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Let $\rho_A,\rho_B$ be two states. Is there any way to characterise the set of bipartite states $\rho$ such that $\mathrm{Tr}_B(\rho)=\rho_A$ and $\mathrm{Tr}_A(\rho)=\rho_B$?

If I assume $\rho$ to be pure, then this is possible iff $\rho_A$ and $\rho_B$ have the same spectrum, and then $\rho=|\psi\rangle\!\langle\psi|$ for any $|\psi\rangle$ of the form $$|\psi\rangle = \sum_k \sqrt{\lambda_k} (|\lambda_k^A\rangle\otimes|\lambda_k^B\rangle),$$ where $\rho_A|\lambda_k^A\rangle=\lambda_k|\lambda_k^A\rangle$ and $\rho_B|\lambda_k^B\rangle=\lambda_k|\lambda_k^B\rangle$. The freedom in choosing $|\psi\rangle$ is the same freedom we have in choosing the eigendecomposition of the marginals: in degenerate eigenspaces any orthonormal basis can be chosen.

On the other extreme, if $\rho$ has maximal entropy, then it is bound to be $\rho=I/d^2$ with $d$ dimension of each space, and the condition is satisfied if and only if $\rho_A=\rho_B=I/d$.

Is there anything similar that can be said for the more general case? Either as a general statement without making any assumption on $\rho$, or as a way to classify the possible $\rho$ with fixed entropy and marginals.


Some related questions are What is the most general quantum operation that preserves the marginal? and Constructing a state with constraints on reduced states. The latter is very similar, although it approaches the problem from a different, more numerical, perspective, and it focuses on finding some solution, rather than characterising the set of possible solutions.

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    $\begingroup$ Given that it's a convex set we really just want to know the boundary and it seems that there have been a few works giving necessary and sufficient conditions to be an extremal point of this set, e.g. see this paper. (disclaimer: I've not read the paper myself). $\endgroup$
    – Rammus
    Mar 27 at 12:40
  • $\begingroup$ @Rammus nice find, thanks for the pointer; I'll have a look at the paper $\endgroup$
    – glS
    Mar 27 at 13:29
  • $\begingroup$ I have the feeling I have seen a question here a while ago which was similar (not the ones you link). $\endgroup$ Mar 27 at 15:15
  • $\begingroup$ I'm not sure I understand the question. What are the conditions you want to have on your global state? Otherwise, $\rho_A\otimes \rho_B$ is always a valid solution. Can you formulate a clear question? $\endgroup$ Mar 27 at 15:16
  • $\begingroup$ @NorbertSchuch I'm asking about a characterisation of the full set of $\rho$ whose marginals are the given $\rho_A,\rho_B$, if such a thing is possible. It is at least possible under some additional assumptions, e.g. asking for a pure state satisfying these criteria. Your example works of course, but is only one state compatible with the given marginals. The linked paper above might contain the (an) answer. I'm not completely sure as I haven't had the time to dig into it further yet, but I think their corollary 1 is pretty much the solution (and is also a surprisingly nice characterisation) $\endgroup$
    – glS
    Mar 27 at 21:29
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The case in which the initial state is bipartite and pure has a geometric solution, please see the original construction by Życzkowski and Kuś and the review: by Bengtsson and Życzkowski. It also appears in section 16.4 of the book: GEOMETRY OF QUANTUM STATES An Introduction to Quantum Entanglement.

The characterization of the bipartite pure states having the same marginals is that they lie on the same fiber in the bundle structure of a space of equally entangled states over the space of unitary equivalent density matrices whose nonvanishing eigenvalues are the squares of the original state Schmidt coefficients. This characterization can be explained as follows:

For a bipartite system, the non-vanishing eigenvalues of the reduced density matrix are the squares of the non-vanishing Schmidt coefficients of the initial pure states. The action of a local unitary transformation $U \otimes V$ on the initial state produces a unitarily displaced reduced density matrix by $U$ on $\mathcal{H}_1$ and $V$ on $\mathcal{H}_2$. Thus, the space unitarily equivalent reduced density matrices with a given set of non-vanishing eigenvalues is obtained by tracing the states belonging to the space of equally entangled bipartite pure states, whose nonvanishing Schmidt coefficients are the square roots of the eigenvalues. By equally entangled states, it is meant that they lie on a local unitary transformation orbit.

