**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.