Can we recover a state on a composite system from a state on the subsystem?

Question

$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right \rvert}$ 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.

Answer 1

$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right \rvert}$

No, because the "decoding" will not in general be unique. Consider the two states $\left( \ket{00} \pm \ket{11} \right)/\sqrt 2$. The density matrices are $$ \rho_\pm = \frac{1}{2} \left( \ket{00}\bra{00} \pm \ket{00}\bra{11} \pm \ket{11}\bra{00} + \ket{11}\bra{11} \right) \, . $$ The reduced density matrices (for either qubit because of the symmetry) are $$ \rho'_\pm = \frac{1}{2} \left( \ket{0}\bra{0} + \ket{1}\bra{1} \right) \, . $$ Therefore we've shown that two different pure states lead to the same reduced density matrices. This is generally the case: the mapping from the set of pure states to the set of reduced density matrices is not one-to-one.

The process of converting a reduced density matrix to a pure state is called purification and it is well known that given a reduced density matrix, the choice of purification is not unique.

Answer 2

I think your problem is an instance of what is called a quantum marginal problem. In the quantum marginal problem, we have a multi-part quantum system with Hilbert space $\mathbb{H} = \mathbb{H}_{A} \otimes \mathbb{H}_{B}\otimes \mathbb{H}_{C} ... $ and we have some density matrices on subsystems, for example $\rho_{AB} \in B(\mathbb{H}_{A} \otimes \mathbb{H}_{B})$, $\rho_{BC} \in B(\mathbb{H}_{B} \otimes \mathbb{H}_{C})$. We want to know if there exists a quantum state $\rho \in B(\mathbb{H})$ which is compatible with all of these reduced density matrixes, in the sense that you get the reduced density matrices when you perform the partial trace. So for example $Tr_{CDE..}[\rho]=\rho_{AB}$, $Tr_{ADE..}[\rho]=\rho_{BC}$.

Importantly, it is not always the case that there exists a state $\rho$ satisfying the conditions. Additionally, if $\rho$ exists, sometimes it is unique, and sometimes it is not.

In its most general form, the quantum marginal problem is known to be a hard problem, even for quantum computers. Indeed it is known to be QMA complete in general

There are two kinds of important restrictions one could place on the problem to try and simplify it. The first kind of restriction is that you can assume the global state $\rho$ has a particular strucuture, for example you could assume that it is pure: $\rho = \vert \psi\rangle \langle \psi \vert$. But even then it can be non-trivial.

A more useful restriction is to assume that the reduced density matrices act on disjoint subsystems. For example $\rho_{A}$ and $\rho_{B}$. A problem of this kind is called a non-overlapping quantum marginal problem, and the solution to this problem is for the most part given in this paper. One important takeaway is that the answer to the problem in this case depends only on the spectra of the reduced density matrices, and the surprising thing is that the conditions all take the form of linear inequalities. But even then, the inequalities can be highly non-trivial, and the number of them grow rapidly with the system size.

One very straightforward example of a quantum marginal problem which is easy to solve is when $\rho$ is guaranteed to be pure, there are two systems $A$ and $B$, and the reduced density matrices are $\rho_A$ and $\rho_B$. In this case there exists such a $\rho$ if and only if $\rho_A$ and $\rho_B$ have the same non-zero eigenvalues. Outside of this case, as you can see, the problem is quite non-trivial.