Why do completely positive maps satisfy ${\rm Tr}[\Psi(\rho)_++\Psi(-\rho)_+]\leq{\rm Tr}[\Psi(\rho_+)]+{\rm Tr}[\Psi((-\rho)_+)]$?

Question

I am studying a paper of M. Plenio, "Logarithmic Negativity: A Full Entanglement Monotone That is not Convex", PRL 2005 [arXiv:quant-ph/0505071].

In the paper, I do not fully understand the Eq.$(7)$. He said that

Employing linearity of operation $\Psi$ and the fact that $\Psi$ maps positive states to positive states.

And he found

$$\text{Tr}\big[\Psi(\rho)_+ +\Psi(-\rho)_+\big]\leq \text{Tr}\big[\Psi(\rho_+)\big]+\text{Tr}\big[\Psi((-\rho)_+)\big]\,,$$

while $\Psi$ is a completely positive PPT-operation that maps $\rho$ to $\sigma=\Psi(\rho)$ deterministically, and $A_+$ ($A_-$) denotes the positive (negative) part of the operator $A$.

I do not understand which property of the positive and linear mapping guarantees this inequality. If anyone answer this, I am really appreciated.

Answer 1

The inequality can be proved as follows.

First, let $P$ and $Q$ be any two positive semidefinite operators, and consider the operator $\Psi(P - Q)$. Because $\Psi$ is positive, this is a Hermitian operator. In particular, by the positivity of $\Psi$, we can write this operator as a difference between two positive semidefinite operators: $$ \Psi(P - Q) = \Psi(P) - \Psi(Q). $$ This expression may, however, not represent a decomposition of $\Psi(P - Q)$ into its positive and negative parts because $\Psi(P)$ and $\Psi(Q)$ might not have orthogonal images (even if $P$ and $Q$ have orthogonal images).

So let's consider such a decomposition, $$ \Psi(P - Q) = R - S, $$ where $R$ and $S$ are positive semidefinite and satisfy $R S = 0$. Alternatively, we can write $$ R= \Psi(P-Q)_+ \quad\text{and}\quad S = \Psi(-(P-Q))_+ $$ using the notation of the question and the paper to which it refers. Let's also define $\Pi$ and $\Delta$ to be the orthogonal projections onto the images of $R$ and $S$, respectively, so $$ R = \Psi(P - Q) \Pi = \Pi \Psi(P - Q) = \Pi \Psi(P - Q) \Pi $$ and similar for $\Delta$ and $S$. These are projections onto orthogonal subspaces, so $\Pi + \Delta \leq \mathbb{I}$.

Now we can prove the following inequality. $$ \begin{aligned} \operatorname{Tr}(R + S) & = \operatorname{Tr}\bigl((\Pi - \Delta)(R - S)\bigr)\\ & = \operatorname{Tr}\bigl((\Pi - \Delta)(\Psi(P) - \Psi(Q))\bigr)\\ & = \operatorname{Tr}(\Pi \Psi(P)) - \operatorname{Tr}(\Delta \Psi(P)) - \operatorname{Tr}(\Pi \Psi(Q)) + \operatorname{Tr}(\Delta \Psi(Q))\\ & \leq \operatorname{Tr}(\Pi \Psi(P)) + \operatorname{Tr}(\Delta \Psi(P)) + \operatorname{Tr}(\Pi \Psi(Q)) + \operatorname{Tr}(\Delta \Psi(Q))\\ & = \operatorname{Tr}\bigl((\Pi + \Delta)(\Psi(P) + \Psi(Q))\bigr)\\ & \leq \operatorname{Tr}(\Psi(P) + \Psi(Q)) \end{aligned} $$ For the first inequality we're using the positivity of $\Psi$ together with the inequality $\operatorname{Tr}(AB)\geq 0$ for $A, B \geq 0$, and for the second inequality we're using $\Pi + \Delta \leq \mathbb{I}$ along with $\Psi(P) + \Psi(Q) \geq 0$ (so again we're using the positivity of $\Psi$).

If we now choose $P = \rho_+$ and $Q = (-\rho)_+$, we obtain $P - Q = \rho$, so $R = \Psi(\rho)_+$ and $S = \Psi(-\rho)_+$. This yields the inequality in the question.

So, the fact that $\rho_+$ and $\rho_-$ are the positive and negative parts of $\rho$ is a bit of a red herring here — it's the decomposition of $\Psi(\rho)$ into its positive and negative parts that is more relevant to this proof.

Answer 2

Edit: As pointed out by John Watrous again in the comments, the "proven" inequality is wrong so this answer is not correct.

$\def\braket#1{\langle{#1}\rangle}$I made a mistake in my first answer that is addressed in John Watrous' answer (wrongly assuming that $\Psi(\rho)_+ = \Psi(\rho_+)$ and $\Psi(\rho)_- = \Psi(\rho_-)$). In this answer, I use the inequality $\Psi(\rho)_+ \leq \Psi(\rho_+)$ instead.

If $\Psi(\rho)_+ \leq \Psi(\rho_+)$ can be proved for Hermitian $\rho$, then applying this inequality to $\rho$ and $-\rho$, taking the traces, and adding the two resulting inequalities results in $$\operatorname{Tr}\big[\Psi(\rho)_+\big] + \operatorname{Tr}\big[\Psi(-\rho)_+\big]\leq \operatorname{Tr}\big[\Psi(\rho_+)\big]+\operatorname{Tr}\big[\Psi((-\rho)_+)\big].$$ To prove $\Psi(\rho)_+ \leq \Psi(\rho_+)$, first prove that if $A \geq 0$ and $B \leq 0$ then $(A + B)_+ \leq A$.

If $\braket{\phi\vert{(A+B)_+}\vert\phi} = 0$, then it is trivially true that $\braket{\phi\vert{(A+B)_+}\vert\phi} \leq \braket{\phi\vert{A}\vert{\phi}}$. Now suppose that $\braket{\phi\vert{(A+B)_+}\vert\phi} > 0$ (and so $\braket{\phi\vert{(A+B)_-}\vert\phi} = 0$). It follows that $$\braket{\phi\vert{(A+B)_+}\vert\phi} = \braket{\phi\vert{(A+B)}\vert\phi} = \braket{\phi\vert{A}\vert{\phi}}+\braket{\phi\vert{B}\vert\phi} \leq \braket{\phi\vert{A}\vert{\phi}},$$ and thus $(A + B)_+ \leq A$.

By positivity of $\Psi$, $\Psi(\rho_+) \geq 0$ and $\Psi(\rho_-) \leq 0$, so taking $A = \Psi(\rho_+)$ and $B = \Psi(\rho_-)$ above results in $\Psi(\rho)_+ = (\Psi(\rho_+) + \Psi(\rho_-))_+ \leq \Psi(\rho_+)$.~~