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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 ]

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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.

This post has been migrated from (A51.SE)
answered Oct 26, 2011 by Peter Shor (790 points) [ no revision ]

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