### Exercise 7.

**Entropy exchange**. The entropy exchange for a state \(\rho\) and a quantum channel \(\Lambda\) is defined as follows: \[
S(\rho, \Lambda):= S(\rho_{RQ}^\prime),
\] where \(\rho_{RQ}^\prime= ({\rm id}_R \otimes \Lambda) \psi^\rho_{RQ}\), with \(\psi^\rho_{RQ}\) being a purification of \(\rho\).

Prove that \(S(\rho, \Lambda) = S(\rho_E^\prime)\), where \(\rho_E^\prime = \operatorname{tr}_{RQ} (\rho_{RQE}^\prime)\) with \[ \rho_{RQE}^\prime = (I_R \otimes U_{QE}) (\psi^\rho_{RQ} \otimes |0_E\rangle \langle 0_E|) (I_R \otimes U_{QE}^\dagger), \] with \(U_{QE}\) being the Stinespring dilation of the channel \(\Lambda\).

*Thus the entropy exchange can be interpreted as the amount of entropy introduced by \(\Lambda\) into an initially pure environment.*- Prove that the entropy exchange can also be written in the form \[ S(\rho, \Lambda) = S(W) = - \operatorname{tr}W \log W, \] where \(W\) denotes a matrix with elements \(W_{ij} = \operatorname{tr}(A_i \rho A_j^\dagger)\), where \(\{A_i\}\) denote a set of Kraus operators of \(\Lambda\).

- We can see that indeed, \(\rho'_{RQ} = \operatorname{tr}_E \rho'_{RQE}\). Moreover, \(\rho'_{RQE}\) is a pure state as \(\psi_{RQ}^\rho\otimes | 0_E \rangle\langle 0_E |\) is pure, and we conjugate by a unitary, \(I_R\otimes U_{QE}\). Thus, the Schmidt decomposition tells us that across any bipartite partition, the reduced density matrices have the same non-zero eigenvalues, and thus the same entropy. That is, \(S(\rho'_{RQ}) = S(\rho'_E)\) as desired.
- We can trace out \(R\) to find \[\begin{aligned} \rho'_E = \operatorname{tr}_Q [U_{QE} (\rho_Q \otimes |0_E\rangle \langle 0_E|) U_{QE}^\dagger]. \end{aligned}\] Now, we can choose our Strinespring representation to act as \(U | \psi_A \rangle\otimes | 0_E \rangle = \sum_i A_i | \psi_A \rangle\otimes | i \rangle\) for any pure state \(| \psi_A \rangle\) (as shown in the notes where we construct \(U\)), where \(\{A_i\}\) is a Kraus representation of \(\Lambda\). Now, let \(\rho_Q = \sum_\alpha \lambda_\alpha | \psi_\alpha \rangle\langle \psi_\alpha |\) be a decomposition into pure states. Then \[\begin{aligned} \rho'_E &=\sum_\alpha \lambda_\alpha \operatorname{tr}_Q [U_{QE} (| \psi_\alpha \rangle\langle \psi_\alpha | \otimes |0_E\rangle \langle 0_E|) U_{QE}^\dagger]\\ &=\sum_\alpha \lambda_\alpha \sum_{i,j} \operatorname{tr}_Q [ A_i| \psi_\alpha \rangle\langle \psi_\alpha |A_j^\dagger \otimes |i_E\rangle \langle j_E|]\\ &=\sum_\alpha \lambda_\alpha \sum_{i,j} \operatorname{tr}[ A_i| \psi_\alpha \rangle\langle \psi_\alpha |A_j^\dagger] |i_E\rangle \langle j_E| \\ &= \sum_{i,j} \operatorname{tr}[ A_i \rho_Q A_j^\dagger] |i_E\rangle \langle j_E|. \end{aligned}\] Therefore, \(W:= \rho'_E\) has a matrix representation where the matrix elements are given by \(\operatorname{tr}[ A_i \rho_Q A_j^\dagger]\). By Step (1), this completes the proof.