Does the standard 3 qubit code scheme admit a transversal implementation of $H$?

The standard 3 code scheme encodes one qubit into 3 by applying 2 $$CNOT$$s targeted on auxiliary qubits set on ground state $$|0\rangle$$.

I am struggling to perform logical operations between two logical qubits encoded this way, so I am starting to suspect that the operation $$H^{\otimes 3}$$ is not a logical operation for such a code.

The logical computational basis states of the $$3$$-qubit repetition code are $$|0_L\rangle=|000\rangle$$ and $$|1_L\rangle=|111\rangle$$ which are unentangled. However, the logical Hadamard sends $$|0_L\rangle$$ to $$|{+}_L\rangle=\frac{1}{\sqrt2}(|000\rangle+|111\rangle)$$ which is entangled. Every transversal gate is a product of local unitaries $$U_1\otimes U_2\otimes U_3$$ which cannot create entanglement between the physical qubits. Therefore, there is no transversal Hadamard in the repetition code.

• I'm not sure I got the reasoning. Isn't it just a different basis? I.e. $|+_L\rangle = |+++\rangle$. Dec 31, 2022 at 22:35
• No, $|{+}_L\rangle\ne|{+++}\rangle$ as you can verify by expressing both sides in the computational basis of the physical qubits. The LHS is $(|000\rangle+|111\rangle)/\sqrt2$ whereas the RHS is $(|000\rangle+|001\rangle+|010\rangle+|011\rangle+\dots+|111\rangle)/\sqrt8$. The LHS is entangled (it's the GHZ state), but the RHS is not. Dec 31, 2022 at 22:42
• May you give an example of the smallest code admitting a transversal implementation of $H$? Jan 1, 2023 at 11:14
• @DanieleCuomo That's probably better asked as its own question (and an interesting one, too!)
– JSdJ
Jan 2, 2023 at 9:01
• Jan 2, 2023 at 11:40

There's another method of reasoning this. Note that $$H^{\otimes 3} |0_{L}\rangle$$ creates the uniform superposition over all computational basis states.

Since the logical codespace is spanned by just $$|000\rangle$$ and $$|111\rangle$$, this is obviously not even a codestate, let alone the logical $$|{+}_{L}\rangle$$ state.

You can use this to immediately rule out many other potential 'logical' operations, too.

• This proves that $H^{\otimes 3}$ is not a logical operator and in particular not the logical Hadamard. However, it doesn't rule out that a different transversal unitary $U_1\otimes U_2\otimes U_3$ happens to implement the logical Hadamard. Jan 2, 2023 at 17:26