Quantum error correction

Remarks

Example: let $H = (\C^2)^{\otimes 3}$, $\mathcal{C} = \operatorname{span}\set{\lvert 000 \rangle, \lvert 111 \rangle}$, and let $\mathcal{N}(\rho) = p_0 \rho + \sum_{i=1}^{3} p_i X_i \rho X_i$ with $p_i \ge 0$ and $\sum p_i = 1$, where $X_i$ flips qubit $i$. The four subspaces $\mathcal{C}$ and $X_i \mathcal{C}$ are orthogonal. The recovery measures which of them contains the state, which reveals $i$ but nothing about the encoded state, and then applies $X_i$. So $\mathcal{C}$ corrects $\mathcal{N}$.

Non-example: the same code does not correct the noise $\mathcal{N}'(\rho) = \frac12 \rho + \frac12 Z_1 \rho Z_1$, where $Z_1 = \operatorname{diag}(1, -1)$ acts on qubit $1$. Let $\lvert \pm_L \rangle = (\lvert 000 \rangle \pm \lvert 111 \rangle)/\sqrt2$; then $Z_1$ swaps $\lvert +_L \rangle$ and $\lvert -_L \rangle$. So $\mathcal{N}'$ sends both states to the same mixed state, and no recovery can return both.

A code that corrects $\mathcal{N}$ also corrects every channel whose Kraus operators are linear combinations of those of $\mathcal{N}$. This is why correcting finitely many errors corrects a continuum of them.