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An $n$-qubit Pauli channel $\mathcal E$ acting on a quantum state $\rho$ is of the form \begin{equation}\label{PauliChannel} \mathcal E(\rho)=\sum_jp_jP_j\rho P_j \end{equation} where $p_j\in[0,1]$ is the Pauli error rate associated with the Pauli operator $P_j$ and $P_j$ is an $n$-fold tensor product of Pauli operators $I$, $\sigma_x$, $\sigma_y$, $\sigma_z$.

When is $\mathcal E$ invertible?

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  • $\begingroup$ Hint: You can represent the channel on the bases of Pauli strings. $\endgroup$
    – narip
    May 29, 2023 at 11:45
  • $\begingroup$ related (the depolarizing channel is an example of a Pauli channel that is invertible): quantumcomputing.stackexchange.com/questions/20677/… $\endgroup$
    – forky40
    May 29, 2023 at 17:10
  • $\begingroup$ @forky40 The maximally depolarizing channel is clearly not invertible and hence a counterexample to the question. $\endgroup$
    – Rammus
    May 29, 2023 at 19:39
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    $\begingroup$ Mathematically invertible or physically invertible? $\endgroup$ May 29, 2023 at 21:40
  • $\begingroup$ Some references may be useful: arxiv.org/abs/2012.10959 arxiv.org/abs/2207.01403 They discuss why some noise is mathematically invertible but physically not. $\endgroup$
    – Yuchen Guo
    May 31, 2023 at 11:21

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TL;DR: A Pauli channel has a mathematical inverse if and only if it doesn't vanish on any Pauli operator. The inverse is physical if and only if the channel is unitary. The former follows from diagonalization, the latter from the fact that physically invertible channel must preserve the volume of the set of density operators.

Mathematical inverse

Let $D_n\subset \mathbb{C}^{2^n\times 2^n}$ denote the set of $n$-qubit density matrices and let $N:=4^n-1$. For any Pauli operator $P_k$ $$ \mathcal{E}(P_k)=\left[\sum_{i\in\mathcal{C}(k)}p_i-\sum_{i\in\mathcal{A}(k)}p_i\right]P_k\tag1 $$ where $\mathcal{C}(k)\subset\{0,\dots,N\}$ is the set of indices of Pauli operators that commute with $P_k$ and $\mathcal{A}(k)\subset\{0,\dots,N\}$ is the set of indices of Pauli operators that anticommute with $P_k$. Define $$ q_k=\sum_{i\in\mathcal{C}(k)}p_i-\sum_{i\in\mathcal{A}(k)}p_i\tag2 $$ so that $\mathcal{E}(P_k)=q_kP_k$. In other words, the matrix of $\mathcal{E}$ in the Pauli basis is diagonal $$ K(\mathcal{E})=\begin{bmatrix} 1&0&\dots&0\\ 0&q_1&\dots&0\\ \vdots&\vdots&&\vdots\\ 0&0&\dots&q_N \end{bmatrix}\tag3 $$ with $q_k\in[-1,1]$. Thus, as a linear map, $\mathcal{E}$ has an inverse if and only if $q_k\ne 0$ for all $k=1,\dots,N$, i.e. if and only if $\mathcal{E}(P_k)\ne 0$ for every Pauli operator $P_k$. This is guaranteed for example when one of the $p_j$ is larger than the sum of all the others.

Physical inverse

Now, let's assume that $\mathcal{E}^{-1}$ exists and let's consider the question when it is physical. This is clearly the case when exactly one of $p_j$ is non-zero, i.e. when $\mathcal{E}$ is unitary, so let's assume that two or more $p_j$ are non-zero. In this case, there is a $q_k$ with $|q_k|<1$. But $K(\mathcal{E})$ is the Jacobian matrix of $\mathcal{E}$, so $\mathcal{E}$ sends $D_n$ to a set $\mathcal{E}(D_n)\subset D_n$ of smaller volume $$ V(\mathcal{E}(D_n))=\left|\det K(\mathcal{E})\right|\cdot V(D_n)=V(D_n)\prod_{k=1}^N|q_k|<V(D_n).\tag4 $$ Similarly, $V(\mathcal{E}^{-1}(D_n))>V(D_n)$. But then, $\mathcal{E}^{-1}(D_n)\setminus D_n$ is non-empty, so $\mathcal{E}^{-1}(D_n)$ includes a point that fails to correspond to a physical state and $\mathcal{E}^{-1}$ is not a quantum channel.

We conclude that $\mathcal{E}^{-1}$ exists and is a quantum channel if and only if exactly one of $p_j$ is non-zero. In this case, both $\mathcal{E}$ and $\mathcal{E}^{-1}$ are unitary.

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    $\begingroup$ Beside the insightful explanation. Could you please explain to me how this differ from my statement? Or maybe why my statement seems considered wrong? $\endgroup$ May 29, 2023 at 23:16

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