Suppose I have two states $\rho_{AB}$ and $\sigma_{AB}$ such that the marginals $\rho_A = \sigma_A$. What is the most general operation that could have acted on $\rho$ to output $\sigma$?

For example, if there was no reduced state constraint, I think one can always find a unitary $U_{ABC}$ where $C$ is a purifying system such that $U_{ABC}\vert\rho_{ABC}\rangle = \vert\sigma_{ABC}\rangle$.

Now enforcing $\rho_A=\sigma_A$, is it true that there exists some unitary $U_{BC}$ that achieves $(I_A\otimes U_{BC})\vert\rho_{ABC}\rangle = \vert\sigma_{ABC}\rangle$? Or some other way to achieve such a transformation?

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    $\begingroup$ There are some fringe cases here that I can think of quickly. For example, for a fixed $\rho_A$, if we have that its support is entirely contained within an eigenspace of a unitary operator $U_A$ with eigenvalue $1$ then $U_A$ will have a trivial action on it. For example $\sigma_z |0\rangle = |0\rangle$ but $\sigma_z \neq I$. $\endgroup$
    – Rammus
    Mar 10, 2021 at 9:21
  • $\begingroup$ @Rammus good point. So I guess the question is whether I can still achieve the transformation using $I_A$ or if there are cases where the transformation requires me to use a unitary $U_A$ with the properties you've pointed out. $\endgroup$
    – JRT
    Mar 10, 2021 at 9:42
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    $\begingroup$ are you asking for the class of maps (or channels?) $\Phi$ such that $\mathrm{Tr}_2[\Phi(\rho)]=\mathrm{Tr}_2[\Phi(\sigma)]$ for all $\rho,\sigma$, or only the class of maps (channels) preserving the marginals of two specific states? $\endgroup$
    – glS
    Mar 10, 2021 at 11:37
  • $\begingroup$ @gIS I'm asking if for any $\rho, \sigma$ such that $\rho_A = \sigma_A$, can we restrict the map $\Phi: \mathcal{H}_{ABC}\rightarrow\mathcal{H}_{ABC}$ that achieves $\Phi(\rho) = \sigma$ to a form like $I_A\otimes U_{BC}$? $\endgroup$
    – JRT
    Mar 11, 2021 at 3:18

1 Answer 1


In the absence of the constraint on the marginal, it is true that there exists $U_{ABC}$ such that $U_{ABC}|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$. Indeed, extend $|\rho_{ABC}\rangle = |\rho_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\rho_{ABC}^{(k)}\rangle$ and $|\sigma_{ABC}\rangle = |\sigma_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\sigma_{ABC}^{(k)}\rangle$ and define $U_{ABC} =\sum_k |\sigma_{ABC}^{(k)}\rangle\langle\rho_{ABC}^{(k)}|$. This construction makes it clear that $U_{ABC}$ is not unique.

Now, enforcing the constraint on the marginal, it is true that for any $\rho_{AB}$ and $\sigma_{AB}$ with $\mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_B(\sigma_{AB})$ there exists $U_{BC}$ that achieves $(I_A\otimes U_{BC})|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$ where $|\rho_{ABC}\rangle$ and $|\sigma_{ABC}\rangle$ denote the respective purifications.

To see this, consider the Schmidt decompositions of the two pure states relative to the partitioning of $ABC$ into $A$ and $BC$

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \mu_i |i_A'\rangle|i_{BC}'\rangle $$

where $|i_A\rangle$, $|i_A'\rangle$, $|i_{BC}\rangle$ and $|i_{BC}'\rangle$ are orthonormal sets of states (not necessarily full bases) and $\lambda_i$ and $\mu_i$ are positive real numbers. Note that

$$ \mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_{BC}(|\rho_{ABC}\rangle\langle\rho_{ABC}|) = \sum_i\lambda_i^2|i_A\rangle\langle i_A| \\ \mathrm{tr}_B(\sigma_{AB}) = \mathrm{tr}_{BC}(|\sigma_{ABC}\rangle\langle\sigma_{ABC}|) = \sum_i\mu_i^2|i_A'\rangle\langle i_A'| $$

and hence $\lambda_i = \mu_i$. Moreover, if $\lambda_i$ are distinct then the eigendecomposition above is unique and hence $|i_A\rangle = |i_A'\rangle$. If they are not, this is not guaranteed, but we can choose $|i_A\rangle = |i_A'\rangle$. In any case, we can write

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle. $$

Finally, set $U_{BC} = \sum_i |i_{BC}'\rangle\langle i_{BC}|$ and observe that

$$ (I_A\otimes U_{BC})|\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle U_{BC}|i_{BC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle = |\sigma_{ABC}\rangle $$

as desired.

I don't know what the most general form of a marginal-preserving unitary is. It is certainly more general than $I\otimes U_{BC}$ or even $U_A\otimes U_{BC}$, because there are examples of entangling unitaries that satisfy the constraint. For example, let $A$ and $B$ denote two qubits and $\rho_{AB} = |00\rangle\langle 00|$ and $\sigma_{AB} = |01\rangle\langle 01|$. Then $\mathrm{tr}_B\rho_{AB} = \mathrm{tr}_B\sigma_{AB} = |0\rangle\langle 0|$, but $\text{CNOT}\,\rho_{AB}\,\text{CNOT}^\dagger = \sigma_{AB}$.


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