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If I have two systems A and B and I want to find the density operator of the joint systems $\rho_{AB}$, is it just the tensor product of the density matrix of the individual systems, $\rho_{AB} = \rho_A\otimes\rho_B$? If not, what am I neglecting? (Maybe information about the correlation of the two system is lost)

Qmechanic
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nicco
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    The existing answer is good. For a concrete counterexample: consider a system of two qubits, each having reduced density matrix $\rho_R=\begin{pmatrix}1/2&\&1/2\end{pmatrix}$ (like a classical coin flip). The joint state might be $\rho=\rho_R\otimes\rho_R=\begin{pmatrix}1/4&&&\&1/4&&\&&1/4&\&&&1/4\end{pmatrix}$ (like two independent coin flips), or it might be $\rho=|\psi\rangle\langle\psi|=\begin{pmatrix}1/2&&&\&0&&\&&0&\&&&1/2\end{pmatrix},$ where the qubits are in the pure (and entangled) state $|\psi\rangle=\frac1{\sqrt2}(|00\rangle+|11\rangle).$ Or something in between. – HTNW May 24 '23 at 02:29
  • @HTNW This basically is the example I mention in the first link. – Tobias Fünke May 24 '23 at 05:38

1 Answers1

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This is not possible in general, as both density matrices could be the reduced density matrices of several joint density operators, see e.g. here and here.

The reduced density matrices determine the joint (bipartite) density operator uniquely if and only if at least one of these two density matrices is pure.

Indeed, consider a bipartite density matrix $\rho$ and suppose that at least one of its reduced density matrices $\rho_A$ is pure. Owing to the convex nature of the space of density matrices/their spectral properties, we can always write $$\rho=\sum\limits_{k} p_k |\psi_k\rangle\langle \psi_k| \tag 1 \quad , $$

which shows that

$$\rho_A=\mathrm{Tr}_B \rho = \sum\limits_k p_k \mathrm{Tr}_B |\psi_k\rangle\langle \psi_k| = \sum\limits_k p_k \rho_{A,k}\tag 2 \quad .$$

But if $\rho_A$ is pure, then $(2)$ implies that $\rho_A=\rho_{A,k}$ for all $k$, due to the extremal properties of the pure states. The Schmidt decompositions of the $|\psi_k\rangle$'s now show that $$|\psi_k\rangle\langle \psi_k| = \rho_{A,k} \otimes \rho_{B,k} = \rho_A\otimes \rho_{B,k} \tag 3 $$

which then gives

$$\rho=\rho_A\otimes \rho_B \tag{4}\quad , $$ with $\displaystyle \rho_B=\sum\limits_k p_k \,\rho_{B,k}$.

Note that $\rho$ can be mixed, which is the case if and only if $\rho_B$ is mixed. Thus, if you have two density matrices $\rho_A$ and $\rho_B$, and at least one is pure, then the joint density matrix is necessarily of the form $(4)$ and thus uniquely specified.

Conversely, if both $\rho_A$ and $\rho_B$ are mixed, there are uncountably many different density matrices $\rho_1$ and $\rho_2$ with $\rho_A$ and $\rho_B$ as reduced density matrices, which shows that in this case the joint density operator is not uniquely specified. The first link above contains some explicit example.

For a proof of this fact, see e.g. Mathematical Foundations of Quantum Mechanics. John von Neumann. New Edition. Princeton University Press. Chapter 2, p. 276-78.

Tobias Fünke
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