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

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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.

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  • $\begingroup$ $M_{ij,kl}=\langle i|\sigma|j\rangle\langle l|\sigma^\dagger|j\rangle$ should be $M_{ij,kl}=\langle i|\sigma|k\rangle\langle j|\sigma^\dagger|l\rangle$ ? And can you tell me why $M_{ij,kl}=\langle i|\sigma|j\rangle\langle l|\sigma^\dagger|j\rangle =\langle i,j|\sigma\otimes\sigma^*|k,l\rangle$ ? $\endgroup$
    – karry
    Commented Oct 11, 2022 at 7:45
  • $\begingroup$ Sorry, my indices got a bit jumbled in the middle there! Should be correct now, and hopefully a little clearer as a result. $\endgroup$
    – DaftWullie
    Commented Oct 11, 2022 at 8:00

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