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Given density operator of a composite system, say $\rho_{AB}$, we can always calculate reduced density operators of individual system i.e. $\rho_{A}=Tr_{B}(\rho_{AB})$ and $\rho_{B}=Tr_{A}(\rho_{AB})$.

Can we calculate density matrix of a composite system, $\rho_{AB}$, given reduced density operators $\rho_{A}$ and $\rho_{B}$ and also knowing that $\rho_{A}$ and $\rho_{B}$ are mixed state density operators? If yes, then how?

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In general, given only the reduced states, you cannot find a unique state for the composite system. But you can certainly find states $\rho_{AB}$ such that their reduced states are $\rho_{A}, \rho_{B}$. As a trivial example, consider the state $\rho_{AB}=\rho_{A} \otimes \rho_{B}$, it is easy to check that it's reduced states give you $\rho_{A}, \rho_{B}$.

The real question is, why can't you find a unique composite state? Also, are there instances where you can find unique extensions?

The answer, like many things in quantum information theory, lies in entanglement. If the state $\rho_{AB}$ is a product state (that is, it is unentangled) then it's marginals (the reduced states) will uniquely define the composite state. For entangled states, this breaks down -- the intuitive reasoning is that the reduced states cannot account for the nonlocal correlations in the system. As an example, consider the four Bell states, all of them have the same marginals (the maximally mixed state $\frac{\mathbb{I}}{2}$) but the four Bell states are orthogonal to each other and hence completely distinguishable.

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No. For example, $$ \rho_{AB}=\frac{I}{4},\qquad \rho_{AB}=\frac{1}{2}(|00\rangle+|11\rangle)(\langle00|+\langle 11|) $$ both have the same reduced density matrices $\rho_A=\rho_B=I/2$.

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  • $\begingroup$ What if $\rho_{AB}$ is pure state density operator? $\endgroup$ – Omkar Jul 30 at 9:56
  • $\begingroup$ Still no. Apply any single-qubit unitaries to the second case. The reduced density matrices don't change. $\endgroup$ – DaftWullie Jul 30 at 9:57

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