# Necessary and sufficient condition to define logical operation (stabilizer code)

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 ?

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.

• 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). Dec 30, 2020 at 16:59
• 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 $P_C = U P_C U^\dagger$. Dec 30, 2020 at 19:06
• Interesting. I will probably dig into this when I will have more time. Thanks ! Dec 30, 2020 at 19:11