Inequalities involving entropy: quantum discord and mutual information

Question

My question is inspired by the following paper of Olivier and Żurek but for this question to be self-contained I will recall all the necessary definitions: for a quantum state $\rho$ we define the entropy by the formula $H(\rho):=-tr(\rho log \rho)$ (by a state we mean in fact normal, possibly mixed state which we identify with the corresponding density matrix). Now consider a quantum system $S$ and an apparatus $A$: we describe both of them in the context of Hilbert spaces $\mathcal{H}_S$ and $\mathcal{H}_A$. Assume that the composite state $\rho_{SA} \in \mathcal{L}^1(\mathcal{H}_S \otimes \mathcal{H}_A)$ is given. Denote by $\rho_S$ the state $tr_A(\rho_{SA})$ obtained from $\rho_{SA}$ by taking the partial trace: similarly define $\rho_A$ and define the so called mutual information $I(S:A):=H(\rho_S)+H(\rho_A)-H(\rho_{SA})$. Quantum discord will be defined as the difference between mutual information $I$ and the other, quantum version of it, called $J$ which is defined as follows: choose a family of one dimensional mutually orthogonal projections $\{\Pi_j^A\}_j$ summing up to identity $I_{\mathcal{H}_A}$: put $$\rho_{S|\Pi_j^A}=\frac{1}{p_j}(I_{\mathcal{H}_S} \otimes \Pi_j^A) \rho_{SA} (I_{\mathcal{H}_S} \otimes \Pi_j^A)$$ where $p_j=tr((I_{\mathcal{H}_S} \otimes \Pi_j^A)\rho_{SA})$ and define $H(\rho_S|\{\Pi_j^A\}_j):=\sum_j p_j H(\rho_{S|\Pi_j^A})$. We define $J(S:A)_{\{\Pi_j^A\}_j}:=H(\rho_S)-H(\rho_S|\{\Pi_j^A\}_j)$. Finally we define the quantum discord as a difference $$\delta(S:A)_{\{\Pi_j^A\}_j}=I(S:A)-J(S:A)_{\{\Pi_j^A\}_j}.$$ Proposition 1 in the paper mentioned above gives another formula for $H(\rho_S|\{\Pi_j^A\}_j)$ as a difference $H(\rho_{SA}^D)-H(\rho_A^D)$ where $\rho_{SA}^D=\sum_j (I \otimes \Pi_j^A)\rho_{SA}(I \otimes \Pi_j^A)$ and $\rho_A^D=\sum_j \Pi_j^A \rho_A \Pi_j^A$. This allows us to express the quantum discord as follows: $$\delta(S:A)_{\{\Pi_j^A\}_j}=H(\rho_A)-H(\rho_A^D)+H(\rho_{SA}^D)-H(\rho_{SA}) $$

Question 1 How to prove that $\delta(S:A)_{\{\Pi_j^A\}_j} \geq 0$?

In the paper authors claim that this follows somehow from concavity of $H(\rho_{SA})-H(\rho_A)$ but I don't see how (they cite some paper which I've browsed but didn't manage to find an answer)

Question 2 How to prove that $\delta(S:A)_{\{\Pi_j^A\}_j}=0$ implies $\rho_{SA}=\rho_{SA}^D$?

The authors give a proof of this fact but I'm not convinced by their argument: they consider another state $\hat{\rho}_{SA}$ obtained by removing off-diagonal terms and claim that $H(\hat{\rho}_{SA}) >H(\rho_{SA})$ while $H(\hat{\rho}_A)=H(\rho_A)$. I don't see why this is true: once we accept this we clearly obtain $H(\rho_{SA})-H(\rho_A)<H(\hat{\rho}_{SA})-H(\hat{\rho}_A)$ and now they claim that also $H(\hat{\rho}_{SA})-H(\hat{\rho}_A) \leq H(\rho_{SA}^D)-H(\rho_A^D)$ which is also not clear for me (they again invoke concavity).

I will be very grateful for any help with this problem

EDIT: There is already one excellent answer below but unfortunately this answer adresses only the first of my questions: still I would like to see how to prove the second part. Maybe the following discussions will be relavant: see here and here but I'm affraid that somewhere we need strong inequality to finish the argument given by Olivier and Żurek.

Answer 1

This is only a partial answer for now. I will sketch the answer of Question 1 in the case when all Hilbert spaces involved are finite-dimensional for simplicity.

The key point how to use concavity is the following: The map $f\colon\rho_{SA}\mapsto H(\rho_{SA})-H(\mathrm{tr}_S(\rho_{SA}))$ is invariant under conjugation with unitaries in $\mathbb C 1\otimes \mathcal L(\mathcal H_A)$. Moreover, if we denote by $\mu$ the normalized Haar measure on $\mathbb T^m$, we have $$ \sum_{j=1}^m (I\otimes\Pi_j^A)x(I\otimes\Pi_j^A)=\int_{\mathbb T^m}\left(I\otimes\sum_{j=1}^m \lambda_j\Pi_j^A\right)x\left(I\otimes\sum_{k=1}^m\lambda_k\Pi_k^A\right)\,d\mu(\lambda) $$ since for $j\neq k$ one has $\int_{\mathbb T^m}\lambda_j\lambda_k\,d\mu(\lambda)=0$. Note that $\sum_{j=1}^m \lambda_j\Pi_j^A$ is a unitary in $\mathcal L(\mathcal H_A)$.

Then concavity and partial unitary invariance of $f$ imply \begin{align*} f(\rho_{SA}^{D})&=f\left(\int_{\mathbb T^m}\left(I\otimes\sum_{j=1}^m \lambda_j\Pi_j^A\right)\rho_{SA}\left(I\otimes\sum_{k=1}^m\lambda_k\Pi_k^A\right)\,d\mu(\lambda)\right)\\ &\geq \int_{\mathbb T^m}f\left(\left(I\otimes\sum_{j=1}^m \lambda_j\Pi_j^A\right)\rho_{SA}\left(I\otimes\sum_{k=1}^m\lambda_k\Pi_k^A\right)\right)\,d\mu(\lambda)\\ &=\int_{\mathbb{T}^m}f(\rho_{SA})\,d\mu(\lambda)\\ &=f(\rho_{SA}). \end{align*} This settles your first question (note that $\mathrm{tr}_S(\rho^D_{SA})=\rho_A^D$).