# Why use conjugate not transpose complex conjugate in superoperator？

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

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.
• $M_{ij,kl}=\langle i|\sigma|j\rangle\langle l|\sigma^\dagger|j\rangle$ should be $M_{ij,kl}=\langle i|\sigma|k\rangle\langle j|\sigma^\dagger|l\rangle$ ? And can you tell me why $M_{ij,kl}=\langle i|\sigma|j\rangle\langle l|\sigma^\dagger|j\rangle =\langle i,j|\sigma\otimes\sigma^*|k,l\rangle$ ? Commented Oct 11, 2022 at 7:45