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.

Helpful background. Schmidt decomposition identifies pure-state entanglement; separable mixed states explains why a mixed-state measure must account for every ensemble decomposition.

Concurrence and two Schmidt coefficients

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 mixed state, averaging pure-state concurrence over an arbitrarily chosen ensemble would depend on the preparation description. Define the convex roof by minimizing over all pure-state ensembles of the same density operator:

$$C(\rho)=\min_{\rho=\sum_k p_k|\psi_k\rangle\langle\psi_k|} \sum_k p_k C(\psi_k).$$

An explicit ensemble supplies an upper bound, not usually the minimum. Wootters solved this optimization for two qubits. 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 $r_i$.

Bell-diagonal states make the formula transparent

Let $\rho=\sum_{k=1}^4 q_k|B_k\rangle\langle B_k|$ in the Bell basis. Each Bell projector is unchanged by the spin flip, so $\widetilde\rho=\rho$. The eigenvalues of $\rho\widetilde\rho=\rho^2$ are $q_k^2$, and their nonnegative square roots are $q_k$. Therefore

$$C(\rho)=\max\{0,2q_{\max}-1\}.$$

The equal mixture of $|\Phi^+\rangle$ and $|\Phi^-\rangle$ has $q_{\max}=1/2$ and zero concurrence, although both listed components are maximally entangled. Expanding it gives the separable state $\tfrac12|00\rangle\langle00|+\tfrac12|11\rangle\langle11|$. The minimization over ensembles is essential.

Werner-State Example

A two-qubit Werner state can be written

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

The singlet has probability $q_{\rm singlet}=(1+3p)/4$ and each other Bell vector has probability $(1-p)/4$. The Bell-diagonal formula therefore gives

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

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,

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

Here $E_F$ is the minimum ensemble-average pure-state entanglement entropy, not the entropy of $\rho_A$ itself. The formula gives $E_F$ in bits. It does not identify concurrence or $E_F$ with the distillable entanglement of an arbitrary mixed state; different mixed-state resource measures need not agree. See entanglement measures.

What mixing and local changes do

Concurrence is invariant under local unitaries and is zero exactly for separable two-qubit states. Its convexity means

$$C\!\left(\sum_k p_k\rho_k\right)\le\sum_k p_k C(\rho_k).$$

This compares the mixture with the average input concurrence; it does not say a mixture has less concurrence than every component. The same four-eigenvalue formula does not apply to higher-dimensional mixed states. In a multipartite system, concurrence of a selected two-qubit reduction describes only that pair, as discussed in monogamy of entanglement.

Numerical and interpretation checks

Use a normalized positive semidefinite two-qubit density matrix, and sort the four nonnegative numbers before forming the difference. For floating-point calculations, the Hermitian square-root construction above avoids mistaking small non-Hermitian eigensolver artifacts for complex physical eigenvalues. Only clip negative values at a tolerance justified by numerical error; a substantially nonpositive input is not a valid density matrix.

The spin flip includes complex conjugation and is a mathematical construction, not a universal physical quantum channel. It also depends on the chosen subsystem split, even though local changes of basis leave the final concurrence unchanged.

References

• S. Hill and W. K. Wootters, “Entanglement of a pair of quantum bits”, Physical Review Letters 78, 5022–5025 (1997).
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum entanglement”, Reviews of Modern Physics 81, 865–942 (2009).
• W. K. Wootters, “Entanglement of formation of an arbitrary state of two qubits”, Physical Review Letters 80, 2245–2248 (1998).

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,

$$\lvert\psi\rangle = \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2} \otimes \frac{2\lvert0\rangle+\lvert1\rangle}{\sqrt5}.$$

2. A complex relative phase. Compute the concurrence of $|\psi\rangle=(|00\rangle+i|01\rangle+|11\rangle)/\sqrt3$, then recover the eigenvalues of $\rho_A$ from it.

Solution

$ad-bc=1/3$, so $C=2/3$. A qubit density matrix has trace one and determinant $C^2/4=1/9$, hence eigenvalues $(1\pm\sqrt{1-C^2})/2=(3\pm\sqrt5)/6$. The relative phase affects matrix entries but leaves these particular Schmidt probabilities unchanged.

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.