Entanglement cost is an entanglement measure that aims to quantify how many ebits are required to prepare a copy of a state using only LOCC operations. Many copies can be prepared at the same time and the entanglement cost therefore quantifies how many ebits are required per copy of the state. The preparation is allowed to be approximate, as long as the approximation can be made arbitrary good by preparing many copies at a time.
Formal definition
Let $P_+$ be the projector onto a Bell state, $P_+ := |\Phi^+\rangle \langle \Phi^+|$, where $|\Phi^+\rangle = (|00\rangle + |11 \rangle)/\sqrt{2}$. The entanglement cost aims to quantify the rate m/n at which it is possible to convert $P_+^{\otimes m}$ into $\rho^{\otimes n}$ with a LOCC operation $\Lambda$. Since it is usually impossible to perform this exactly, we settle for $\Lambda(P_+^{\otimes m}) \approx \rho^{\otimes n}$ and let the quality of the approximation be quantified by a distance measure $D(\Lambda(P_+^{\otimes m}), \rho^{\otimes n})$ which can be either the Bures distance, the trace distance or another suitable distance. The entanglement cost EC is then the infimum of all possible rates m/n such that the approximation can be made arbitrarily good by choosing m and n large enough. This can be formulated mathematically as [quant-ph/0008134]:
- $E_c(\rho) = \inf \{ E \mid \forall \epsilon > 0, \delta > 0, \exists m, n, \Lambda, |E-\frac{m}{n}| \leq \delta \text{ and } D(\Lambda(P_+^{\otimes m}), \rho^{\otimes n}) \leq \epsilon\}.$
Relations to other entanglement measures
The entanglement cost has been show to be equal to the regularization of the entanglement of formation,
$$ E_C(\rho) = \lim_{n \to \infty} \frac{1}{n} E_f(\rho^{\otimes n}). $$ If the entanglement of formation turns out to be additive, the entanglement cost will be equal to the entanglement of formation.