# Exercise 7

## from Example Sheet 4

### 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 $$\rho$$.

1. 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.

2. 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 $$\Lambda$$.
1. 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.
2. 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.