state entanglement problem.

Pure-State Concurrence

Write a normalized two-qubit pure state in the computational product basis:

$$\lvert\psi\rangle = a\lvert00\rangle + b\lvert01\rangle + c\lvert10\rangle + d\lvert11\rangle.$$

The pure-state concurrence is

$$C(\psi) = 2\lvert ad-bc\rvert.$$

Equivalently, if the coefficients are arranged as the matrix

$$M_\psi = \begin{pmatrix} a & b\\ c & d \end{pmatrix},$$

then

$$C(\psi) = 2\lvert\det M_\psi\rvert.$$

This makes the product-state criterion transparent. A two-qubit vector is product exactly when the coefficient matrix has rank one, which is equivalent to

$$\det M_\psi=0.$$

Thus $C(\psi)=0$ for product states and $C(\psi)>0$ for entangled pure states.

Relation to Reduced States

Let

$$\rho_A = \operatorname{Tr}_B \lvert\psi\rangle\langle\psi\rvert.$$

For a pure two-qubit state,

$$C(\psi) = 2\sqrt{\det\rho_A}.$$

Since $\rho_A$ is a $2\times2$ density matrix, this can also be written as

$$C(\psi) = \sqrt{ 2\left(1-\operatorname{Tr}\rho_A^2\right) }.$$

If the Schmidt coefficients are $s_0$ and $s_1$, then

$$C(\psi) = 2s_0s_1.$$

The value lies in the interval

$$0\le C(\psi)\le 1.$$

The lower endpoint means product; the upper endpoint means maximally entangled.

Spin-Flip Form

The standard compact definition uses the two-qubit spin flip. Let complex conjugation be taken in the computational basis and define

$$\lvert\widetilde\psi\rangle = (\sigma_y\otimes\sigma_y) \lvert\psi^*\rangle.$$

Then

$$C(\psi) = \left\lvert \langle\psi\vert\widetilde\psi\rangle \right\rvert.$$

For the coefficient convention above, this equals $2\lvert ad-bc\rvert$.

The spin-flip definition is basis-dependent in its intermediate notation, but the final concurrence is invariant under local unitaries. It is measuring entanglement across the fixed two-qubit split, not a preferred computational basis.

Examples

A product state such as

$$\lvert00\rangle$$

has $a=1$ and $b=c=d=0$, so

$$C=0.$$

The Bell state

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \left( \lvert00\rangle+\lvert11\rangle \right)$$

has $a=d=1/\sqrt2$ and $b=c=0$, so

$$C(\Phi^+)=1.$$

For the partially entangled state

$$\lvert\psi_\theta\rangle = \cos\theta\,\lvert00\rangle + e^{i\phi}\sin\theta\,\lvert11\rangle, \qquad 0\le\theta\le\frac{\pi}{4},$$

the concurrence is

$$C(\psi_\theta) = \sin(2\theta).$$

The phase $\phi$ does not change the concurrence, because it can be removed by a local phase rotation.

Mixed-State Concurrence

For a two-qubit density operator $\rho$, define the spin-flipped density operator

$$\widetilde\rho = (\sigma_y\otimes\sigma_y) \rho^* (\sigma_y\otimes\sigma_y),$$

where complex conjugation is again taken in the computational basis.

Let $r_1,r_2,r_3,r_4$ be the square roots of the eigenvalues of

$$\rho\widetilde\rho$$

arranged in nonincreasing order:

$$r_1\ge r_2\ge r_3\ge r_4\ge 0.$$

Wootters’ two-qubit concurrence is

$$C(\rho) = \max\{0,\ r_1-r_2-r_3-r_4\}.$$

The matrix $\rho\widetilde\rho$ is not generally Hermitian, but its eigenvalues entering this formula are nonnegative real numbers for physical two-qubit states. For numerical work, it is often more stable to use the Hermitian positive matrix

$$\sqrt{\sqrt{\rho}\,\widetilde\rho\,\sqrt{\rho}},$$

whose eigenvalues are the same numbers $r_i$.

Werner-State Example

A two-qubit Werner state can be written

$$\rho_W(p) = p\lvert\Psi^-\rangle\langle\Psi^-\rvert + \frac{1-p}{4}I, \qquad 0\le p\le 1.$$

Its concurrence is

$$C(\rho_W) = \max\left\{ 0,\frac{3p-1}{2} \right\}.$$

Thus the state is detected as entangled exactly when

$$p>\frac13.$$

At $p=1$, the state is the singlet Bell state and has concurrence $1$. At $p=0$, the state is maximally mixed and has concurrence $0$.

Relation to Entanglement of Formation

For two qubits, concurrence is closely tied to entanglement of formation. If logarithms are base $2$, define the binary entropy

$$h_2(x) = -x\log_2 x -(1-x)\log_2(1-x).$$

For a two-qubit state,

$$E_F(\rho) = h_2\left( \frac{1+\sqrt{1-C(\rho)^2}}{2} \right).$$

This relation is important historically and operationally, but this page uses it only as orientation. Entanglement of formation and other resource measures require a broader quantum-information treatment.

