Exercise 10.8

Consider a three qubit phase flip code with logical qubit as \(|0_{L}\rangle = |+++\rangle\) and \(|1_{L}\rangle = |---\rangle\), the projector onto the codespace is

(1)\[ P = |+++\rangle\langle +++|+ |---\rangle\langle ---| \]

Since \(Z|+\rangle = |-\rangle\) and \(Z|-\rangle = |+\rangle\), we have

(2)\[\begin{split} \begin{align} Z_1P = |-++\rangle\langle +++|+ |+--\rangle\langle ---|\\ Z_2P = |+-+\rangle\langle +++|+ |-+-\rangle\langle ---|\\ Z_3P = |++-\rangle\langle +++|+ |--+\rangle\langle ---|\\ PZ_1 = |+++\rangle\langle -++|+ |---\rangle\langle +--|\\ PZ_2 = |+++\rangle\langle +-+|+ |---\rangle\langle -+-|\\ PZ_3 = |+++\rangle\langle ++-|+ |---\rangle\langle --+|\\ \end{align} \end{split}\]

If \(E = \{E_1 = I, E_2 = Z_1, E_3 = Z_2, E_4 = Z_3\}\)​, we could compute that

(3)\[\begin{split} PE^{\dagger}_1E_1 P = PII P = P^2 = P \iff \alpha_{11} = 1\\ PE^{\dagger}_2E_2 P = PZ_1Z_1 P =P^2 = P \iff \alpha_{22} = 1\\ PE^{\dagger}_3E_3 P = PZ_2Z_2 P = P^2 = P \iff \alpha_{33} = 1\\ PE^{\dagger}_4E_4 P = PZ_3Z_3 P = P^2 = P \iff \alpha_{44} = 1\\ \end{split}\]

Also, we could find that for \(i\neq j\) we have \(PE^{\dagger}_iE_j P = 0\) since we have \(PE^{\dagger}_i \neq E_j P\) when \(i\neq j\), and the product of states with orthonormal basis can only be zero. Therefore, we could find that \(\alpha_{ij} = \delta_{ij}\) where \(\delta_{ij} = 1\) if \(i=j\) and \(\delta_{ij} = 0\) when \(i\neq j\). So the error set \(E = \{E_1 = I, E_2 = Z_1, E_3 = Z_2, E_4 = Z_3\}\) satisfies the quantum error-correction conditions and matrix \(\alpha\) is Hermitian matrix (identity matrix).