I have come across an article which talks about quantum discord (Observing the operational significance of discord consumption. M. Gu et al. Nature Physics 8, 671–675 (2012) doi:10.1038/nphys2376), but I am struggling with the abstract, which says:

The amount of extra information recovered by coherent interaction is quantified and directly linked with the discord consumed during encoding. No entanglement exists at any point of this experiment. Thus we introduce and demonstrate an operational method to use discord as a physical resource.

So how can this happen? Usually quantum computations are thought to consume entanglement to perform calculations, and only pure states are involved. However, when quantum discord is discussed, the mixed state description makes it hard to understand where 'discordance' comes from.

Are there explicit (mathematical) examples that demonstrate whether it is possible to have

  • (a) zero entanglement, non-zero discord;
  • (b) non-zero entanglement, zero discord; and
  • (c) non-zero entanglement, non-zero discord?

In classical information theory, the (Shannon) entropy is defined by $ H(A) = -\sum p_i \log(p_i) $ for a given probability distribution $p_i$. The mutual information can be defined as

$$\mathcal{I}(A:B) = H(A)+H(B)-H(A,B)$$ or $$\mathcal{J}(A:B) = H(A)+H(B|A),$$

where $H(A,B)$ and $H(B|A)$ are the entropy of joint system and the conditional entropy, respectively. These two expressions are equivalent in the classical case, and the second one can be derived from the first one using Bayes' rule.

Similar expressions can be defined in the quantum system, with the entropy replaced by the Von Neumann entropy, $$H(\rho)=-Tr(\rho \ln \rho).$$

However, the expressions $\mathcal{I}(A:B)$ and $\mathcal{J}(A:B)$ are not always equivalent in quantum systems and the difference is called the quantum discord:

$$ \mathcal{D}(A:B) = \mathcal{I}(A:B) - \mathcal{J}(A:B).$$

Therefore, it can be treated as a measure for the non-classical correlations in a quantum system.


2 Answers 2


Look at the paper where it is introduced:

They analyze Werner states $$\rho = \tfrac{1-z}{4}\mathbb{I} + z |\psi\rangle\langle\psi|$$ for $0\leq z \leq 1$ and $|\psi\rangle = (|00\rangle+|11\rangle)/\sqrt{2}$.

It can be seen that discord is greater than $0$ in any basis when $z\geq 0$, which contrasts with the well-known separability of such states when $z \leq 1/3$.


  • (a) zero entanglement, non-zero discord - YES, e.g. Werner state $z=1/3$
  • (b) non-zero entanglement, zero discord - I THINK NO - Discord is amount of information destroyed by classical measurement (and classical measurement destroys all entanglement),
  • (c) non-zero entanglement, non-zero discord - YES, e.g. Werner state $z=1$.
  • $\begingroup$ Thanks. I have found that paper, but I havent read it yet. The situation (b) is not possible, but I dont know the proof yet. I am more concern with the situation (a) as I dont really understand what it means physically. That is how can two system no entangled together but with non-trivial classical correlation. Do you have some idea about that? I will read the paper soon and give you bounty if it is correct $\endgroup$
    – unsym
    Commented Dec 20, 2012 at 17:43
  • $\begingroup$ I'm not that deep into discord, but AFAIR some separable states with non-zero discord can be used for entanglement activation or something. BTW: indico.cern.ch/getFile.py/… When it comes to (a) - IMHO the intuition is the following: for mixed states, not all non-classical effects are related to entanglement; there are systems, where with a fixed set of measurements, you can 'transmit' less information than from general quantum-mechanical bound. $\endgroup$ Commented Dec 21, 2012 at 14:54
  1. (zero entanglement, zero discord): aside from the standard example of a Werner state, another very simple one is the state $\frac12(\mathbb{P}_0\otimes\mathbb{P}_0+\mathbb{P}_1\otimes\mathbb{P}_+)$ where $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$. This is clearly separable, but has nonzero discord from right to left (though zero discord from left to right).

  2. (non-zero entanglement, zero discord): This is not possible. A states has zero discord (left-to-right) if and only if it can be written as $\sum_k p_k \mathbb{P}_k\otimes\rho_k$ for some collection of projections $\mathbb{P}_k$, and states $\rho_k$. Such a state is separable, i.e. has no entanglement.

  3. (non-zero entanglement, non-zero discord): most cases are like this. For example, any pure state $|\psi\rangle$ gives $I(X:Y)=2 S(X)$, meaning that its discord equals twice its entanglement entropy. So in particular any maximally entangled state has unit discord.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.