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?

  • $\begingroup$ The Pauli matrices (along with the identity) are a basis for the vector space of $2\times 2$ matrices. An error-correcting code is a subspace so it must be closed under linear combinations. Remember that errors may come from unitaries in the larger Hilbert space but to the code, they just look like arbitrary linear transformations. $\endgroup$
    – Condo
    Dec 24, 2020 at 16:18
  • $\begingroup$ @Condo I know that the trick is to use the Fact Pauli matrix form a basis for linear operator. But it is possible for me to prove that if all the Pauli matrices "necessary to describe the problem" (i.e used in the decomposition of any $M_{\mu}$, then Knill Laflamme on those Pauli matrices imply Knill Laflamme for Kraus of the map. But the reciprocal (which seems to be the point of Preskill notes) is not obvious for me. $\endgroup$ Dec 24, 2020 at 16:29
  • $\begingroup$ While writing my comment I realize that what I said is enough to show that correcting error for Pauli errors is enough to correct for an arbitrary error. However, I am still confused by what the document seem to say (the other direction of the logic: correcting for those Kraus means you can correct for arbitrary Pauli on which the Kraus are expanded). $\endgroup$ Dec 24, 2020 at 16:31

1 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.

  • $\begingroup$ Thank you very much for your answer. Even though I see what you mean (using the syndrome to correlate the $X$, $Y$ or $Z$ error to an ancilla), as it is restricted to a particular case (unitary $U$) I don't see why it works "in general". My example about the $U$ was more to provide what I thought was a counter example to Preskill notes. I think I would be satisfied by an answer simply confirming me that if Knill-Laflamme is verified for the set of Pauli error that are required to decompose the Kraus of the map on, then you can correct for the "real" map (i.e based on the non Pauli Kraus). $\endgroup$ Dec 27, 2020 at 22:52
  • $\begingroup$ But the other way around is not necesseraly true as suggested by Preskill (if Knill Laflamme are verified for the Kraus of the map it doesn't imply they would be verified for the Pauli operator on which those Kraus can be decomposed on). I guess it is a mistake in his note but I would like to be sure I don't make a mistake saying so. $\endgroup$ Dec 27, 2020 at 22:53
  • $\begingroup$ Nothing about my answer uses unitarity of the error operator. You could replace $U$ with any $E_k$. The point is that whatever operator can be decomposed in terms of Pauli operators. If you can correct for each of those terms in the sum, you can correct for the whole thing, in exactly the way I demonstrated by example. As you say, the opposite is not true. There are some errors that can be corrected even if their individual terms cannot all be corrected. $\endgroup$
    – DaftWullie
    Dec 28, 2020 at 12:20
  • 1
    $\begingroup$ There is an easy counter-example to this. Think of the 3-qubit majority vote code, but change the basis by rotating each qubit by the same unitary $U$. This code is protected against any single-qubit error $UXU^\dagger$, but is not protected against $X$, $Y$ or $Z$ errors. $\endgroup$
    – DaftWullie
    Dec 28, 2020 at 12:22
  • $\begingroup$ Thank you very much for your help. $\endgroup$ Dec 28, 2020 at 12:35

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