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Let's say we have a pure tripartite state $\rho_{ABE}$ and a completely positive map $\mathcal{R}$, which is defined as:

$$ \mathcal{R} : \rho \rightarrow \sum_j \langle\psi_j|\rho |\psi_j \rangle |\psi_j\rangle\langle\psi_j|, $$ for some density operator $\rho$. Now we apply this map $\mathcal{R}$ to the subsystem $A$ only. So the resulting state is:

$$ \rho'_{ABE} = (\mathcal{R} \otimes \mathcal{I} \otimes \mathcal{I}) (\rho_{ABE}), $$ where $\mathcal{I}$ is the identity map. Now, in this state $\rho'_{ABE}$, how do I prove that:

\begin{equation} H(RB) = H(RE) \tag{1} \end{equation} where, $R$ is the classical state that we get when we apply map $\mathcal{R}$ on subsystem $A$. I know that for pure state $\rho_{ABE}$, any bipartite cut would produce the same entropy, i.e. :

$$ H(AB) = H(E), H(AE) = H(B), H(A) = H(BE). $$ However, I don't think it is the case that, $H(AB) = H(AE)$, isn't it? If so, then how come equation ($1$) is true?

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As $\rho_{ABE}$ is pure we have $\rho_{ABE} = |\psi\rangle \langle \psi|$. We'll rewrite the output of the channel $\mathcal{R}$ as $$ \rho_{ABE}' = \sum_j (P_j \otimes I_{BE}) |\psi\rangle \langle \psi| (P_j \otimes I_{BE}) $$ where $P_j = |\psi_j\rangle \langle \psi_j|$ are a collection of orthogonal rank one projections. Let $\rho_{ABE}^j = (P_j \otimes I_{BE}) |\psi\rangle \langle \psi| (P_j \otimes I_{BE})$. Note that $\rho_{ABE}^j$ is rank one as $\rho_{ABE}^j = |\phi^j\rangle \langle \phi^j|$ with $|\phi^j\rangle = (P_j \otimes I_{BE})|\psi\rangle$.

The important thing to notice is that because the projections are rank one we also have that $\rho_{BE}^j$ is pure. Indeed, $\rho_{BE}^j = |\Phi^j\rangle \langle \Phi^j|$ with $|\Phi_j\rangle = ( \langle\psi_j|\otimes I_{BE})|\psi\rangle$. This means that $$ \rho_B^j = \mathrm{tr}_{AE}[\rho_{ABE}^j] \qquad \text{and} \qquad \rho_E^j = \mathrm{tr}_{AB}[\rho_{ABE}^j] $$ have the same eigenvalues (this can be seen by applying the schmidt decomposition to $\rho_{BE}^j$).

The next thing to show is that this implies that $$ \rho_{AB}' = \sum_j (P_j \otimes I_B) \rho_{AB} (P_j \otimes I_B) \quad \text{and} \quad \rho_{AE}' = \sum_j (P_j \otimes I_E) \rho_{AE} (P_j \otimes I_E) $$ have the same eigenvalues. Note that we can write $$ \rho_{AB}' = \sum_j P_j \otimes \rho_{B}^j. $$ One can now see from the spectral theorem that the eigenvalues of $\rho_{AB}'$ will be the union of the eigenvalues of the different $\rho_{B}^j$. Moreover each eigenvector will be of the form $|\psi_k\rangle \otimes |v^k\rangle$ where $|v^k\rangle$ is an eigenvector of $\rho_{B}^k$ for some $k$. Repeating this for $\rho_{AE}'$ we find that the eigenvalues of $\rho_{AB}'$ and $\rho_{AE}'$ must be equal and hence we also have $H(RB) = H(RE)$.

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  • $\begingroup$ Your answers are always great, Rammus. Thanks a lot! $\endgroup$ Jul 14 at 18:29
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    $\begingroup$ No problem, it's a cute little lemma that I'd never seen before so thanks for the question. Seems like it could be useful in cryptography. $\endgroup$
    – Rammus
    Jul 14 at 18:38
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    $\begingroup$ Indeed! It appeared in this paper: arxiv.org/pdf/0909.0950.pdf , page 6. The result from the paper (H(R|E) + H(S|B) >= some amount)) is used to lower bound Eve's(E) information about Alice's measurement outcomes and consequently proving security of QKD protocols in the crypto community. $\endgroup$ Jul 14 at 18:44

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