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What is the Holevo-Schumacher-Westmoreland capacity of a Pauli channel?

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Suppose you are given an $n$-qubit quantum channel defined as $\mathcal{E}(\rho) = \sum_{i} p_i X_i \rho X_i^\dagger$, where $X_i$ denotes an $n$-fold tensor product of Pauli matrices and $\{p_i\}$ is a probability distribution. The Holevo-Schumacher-Westmoreland capacity of the channel is defined by $$ \chi(\mathcal{E}) = \max_{\{q_j, \rho_j\}} \left[S\left(\sum_j q_j \rho_j\right) -\sum_j q_j S\left(\rho_j\right) \right], $$ where $S$ denotes the von Neumann entropy of a density matrix (see, for example, http://theory.physics.helsinki.fi/~kvanttilaskenta/Lecture13.pdf). Is it known how to calculate this number as a function of $p_i$ and $n$?

This post has been migrated from (A51.SE)
asked Oct 26, 2011 in Theoretical Physics by Kernel (125 points) [ no revision ]

1 Answer

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Finding the HSW capacity is an optimization problem which I believe is moderately tractable. There is an iterative numerical method outlined in this paper of mine ("Capacities of quantum channels and how to find them."). A different, although somewhat similar, method was detailed in the paper "Qubit channels which require four inputs to achieve capacity: implications for additivity conjectures" by Masahito Hayashi, Hiroshi Imai, Keiji Matsumoto, Mary Beth Ruskai and Toshiyuki Shimono. If the number of qubits $n$ is not quite small, however, the high dimensionality of the space is going to keep these techniques from working.

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answered Oct 26, 2011 by Peter Shor (780 points) [ no revision ]

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