What is the difference between classical-quantum and completely classical states?

Question

States that are completly classical : $$ \begin{aligned} \tilde\rho_{A B} & =\sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} p_{X, Y}(x, y)(|x\rangle \otimes|y\rangle)(\langle x| \otimes\langle y|)_{A B} \\ & =\sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} p_{X, Y}(x, y)|x\rangle\left\langle\left. x\right|_A \otimes \mid y\right\rangle\left\langle\left. y\right|_B\right. \end{aligned} \tag1 $$

• The states $\left\{|x\rangle_A\right\}_{x \in \mathcal{X}}$ and $\left\{|y\rangle_x\right\}_{y \in \mathcal{Y}}$ form an orthonormal basis for the respective systems $x$ and $y$.
• State has only classical correlations because can be prepared by LOCC.

Classical-Quantum state : $$ \begin{aligned} \rho^{A B} & =\sum_{x, y} p(x, y)|x\rangle\langle x| \otimes \pi_y \\ &\equiv \rho^{A B}=\sum_x p(x)|x\rangle\langle x| \otimes \rho_x^B \end{aligned} $$ with $p(x)=\sum_y p(x, y)$ and $\rho_x^B=\frac{1}{p(x)} \sum_y p(x, y) \pi_y$

Let the spectral decomposition of $\rho_x^B$ be: $$ \rho_x^B=\sum_{y \in \mathcal{Y}} p_{Y \mid X}(y \mid x) \left|y_x\right\rangle\left\langle\left. y_x\right|_B\right. $$ where $\left\{\left|y_x\right\rangle\right\}$ forms orthonormal basis for $B$.

$$ \begin{aligned} \Rightarrow \rho^{A B} & =\sum_x p_X(x)|x\rangle\left\langle x\right|_A \otimes \sum_y p_{Y \mid X}(y \mid x) \mid y_x\rangle\left\langle y_x\right|_B \\ & =\sum_x \sum_y p_X(x) p_{Y \mid X}(y \mid x)|x\rangle\left\langle x\right|_A \otimes \mid y_x\rangle\left\langle y_x\right|_B \end{aligned} $$ $$ =\sum_x \sum_y p_{X, Y}(x, y)|x\rangle\left\langle\left. x\right|_A \otimes \mid y_x\right\rangle\left\langle y_x\right|_B \tag2 $$

Question: From equation 1 and 2, we see $\rho$ and $\tilde \rho$ have similar form, expressed in terms of orhtonormal basis of both systems. I would be greateful if someone properly explain (in detail) the difference between Classical-Quantum states and States that are completely classical, possibly in terms of the expresssions of $\rho$ and $\tilde \rho$.

Answer 1

Your expressions give a pretty clear distinction: in the classical-quantum state, the eigenbases $\left|y_x\right>$ can be different for different states $x$ of the classical register $A$, and in the completely classical state the basis $\left|y\right>$ is fixed and does not depend on $x$.

We can also phrase it lake this: choose a basis $\{\left|y\right>\}_{y \in \mathcal{Y}}$ and say that classical states of the system $B$ are convex linear combinations of $\{\left|y\right>\left<y\right|\}_{y \in \mathcal{Y}}$ (so they are given by classical probability distributions on $\mathcal{Y}$). Then a classical-quantum state has the form $$\rho^{AB} = \sum_{x} \left|x\right>\left<x\right| \otimes \rho_x^B$$ and it is completely classical exactly when all $\rho_x^B$ are classical states.