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

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

• Think about, if there are two non-identical states $\rho$ that would have the same reduced density matrix $\rho’$ of the subsystem. Mar 20, 2019 at 5:42
– user225954
Mar 20, 2019 at 6:14
• This sentence: "Equivalently, can we determine the initial state $|\phi\rangle = \alpha |0\rangle + \beta |1\rangle \in \mathbb{C}_2$ to any state to solve this problem?" does not make sense. Can you please clarify what you're trying to say? Mar 20, 2019 at 7:14
• Sorry. That's a mistake. I meant "you can encode the initial state $|\psi>$ to any state that you want to solve the problem." But any operation to the state has to be represented by a unitary matrix.
– user225954
Mar 20, 2019 at 7:32

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

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.

• Decoding doesn't have to be unique "in general". I want to know if or not there exist non-trivial states that satisfy my problem.
– user225954
Mar 20, 2019 at 7:37
• @John Then I think the question needs to be clarified. It's not clear what your "problem" is. Mar 20, 2019 at 7:43
• OK, let me clarify my problem. Do you know about deletion error-correcting codes in the classical coding theory? The famous codes are Levenshtein codes, VT codes and so on. I'm wondering if we construct such codes in quantum coding theory. Then, my problem is that whether or not there exists an operation satisfing the followings:
– user225954
Mar 20, 2019 at 14:47
• 1.$|\psi>\in \mathbb{C}^2$ is an entangled state.$\$2.You can prepare some initialized states $|0>$ and can operate these qubits using unitary matrices (ex. if $|\psi>=\alpha |0>+\beta |1>$, then $CNOT(|\psi>\otimes |0>)=\alpha |00>+\beta |11>$). $\$3. Let the state after the operations be $|\Psi>\in\mathbb{C}^{2\otimes n}(n\in \mathbb{Z}_{\ge2})$.Then, you send $n-1$ qubits to Alice (so $\rho=|\Psi><\Psi|$ and the sent qubits are represented $Tr_k (\rho)(k\in\{1,...,n\})$).$\$4.Doing proper operations, Alice wants to get $|\Psi>$.
– user225954
Mar 20, 2019 at 14:47
• In this case, you don't know the state $|\psi>$ in 1 (so $|\psi>=\alpha |0>+\beta |1>$ and you can't know $\alpha$ and $\beta$.
– user225954
Mar 20, 2019 at 14:49

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.

• I appreciate your answer. I did not know "quantum marginal problem", but it sounds interesting. In my problem, one can only obtain a subsystem of a composite system and wants to know the state on the composite system from the state of its subsystem. Is it possible for some cases, or impossible for any cases.
– user225954
Mar 21, 2019 at 7:53
• Hi John, it is not clear to me how the case you are describing is different from what I described in my answer. Mar 21, 2019 at 14:52
• Sorry, I don't understand your answer that much so that I'm reading it repeatedly. Could you give me an example that I want to know?
– user225954
Mar 21, 2019 at 16:56
• Hi Joel. I read your answer and understand it except some proofs in the propositions. Could you prove the last proposition, please? You said "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".
– user225954
Mar 24, 2019 at 7:46
• Hi John, A bipartite pure state always admits a decomposition of the form: $\vert \psi \rangle = \sum_i \sqrt{\lambda_i}\vert a_i \rangle \vert b_i \rangle$ where the sets $\{\vert a_i \rangle \}$ and $\{ \vert b_i \rangle\}$ are each sets of orthonormal basis states. This is related to the singular value decomposition. Mar 25, 2019 at 19:05