Why use conjugate not transpose complex conjugate in superoperator？

Question

For the n-qubit depolarizing noise, I want to know why it uses $\sigma_{0}^{i*}$ instead of $\sigma_{0}^{i}$ or $\sigma_{0}^{i\dagger}$.

Answer 1

This is just the standard way that you write down the superoperator. Think about the following:

If I have $$ \rho\mapsto \sigma\rho \sigma^\dagger $$ then if I want to write this as a superoperator, I start with $$ \rho=\sum_{ij}\rho_{ij}|i\rangle\langle j| $$ and rewrite it as a vector $$ |\rho\rangle=\sum_{i,j}\rho_{i,j}|i,j\rangle. $$ The the superoperator is the matrix $M$ such that $$ M|\rho\rangle=|\sigma\rho \sigma^\dagger\rangle. $$ Thus, $M$ is written $$ M=\sum_{i,j,k,l}M_{ij,kl}|i,j\rangle\langle k,l|. $$ We have $$ \sum_{kl}M_{ij,kl}\langle k,l|\rho\rangle=\langle i,j|\sigma\rho\sigma^\dagger\rangle=\langle i|\sigma\rho\sigma^\dagger|j\rangle. $$ Hence, $$ M_{ij,kl}=\langle i|\sigma|k\rangle\langle l|\sigma^\dagger|j\rangle. $$ Now, transpose the second term, $$ M_{ij,kl}=\langle i|\sigma|k\rangle\langle j|\sigma^\star|l\rangle. $$ You can verify that this is the same as $$ =\langle i,j|\sigma\otimes \sigma^\star|k,l\rangle. $$ (If in doubt, work backwards!) Thus, $$ M=\sigma\otimes\sigma^\star, $$ as required.