I have a question about the decomposition of Pauli errors. Pauli error $E \in \{I,X,Y,Z\}^{{\bigotimes}n}$ that satisfies the syndrome $S$ can be decomposed into a product of pure error $T(S)$, Logical operator $L$, stabilizer operator $G$ as follows [1]
\begin{equation} E = T(S)LG \end{equation}
I have two questions: First, what is "pure error"? Second, why is this decomposition possible?
For stabilizer codes, a pure error is basically a correctable error that is not a stabilizer. This means that a pure error takes the code to a space orthogonal to the code (i.e. a space corresponding to a non-identity syndrome). As convention, the identity is considered to be a pure error. The idea is that a pure error corresponds uniquely to an error syndrome.
For the second question, we first take a look at a special case. Consider the code $C \subseteq (\mathbb{C}^2)^{\otimes n}$ consisting of codewords of the form $\vert{\psi}\rangle\vert{0}\rangle^{\otimes (n - k)}$ where $\vert{\psi}\rangle$ is a state of the first $k$ qubits. There are three types of Pauli errors we can specify in relation to $C$:
All Pauli errors are a product of these three types of errors (along with multiplication by imaginary $i$), so the mentioned decomposition can be realized using the anticommuting property of Pauli errors.
Now for a general $n$-qubit stabilizer code $C'$ with $n - k$ stabilizer generators, there is an encoding unitary $U$ that takes $C$ to $C'$. $U$ is in the Clifford group, so if $E$ is a Pauli error then $UEU^\dagger$ is also a Pauli error. Also, if $E$ is a Pauli error that is a pure error/logical operator/stabilizer for $C$, then $UEU^\dagger$ is a pure error/logical operator/stabilizer for $C'$. This gives the mentioned decomposition of Pauli errors for $C'$.
