Show that all extensions of $\rho$ can be obtained as a channel applied to its purification

Question

I am struggling with this exercise here:

Let $H:A, H_E$ and $H_{E′}$ denote complex Euclidean spaces. Consider a purification $|ψ_{AE}⟩⟨ψ_{AE}| ∈ D(H_A ⊗ H_E)$ of a quantum state $ρ_A ∈ D(H_A)$ and a quantum state $σ_{AE′} ∈ D(H_A ⊗ H_{E′})$ such that $Tr_{E′} [σ_{AE′}] = ρ_A$.

Show that there exists a quantum channel $S : B(H_E) → B(H_{E′})$ such that $(id_A ⊗ S) (|ψ_{AE}⟩⟨ψ_{AE}|) = σ_{AE′}$.

Attempt: I know that since pure states are the extreme points of the set of quantum states (which is convex), it suffices to show this for pure states.

To show for $ σ_{AE′}$ being a pure state$(σ_{AE′}=|ψ_{AE'}⟩⟨ψ_{AE'}|)$. I try using the fact that for different purifications of the same state, we have an isometry V s.t. $(id_A \otimes V)|ψ_{AE}⟩=|ψ_{AE'}⟩$ but i get stuck at: $$ σ_{AE′}=(id_A \otimes V)|ψ_{AE'}⟩⟨ψ_{AE'}|(id_A \otimes V^{\dagger}) $$

Answer 1

Changing your notation for clarity, for a fixed state $\rho_A$, we are given $\sigma_{AE'}\in \text{D}(\mathcal{H}_{AE'})$ and $|\psi_{AE}\rangle \in \mathcal{H}_{AE}$ such that \begin{equation} \text{Tr}_{E'}(\sigma_{AE'}) = \rho_A = \text{Tr}_{E}|\psi_{AE}\rangle \langle \psi_{AE}| \tag{1} \end{equation} You've already found the answer for when $\sigma_{AE'}$ is a pure state $|\phi_{AE'}\rangle \langle \phi_{AE'}|$: Since purifications are isometrically related, then implies that there is some isometry $V: \mathcal{H}_E \rightarrow \mathcal{H}_{E'}$ such that \begin{equation} |\phi_{AE'}\rangle = (I_A \otimes V)|\psi_{AE}\rangle, \tag{2} \end{equation} and the channel you want is just an isometry channel $\mathcal{V}: \mathcal{L}(\mathcal{H}_E) \rightarrow \mathcal{L}(\mathcal{H}_{E'})$ defined by conjugation $\mathcal{V}(\rho):= V\rho V^\dagger$. Then, from Eq. (2) we have \begin{equation} (I_A\otimes \mathcal{V})(|\psi_{AE}\rangle \langle \psi_{AE}|) = \sigma_{AE'}. \tag{3} \end{equation}

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More generally, for a mixed state $\sigma_{AE'}$ you can just follow the same procedure but use a purification of $\sigma_{AE'}$: Let $R$ be a system purifying $\sigma_{AE'}$, containing a state $|\phi_{AE'R}\rangle$ satisfying \begin{equation} \sigma_{AE'} = \text{Tr}_R |\phi_{AE'R}\rangle \langle \phi_{AE'R} | \tag{4} \end{equation} From Eq. (1), we know that $|\phi_{AE'R}\rangle$ is also a purification for $\rho_A$: \begin{equation} \text{Tr}_{E'R}|\phi_{AE'R}\rangle \langle \phi_{AE'R} | = \text{Tr}_{E'}(\sigma_{AE'}) = \rho_A \tag{5} \end{equation} Since purifications are isometrically related, there is some $W: \mathcal{H}_E \rightarrow \mathcal{H}_{E'R}$ such that \begin{equation} |\phi_{AE'R}\rangle = (I_A \otimes W)|\psi_{AE}\rangle. \tag{6} \end{equation} Then, defining $\mathcal{W}: \mathcal{L}(\mathcal{H}_E) \rightarrow \mathcal{L}(\mathcal{H}_{E'R})$ according to $\mathcal{W}(\rho):= W\rho W^\dagger$, we arrive at \begin{equation} (I_A\otimes \mathcal{S})(|\psi_{AE}\rangle \langle \psi_{AE}|) = \sigma_{AE'} \tag{7} \end{equation} with $\mathcal{S} = \text{Tr}_R \circ \mathcal{W}$.

