2
$\begingroup$

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.

$\endgroup$

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ I'm not sure I got the reasoning. Isn't it just a different basis? I.e. $|+_L\rangle = |+++\rangle$. $\endgroup$ Dec 31, 2022 at 22:35
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 31, 2022 at 22:42
  • $\begingroup$ May you give an example of the smallest code admitting a transversal implementation of $H$? $\endgroup$ Jan 1 at 11:14
  • 2
    $\begingroup$ @DanieleCuomo That's probably better asked as its own question (and an interesting one, too!) $\endgroup$
    – JSdJ
    Jan 2 at 9:01
  • 1
    $\begingroup$ Here it is! quantumcomputing.stackexchange.com/questions/29531/… $\endgroup$ Jan 2 at 11:40
1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 2 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.