# Why can Pauli errors $E$ be decomposed as $E=T(S)LG$ with $T(S)$ "pure errors"?

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]

$$$$E = T(S)LG$$$$

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 $$C$$:

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 $$C'$$.

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