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

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 $$\tilde \rho$$.

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 $$\{\left|y\right>\left (so they are given by classical probability distributions on $$\mathcal{Y}$$). Then a classical-quantum state has the form $$\rho^{AB} = \sum_{x} \left|x\right>\left and it is completely classical exactly when all $$\rho_x^B$$ are classical states.