Consider the single qubit depolarizing noise channel given by

$$\Phi(\rho) = \frac{\lambda}{d} \mathbb{I} + (1- \lambda) \rho.$$

What might be the adjoint $\Phi^{*}(\cdot)$ of this channel? In particular, I am trying to find how the adjoint acts on standard basis states $|0\rangle\langle 0|$ and $|1\rangle\langle 1|$.


1 Answer 1


TL;DR: $\Phi^*=\Phi$.

If quantum channel $\Psi:\mathcal{X}\to\mathcal{Y}$ has a Kraus representation $\Psi(X)=\sum_iK_iXK_i^\dagger$ then its adjoint $\Psi^*:\mathcal{Y}\to\mathcal{X}$ has a Kraus representation $\Psi^*(Y)=\sum_iK_i^\dagger YK_i$, because $$ \begin{align} \langle Y,\Psi(X)\rangle&=\mathrm{tr}\left[Y^\dagger \left(\sum_iK_iXK_i^\dagger\right)\right]\tag1\\ &=\mathrm{tr}\left[\left(\sum_iK_i^\dagger Y^\dagger K_i\right)X\right]\tag2\\ &=\mathrm{tr}\left[\left(\sum_iK_i^\dagger YK_i\right)^\dagger X\right]\tag3\\ &=\langle\Psi^*(Y),X\rangle.\tag4 \end{align} $$ Therefore, every channel with Hermitian or anti-Hermitian Kraus operators is self-adjoint. In particular, all Pauli channels including depolarizing channel are self-adjoint.


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