Generalized "purification" scheme

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 $${\mathcal H_{C}}$$)?

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.

• Entropy is not a correct indicator here, consider the case when $\rho$ is maximally mixed (then $\sigma$ is also maximally mixed but in a lower dimension, so $S(\sigma) < S(\rho)$). It has to be $\Lambda(\sigma) \prec \Lambda(\rho)$, i.e. the spectrum of $\sigma$ majorized by the spectrum of $\rho$. And I think this is a sufficient condition. Oct 15 '20 at 8:57
• Thanks! You are correct, the entropy depends on the dimension ... I was thinking about the spectrum in the first place and this seems very plausible! Oct 15 '20 at 9:06
• $\Lambda(\sigma) \prec \Lambda(\rho)$ is not quite correct too, since they are also in different dimensions. But it's definitely just some condition on both spectrums. I'll check this later. Oct 15 '20 at 9:09
• It turned out to be harder than I thought. A known majorization relations (arxiv.org/abs/quant-ph/0008073) are not directly applicable in this case. Oct 16 '20 at 6:04

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 $$\sigma \in D(\mathcal{H_B})$$.

• Unfortunately, this argumentation is not correct. The tensor product structure used to define $\rho$ can be different from the one used to factor $A$ into $B$ and $C$. I'll try to comment on this below Oct 15 '20 at 8:03
• Just read your answer but still can't see how my argument does not provide a valid "counter-example". Could you please provide more details? Oct 15 '20 at 8:48
• The choice of a tensor product structure for a Hilbert space $\mathcal H_A$ is the choice of an explicit unitary isomorphism $f:\, \mathcal H_A \rightarrow \mathcal H_B \otimes \mathcal H_C$. In your case, set for simplicity $\mathcal H_A:=\mathbb C^d \otimes \mathbb C^d$. Then, we can arbitrarily "twist" the space by a unitary, e.g. choose $f = CX\cdot H^{\otimes 2}$. Then, the image of $\rho$ under $f$ is the maximally entangled state and the reduced state will be maximally mixed. The question is under which circumstances can we find a suitable $f$ for given $\rho$ and $\sigma$. Oct 15 '20 at 11:03
• Great! Thank you so much for clarifying this! Oct 15 '20 at 11:46
• Is it like we are allowed to evolve our "initial" state $\rho$ with arbitrary unitary such that in the final state a part of the system is in $\sigma$? If so, can we restate the problem as finding a quantum channel $\Phi \in C(\mathcal{H}_A, \mathcal{H}_B)$ such that $\Phi(\rho) = \sigma$ and $rank(J(\Phi)) = \frac{d_A}{d_B}$ ? Oct 16 '20 at 10:52