I have three qubits in subsystems $A$, $B$, $C$. System $A$ initially contains some state $\rho_A$, and $BC$ contains a bipartite pure state $|\psi\rangle_{BC}$. I apply a unitary operation $U$ acting on $\mathcal{H}_A\otimes \mathcal{H}_B$ and then I trace out system $B$, resulting in the state \begin{equation} \rho_{AC}' = \text{Tr}_B \left((U \otimes I_C)(\rho_A \otimes |\psi\rangle \langle \psi|_{BC})(U^\dagger \otimes I_C)\right) \end{equation} I would like to know, for what conditions on the choice of unitary $U$ will there necessarily exist $\rho_A$ and $|\psi\rangle$ such that the final state $\rho_{AC}'$ is not separable in the space $\mathcal{H}_A\otimes \mathcal{H}_C$?

  • 2
    $\begingroup$ maybe a trivial solution but, what about if $U$ swaps $A$ and $B$. That way you effectively trace out $\rho_B$ at the end and the final state on $AC$ is precisely $|\psi\rangle$, which can be entangled $\endgroup$
    – glS
    Commented Sep 9, 2023 at 2:50
  • $\begingroup$ yes thats a good example - I am hoping for a set of conditions on $U$ such that $AC$ can be entangled. I'll edit to clarify $\endgroup$
    – forky40
    Commented Sep 9, 2023 at 3:13
  • $\begingroup$ is $\rho_{A}$ an arbitary density matrix, or a representation of a pure state? $\endgroup$ Commented Sep 9, 2023 at 8:39
  • $\begingroup$ $\rho_A$ can be an arbitrary qubit state, mixed or pure. $\endgroup$
    – forky40
    Commented Sep 9, 2023 at 15:14

2 Answers 2


This isn't a full solution. More a way to reduce the problem in a more general one, and possibly pinpoint the difficult question that needs answering.

Let's assume $\rho_A$ is pure, $\rho_A\equiv |\phi\rangle\!\langle \phi|$, and write the Schmidt decomposition of $|\psi\rangle_{BC}$ as $$|\psi\rangle = \sum_k \sqrt{p_k} |k,k\rangle.$$ We can always write it this way in some suitable basis for the spaces $\mathcal H_B$ and $\mathcal H_C$. Before the partial trace on $B$ the state is therefore $$|\Psi_f\rangle\equiv (U_{AB}\otimes I_C)(|\phi\rangle\otimes |\psi\rangle) = \sum_k \sqrt{p_j} \big[ (U_{AB}(|\phi\rangle_A\otimes|j\rangle_B))\otimes |j\rangle_C\big].$$ Define $V:\mathcal H_B\to \mathcal H_A\otimes \mathcal H_B$ as the isometry with action $V|j\rangle_B\equiv U_{AB}(|\phi\rangle_A\otimes|j\rangle_B)$. This is purely for notational conciseness, it's not strictly required, but allows to write the state concisely as $$|\Psi_f\rangle = \sum_k \sqrt{p_k} (V|j\rangle\otimes |j\rangle).\tag X$$ The rationale is that asking "is there a unitary $U_{AB}$ such that there are $|\phi\rangle_A$ and $|\psi\rangle_{BC}$ such that XYZ?" is the same as asking "is there an isometry $V$ such that there is $|\psi\rangle_{BC}$ such that XYZ?", and we can reduce the "elementary objects" involved in the question.

We thus translate the question in the following on: given a generic isometry $V$, are there bases for $\mathcal H_B,\mathcal H_C$ and weights $p_k\ge0, \sum_k p_k=1$ such that the $|\Psi_f\rangle$ in (X) is entangled after partial tracing $\mathcal H_B$? In this terms, this seems tightly related to the following more general question: given a tripartite state $|\Psi_{ABC}\rangle$ that is entangled in the bipartition $AB:C$, when is the reduced state $\operatorname{tr}_B[|\Psi_{ABC}\rangle\!\langle\Psi_{ABC}|]$ still entangled? I don't have a good answer for this. I would have guessed that you get a separable state whenever you have maximal entanglement in the bipartition $B:AC$, but this simple condition appears to not be quite right: in such situations you get a balanced mixture of projection onto orthogonal states in $AC$, but this can still be entangled. See eg examples below:

Consider as a toy example $|\Psi_f\rangle = |000\rangle+|111\rangle$. Then partial tracing $B$ we get $\mathbb{P}_{00}+\mathbb{P}_{11}$ where $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$, which is separable.

If instead $$|\Psi_f\rangle\equiv (|00\rangle+|11\rangle)|0\rangle+(|00\rangle-|11\rangle)|1\rangle$$ then partial tracing we get $\mathbb{P}_0\otimes\mathbb{P}_- + \mathbb{P}_1\otimes\mathbb{P}_-$ which is again separable.