Said differently, a space of equally entangled bipartite pure states is a fiber bundle over the unitary orbit of the reduced density matrix whose non-vanishing eigenvalues is equal to the initial state Schmidt vector.

The fibers are isomorphic to $U(N)/U(m_0)\times U(1)$, where $N$ is the dimension of the Hilbert space which was traced on and $m_0$ is the number of vanishing Schmidt coefficients. The fibers corresponds to spaces of states having the same reduced density matrix. Please see table 16.1 in the third reference for the various orbit structures in the case of 2 qubits and 2 qutrits.

Update

Given a pure state $|\Psi\rangle$ living in $\mathcal{H}_1\otimes\mathcal{H}_2=\mathbb{C}^n\otimes\mathbb{C}^n $, then its local unitary orbit is

$$\mathcal{O} = \mathrm{Span} \left(U(n)\otimes U(n) \circ \Psi \right)$$

This is the total space of the bundle: $\mathcal{F}\rightarrow \mathcal{O}\rightarrow \mathcal{M}$. Suppose that the Schmidt coefficients of this state are given by: $\mathbf{v_s} = \{\underbrace{0,…}_{m_0 \mathrm{times}}, \underbrace{\lambda_1,…}_{m_1\mathrm{times}}, \underbrace{\lambda_2,…}_{m_2\mathrm{times}}, …\}$

Then (after tracing on the second factor) base space of the bundle is:

$$\mathcal{M} = \mathrm{Span} \left(U(n) \circ \mathrm{tr}_2\left(|\Psi\rangle\langle|\Psi|\right) \right)= \mathrm{Ad}\left(U(n)\right) \mathrm{diag}(|\mathbf{v_s}|^2)= U(n)/U(m_0)\times U(m_1)\times U(m_2)…$$ The fiber of the bundle is: $$\mathcal{F} = U(n)/U(m_0)$$

(The reason that $U(m_0)$ appears in the isotropy group of both the base space and the fiber is that the isotropy group of a unit matrix $\mathbf{1}$ is $U(m_1)$: $\mathbf{1} = U \mathbf{1}U^{\dagger} $, while the isotropy group of a zero matrix is $U(m_0) \times U(m_0)$: $\mathbf{0} = U \mathbf{0}V^{\dagger} $)

The span in the base space is due to the first $U(n)$ factor of the local unitary group $ U(n)\otimes U(n)$, i.e., the factor acting on $\mathcal{H}_1$, while the span in the fiber is the second $U(n)$ factor acting on $\mathcal{H}_2$. The base space $M$ is the unitary orbit of the reduced state, thus, if we wish to characterize states descending to the same reduced state, we have to fix the second $U(n)$ factor, thus remain with a space isomorphic to the fiber $\mathcal{F}$.

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  • $\begingroup$ so if I'm understanding this correctly, you are considering the fiber bundle corresponding to the projection $|\Psi\rangle\mapsto\mathrm{Tr}_B|\Psi\rangle\!\langle\Psi|$, whose base space is the orbit $\{U\rho U^\dagger: U \text{ unitary}\}$ induced by any density matrix $\rho$ with the correct eigenvalues? Doesn't this not take into account the addition freedom you have in also choosing the eigenbasis for the other reduced state? Shouldn't you have a structure with two orbits? $\endgroup$
    – glS
    Apr 9 at 10:11
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    $\begingroup$ The local unitary group is a direct product of the unitary groups of the first and the second Hilbert spaces. In the bundle picture, the base space is a unitary orbit of the first factor, while the fiber is a homogeneous space of the second factor. I have added an update to the answer with a more detailed explanation. $\endgroup$ Apr 13 at 8:50

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