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The twirled operation of a quantum channel $\mathcal E$ is defined as \begin{align} \mathcal E_T(\rho) &= \int dU U^\dagger \mathcal E(U \rho U^\dagger)U, \end{align} where the integral is over the normalized Haar measure $dU$.

Let $P$ be a one-dimensional projector, and $Q \equiv I − P$ be the projector onto the orthocomplementary space. Letting $V$ be block diagonal with respect to the spaces onto which $P$ and $Q$ project, we see that $V P V^\dagger = P$ and thus $V \mathcal E_T (P )V^\dagger = \mathcal E_T (P)$. How to prove that $\mathcal E_T(P) = \alpha P+ \beta Q$ for some $\alpha$ and $\beta$?

This result was in this article by Nielsen. enter image description here

glS
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Michael.Andy
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    please try to avoid using screenshots of text. The hinder readability and searchability of the post. You can use > quote blocks to quote text – glS Mar 27 '23 at 12:25
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    Roughly speaking, $V\mathcal{E} _T\left( P \right) V^{\dagger}=\mathcal{E} _T\left( P \right) $ means $\mathcal{E} _T\left( P \right) $ commutes with $V$. Since $V$ can be any unitary as long as it's block diagonal w.r.t. spaces onto which $P$ and $Q$ project, $\mathcal{E} _T\left( P \right) $ can only be an identity in spaces which $P$ and $Q$ project. Think of this simple example: the only possibility that a matrix commutes with any unitary matrices is when it is proportional to the identity. – narip Mar 27 '23 at 13:43
  • possible duplicate of https://quantumcomputing.stackexchange.com/q/23845/55, and links therein – glS Mar 27 '23 at 14:12

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