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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.

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 $\phi$:

  • 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.

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.

Source Link
glS
glS
  • 15.2k
  • 5
  • 41
  • 109

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 $\phi$:

  • 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.