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 $D(\mathcal{E}(\rho), \mathcal{E}(\sigma)) \leq D(\rho,\sigma)$

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 $D(\mathcal{E}(\rho), \mathcal{E}(\sigma))=Tr(P(\mathcal{E}(\rho)-\mathcal{E}(\sigma)))$
[itex] [/itex]

The Attempt at a Solution

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

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