What Concurrence Does and Does Not Tell You

Concurrence is an entanglement measure for two qubits. In that setting:

• $C(\rho)=0$ exactly for separable states;
• $C(\rho)=1$ for maximally entangled Bell states;
• local unitaries do not change $C$;
• mixing generally reduces $C$;
• $C$ gives a closed route to two-qubit entanglement of formation.

But concurrence should not be overread:

• it is not a measure of total correlation;
• it is not defined by the same simple formula for qubit-qutrit or higher-dimensional mixed states;
• it does not replace separability criteria such as partial transpose;
• it does not by itself describe an experimental measurement protocol;
• it is not the canonical language for multipartite entanglement, where inequivalent entanglement types appear.

Use concurrence when the system is genuinely a two-qubit bipartite system and a scalar mixed-state entanglement measure is wanted.

Common Mistakes

• Applying the two-qubit Wootters formula to larger bipartite systems.
• Confusing concurrence with mutual information. A separable mixed state can have nonzero mutual information but zero concurrence.
• Forgetting to order the four numbers $r_i$ before using the mixed-state formula.
• Treating the spin-flip operation as a physical time evolution rather than a mathematical construction.
• Assuming that a large classical mixture of Bell-state preparations must have large concurrence.
• Ignoring the subsystem decomposition before calling a pair of degrees of freedom “two qubits.”

Cross-Links

• Entangled States
• Separable Mixed States
• Entanglement Measures
• Bell States
• Local Unitary Equivalence
• Classical Correlation versus Entanglement
• Entanglement Depends on a Decomposition
• Reduced Density Operators
• Schmidt Decomposition
• Schmidt Rank
• Entanglement Entropy
• Renyi Entropies
• Mutual Information
• Negativity and PPT Criterion
• Entanglement Witnesses
• LOCC Preview
• Monogamy of Entanglement
• Density Operators
• Pauli Matrices

References

• S. Hill and W. K. Wootters, “Entanglement of a Pair of Quantum Bits,” Physical Review Letters 78, 5022-5025, 1997.
• W. K. Wootters, “Entanglement of Formation of an Arbitrary State of Two Qubits,” Physical Review Letters 80, 2245-2248, 1998.
• C. H. Bennett, D. P. DiVincenzo, J. A. Smolin, and W. K. Wootters, “Mixed-State Entanglement and Quantum Error Correction,” Physical Review A 54, 3824-3851, 1996.
• V. Coffman, J. Kundu, and W. K. Wootters, “Distributed Entanglement,” Physical Review A 61, 052306, 2000.
• 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, Cambridge University Press, 2010.

Exercises

1. Coefficient determinant test. Compute the concurrence of

$$\lvert\psi\rangle = \frac{1}{\sqrt{10}} \left( 2\lvert00\rangle + \lvert01\rangle + 2\lvert10\rangle + \lvert11\rangle \right).$$

Is the state product or entangled?

Solution

Here

$$a=\frac{2}{\sqrt{10}}, \quad b=\frac{1}{\sqrt{10}}, \quad c=\frac{2}{\sqrt{10}}, \quad d=\frac{1}{\sqrt{10}}.$$

Thus

$$ad-bc = \frac{2}{10} - \frac{2}{10} = 0.$$

Therefore $C=0$, and the state is product. Indeed,

$$\lvert\psi\rangle = \frac{1}{\sqrt5} \left(2\lvert0\rangle+\lvert1\rangle\right) \otimes \frac{1}{\sqrt2} \left(\lvert0\rangle+\lvert1\rangle\right).$$

2. Partially entangled state. For

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

show that $C=\sin(2\theta)$ for $0\le\theta\le\pi/4$.

Solution

The only nonzero coefficients are

$$a=\cos\theta, \qquad d=\sin\theta.$$

Therefore

$$C = 2\lvert ad-bc\rvert = 2\cos\theta\sin\theta = \sin(2\theta).$$

3. Werner-state threshold. Use

$$C(\rho_W) = \max\left\{ 0,\frac{3p-1}{2} \right\}$$

to determine the concurrence for $p=1/4$, $p=1/2$, and $p=1$.

Solution

For $p=1/4$,

$$\frac{3p-1}{2} = \frac{3/4-1}{2} = -\frac18,$$

so $C=0$.

For $p=1/2$,

$$\frac{3p-1}{2} = \frac{3/2-1}{2} = \frac14,$$

so $C=1/4$.

For $p=1$,

$$C=1.$$

4. Concurrence versus total correlation. Can a two-qubit state have zero concurrence but nonzero mutual information?

Solution

Yes. The separable classically correlated state

$$\rho = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert$$

has zero concurrence because it is separable. It has nonzero mutual information because measurements in the computational basis are perfectly correlated.