# When is $\rho_{AB} = \rho_{A} \otimes \rho_{B}$ true?

Say $$\rho_{AB}$$ is the density matrix of the joint system AB, and $$\rho_{A}$$ ($$\rho_{B}$$) is the density matrix of system A (system B).

When can one write $$\rho_{AB} = \rho_{A} \otimes \rho_{B}$$. Is this always true?

A pure or mixed density operator of the form $$\rho = \rho_1 \otimes \rho_2 \tag{1}$$ on a bipartite Hilbert space $$H=H_1 \otimes H_2$$ is called a product state. Not all states are product states; easy counter examples for a two-qubit system are the so-called Bell states.
Consider two observables $$A$$ and $$B$$, on $$H_1$$ and $$H_2$$ respectively, in their spectral representation: \begin{align} A&:=\sum\limits_i a_i \, P^{A}_i\\ B&:=\sum\limits_j b_j \, P^{B}_j \quad . \end{align} We say that a state $$\rho$$ has no correlations if for all joint projective measurements $$P^{AB}_{ij}:=P_i^{A} \otimes P_j^{B}$$ and for all possible pairs of observables $$A$$ and $$B$$ the corresponding joint probability distribution $$p^{AB}_\rho(i,j):=\mathrm{Tr}(P^{AB}_{ij}\rho)$$ has no correlations.
One can now show (assume that $$H$$ is finite-dimensional) that a density operator $$\rho$$ has no correlations if and only if it is a product state, which in turn is equivalent to the fact that for all observables $$A, B$$ on the respective Hilbert spaces it holds that $$C_\rho(A,B)=0$$, where $$C_\rho(A,B):=\langle A \otimes B \rangle_\rho -\langle A\otimes \mathbb I_2\rangle_\rho\langle \mathbb I_1\otimes B\rangle_\rho \quad . \tag{2}$$ The expectation value is defined as $$\langle \cdot \rangle_\rho :=\mathrm{Tr}(\cdot\rho)$$.
The upshot is that when you perform joint (projective) measurements on your system, then for a product state $$(1)$$ the measurement results are not correlated. In contrast, there are the so-called separable states (of which product states are a special case), which, in general, allow for classical correlation. The Bell states mentioned above are examples of entangled states. In fact, entanglement is one seminal (non-classical) feature of quantum mechanics: It allows and predicts correlations which cannot be reproduced by a classical (what exactly this means can be defined precisely) theory. Moreover, these predictions by quantum mechanics were found to agree on experimental results. There are many questions and answers here on physics stack exchange regarding entanglement or the Bell inequalities and related topics.
• I would like just to add that given $\rho_1$ and $\rho_2$ it is easy to compute $\rho=\rho_1\otimes\rho_2$. However, given a $\rho$, it is much more difficult to know if $\rho$ can be separated. Jun 22 at 12:33
• @Mauricio Yes, but note that given any two reduced density matrices $\rho_1$ and $\rho_2$, there is always a separable density operator $\rho$ with $\rho_1$ and $\rho_2$ as reduced density matrices, namely $\rho=\rho_1 \otimes \rho_2$. However there could be several density matrices which would yield $\rho_1$ and $\rho_2$ as reduced density matrices, which, in general (depending on $\rho_1$ and $\rho_2$), can differ in their purity and separability, cf. this. Jun 22 at 12:38