# 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}