Are $n$-qubit Pauli channels $\mathcal E(\rho)=\sum_j p_j P_j \rho P_j$ invertible?

Question

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?

Answer 1

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.