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I'm currently working with stabilizer codes in quantum error correction and have a question regarding the generation of all logical Clifford operators.

Given a stabilizer code, I understand that for logical Paulis like $X$, $Y$, and $Z$, any element of the stabilizer group multiplied by a representative of these logical operators results in another valid logical operator, forming a coset of the stabilizer group.

My query is specifically about the logical Hadamard operators in this context:

Does this coset property extend to logical Clifford operators (for example logical Hadamard)? In other words, if I multiply a representative logical Hadamard operator by elements of the stabilizer group, will this process yield all possible logical Hadamard operators for the code, or just a subset?

UPDATE:

Answer to the Question: The conclusion is that the number of possible logical Clifford operations is greater than initially expected. The synthesis of logical Cliffords can be achieved using the symplectic method, which involves mapping each Clifford to a symplectic matrix. For an [๐‘›,๐‘›โˆ’๐‘˜] stabilizer code, each logical Clifford can have $2^{๐‘˜(๐‘˜+1)/2}$ symplectic solutions. Furthermore, each of these solutions, when multiplied by the entire stabilizer group, remains a valid logical Clifford. This situation is distinct from that of the logical Pauli.For more comprehensive information, please refer to https://arxiv.org/abs/1803.06987. The specific details I mentioned correspond to Theorem 27 in the paper.

In terms of distinction: each logical Pauli corresponds to a single symplectic solution, while each non-Pauli logical Clifford corresponds to $2^{k(k+1)/2}$ symplectic solutions for an $[n, n-k]$ stabilizer code. Each symplectic solution, when multiplied by the stabilizer, remains an equivalent logical Clifford (note that multiplying by a Pauli does not change the symplectic form of a Clifford gate, up to a sign). Therefore, the number of valid logical Cliffords is $2^{k(k+1)/2}$ times that of logical Paulis.

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Yes, the coset property will extend. You can look at the stabilizer elements as the "Logical Identity" operation on your codespace, i.e., operations that leave the codespace invariant. So if you multiply your logical Hadamard, say $\bar{H}$, with any stabilizer element $S_i \in \mathcal{S}$ where $\mathcal{S}$ is the set of your stabilizers, then you can see that every operation $\bar{H} S_i$ is "logical identity" on the codespace followed by logical Hadamard. So every $\bar{H} S_i$ will act as the logical Hadamard. This is true for any logical operation.

Let's confirm via some algebra. Let $|\psi\rangle$ be a state in your codespace. Therefore, by the definition of stabilizers, we know

$$S_i |\psi \rangle = |\psi\rangle \tag{1}\,.$$

Now, let $\bar{H}$ be your logical Hadamard or any logical operation on the codespace $\mathcal{C}$. It transforms your state as follows -

$$\bar{H}|\psi\rangle = |\phi\rangle \,.\tag{2}$$

Now let's check the action of $\bar{H} S_i$.

$$\begin{align} \bar{H} \underbrace{S_i|\psi\rangle}_{\text{from (1)}} =\underbrace{\bar{H}|\psi\rangle}_{\text{from (2)}}= |\phi\rangle \,.\tag{3} \end{align}$$

Therefore,

$$\bar{H}S_i |\psi\rangle = \bar{H} |\psi\rangle\,, \tag{4}$$ holds true $\forall S_i \in \mathcal{S}$ and $\forall |\psi\rangle \in \mathcal{C}\,.$ Hence, we have confirmed our original argument.

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  • $\begingroup$ Thank you for your detailed answer. I am still wondering if it's possible for some logical (\overline{H}') to exist such that it cannot be written as (\overline{H} S_i) for some (S_i) in the stabilizer group. $\endgroup$
    – Podrick
    Dec 18, 2023 at 6:09
  • $\begingroup$ No. Because, like I said, $S_i$ are the elements which act like identity on your codespace. It's exactly the same as for any matrix $M$, $M \cdot I = M$ where $I$ is the identity matrix. $\endgroup$
    – FDGod
    Dec 18, 2023 at 6:36
  • $\begingroup$ For every logical operator, say $X$, $X$ and $X S_i$ both will have the same exact effect within the codespace. They may act differently outside the codespace. $\endgroup$
    – FDGod
    Dec 18, 2023 at 6:37
  • $\begingroup$ I kind of get your point. Thank you. $\endgroup$
    – Podrick
    Dec 18, 2023 at 10:00

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