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?


1 Answer 1


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 $C$:

  1. $X$ operators acting only on the latter $n - k$ qubits. These take the code to a unique space orthogonal to the code, so these are pure errors.
  2. Pauli errors that acting only on the first $k$ qubits. These are logical operators.
  3. $Z$ operators acting only on the latter $n - k$ qubits. These act as the identity on $C$, so these are stabilizers.

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 $C'$.

  • $\begingroup$ Thank you for your response. Your detailed explanation was very helpful and informative. $\endgroup$
    – Kmai
    Commented Jun 25 at 8:50

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