Controlled Quantum Dynamics Quantum lost property: An operational meaning for the Hilbert-Schmidt product? Matthew F. Pusey, Terry Rudolph Introduction Suppose you are given a quantum system prepared, with equal probability, in state ρ or σ. After a measurement, your probability of guessing the correct state depends on the trace distance 1 δ = tr |ρ − σ| . 2 Suppose ρ is prepared by sampling from a distribution {pi} and P P preparing ρi, such that ρ = i piρi and similarly σ = j q j σj . If in each run you are given the classical information (i, j), your average probability of success depends on X ∆= piqj δ(ρi, σj ). i,j We prove that ∆≤ p 1 − tr(ρσ) and conjecture that for all ρ and σ, there exist decompositions that saturate this bound. tr(ρσ) is known as the Hilbert-Schmidt inner product. To our knowledge, this is the first time this inner product between two states has appeared in a simple operational setting. arXiv:1208.2550 An equivalent conjecture: existence of unbiased decompositions A decomposition into pure states ρ = |ψ i hψ | i i i and σj = φj φj saturates the bound if and only if ψi φj is independent of i, j, so we conjecture that such a decomposition exists for all ρ, σ. Such decompositions may be useful in other settings. The evidence Numerics indicate that such decompositions can always be found. We have analytic proofs for when the rank of ρ is at most 2, and for when ρ is the maximally mixed state (in the latter case the unbiased decompositions are simply mutually unbiased bases). Ideas for a proof or a counter-example are very welcome! Special case: qubit states Rotate so that ρ~ = (0, 0, r) and ~σ = (sx, 0, sz ). Then can take p ρ~1 = 0, 1 − r2, r , p ρ~2 = 0, − 1 − r2, r , q ~σ1 = 1 − s2z , 0, sz and q ~σ2 = − 1 − s2z , 0, sz . It’s easy to check that these are valid decompositions into pure states, and ρ~i · ~σj = rsz which is independent of i and j as required.
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