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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.
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.
See also
Last modified:
Saturday, February 4, 2017 - 00:16
