Universal finite truncation of quantum and private capacities

Problem

Do there exist channel-independent finite integers \(m_Q\) and \(m_P\) that determine, respectively, the quantum capacity and the private classical capacity of every finite-dimensional quantum channel? Let \(V_{A\to BE}\) be a Stinespring isometry for a channel and its complement, and define coherent information by

\begin{equation} \mathcal N_{A\to B}(\rho):=\operatorname{Tr}_{E}[V\rho V^\dagger], \qquad \mathcal N^c_{A\to E}(\rho):=\operatorname{Tr}_{B}[V\rho V^\dagger], \qquad I_{\rm c}(\rho,\mathcal N):= S(\mathcal N(\rho))-S(\mathcal N^c(\rho)). \tag{1} \end{equation}

For each \(m\geq1\), use Eq. (1) to set

\begin{equation} Q^{(m)}(\mathcal N) :=\frac1m\max_{\rho_{A^m}} I_{\rm c}(\rho_{A^m},\mathcal N^{\otimes m}), \qquad Q(\mathcal N):=\sup_{m\geq1}Q^{(m)}(\mathcal N). \tag{2} \end{equation}

For an ensemble \(\{p_x,\rho_x^{A^m}\}\), let its joint channel output be

\begin{equation} \omega^{XB^mE^m} :=\sum_x p_x|x\rangle\!\langle x|^X\otimes V^{\otimes m}\rho_x^{A^m}(V^\dagger)^{\otimes m}. \tag{3} \end{equation}

In terms of the state in Eq. (3), define

\begin{equation} P^{(m)}(\mathcal N) :=\frac1m\max_{\{p_x,\rho_x^{A^m}\}} \bigl[I(X;B^m)_\omega-I(X;E^m)_\omega\bigr], \qquad P(\mathcal N):=\sup_{m\geq1}P^{(m)}(\mathcal N). \tag{4} \end{equation}

The question is whether there exist finite integers \(m_Q\) and \(m_P\), independent of the channel dimensions and of \(\mathcal N\), such that

\begin{equation} Q(\mathcal N)=Q^{(m_Q)}(\mathcal N) \quad\text{and}\quad P(\mathcal N)=P^{(m_P)}(\mathcal N) \quad\text{for every finite-dimensional }\mathcal N. \tag{5} \end{equation}

Source

Wilde asks whether an entropic formula evaluated on some finite tensor power could replace the regularizations in Eqs. (2) and (4). Equation (5) records the uniform, channel-independent interpretation of that question [Wil17].

Progress

Cubitt et al. proved that for every positive integer \(n\) there is a finite-dimensional channel \(\mathcal M_n\) such that

\begin{equation} Q^{(n)}(\mathcal M_n)=0 \quad\text{while}\quad Q(\mathcal M_n)>0. \tag{6} \end{equation}

Equation (6) rules out every proposed universal value of \(m_Q\) [CEM+15].
Elkouss and Strelchuk proved that for every \(n\) there is a channel \(\mathcal N_n\) for which, for all \(1\leq k<n\),

\begin{equation} P^{(k)}(\mathcal N_n) <Q^{(k+1)}(\mathcal N_n) \leq P(\mathcal N_n). \tag{7} \end{equation}

Choosing \(n>m_P\) in Eq. (7) rules out every universal private-information truncation \(m_P\) [ES15].

Comment

The answer to Eq. (5) is negative for both capacities. These counterexamples exclude only a channel-independent block length for the standard coherent- and private-information regularizations; they do not exclude finite stabilization for a particular channel or a different exact capacity formula.

References

[Wil17]: M. M. Wilde, Quantum Information Theory, 2nd ed., Cambridge University Press (2017), Sec. 26.6.DOI arXiv
[CEM+15]: T. S. Cubitt, D. Elkouss, W. Matthews, M. Ozols, D. Pérez-García, and S. Strelchuk, “Unbounded Number of Channel Uses May Be Required to Detect Quantum Capacity,” Nature Communications 6, 6739 (2015).DOI arXiv
[ES15]: D. Elkouss and S. Strelchuk, “Superadditivity of Private Information for Any Number of Uses of the Channel,” Physical Review Letters 115, 040501 (2015).DOI arXiv

Universal finite truncation of quantum and private capacities

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