# What is the most general quantum operation that preserves the marginal?

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?

• 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 $\sigma_z \neq I$. Mar 10 at 9:21
• @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.
– JRT
Mar 10 at 9:42
• 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?
– glS
Mar 10 at 11:37
• @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 $I_A\otimes U_{BC}$?
– JRT
Mar 11 at 3:18

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 $$\text{CNOT}\,\rho_{AB}\,\text{CNOT}^\dagger = \sigma_{AB}$$.