# Prove that for a pure tripartite state $\rho_{ABE}$, $H(RB) = H(RE)$

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?

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 $$H(RB) = H(RE)$$.