Arbitrary error correctable iff Pauli error are: misunderstanding from Preskill notes

Question

I am following Preskill notes.

What I want to understand is why it is in general enough to be able to correct n-qubit Pauli errors to say that an arbitrary error can be corrected.

I call: $\mathcal{M}(\rho)=\sum_a M_a \rho M_a^{\dagger}$ an error map, $C$ the code space.

A CPTP recovery operation exists if and only if, the Kraus operator of the error map verify the Knill-Laflamme condition:

$$\forall (i,j) \in C : \langle \overline{i} | M_{\delta}^{\dagger} M_{\mu} | \overline{j} \rangle = C_{\delta \mu} \delta_{ij}$$

With $C_{\delta \mu}$ an Hermitian matrix and $\langle \overline{i} | \overline{j} \rangle=\delta_{ij}$ (the family $| \overline{i} \rangle$ forms an orthonormal basis of $C$).

My question

From this condition, and the fact any Kraus operator can be decomposed as a sum of n-qubit Pauli matrices, Preskill seem to say that it shows that if one is able to correct against Pauli error it is able to correct against an arbitrary error (that verifies Knill Laflamme condition of course).

I call $\mathcal{E}$ the set of n-qubit Pauli operators on which we can decompose any of the $M_a$'s.

He says on page 10, just below (7.26)

"since each $E_a$ in $\mathcal{E}$ is a linear combination of $M_{\mu}$'s, then"

$$\forall (i,j) \in C: \langle \overline{i} | M_{\delta}^{\dagger} M_{\mu} | \overline{j} \rangle = C_{\delta \mu} \delta_{ij} \Rightarrow \forall (i,j) \in C: \langle \overline{i} | E_{a}^{\dagger} E_{b} | \overline{j} \rangle = C_{ba} \delta_{ij}$$

And I don't understand this. I could imagine $\mathcal{M}(\rho)=U \rho U^{\dagger}, U \notin \mathcal{E}$ (a unitary error that is not strictly a Pauli one). Wouldn't this be a counter example in which the Pauli matrix on which the error can be decomposed cannot in turn be expressed in function of Kraus operator.

To say things more shortly: is there an easy way to show that if one is able to correct Pauli error we can show it is equivalent to Knill Laflamme condition?

Answer 1

I think it probably helps to explore the specific example you asked about. Let $U$ be the unitary error $$ U=\alpha 1+\beta X+\gamma Y+\delta Z $$ for some complex parameters $\alpha,\beta,\gamma,\delta$, acting on the first qubit of a multi-qubit system. We could constrain them a bit, but we don't need to. Now imagine you have a state encoded into an error correcting code: $$ |\psi\rangle=a|0_L\rangle+b|1_L\rangle. $$ If I apply the $U$, then obviously I have the state $$ U|\psi\rangle=\alpha|\psi\rangle+\beta X|\psi\rangle+\gamma Y|\psi\rangle+\delta Z|\psi\rangle. $$ Now you perform syndrome extraction for you code. You'll have some ancillas that you end up measuring which will tell you what error you have (type and location) assuming no more than one error has happened anywhere. So, effectively, we'd have some state that looks like $$ \alpha|\psi\rangle|\text{none}\rangle+\beta X|\psi\rangle|X,1\rangle+\gamma Y|\psi\rangle|Y,1\rangle+\delta Z|\psi\rangle|Z,1\rangle. $$ When we measure the ancilla, it will return one of those values. Perhaps $|Y,1\rangle$. In which case, you know your system is in the state $Y|\psi\rangle$ after measurement, and you can correct it by applying $Y$ on qubit 1 as the syndrome extraction told you to.

Hopefully you can now see, thanks to linearity + measurement, how the decomposition of any error map in terms of Pauli errors works.