# How to write the general state of a corrupted n-qubit codeword in quantum error correction?

I am reading the chapter on quantum error correction in Quantum Computer Science by David Mermim. Which says, for a single qubit in a random state $$|\Phi\rangle$$ of the superposition of $$|0\rangle$$ and $$|1\rangle$$, the general form of this corrupted qubit, when interacting with the environment in state $$|e\rangle$$ is: $$$$\tag{1} |e\rangle|\Psi\rangle \to (|a\rangle\textbf{X}+|b\rangle\textbf{Y}+|c\rangle\textbf{Z}+|d\rangle\textbf{1})|\Psi\rangle$$$$

where $$|a\rangle,|b\rangle,|c\rangle,|d\rangle$$ are four random state of the environment, and $$\textbf{X}, \textbf{Y}, \textbf{Z}$$ mean a flip error, a combined flip error and phase error, and a phase error, respectively. $$\textbf{1}$$ is the identity operator representing no error.

From this, the state of a corrupted n-qubit codeword is:

$$$$\tag{2} |e\rangle|\Psi\rangle \to \sum_{\mu_{1}=0}^{3}...\sum_{\mu_{n}=0}^{3}|e_{\mu_{1}...\mu_{n}}\rangle \textbf{X}^{\mu_1}\otimes...\otimes\textbf{X}^{\mu_n}|\Psi\rangle$$$$

I can understand the above two equations. But when constructing the general form of the state of a corrupted n-qubit codeword with the assuming that at most one single qubit in the codeword suffers an error, the book says that 'the terms in (2) are dominated by those in which only a single one of the $$X^{\mu_i}$$ differs from $$\textbf{1}$$, and the general form of (2) becomes a superposition of terms in which each individual Qbit sufferers an error.

$$$$\tag{3} |e\rangle|\Psi\rangle \to [\sum_{i=0}^{n-1}(|a_i\rangle\textbf{X}_i+|b_i\rangle\textbf{Y}_i+|c_i\rangle\textbf{Z}_i)+|d\rangle\textbf{1}]|\Psi\rangle$$$$

What I do not understand is that since we assume only one qubit has error, then only one of the $$X^{\mu_i}$$ in $$X^{\mu_1}\otimes...\otimes X^{\mu_n}$$ in equation (5.16) looks like $$|d\rangle\textbf{1} + |a\rangle\textbf{X} + |b\rangle\textbf{Y} + |c\rangle\textbf{Z}$$. As a result, the state $$\Psi$$ should only suffer error in this ith qubit. How do the auther get from (2) to (3) where all the qubits seem to have an error?