Exercise 5

from Example Sheet 3

Exercise 5.

Show that the fidelity is multiplicative under tensor products. That is, \[ F(\rho_1 \otimes \rho_2 , \sigma_1\otimes \sigma_2) = F(\rho_1,\sigma_1) F(\rho_2, \sigma_2). \]

Hint: first show that \(\|A\otimes B\|_1 = \|A\|_1\|B\|_1\).

First, we have that \[\begin{aligned} \| A\otimes B\|_1 &= \operatorname{tr}[ \sqrt{(A^\dagger\otimes B^\dagger)(A\otimes B)} ]\\ &= \operatorname{tr}[ \sqrt{A^\dagger A \otimes B^\dagger B}]\\ &= \operatorname{tr}[ \sqrt{A^\dagger A} \otimes \sqrt{B^\dagger B}]\\ &= \operatorname{tr}[ \sqrt{A^\dagger A}] \operatorname{tr}[ \sqrt{B^\dagger B} ]\\ &= \|A\|_1 \|B\|_1. \end{aligned}\] We used that the square root of the tensor product of two operators \(X,Y\geq 0\) has \(\sqrt{X\otimes Y} = \sqrt{X}\otimes \sqrt{Y}\). We can see this by using their spectral decompositions: let \(X =\sum_i \lambda_i P_i\) where \(\lambda_i\) are the eigenvalues of \(X\) and \(P_i\) the associated eigenprojections, and similarly \(Y = \sum_i \mu_i Q_i\). Then \(X\otimes Y = \sum_{ij} \lambda_i \mu_j P_i \otimes Q_j\) is the spectral decomposition of \(X\otimes Y\). Therefore \[ \sqrt{X \otimes Y} = \sum_{ij} \sqrt{\lambda_i \mu_j} P_i \otimes Q_j = \sum_i \sqrt{\lambda_i} P_i \otimes \sum_j \sqrt{\mu_j} Q_j = \sqrt{X}\otimes \sqrt{Y}. \] Then, \[\begin{aligned} F(\rho_1 \otimes \rho_2 , \sigma_1\otimes \sigma_2) &= \| \sqrt{\rho_1 \otimes \rho_2} \sqrt{\sigma_1\otimes \sigma_2}\|_1\\ &=\| (\sqrt{\rho_1} \otimes \sqrt{\rho_2}) (\sqrt{\sigma_1}\otimes \sqrt{\sigma_2})\|_1\\ &=\| (\sqrt{\rho_1}\sqrt{\sigma_1} \otimes \sqrt{\rho_2} \sqrt{\sigma_2})\|_1\\ &=\| (\sqrt{\rho_1}\sqrt{\sigma_1}\|_1\|\sqrt{\rho_2} \sqrt{\sigma_2})\|_1\\ &= F(\rho_1,\sigma_1) F(\rho_2,\sigma_2). \end{aligned}\]