Yet another example is $$|\Psi_f\rangle\equiv (|00\rangle+|11\rangle)|0\rangle+(|01\rangle+|10\rangle)|1\rangle$$ then partial tracing we get $\mathbb{P}(|00\rangle+|11\rangle)+\mathbb{P}(|01\rangle+|10\rangle)$. This is an $X$ state, which is also separable (see e.g. https://quantumcomputing.stackexchange.com/a/5565/55).

Note how in all of these examples, $B$ is maximally entangled with $AC$. As an example of this not being the case, say $$|\Psi_f\rangle \equiv |00\rangle|0\rangle+\left(\frac{|01\rangle+|10\rangle)}{\sqrt2}\right) |1\rangle = |0\rangle_B\left(|00\rangle+\frac{1}{\sqrt2}|11\rangle\right)_{AC} + \frac{1}{\sqrt2}|1\rangle_B |01\rangle_{AC},$$ where the second identity is a rewriting to highlight the entanglement features in the bipartition $B:AC$. Properly renormalising the above you see that the Schmidt coefficients in this bipartition are $(\sqrt3/2,1/2)$, and thus the entanglement is not maximal. And indeed, you can easily observe eg via partial transpose that the reduced state is entangled.

For a state that violates the pattern above, consider instead $$|\Psi_f\rangle = (V\otimes I_C) (\sqrt3/2 |00\rangle+1/2|11\rangle), \\ V|0\rangle\equiv (3/2)^{-1/2} (|00\rangle+\frac{1}{\sqrt2}|11\rangle), \qquad V|1\rangle \equiv |01\rangle. $$ Isolating $B$ you get the maximally entangled state $$|0\rangle_B\otimes|00\rangle_{AC} + |1\rangle_B\otimes\left(\frac{|01\rangle+|10\rangle}{\sqrt2}\right)_{AC}.$$ However, tracing out $B$ you still get an entangled state, as again verifiable eg via partial transpose.

The question seems related to the topic of robustness of entanglement, so a look at the topical literature there might help. Eg https://arxiv.org/abs/1809.00622 is a start.


This is not a fully characterised solution, but perhaps you will find it helpful...

We start by writing the initial $|\psi\rangle$ in its Schmidt basis, $$ |\psi\rangle=\alpha|00\rangle+\beta|11\rangle. $$ I'm going to assume that the input of the other qubit is pure for simplicity, though I don't think it makes much difference. Now act the unitary on it. It must produce $$ |\Psi\rangle=\alpha|0\rangle_C|\psi_0\rangle_{AB}+\beta|1\rangle|\psi_1\rangle. $$ We take the partial trace over $B$, leaving $$ \rho=|\alpha|^2|0\rangle\langle 0|\otimes A+|\beta|^2|1\rangle\langle 1|\otimes D+\alpha \beta^\star|0\rangle\langle 1|\otimes B+\alpha^\star \beta|1\rangle\langle 0|\otimes C $$ where $A,B,C,D$ are $2\times 2$ matrices relating to partial trace of $|\psi_i\rangle\langle\psi_j|$ for different $i,j$. This is just a block matrix $$ \rho=\left(\begin{array}{cc} |\alpha|^2 A & \alpha\beta^\star B \\ \alpha^\star\beta C & |\beta|^2D \end{array}\right). $$ We know that this is a density matrix, so its positive. If $A$ is invertible and $\alpha\neq 0$, then $$ \text{det}(\rho)=|\alpha|^2|\beta|^2\text{det}(A)\text{det}(D-CA^{-1}B)\geq 0. $$ Now, you can take the partial transpose, $$ \rho^{T_C}=\left(\begin{array}{cc} |\alpha|^2 A & \alpha^\star\beta C\\ \alpha\beta^\star B & |\beta|^2D \end{array}\right). $$ The state is entangled if and only if this has a negative eigenvalue. Moreover, it can have at most one negative eigenvalue (according to https://arxiv.org/abs/1304.6775), so its determinant would have to be negative. $$ \text{det}(\rho^{T_C})=|\alpha|^2|\beta|^2\text{det}(A)\text{det}(D-BA^{-1}C). $$ So, aside from the special cases identified (i.e. non-invertible $A$), the conditions on $U$ are that it creates $A,B,C,D$ such that $\text{det}(D-CA^{-1}B)<0$ even though $\text{det}(D-BA^{-1}C)\geq 0$. Quite what that means in terms of $U$, I'm not sure (yet), but it is interesting that it does not depend on the Schmidt coefficients of the initially entangled state!


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