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

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?

• How do you define Alice-discord? May 29 at 22:02
• @Rammus I simply mean the discord with respect to Alice's measurements. I included the explicit definition
– glS
May 29 at 22:50
• The only thing that comes to mind is that, due to this being a CQ state, the maximisation over the measurement for the asymmetric mutual information will just be the set of orthogonal projectors that Alice's system is prepared in, so the two measures should be equivalent. So maybe he is making the assumption of it being "well known" based on well known information about mutual information of CQ states? May 31 at 11:32
• arxiv.org/pdf/1611.01959.pdf might be what you are looking for. Specifically the paragraph under equation 15 on page 8. May 31 at 11:35

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

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

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

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.

• to be clear, you are saying that there is some projective measurement such that this is true, not that any $\{\Pi_i\}$ will do, right? In other words, if one uses the projective measurement corresponding to $\pi_j$ you have this, otherwise the entropy might be different
– glS
Oct 10 at 18:20
• Apologies for the late reply. Correct. It has been shown that projections minimize discord, though which set of projectors depends on the state in question. Nov 2 at 17:18