Why is the quantum discord of $\rho$ zero iff $\rho=\sum_j p_j \pi_j\otimes \rho_j$ for mutually orthogonal projections $\pi_j$?

Question

In (Wiseman 2012), the author mentions (equation (4), page 6), that a state $\rho$ has zero quantum discord (more precisely, zero Alice-discord) if and only if it can be written in the form $$\rho = \sum_j p_j \pi_j\otimes \rho_j,$$ for some probability distribution $p_j$, some collection of orthogonal projections $\{\pi_j\}_j$ on Alice's system, and some collection of states $\rho_j$ on Bob's system.

By "Alice-discord" I mean here the discord with respect to measurements performed by Alice (the first system). More precisely, the discord is defined here $$\delta_A(\rho) =I(\rho) - \max_{\{\Pi^A_j\}} J_{\{\Pi^A_j\}}(\rho) = S(\rho_A) - S(\rho) + \min_{\{\Pi^A_j\}} S(\rho_{B|\{\Pi^A_j\}}),$$ where $I$ and $J$ are the two classically equivalent forms of the mutual information, and the maximisation and minimisations are performed with respect to possible measurements performed by Alice.

The author mentions this as "well known" and does not provide a reference. Is there an easy way to see why this is the case? Alternatively, what is a reference discussing this fact?

Answer 1

$$I(A:B)=S(A)+S(B)-S(AB)$$

$$J(A_{\{\Pi_{i}\}}:B)=S(A_{\{\Pi_{i}\}})+S(B)-S(A_{\{\Pi_{i}\}}B)$$

$$I(A:B)-J(A_{\{\Pi_{i}\}}:B)=S(A)-S(AB)-S(A_{\{\Pi_{i}\}})+S(A_{\{\Pi_{i}\}}B)$$

Since $$\rho = \sum_j p_j \pi_j\otimes \rho_j,$$ it is invariant to the measurements performed on the subsystem A, after the results are forgotten, so $$S(A)=S(A_{\{\Pi_{i}\}})$$ and $$S(AB)=S(A_{\{\Pi_{i}\}}B)$$

resulting in a discord (for alice) of 0.

If $\rho$ were any other state, wherein it was not invariant to a set of measurements after maximisation, then the measurements on it would lead to an increase in entropy, and a non-zero discord.

Both Quantum Discord: A Measure of the Quantumness of Correlations and Studies on the Role of Entanglement in Mixed-state Quantum Computation go over this in greater detail.