I'm looking for a relatively simple, nontrivial example of computing quantum discord (if such an example exists). Could anyone provide such an example?
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2 Answers
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Perhaps first consider the two-qubit case. It is nontrivial enough that it was published in PRA, but simple enough that it can be handled analytically since the derivations just involve 4x4 matrices. The tediousness of the derivations depend on what type of states you're working with. See for example
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Define the discord as $\newcommand{\Iacc}{I_{\rm acc}}\delta(\rho) = I(\rho)-I_{\rm acc}(\rho)$, with $I(\rho)$ quantum mutual information, and $\Iacc(\rho)$ the accessible mutual information, defined as the mutual information of the joint probability distribution corresponding to a choice of local measurements for $\rho$, maximised over all such choices. Note that this defines a symmetric quantum discord. See this other answer of mine for more details.
For an arbitrary pure state, $\rho=|\Psi\rangle\!\langle\Psi|$, one has $I(\rho)=2\Iacc(\rho)$. In particular, the discord is symmetric, and equals the accessible information. This is simple to prove: let $|\Psi\rangle=\sum_k \sqrt{p_k}|k,k\rangle$ be a Schmidt decomposition. Then $$I(\rho)= S(\rho_A)+S(\rho_B)-S(\rho) = 2S(\rho_A) = 2 H(\mathbf p),$$ where $H(\mathbf p)$ denotes the Shannon entropy of the probability vector with elements $p_k$. On the other hand, a projective measurement would give outcome probabilities $$p(a,b)=\lvert \langle u_a,v_b|\Psi \rangle \rvert^2.$$ You can prove that the mutual information of this $p$ is maximised if the measurement basis is the Schmidt basis, and thus $p(a,b) = \delta_{a,b} p_a$. We now need to compute the (classical) mutual information of this quantity, which this time reads $$\Iacc(\rho) = I(A:B)_{\bf p} = H(\mathbf p_A) = H(\mathbf p).$$
Consider the two-qubit state $$\rho = \frac12 ( \mathbb{P}_0\otimes \mathbb{P}_0 + \mathbb{P}_1\otimes \mathbb{P}_+), \qquad \mathbb{P}_v\equiv |v\rangle\!\langle v|.$$ This is classical-quantum wrt measurements on $A$, thus has zero discord left-to-right. But it also has nonzero discord wrt measurements on $B$. Note how this is also a separable state, showcasing how discord is a form of nonclassicality that is different from entanglement.
To compute the accessible mutual information in this case, consider a pair of projective measurements and the corresponding outcome probabilities, which in this case will read $$p(a,b) \equiv \langle \mathbb{P}_{u_a}\otimes\mathbb{P}_{v_b},\rho\rangle = \frac12( |\langle u_a|0\rangle|^2 |\langle v_b|0\rangle|^2 + |\langle u_a|1\rangle|^2 |\langle v_b|+\rangle|^2 ).$$ I'm not sure if there's an analytic way to maximise the mutual information of this with respect to all measurements. Numerically, I find the maximal mutual information to be $\Iacc\simeq 0.39912$, corresponding to $|u_a\rangle=|a\rangle$ (first space measured in the computational basis, as expected) and $|v_0\rangle=\sqrt p|0\rangle+\sqrt{1-p}|1\rangle$, $|v_1\rangle=\sqrt{1-p}|0\rangle-\sqrt{p}|1\rangle$ with $p\simeq 0.146$. Here's a contour plot of the mutual information as a function of the parameters of the second measurement, allowing also a phase $\phi$:
See also this other question on qc.SE about the optimal measurement basis for CQ states.
For an example of a two-qubit separable discordant state that is not classical-quantum, one can consider the standard example given in Ollivier and Zurek's original paper: a Werner state of the form $$\rho_z = \frac{1-z}{4}I +z |\Phi^+\rangle\!\langle\Phi^+| =\frac14\begin{pmatrix}1+z & 0&0& 2z \\ 0& 1-z & 0 & 0 \\ 0&0&1-z&0 \\ 2z & 0 & 0 & 1+z\end{pmatrix}, \\ |\Phi^+\rangle\equiv\frac{1}{\sqrt2}(|00\rangle+|11\rangle),$$ which is separable for $z\le 1/3$ but has nonzero discord in both directions.
It's worth noting that I also only addressed choices of projective measurements here. Arguably one might want to also consider more general non-projective measurements in the maximisation, as it is known that at least in some cases the discord is achieved by those. I don't think that's the case for the examples here, but I'm not entirely sure what's a simple way to prove it either. See this post on qc.SE for more info about such scenarios.
