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My question is highly related to this topic

It is about defining logical operation on a Stabilizer code.

I call $S$ the stabilizer group of a code space $C$, and I assumed it is generated by a family $S=\langle s_1,...,s_p \rangle$. I call $G_n$ the $n$-Pauli matrix group ($n$ being the dimension of the full Hilbert space).

A definition of logical operation is as follow:

$U_L$ is a logical operation if $\forall |\psi \rangle \in C$, $U_L | \psi \rangle \in C$

And, we realize that if $|\psi \rangle$ is stabilized by $g$, $U_L |\psi \rangle$ will be stabilized by $U_L g U_L^{\dagger}$.

Questions: which condition to ensure $U_L$ is a logical operation

A sufficient condition is to have $U_L S U_L^{\dagger} = S$, which means that $U_L \in N(S)$ (where $N(S)$ is the normalizer of $S$).

Indeed, this way we would be certain that $U_L |\psi\rangle$ will be stabilized by $S$ and thus be in the codespace.

What disturbs me is that according to the comments here (and some of the sources attached), the logical operation are actually exactly elements of $N(S)$. I see the sufficient condition but not the necessary one.

For instance, if $U_L$ is non clifford, for $s \in S$, $U_L s U_L^{\dagger}$ might not even be an n-Pauli matrix, thus $U_L S U_L^{\dagger} \neq S$ as $S \subset G_n$. In this case obviously $U_L$ wouldn't be in the normalizer of $S$. But wouldn't it be possible to have a non n-Pauli matrix that still stabilize appropriately $C$ ?

So my question is: Why is it sufficient and necessary to have $U_L \in N(S)$ so that $U_L$ is a logical operation ?

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You are correct that there are logical operators (i.e. ones that preserve $C$) outside $N(S)$. Also, your argument is sound and can be stated rigorously as follows. Let $U \in N(S)$. Then for any $s \in S$ we have $UsU^\dagger \in S$. Now, diagonalize $s$

$$ s = VdV^\dagger $$

and define

$$ V(\theta) = V \, \mathrm{diag}(1, \dots, 1, e^{i\theta}) \, V^\dagger. $$

Note that $V(\theta)$ commutes with $s$. Now, define

$$ U(\theta) = U V(\theta) $$

and note that

$$ U(\theta) s U(\theta)^\dagger = UV(\theta)sV(\theta)^\dagger U^\dagger = UsU^\dagger. $$

Thus, we found a continuous, one-parameter group of operators that preserve $C$. It cannot be a subset of $N(S)$ because $N(S)$ is discrete.


The requirement that $U\in N(S)$ is sufficient, but not necessary for $U$ to be a logical operator. However, when one restricts their consideration to the Pauli group $G_n$ the requirement is both necessary and sufficient.

In quantum error correction one sometimes restricts consideration to $G_n$, because of discretization of quantum errors. This result says that if an error correction operation $\mathcal{R}$ recovers from errors in a set $\{E_i\}$ then it recovers from errors $\{F_j\}$ which are linear combinations of $\{E_i\}$. Thus, for the purposes of establishing whether an error is a logical operator (and thus uncorrectable) it is sufficient to consider errors in $G_n$ since every operator can be written as a linear combination of elements of $G_n$.

See for example theorem 10.2 on p.438 in section 10.3.1 of Nielsen & Chuang and the discussion following the proof. A good summary is also section IX "Digitization of quantum errors" in https://arxiv.org/abs/0905.2794.

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  • $\begingroup$ Thank you very much. So in the end is there an "easy" necessary and sufficient condition equivalent to the definition to know if the operation is a logical one "in general" or this is something not so easy to find out ? (I mean if we do not assume $U \in G_n$, in this case the equivalent property based on the normalizer is clear). $\endgroup$ – StarBucK Dec 30 '20 at 16:59
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    $\begingroup$ Yes, there is! Mathematically, a unitary $U$ preserves a subspace $C$ if and only if it can be written as a direct sum $U = V \oplus U_L$ where $U_L$ is a unitary on $C$. In terms of matrices this means that in any basis of $\mathcal{H}$ that extends a basis of $C$, $U$ is block diagonal with blocks $V$ and $U_L$. In practice, you can verify that $U$ preserves $C$ by checking whether it commutes with the projector $P_C$ onto $C$. Consequently, $U$ is a logical operator if and only if $P_C = U P_C U^\dagger$. $\endgroup$ – Adam Zalcman Dec 30 '20 at 19:06
  • $\begingroup$ Interesting. I will probably dig into this when I will have more time. Thanks ! $\endgroup$ – StarBucK Dec 30 '20 at 19:11

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