Given $\rho\in L(\mathcal H_A)$, can we find $\mathcal H_B\le \mathcal H_A$ such that ${\rm tr}_C(\rho)=\sigma$, for fixed $\sigma$?

Question

Given an arbitrary density matrix $\rho \in L({\mathcal H_{A}})$, can one always find a subspace ${\mathcal H_{B}}$ of ${\mathcal H_{A}}$ such that ${\mathcal H_{A}}={\mathcal H_{B}}\otimes{\mathcal H_{C}}$ and ${\rm tr}_{C}(\rho)=\sigma$ for a fixed density matrix $\sigma \in L({\mathcal H_{B}})$?

While a purification of $\sigma$ yields a pure state $| \Psi \rangle \langle \Psi |$ with the property ${\rm tr}_{C}(| \Psi \rangle \langle \Psi |)=\sigma$, I am wondering whether one can do the same given a fixed (possibly mixed) state $\rho$.

If this is possible, what would be the restrictions on the dimension of ${\mathcal H_{A}}$ (in relation to the size of ${\mathcal H_{C}}$)?

Answer 1

First of all, your question should be more carefully formulated, since it is not even possible to always find a non-trivial subsystem (not subspace) of $\mathcal H_A$, see also my comment here Can a single qutrit in superposition be considered entangled?

Thus, let us assume that $\mathcal H_A$ does not have prime dimension, $d_A=\dim\mathcal H_A$. The question should then be the following: Given a Hilbert space $\mathcal H_B$ where $d_B=\dim\mathcal H_B$ divides $d_A$, and a density matrices $\rho\in\mathcal H_A$ and $\sigma\in\mathcal H_B$, is there a choice of tensor product $\mathcal H_A \simeq \mathcal H_B\otimes\mathcal H_C$, such that $\mathrm{tr}_C(\rho)=\sigma$.

I'm not so sure why you would call this a "purification", though.

This can sometimes be true. For example, take for simplicity $\mathcal H_B=\mathcal H_C=\mathbb C^d\otimes\mathbb C^d$ and define $\mathcal H_A := \mathcal H_B\otimes \mathcal H_C$. Let $\phi^+$ be the standard maximally entangled state with respect to the bipartition of $\mathcal H_C=\mathcal H_B$. Consider the product state $\rho=|\phi^+\rangle\langle\phi^+|\otimes|\phi^+\rangle\langle\phi^+|$, and $\sigma = \mathbb{I}/d^2$. Then there is always a different bipartition $A=B'|C'$ of $A$, such that $\mathrm{tr}_{C'}(\rho) = \sigma$. You can explicitly verify this by writing down the state and swapping the second tensor factors of $B$ and $C$.

Update: However, in general, this is false since the partial trace make the state $\rho$ "more mixed". As pointed put by Danylo, the to states should fulfill some majorisation condition about their spectrum.

Here's a counter-example. Take $\rho = \mathbb{I}/d_A$. Then, for any choice of bipartition, $\mathrm{tr}_C(\rho) = \mathbb{I}/d_B$. Take $\sigma$ to be any other state, e.g. a pure one.

Answer 2

If i understand your question correctly, no, it's not possible. Take for example $$ \rho = | 0 \rangle \langle 0|^{\otimes |A|} \hspace{0.2em} \in D(\mathcal{H_A}) $$ For every choice $ \mathcal{H_B}, \mathcal{H_C} $ such that $ \mathcal{H_A} = \mathcal{H_B} \otimes \mathcal{H_C} $ we have $$ \text{Tr}_C [ \rho ] = | 0 \rangle \langle 0|^{\otimes |B|} $$ so it can not be the case that $ \text{Tr}_C [ \rho ] = \sigma $ for an arbitary fixed $ \sigma \in D(\mathcal{H_B}) $.