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

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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.
Tang, Zhu, Bai, and Wang determine the entanglement cost for the full damping range in their preprint of 23 September 2026 [TZBW26], Theorem 6.3, Eqs. (6.12)–(6.13). In the notation of Eq. (3), their result is

\begin{equation} \begin{aligned} E_C(\omega_p)&=E_F(\omega_p) =h_2\!\left(\frac{1+\sqrt p}{2}\right), &&0\le p\le1,\\ E_F(\omega_p^{\otimes n})&=nE_F(\omega_p), &&n\ge1. \end{aligned} \tag{6} \end{equation}

Here \(n\) is an integer, and the cost is measured in ebits per copy under LOCC with vanishing preparation error. Equation (6) shows that regularization in Eq. (5) does not lower the single-copy value. The proof uses their strong superadditivity criterion (Theorem 4.1 and Corollary 4.4): a two-qubit state with a product vector in its kernel has additive entanglement of formation with every finite-dimensional bipartite partner. The Choi state here satisfies \(\omega_p\lvert01\rangle=0\).

Comment

Solved by the Choi-state entanglement-cost result reported in [TZBW26], Theorem 6.3, as summarized in Eq. (6). The cited source is a preprint. The former regularization gap is closed for every \(0\le p\le1\); the endpoint costs are \(E_C(\omega_0)=1\) and \(E_C(\omega_1)=0\) ebits per copy.

The problems of capacity-achieving codes for amplitude damping and two-way quantum capacity of the amplitude-damping channel 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
[TZBW26]: Z. Tang, C. Zhu, G. Bai, and X. Wang, “Classical Capacity and Entanglement Cost of the Amplitude Damping Channel,” arXiv preprint (2026), version 1, 23 September 2026.arXiv

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