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Proof of "Entropy of preparation" in Von Neumann entropy

  1. Jan 23, 2015 #1
    How should I prove this?

    From John Preskill's quantum computation & quantum information lecture notes(chapter 5)

    If a pure state is drawn randomly from the ensemble{|φx〉,px}, so that the density matrix is ρ = ∑pxx〉<φx|
    Then, H(X)≥S(ρ)
    where H stands for Shannon entropy of probability {px} and S stands for Von Neumann entropy.
     
  2. jcsd
  3. Jan 23, 2015 #2

    jfizzix

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    If you can find it, on page 519 in Nielsen and Chuang's (Quantum Computation and quantum information, 10th anniversary edition), they show how to do it.

    In short, you express [itex]|\phi_{x}\rangle[/itex] in terms of the eigenstates of the density matrix [itex]\rho[/itex]. Since the entropy of a mixture of states is larger than the corresponding mixture of entropies, you can eventually find:
    [itex]S(\rho)\leq H(X)+\sum_{x}p_{x}S(\rho_{x})[/itex]
    Since the states in the ensemble are pure, this would give you your result.
     
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