Answer 2

1. Any purification of a state $\rho$ can be written as $\operatorname{vec}(\sqrt\rho V^\dagger)$ for some partial isometry $V$. More specifically, $V$ is an isometry on the support of $\rho$, so an isometry if $\rho$ is full rank, and a partial isometry in the general case. This is an equivalent way to say that purifications all have the form $$\sum_k \sqrt{p_k} \,|u_k\rangle\otimes |v_k\rangle$$ for some orthonormal set of vectors $|v_k\rangle$ (possibly living in a larger space than $\rho$). Here $\rho=\sum_k p_k |u_k\rangle\!\langle u_k|$ is the eigendecomposition of $\rho$.

   A concise way to show the equivalence of these two formulations is to write $$\sum_k \sqrt{p_k}|u_k\rangle\otimes|v_k\rangle=\operatorname{vec}\left(\sum_k \sqrt{p_k} |u_k\rangle\!\langle \bar v_k|\right) = \operatorname{vec}(\sqrt\rho V^\dagger),\\ \text{where } V\equiv \sum_k |\bar v_k\rangle\!\langle u_k|.$$

2. It follows that any extension of $\rho$ can be obtained as the partial trace of the form above. This is because given any (possibly non-pure) extension, you can purify that extension, and then suitably partial tracing you get back $\rho$. In other words, all extensions $\tilde\rho$ can be written as $$\tilde\rho = \operatorname{tr}_S[ \mathbb{P} (\operatorname{vec}(\sqrt\rho V^\dagger))], \tag2$$ for some partial isometry $V$ and some subset $S$ of the ancillary degrees of freedom to trace out. Here I'm using the shorthand notation $\mathbb{P}(v)\equiv vv^\dagger$, or $\mathbb{P}(|v\rangle)\equiv |v\rangle\!\langle v|$ in bra-ket notation.

   Equation (2) is pretty much the answer to the question, because you can take as channel $\Phi(\rho)\equiv \operatorname{tr}_S \rho$. You thus just showed that there's always a purification such that the extension can be written as the partial trace (ie "some channel") of that purification. If the question fixed a specific choice of purification, then you can get the one above via some partial isometry applied to that purification, composed with the partial trace, and you reach the same conclusion.

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Some additional possibly relevant information:

• A closely related question about how different purifications are related is Prove that different purifications of a state can be mapped into one another via local unitaries, and this other answer of mine on math.SE.
• A related result is the following: if $|\psi\rangle$ and $|\phi\rangle$ are bipartite pure states such that $\operatorname{tr}_2\mathbb{P}_\psi=\operatorname{tr}_2\mathbb{P}_\phi$, then there's a unitary $U$ such that $|\psi\rangle=(I\otimes U)|\phi\rangle$. This follows immediately observing that $\operatorname{tr}_2\mathbb{P}_\psi=\psi\psi^\dagger$ with $|\psi\rangle=\operatorname{vec}(\psi)$, and the characterisation of matrices $A,B$ such that $AA^\dagger=BB^\dagger$, discussed e.g. here.
• A slight generalisation of the above is the following: if $\operatorname{tr}_2\mathbb{P}_\psi=\operatorname{tr}_{2,3}\mathbb{P}_\phi$ (which means now $|\phi\rangle$ lives in a higher-dimensional space), then $|\phi\rangle=(I\otimes V)|\psi\rangle$ for some isometry $V$. The proof is based on essentially the same ideas used above. As a toy example you can consider $|\phi\rangle=|000\rangle+|111\rangle$ and $|\psi\rangle=|00\rangle+|11\rangle$.
• If a generic bipartite state $\rho$ and a pure bipartite state $|\psi\rangle$ are such that $\operatorname{tr}_2\rho=\operatorname{tr}_2\mathbb{P}_\psi$, then there is some channel $\Phi$ such that $\rho=(I\otimes \Phi)\mathbb{P}_\psi$. This result is also discussed in chapter 2 of Watrous' book. It also follows somewhat trivially from the result above because there's an extension $|\phi\rangle$ such that $\rho=\operatorname{tr}_3\mathbb{P}_\phi$, thus we have $\operatorname{tr}_{2,3}\mathbb{P}_\phi=\operatorname{tr}_2\mathbb{P}_\psi$, thus from the result above there's an isometry $V$ such that $|\phi\rangle=(I\otimes V)|\psi\rangle$, and thus $$\rho=\operatorname{tr}_2\mathbb{P}_\phi = \operatorname{tr}_2[ (I\otimes V)\mathbb{P}_\psi(I\otimes V^\dagger)] = (I\otimes\Phi)\mathbb{P}_\psi$$ where $\Phi\equiv \operatorname{tr}_2\circ \Phi_V$ is the channel with Stinespring dilation $\Phi(X)=\operatorname{tr}_2[VXV^\dagger]$.