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I know that for Steane code, we can implement transversally some gates like cNOT, Hadamard and Pauli.

What I am looking for is a resource in which it is explained why implementing those gate give rise to the good logical operation.

If this is a result more general than Steane I would be interested in a "general enough" approach showing it.

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Let $\mathcal{H}$ be the Hilbert space of a set of physical qubits and let $S$ be the stabilizer group of a stabilizer code $\mathcal{G} \subset \mathcal{H}$.

A transversal operator $U$ on $\mathcal{H}$ implements a logical operator on $\mathcal{G}$ if it maps $\mathcal{G}$ back to itself. This can be established by showing that $U$ does not change the stabilizer group $S$ of $\mathcal{G}$. In order to do this, we need to know how stabilizers transform under unitary gates. Suppose that $g$ is an operator that stabilizes $|\psi\rangle$, i.e. $g|\psi\rangle = |\psi\rangle$. Then

$$ UgU^\dagger U|\psi\rangle = Ug|\psi\rangle = U|\psi\rangle\tag1 $$

and we see that $UgU^\dagger$ stabilizes $U|\psi\rangle$. We will use this fact to demonstrate that certain transversal operators do not change the stabilizer group of stabilizer codes that meet certain criteria. We will also use $(1)$ to determine the action of those transversal operators within the code subspace by analyzing their effect on the logical Pauli operators.

Controlled-NOT

Claim: If $\mathcal{G}=CSS(C_1, C_2)$ is a Calderbank-Shor-Steane code for classical linear codes $C_1$ and $C_2$, $C_2^\perp \subset C_1$ then transversal CNOT is a logical CNOT on $\mathcal{G}$.

Remark 1: There is ambiguity in the use of the $CSS(C_1, C_2)$ notation in the literature. For example, Wikipedia and Nielsen & Chuang put $C_2^\perp$ in the second position rather than $C_2$.

Remark 2: The condition $C_2^\perp \subset C_1$ is not an additional restriction on $\mathcal{G}$. It is part of the definition of a CSS code necessary to make sure that the $X$ type and $Z$ type stabilizer generators commute. Consequently, the claim says that CNOT admits transversal implementation for any CSS code.

Proof sketch: Let $g_x$ be a tensor product of identity $I$ and Pauli $X$ operators and similarly for $g_z$. Since $\mathcal{G}$ is a CSS code, we can choose stabilizer generators that are either of the form $g_x$ or of the form $g_z$. Calculate that

$$ \mathrm{CNOT} \circ (g_x \otimes I) \circ \mathrm{CNOT} = g_x \otimes g_x \\ \mathrm{CNOT} \circ (I \otimes g_x) \circ \mathrm{CNOT} = I \otimes g_x \\ \mathrm{CNOT} \circ (g_z \otimes I) \circ \mathrm{CNOT} = g_z \otimes I \\ \mathrm{CNOT} \circ (I \otimes g_z) \circ \mathrm{CNOT} = g_z \otimes g_z. $$

Note that if $g_x, g_z \in S$ then all operators on the right hand sides of the four equations above are in $S \times S$. Moreover, every operator in $S \times S$ can be obtained as a composition of operators of the form $g_x \otimes I$, $g_z \otimes I$, $I \otimes g_x$ and $I \otimes g_z$. We conclude that transversal CNOT preserves $S \times S$.

By analyzing the action of transversal CNOT on the logical $X$ and $Z$ operators along the same lines as above, we see that the logical operator effected by performing transversal CNOT on the physical qubits is in fact the logical CNOT. $\square$

Hadamard

Claim: If $\mathcal{G}=CSS(C_1, C_2)$ is a Calderbank-Shor-Steane code where the two classical linear codes are the same $C_1 = C_2$ then transversal Hadamard is a logical Hadamard on $\mathcal{G}$.

Remark: The definition of the CSS code requires that $C_2^\perp \subset C_1$, so $C_1$ cannot be arbitrary. It necessarily contains its own dual.

Proof sketch: As before, let $g_x$ be a tensor product of identity $I$ and Pauli $X$ operators. We see that

$$ H g_x H = g_z \\ H g_z H = g_x. $$

where $g_z$ is obtained from $g_x$ by replacing $X$ operators in the tensor product with $Z$ operators. Note that since $\mathcal{G}$ is a CSS code, we can choose stabilizer generators that are either of the form $g_x$ or of the form $g_z$. Moreover, since $C_1 = C_2$, $g_x \in S$ if and only if $g_z \in S$. Consequently, transversal Hadamard preserves the stabilizer.

As before, the analysis extends to logical $X$ and $Z$ operators and therefore transversal $H$ on physical qubits implements the logical Hadamard gate. $\square$

Phase gate

(For completeness we include the phase gate even though it is not mentioned in the question.)

