$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right \rvert}$ I'm wondering if we can recover the state of a composite system from the information of its subsystem. For instance denote the state of the composite system as $$ \ket{\psi} = \alpha \ket{00} + \beta \ket{11} \in \mathbb{C}^2 \otimes \mathbb{C}^2 \, .$$ We can denote the state using the density matrix \begin{align} \rho &= \ket{\psi}\bra{\psi} \\ &= |\alpha|^2 \ket{00}\bra{00} + \alpha \bar{\beta} \ket{00}\bra{11} + \bar{\alpha} \beta \ket{11}\bra{00} + |\beta|^2 \ket{11}\bra{11} \, . \end{align} The state $\rho'$ of either subsystem consisting of either just the first or just the second qubit is $$\rho' \equiv \text{Tr}_2(\rho) = |\alpha|^2 \ket{0}\bra{0} + |\beta|^2 \ket{1}\bra{1} \, .$$ Can we recover $\rho$ if we know $\rho'$? Equivalently, can we determine the initial state $\ket{\phi} = \alpha \ket{0} + \beta \ket{1} \in \mathbb{C}^2$ to any state to solve this problem?
When the initial state is a separable state $\ket{\xi} = \ket{00} \in \mathbb{C}^2\otimes \mathbb{C}^2$, the density matrix is $\tilde{\rho} = \ket{00} \bra{00}$, so that the state of either of its subsystems is $\tilde{\rho'} = \text{Tr}_1(\tilde{\rho}) = \text{Tr}_2(\tilde{\rho}) = \ket{0}\bra{0}$. In this case, we can recover $\tilde{\rho}$ from $\tilde{\rho'}$ since $\tilde{\rho'} \otimes \ket{0}\bra{0}=\tilde{\rho}$. Unlike this, I'm looking for the solution in non-trivial cases.