# Classical versus quantum correlations and partial traces

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 $$X$$.

• Close voters, what detail/clarity do you want? Help us out here. Also I see that the answer by Rasmus already tells the user that the two systems are not correlated, so the question has been answered. Jul 25 '21 at 13:45
• @user1271772 The question is still very vague, it is quite hard to understand what the OP wants. Jul 25 '21 at 14:06
• @Rammus Sorry for mis-spelling your name accidentally! (can't edit comment now). Also, thanks for clarifying why you voted to close, though I wish the person who originally voted to close would say why they did too. Also, if you think it should be closed (i.e. not accept answers), it's a bit strange that you would answer it. Your answer could have been a comment! Jul 25 '21 at 14:17
• @user1271772 No worries. I answered it a few days ago and only voted to close it today after I saw it in the queue and thought about it. Jul 25 '21 at 15:08
• Guess I should enter this conversation :-) as I said in another comment a more focused version of my question would be, 'is there any relationship between $\rho_A$ and $\rho_B$ obtained as partial traces of a non-entangled state that exhibits classical correlations?' Jul 25 '21 at 16:26

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.
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.