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.