Classical versus quantum correlations and partial traces

Question

Given a bipartite state $\rho_{AB}$ living in the Hilbert space $\mathcal H(A\otimes B)$ we can always define two local states on $A$ and $B$ respectively by taking the appropriate partial traces: $$\rho_A=\mathrm{tr}_B[\rho_{AB}], \quad \rho_B=\mathrm{tr}_A[\rho_{AB}]. $$ If the state $\rho_{AB}$ is separable, then $\rho_{AB}=\rho_A\otimes \rho_B$. In the more general case of an entangled state, since taking the partial trace corresponds to discarding part of the system any correlations between such a subsystem and its complement are lost, so $\rho_{AB}\ne \rho_A\otimes \rho_B$ and more generally there can be no pair of local states $\sigma_A,\tau_B$ such that $\rho_{AB}=\sigma_A\otimes \tau_B$.

But what about states that have classical correlations? I give an example below.

---

Example: Given an ensemble $\mathcal E=\{p_j,\rho_Q(j)\}$ consider the state $$\rho_{QX}=\sum_j p_j\rho_Q(j)\otimes |j\rangle\langle j|_X $$ where $\{|j\rangle_X\}$ is an orthonormal basis of $\mathcal H(Y)$. This state is correlated as the index $j$ connects the states on $Q$ and on $X$, but it is not entangled since it can be written as $$\rho_{QX}=\sum_j p_j\rho_Q(j)\otimes \rho_X(j) . $$

The partial traces give $$\rho_Q=\sum_j p_j\rho_Q(j), \quad \rho_X=\sum_j p_j|j\rangle\langle j|_X $$ and so $\rho_Q\otimes \rho_X=\sum_j p_j^2\rho_Q(j)\otimes |j\rangle\langle j|_X$. This looks almost like $\rho_{QX}$ apart from that coefficient $p_j^2$ that seems to reflect the fact that when we separate $\rho_{QX}$ we can choose to 'put' $p_j$ either in local state on $Q$ or on the on $X$.

Answer 1

I'm not sure exactly what the question is, but I can expand a bit about these states.

The states you mention are sometimes referred to as "one-way quantum-classical correlated states" (eg here and arxiv version) to refect the properties you describe. They differ from "strictly classical-classical states" of the form $\sum_j p_j|j\rangle \langle j|_Q\otimes |j\rangle\langle j|_X$ in that only the latter have vanishing quantum discord, so the quantum-classical states still have nonzero discord even though they are separable. This means that the total amount of correlations encoded in the states is more than the classical correlations, so there is still something quantum going on without entanglement here.

Answer 2

Note that the last formula is incorrect, it should read $$ \rho_{Q} \otimes \rho_X = \sum_{ij} p_i p_j \rho_Q(i) \otimes |j \rangle \langle j| $$ and then you see that in fact the two systems are not correlated.