# How is a quantum error correcting code $C(E)=\langle\psi|E^\dagger E|\psi\rangle$ diagonalized?

UIn page 6 of the following paper: https://arxiv.org/pdf/0904.2557.pdf, in the proof of theorem 3:"Suppose equation (30) holds. We can diagonalize $$C_{ab}$$. This involves choosing new basis $$\{F_a\}$$ for $$\mathcal{E}$$...".

Does this mean the matrix representation of the bilinear form $$b(E)=\langle\psi|E^{\dagger}E|\psi\rangle$$ with respect to the basis that diagonalizes it? (Here, $$\psi$$ is a fixed codeword.)

• It's unclear to me what your first question means. Could you try rephrasing it? Aug 24, 2023 at 9:39
• How can we formulate diagonalizability of $[C_{ab}]$ in terms of new basis? Aug 24, 2023 at 10:59

$$[C_{ab}] = \langle \psi | F_a^{\dagger} F_b | \psi \rangle$$
$$\langle \overline{0} | F_a^{\dagger} F_b | \overline{1}\rangle = 0$$. Similairly in Equations 27 and 28 the error terms are replaced by elements of the new basis $$\{ F_a \}$$
If we assume the outcome of $$\langle \psi | E^\dagger E |\psi \rangle$$ only depends on the error and not on the state, (i.e the outcome C is only a function of $$E$$ not $$\psi$$), this means that $$\langle \psi | E^\dagger E |\psi \rangle = \langle \phi | E^\dagger E |\phi \rangle$$.