# Correlation vs. entanglement for composite quantum system

Some authors exclusively use "Correlation" to classify composite quantum states, whereas most only speak of "Entanglement".

Correlation basically means that measurements on the subsystems are stochastically dependent and entanglement means non-separability of the composite state.

I am wondering, are those classifications equivalent, or is there any hierarchy (e.g. if a composite state is non entangled, it is always uncorrelated). Does the entropy of entanglement (in some cases) predict whether a state is (un)correlated?

References to a proof would be much appreciated!

Feel free to criticize me on the casual definitions given above as well.

• DId you mean to write "stochastically dependent" instead of "stochastically independent"? States of the form $A\otimes B$ can give correlated results for many measurements, but are unentangled. – WillO Dec 30 '19 at 18:25
• @WillO Not for connected correlation functions. – Norbert Schuch Jan 4 '20 at 23:04
• @NorbertSchuch: The phrase "connected correlation function" is new to me. I'm off to learn about it. – WillO Jan 4 '20 at 23:13
• @WillO <XY>-<X><Y>. – Norbert Schuch Jan 5 '20 at 22:11

On the other hand, entanglement is a property of a state, with respect to some partition on the underlying space. A bipartite state $$\rho$$ is said to be entangled if it cannot be written as a convex combination of product states, that is, if it cannot be written in the form $$\rho=\sum_k p_k \rho_k^A\otimes\rho_k^B$$ for some $$p_k\ge0, \sum_k p_k=1$$ and states $$\rho_k^A,\rho_k^B$$. The bipartite structure is usually, although not necessarily, taken to refer to degrees of freedom of spatially separated particles. It can however refer to any pair of degrees of freedom of a quantum system.
Entangled states can produce nonclassical correlations, but this is not necessarily the case. For example, not all entangled states can produce Bell violations. On the other hand, entangled states always display some form of correlation: given a pure entangled state $$|\psi\rangle$$, write it in its Schmidt decomposition as $$|\psi\rangle=\sum_k \sqrt{p_k} |u_k\rangle\otimes|v_k\rangle$$. Then, measuring in the $$\{|u_k\rangle\}_k$$ basis on the first space and in the $$\{|v_k\rangle\}_k$$ basis in the second will give correlated outcomes (the $$k$$-th outcome for the first party implies that the second party also will have measured its $$k$$-th outcome).