Proof - Trace-preserving quantum operations are contractive

[Emil_M]

Homework Statement

Let \mathcal{E} be a trace-preserving quantum operation. Let \rho and \sigma
be density operators. Show that <br /> D(\mathcal{E}(\rho), \mathcal{E}(\sigma)) \leq D(\rho,\sigma)<br />

Homework Equations

D(\rho, \sigma) := \frac{1}{2} Tr \lvert \rho-\sigma\rvert
We can write \rho-\sigma=Q-S where Q and S are positive matrices with orthogonal support. We choose a projector P, such that <br /> D(\mathcal{E}(\rho), \mathcal{E}(\sigma))=Tr(P(\mathcal{E}(\rho)-\mathcal{E}(\sigma)))<br />

The Attempt at a Solution

<br /> \begin{align*}<br /> D(\rho, \sigma) &amp;=\frac{1}{2} Tr \lvert \rho-\sigma\rvert \\<br /> &amp;=\frac{1}{2} Tr \lvert Q-S\rvert \\<br /> &amp;=\frac{1}{2}(Tr(Q)+Tr(S))\\<br /> &amp;=\frac{1}{2}(Tr(\mathcal{E}(Q)+\mathcal{E}(S))\\<br /> &amp;=Tr(\mathcal{E}(Q))\;\; \Big(\text{since } Tr(Q)=Tr(S) \Big) \\<br /> &amp;\geq Tr(P\mathcal{E}(Q))<br /> \end{align*}<br />

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

Thanks for you help!

 

[Fightfish]

Let's say we have an operator \hat{H} with eigenbasis \{|i \rangle\}. Then we can conveniently express both the trace and the projector in terms of this eigenbasis: \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
Inserting these expressions into \mathrm{Tr}[\hat{P} \hat{H}] and using the fact that \hat{H} is diagonal in this basis should enable you to prove that \mathrm{Tr}[\hat{P} \hat{H}] \leq \mathrm{Tr}[\hat{H}]

 

[Emil_M]

Thanks