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Proof - Trace-preserving quantum operations are contractive

  1. Dec 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex] \mathcal{E}[/itex] be a trace-preserving quantum operation. Let [itex] \rho [/itex] and [itex]\sigma [/itex]
    be density operators. Show that [itex]
    D(\mathcal{E}(\rho), \mathcal{E}(\sigma)) \leq D(\rho,\sigma)
    [/itex]
    2. Relevant equations
    [itex]D(\rho, \sigma) := \frac{1}{2} Tr \lvert \rho-\sigma\rvert [/itex]
    We can write [itex] \rho-\sigma=Q-S [/itex] where [itex] Q [/itex] and [itex] S[/itex] are positive matrices with orthogonal support. We choose a projector [itex] P [/itex], such that [itex]
    D(\mathcal{E}(\rho), \mathcal{E}(\sigma))=Tr(P(\mathcal{E}(\rho)-\mathcal{E}(\sigma)))
    [/itex]
    [itex] [/itex]
    3. The attempt at a solution
    [itex]
    \begin{align*}
    D(\rho, \sigma) &=\frac{1}{2} Tr \lvert \rho-\sigma\rvert \\
    &=\frac{1}{2} Tr \lvert Q-S\rvert \\
    &=\frac{1}{2}(Tr(Q)+Tr(S))\\
    &=\frac{1}{2}(Tr(\mathcal{E}(Q)+\mathcal{E}(S))\\
    &=Tr(\mathcal{E}(Q))\;\; \Big(\text{since } Tr(Q)=Tr(S) \Big) \\
    &\geq Tr(P\mathcal{E}(Q))
    \end{align*}
    [/itex]

    Why is the last step valid? Why can a projector never increase the trace?

    Thanks for you help!
     
  2. jcsd
  3. Dec 30, 2015 #2
    Let's say we have an operator [itex]\hat{H}[/itex] with eigenbasis [itex]\{|i \rangle\}[/itex]. Then we can conveniently express both the trace and the projector in terms of this eigenbasis: [tex]\mathrm{Tr}[\hat{O}] = \sum_{i} \langle i | \hat{O} | i \rangle \qquad \qquad \hat{P} = \sum_{i,j} a_{i}a_{j}^{*} |i\rangle \langle j| \quad \mathrm{where } \sum_{i} |a_{i}|^{2} = 1[/tex]
    Inserting these expressions into [itex]\mathrm{Tr}[\hat{P} \hat{H}][/itex] and using the fact that [itex]\hat{H}[/itex] is diagonal in this basis should enable you to prove that [tex]\mathrm{Tr}[\hat{P} \hat{H}] \leq \mathrm{Tr}[\hat{H}] [/tex]
     
  4. Dec 30, 2015 #3
    Thanks
     
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