Claim: If $\mathcal{G}=CSS(C_1, C_2)$ is a Calderbank-Shor-Steane code for classical linear codes $C_1$ and $C_2$, $C_2^\perp \subset C_1$ where $C_2^\perp$ is doubly-even (i.e. all its codewords have Hamming weight divisible by four) then transversal phase gate $P$ preserves the stabilizer $S$.

Proof sketch: Since the phase gate commutes with the Pauli $Z$ operator it is clear that all $g_z$ stabilizers are preserved. For any $g_x$, we have

$$ P g_x P^\dagger = i^{w(g_x)} g_x g_z $$

where $w(g_x)$ is the weight of stabilizer $g_x$, i.e. the number of non-identity factors in it. Since $C_2^\perp$ is doubly-even, we see that

$$ P g_x P^\dagger = g_x g_z $$

and thus $P g_x P^\dagger \in S$. $\square$

Note that the transversal phase operator is not necessarily the logical phase gate. However, $C_2^\perp$ is an even code so the transversal $Z$ belongs to the normalizer $N(S)$. Thus, as long as transversal $Z$ is not in the stabilizer we can choose it to play the role of the logical $Z$. Under this choice transversal phase operator commutes with the logical $Z$ and therefore its action on $\mathcal{G}$ is a diagonal gate. Moreover, applying transversal phase operator twice yields the logical $Z$. We conclude that in this case the transversal phase operator is either a logical phase gate or its inverse.

Pauli operators

Claim: If $\mathcal{G}$ is a stabilizer code, then logical Pauli operators are transversal.

Proof sketch: This follows immediately from the fact that logical Pauli operators are chosen from the normalizer $N(S)$ of the stabilizer group $S$ in the $n$-qubit Pauli group $\mathcal{P}_n$ since all operators in $\mathcal{P}_n$ are transversal by definition. $\square$

Steane code

Steane code is $CSS(C_1, C_2)$ where $C_1 = C_2$ is the Hamming $[7, 4, 3]$ code.

Since it is a stabilizer code, Pauli operators are transversal. Since it is a CSS code, CNOT is transversal. Since $C_1 = C_2$, Hadamard is transversal. Since $C_1$ is doubly-even, the phase gate is transversal.

Thus, the entire Clifford group admits a transversal implementation in the Steane code. By Eastin-Knill theorem, we cannot extend the set of transversal gates to a universal gateset. In particular, the $T$ gate does not have a transversal implementation in the Steane code.

References

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  • $\begingroup$ Thank you very much for your very complete answer. I have a question to understand the CNOT part. Do you agree with the following: Basically, need to prove that $U |\psi \rangle$ is still in the code space. You know that this state will be stabilized by $UgU^{\dagger}$ for any $g \in S$. Thus what is important to prove is that $\{U g U^{\dagger}, g \in S \}=S$. Referring to your text the important ingredient is then that you can map any $g' \in S$ in the rhs of your equations (by combining appropriately the lhs by product of Pauli). Do you agree ? $\endgroup$ – StarBucK Dec 28 '20 at 15:43
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    $\begingroup$ Re your first comment: Yes, I agree with what you said. Note that there is a minor caveat in the CNOT case: since we're dealing with two logical qubits the stabilizers that must be preserved by the gate are of the form $g_1 \otimes g_2$ where $g_1, g_2 \in S$. Then $U$ is a logical operation on two logical qubits iff $\{U(g_1\otimes g_2)U^\dagger, g_1, g_2 \in S\} = S \otimes S$ where the tensor product of sets of operators is defined elementwise, i.e. $A \otimes B = \{a\otimes b | a\in A, b\in B\}$. You can convince yourself of this by writing equation $(1)$ in two-logical-qubit case. $\endgroup$ – Adam Zalcman Dec 28 '20 at 18:50
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    $\begingroup$ Re your second and third comments: Yes. Logical operators are elements of $N(S)$. Among them, those that are also in $S$ are the logical identity (IOW, their restriction to the code subspace is the logical identity). So you are correct that non-trivial logical operators live in $N(S) - S$. And finally any two operators $a, b \in N(S)$ such that $a^\dagger b \in S$ agree on the code subspace and therefore we can treat them as the same logical operator. This corresponds to taking the quotient $N(S)/S$. Thus, we identify different logical operators with the cosets of $S$ in $N(S)$. $\endgroup$ – Adam Zalcman Dec 28 '20 at 18:58
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    $\begingroup$ Your answer is really nice, but I need some time to go through all the details (and checking some aspect with litterature to really understand). Once I will agree with everything I will validate it ! $\endgroup$ – StarBucK Dec 29 '20 at 18:13
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    $\begingroup$ I see. Yes, I am making the assumption that we have two codeblocks, each separately encoded with the same type of CSS code. IOW, $\dim \mathcal{G} = 2$ on each codeblock. $\endgroup$ – Adam Zalcman Dec 30 '20 at 19:39

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