Entanglement of Formation

For pure states the entanglement entropy answered everything. The entanglement of formation $E_F$ extends that answer to mixed states by asking: of all the ways to build $\rho$ out of pure states, what is the cheapest, measured in average pure-state entanglement?

The convex-roof construction

A mixed state $\rho$ has infinitely many pure-state ensembles that realize it:

\rho = \sum_i p_i\,|\psi_i\rangle\langle\psi_i|, \qquad p_i \geq 0,\ \sum_i p_i = 1.

Each pure member $|\psi_i\rangle$ has a well-defined entanglement, its entanglement entropy $E(|\psi_i\rangle) = S(\operatorname{Tr}_B |\psi_i\rangle\langle\psi_i|)$ . The average entanglement of a particular ensemble is $\sum_i p_i E(|\psi_i\rangle)$ . But the same $\rho$ can be decomposed in many ways, and these averages differ. The entanglement of formation takes the most economical decomposition — the infimum over all ensembles:

E_F(\rho) = \min_{\{p_i,\,|\psi_i\rangle\}} \sum_i p_i\,E(|\psi_i\rangle),

with the minimum over all pure-state ensembles consistent with $\rho$ . This is called the convex roof of the pure-state entanglement: $E_F$ is the largest convex function that agrees with the entanglement entropy on pure states.

Why this is the right notion of cost

$E_F$ has a clean operational reading. Suppose Alice and Bob want to manufacture many copies of $\rho$ using only shared Bell pairs and LOCC. The number of Bell pairs consumed per copy, in the asymptotic limit, is the entanglement cost $E_C(\rho)$ , and it is given by the regularized entanglement of formation,

E_C(\rho) = \lim_{n\to\infty} \frac{1}{n}\,E_F\!\big(\rho^{\otimes n}\big).

If $E_F$ were additive — $E_F(\rho^{\otimes n}) = n\,E_F(\rho)$ — then the cost would be just $E_F$ itself. Additivity was long conjectured but is now known to fail in general (Hastings' counterexample to additivity of minimum output entropy). So $E_F$ is an upper bound on the cost per copy, and the true cost is its regularization.

The two-qubit miracle: a closed form

In general the convex-roof minimization is intractable. The remarkable exception is two qubits, where Wootters found a closed formula. Define the concurrence $C(\rho)$ (built in the next two lessons). Then

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),

where $h$ is the binary entropy. $E_F$ is a monotonically increasing function of $C$ on $[0,1]$ , so the two carry the same ordering information: $C = 0 \Rightarrow E_F = 0$ (separable), and $C = 1 \Rightarrow E_F = 1$ bit (maximally entangled). This is why concurrence is so useful — it is an algebraic stand-in for the otherwise hard-to-compute $E_F$ .

Properties at a glance

$E_F(\rho) = 0$ iff $\rho$ is separable (it is faithful).
$E_F$ is convex and LOCC-monotone, so it is a bona fide entanglement measure.
On pure states it reduces to the entanglement entropy, as required.
It upper-bounds the distillable entanglement: $E_D(\rho) \leq E_C(\rho)$ , and the gap can be strictly positive — the hallmark of bound entanglement.

The next lesson makes concurrence concrete and computable, turning this abstract minimization into an arithmetic you can run by hand.