Entanglement cost of an amplitude-damping-channel Choi state

Problem

What is the entanglement cost of the Choi state of the qubit amplitude-damping channel

\begin{equation} \mathcal A_p(\rho)=A_0\rho A_0^\dagger+A_1\rho A_1^\dagger, \qquad 0\le p\le1? \tag{1} \end{equation}

The Kraus operators in Eq. (1) are

\begin{equation} \begin{aligned} A_0&=\lvert0\rangle\!\langle0\rvert +\sqrt{1-p}\,\lvert1\rangle\!\langle1\rvert =\begin{pmatrix}1&0\\0&\sqrt{1-p}\end{pmatrix},\\ A_1&=\sqrt p\,\lvert0\rangle\!\langle1\rvert =\begin{pmatrix}0&\sqrt p\\0&0\end{pmatrix}. \end{aligned} \tag{2} \end{equation}

In Eq. (2), \(p\) is the decay probability of the excited state. Let \(\lvert\Phi^+\rangle_{RA}=(\lvert00\rangle+\lvert11\rangle)/\sqrt2\). The normalized Choi state of the channel in Eq. (1) is

\begin{equation} \begin{aligned} \omega_p^{RB} &:=(\operatorname{id}_R\otimes\mathcal A_p) (\lvert\Phi^+\rangle\!\langle\Phi^+\rvert_{RA})\\ &=\frac12\Bigl[ \lvert00\rangle\!\langle00\rvert +\sqrt{1-p}\bigl(\lvert00\rangle\!\langle11\rvert +\lvert11\rangle\!\langle00\rvert\bigr) +(1-p)\lvert11\rangle\!\langle11\rvert +p\lvert10\rangle\!\langle10\rvert \Bigr]. \end{aligned} \tag{3} \end{equation}

Here the first and second entries in each ket in Eq. (3) label \(R\) and \(B\), respectively. Thus the question is to determine \(E_C(\omega_p)\), the asymptotic number of ebits per copy required to prepare many copies of \(\omega_p\) by local operations and classical communication.

Source

The question is implicit in the identity between entanglement cost and regularized entanglement of formation, together with Wootters’ single-copy formula applied to this Choi state [HHT01], [Woo98].

Progress

Wootters’ two-qubit formula, together with the concurrence of \(\omega_p\), gives the exact single-copy entanglement of formation

\begin{equation} C(\omega_p)=\sqrt{1-p}, \qquad E_F(\omega_p)=h_2\!\left(\frac{1+\sqrt p}{2}\right), \tag{4} \end{equation}

where \(h_2(x):=-x\log_2x-(1-x)\log_2(1-x)\), with \(0\log_2 0:=0\), [Woo98]. Equation (4) determines one copy exactly, but it does not by itself determine the asymptotic entanglement cost.
The entanglement cost equals the regularized entanglement of formation

\begin{equation} E_C(\omega_p) =\lim_{n\to\infty}\frac1n E_F(\omega_p^{\otimes n}) \le E_F(\omega_p) =h_2\!\left(\frac{1+\sqrt p}{2}\right) \tag{5} \end{equation}

[HHT01]. Consequently, Eq. (5) reduces the problem to evaluating the regularization for this particular family of Choi states.

Comment

The unresolved step is to decide whether regularization lowers the single-copy value in Eq. (4); equivalently, one must evaluate the entanglement of formation on tensor powers of this Choi state. Problems 2 and 5 concern the same channel family but ask about channel capacities, whereas the present problem asks about the entanglement cost of a bipartite state associated with the channel.

References

[Woo98]: W. K. Wootters, “Entanglement of Formation of an Arbitrary State of Two Qubits,” Physical Review Letters 80, 2245–2248 (1998).DOI arXiv
[HHT01]: P. M. Hayden, M. Horodecki, and B. M. Terhal, “The Asymptotic Entanglement Cost of Preparing a Quantum State,” Journal of Physics A: Mathematical and General 34, 6891–6898 (2001).DOI arXiv

Entanglement cost of an amplitude-damping-channel Choi state

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