Concurrence

Concurrence is the algebraic engine behind two-qubit entanglement of formation. It turns the intractable convex-roof minimization into a finite computation involving a "spin-flip" operation and four eigenvalues — and for pure states it collapses to a one-line determinant.

The spin-flip and the tilde state

Concurrence is built from time reversal of spins. The single-qubit spin flip is the Pauli $Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$ , which sends $|0\rangle \leftrightarrow |1\rangle$ with a phase. For a two-qubit density operator $\rho$ , define the tilde (spin-flipped) state

\tilde\rho = (Y \otimes Y)\,\rho^{*}\,(Y \otimes Y),

where $\rho^{*}$ is the complex conjugate in the computational basis. Both qubits are flipped and the amplitudes conjugated.

Wootters' formula

Form the (non-Hermitian) product $R = \rho\,\tilde\rho$ . Although $R$ is not Hermitian, its eigenvalues are real and non-negative. Let $\lambda_1 \geq \lambda_2 \geq \lambda_3 \geq \lambda_4$ be the square roots of those eigenvalues, sorted in decreasing order. The concurrence is

C(\rho) = \max\{0,\ \lambda_1 - \lambda_2 - \lambda_3 - \lambda_4\}.

The $\max$ with $0$ is what makes separable states report exactly zero: for such states the largest square-root eigenvalue never exceeds the sum of the other three. Concurrence ranges over $[0,1]$ , with $C = 1$ for the Bell states.

Once $C$ is known, the entanglement of formation follows from the closed form of the previous lesson,

E_F(\rho) = h\!\left(\frac{1 + \sqrt{1 - C^2}}{2}\right), \qquad h(x) = -x\log_2 x - (1-x)\log_2(1-x).

The pure-state shortcut

For a pure state $|\psi\rangle = a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$ the machinery collapses dramatically. The spin-flipped ket is $|\tilde\psi\rangle = (Y\otimes Y)|\psi^{*}\rangle$ , and the concurrence equals the modulus of the overlap,

C(|\psi\rangle) = \big|\langle\psi|\tilde\psi\rangle\big| = 2\,|ad - bc|.

The quantity $ad - bc$ is the determinant of the $2\times 2$ matrix of amplitudes $\begin{psmallmatrix} a & b \\ c & d \end{psmallmatrix}$ . A pure state is a product state precisely when this matrix has rank $1$ , i.e. determinant $0$ — and then $C = 0$ . So concurrence directly measures how far the amplitude matrix is from being a rank-one (separable) matrix.

Worked example

Take $|\psi\rangle = \cos\theta\,|00\rangle + \sin\theta\,|11\rangle$ . Here $a = \cos\theta$ , $d = \sin\theta$ , and $b = c = 0$ , so

C = 2|ad - bc| = 2\cos\theta\sin\theta = |\sin 2\theta|.

At $\theta = \pi/4$ this is $C = 1$ : the state is the Bell state $\tfrac{1}{\sqrt2}(|00\rangle + |11\rangle)$ , maximally entangled. At $\theta = 0$ it is the product state $|00\rangle$ with $C = 0$ . The exercise evaluates this at $\theta = \pi/5$ .

Try it

Run your code to see the quantum state.