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The Knill-Laflamme condition for a stabilizer $\mathcal{M}$ is

An error with Kraus operators $\{E_k\}$ is correctable if either $$E^\dagger_kE_l\in\mathcal{M}\quad\forall\, k,l $$ or there exists $M\in\mathcal{M}$ such that $$\{M,E_k^\dagger E_k\}=0\quad\forall \,k $$

But consider a unitary error $U$, then $U^\dagger U=I\in \mathcal{M}$. Does this mean that all unitary errors are always correctable by any stabilizer? It shouldn't, because for example Shor's code doesn't correct double bit flips. What am I missing?

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  • $\begingroup$ Why would it mean that? The second condition is never satisfied. Why would the first be? (And what is $\mathcal M$?) $\endgroup$ – Norbert Schuch Jul 28 at 20:21
  • $\begingroup$ $\mathcal{M}$ is a stabilizer (I think they're called like that, an abelian subgroup of the Pauli group, and we use as codewords eigenstates of matrices in $\mathcal{M}$ with eigenvalue $1$) the first condition, since there is only one Kraus operator, is always satisfied, since $U^\dagger U=I\in\mathcal{M}$ $\endgroup$ – user2723984 Jul 28 at 20:26
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    $\begingroup$ But is has to be satisfied for all error pairs $E_k^\dagger E_l$. Unless your error is always a fixed unitary. In that case, it can obviously be corrected, because you know what happened to your system. $\endgroup$ – Norbert Schuch Jul 28 at 20:29
  • $\begingroup$ I don't understand, $\{E_k\}$ are the Kraus operators of one error, not of all possible errors that can happen to the system, or have I misunderstood? $\endgroup$ – user2723984 Jul 28 at 20:30
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    $\begingroup$ $\{E_k\}$ are the Kraus operators of the channel which describes the error - for instance, for unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $\rho\mapsto p_k U_k\rho U_k^\dagger + q\rho$, where $q$ is the probability that no error occurs. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$. $\endgroup$ – Norbert Schuch Jul 28 at 20:32
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If you only have one $E_k$ (i.e., $k=1$ can only take one value), and this $E_k=U$ is unitary, then - as you point out in the comments - the first condition is always satisfied, and the error can be corrected.

However, this also means that your "error" is the deterministic application of $U$. So after applying the "error" map, you just have to undo $U$, because you know which error has been applied.


On the other hand, if you have several unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $\rho\mapsto p_k U_k\rho U_k^\dagger + q\rho$, where $q$ is the probability that no error occurs. Thus, we have $E_k=\sqrt{p_k}U_k$, and additionally $E_0=\sqrt{q}I$. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.

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  • $\begingroup$ Then the condition is worded a bit weirdly I think. To me a double bit flip is "an error" and following this condition it is correctable. Should I take the condition to mean that an ensemble of errors is correctable only if the map induced by the most general linear combination of them has Kraus operators that satisfy the Knill-Laflamme condition? $\endgroup$ – user2723984 Jul 28 at 21:43
  • $\begingroup$ @user2723984 In QECC, you look at the map which corresponds to an average error (including no error) during a given evolution. There is nothing weird about that. Indeed, the right formulation for that is a CPTP map with Kraus operators $E_k$, which are not unique -- an unknown error is nothing but a noisy evolution of the quantum system, nothing more. $\endgroup$ – Norbert Schuch Jul 28 at 22:01
  • $\begingroup$ thank you, I misinterpreted the condition as I was considering each possible flip an individual error that could either be correctable or not, rather than a general error. So I expected the condition to be fulfilled individually by single flip errors and not by double flips in the bit flip code for example. $\endgroup$ – user2723984 Jul 28 at 22:05

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