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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: $$ σ_{AE′}=(id_A \otimes V)|ψ_{AE'}⟩⟨ψ_{AE'}|(id_A \otimes V^{\dagger}) $$

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2 Answers 2

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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}


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 $\mathcal{S} = \text{Tr}_R \circ \mathcal{W}$.

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  • $\begingroup$ Could you use the fact that the pure state are the extreme points of the set of states, and that this set is convex to then write S as a convex combination of V? $\endgroup$ Mar 31 at 7:55
  • $\begingroup$ I'm not quite sure what you mean. But something that comes to mind is, if a mixed state satisfies $\text{Tr}_{E'}\sigma_{AE'} =\rho_A$ this does not imply that $\sigma_{AE'}$ is a convex combination of purifications of $\rho_A$. $\endgroup$
    – forky40
    Mar 31 at 15:05
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  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.


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 $\Phi(X)=\operatorname{tr}_2[VXV^\dagger]$.
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