# All unitary errors are correctable

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?

• Why would it mean that? The second condition is never satisfied. Why would the first be? (And what is $\mathcal M$?) – Norbert Schuch Jul 28 '19 at 20:21
• $\mathcal{M}$ 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 $U^\dagger U=I\in\mathcal{M}$ – user2723984 Jul 28 '19 at 20:26
• 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. – Norbert Schuch Jul 28 '19 at 20:29
• 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? – user2723984 Jul 28 '19 at 20:30
• $\{E_k\}$ 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 $E_k$, $E_l$. – Norbert Schuch Jul 28 '19 at 20:32

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 $$E_k$$, $$E_l$$.
• @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. – Norbert Schuch Jul 28 '19 at 22:01