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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.

Adam Zalcman
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Daniele Cuomo
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2 Answers2

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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.

Adam Zalcman
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  • I'm not sure I got the reasoning. Isn't it just a different basis? I.e. $|+_L\rangle = |+++\rangle$. – Daniele Cuomo Dec 31 '22 at 22:35
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    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. – Adam Zalcman Dec 31 '22 at 22:42
  • May you give an example of the smallest code admitting a transversal implementation of $H$? – Daniele Cuomo Jan 01 '23 at 11:14
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    @DanieleCuomo That's probably better asked as its own question (and an interesting one, too!) – JSdJ Jan 02 '23 at 09:01
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    Here it is! https://quantumcomputing.stackexchange.com/questions/29531/examples-of-codes-admitting-a-transversal-implementation-of-the-hadamard-gate – Daniele Cuomo Jan 02 '23 at 11:40
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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.

JSdJ
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    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. – Adam Zalcman Jan 02 '23 at 17:26