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

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

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 $$$$\text{Tr}_{E'}(\sigma_{AE'}) = \rho_A = \text{Tr}_{E}|\psi_{AE}\rangle \langle \psi_{AE}| \tag{1}$$$$ You've already found the answer for when $$\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'}$$ such that $$$$|\phi_{AE'}\rangle = (I_A \otimes V)|\psi_{AE}\rangle, \tag{2}$$$$ and the channel you want is just an isometry channel $$\mathcal{V}: \mathcal{L}(\mathcal{H}_E) \rightarrow \mathcal{L}(\mathcal{H}_{E'})$$ defined by conjugation $$\mathcal{V}(\rho):= V\rho V^\dagger$$. Then, from Eq. (2) we have $$$$(I_A\otimes \mathcal{V})(|\psi_{AE}\rangle \langle \psi_{AE}|) = \sigma_{AE'}. \tag{3}$$$$

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 $$$$\sigma_{AE'} = \text{Tr}_R |\phi_{AE'R}\rangle \langle \phi_{AE'R} | \tag{4}$$$$ From Eq. (1), we know that $$|\phi_{AE'R}\rangle$$ is also a purification for $$\rho_A$$: $$$$\text{Tr}_{E'R}|\phi_{AE'R}\rangle \langle \phi_{AE'R} | = \text{Tr}_{E'}(\sigma_{AE'}) = \rho_A \tag{5}$$$$ Since purifications are isometrically related, there is some $$W: \mathcal{H}_E \rightarrow \mathcal{H}_{E'R}$$ such that $$$$|\phi_{AE'R}\rangle = (I_A \otimes W)|\psi_{AE}\rangle. \tag{6}$$$$ Then, defining $$\mathcal{W}: \mathcal{L}(\mathcal{H}_E) \rightarrow \mathcal{L}(\mathcal{H}_{E'R})$$ according to $$\mathcal{W}(\rho):= W\rho W^\dagger$$, we arrive at $$$$(I_A\otimes \mathcal{S})(|\psi_{AE}\rangle \langle \psi_{AE}|) = \sigma_{AE'} \tag{7}$$$$ with $$\mathcal{S} = \text{Tr}_R \circ \mathcal{W}$$.

• 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? Mar 31 at 7:55
• 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 $\rho_A$. Mar 31 at 15:05
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.

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