The logical basis states of the Shor’s 9-qubit code are given by

$|0_L\rangle = \frac{(|000\rangle+|111\rangle)(|000\rangle+|111\rangle)(|000\rangle+|111\rangle)}{2\sqrt{2}}, |1_L\rangle = \frac{(|000\rangle-|111\rangle)(|000\rangle-|111\rangle)(|000\rangle-|111\rangle)}{2\sqrt{2}}$

What's the logical bit-flip $\hat{X}_L$ and phase-flip operators $\hat{Z}_L$


The logical $Z_{L}$ can be performed by flipping every bit in either of the three 'blocks': $$ Z_{L} \hat{=} X_{1}X_{2}X_{3} \hat{=} X_{4}X_{5}X_{6} \hat{=} X_{7}X_{8}X_{9} $$ or, for something more symmetric, just flip all bits: $$ Z_{L} \hat{=} X_{1}X_{2}X_{3}X_{4}X_{5}X_{6}X_{7}X_{8}X_{9} $$

Here, the $\hat{=}$ should be read as 'can be implemented as'. These operators are not necessarily the same operators, but they all have the same effect on the codespace (so for any logical state they are the 'same').

The logical $X_{L}$ can be performed by applying $3$ different single-qubit $Z$ operations, where the first of those three is on either of the first three qubits (i.e. $Z_{1}$, $Z_{2}$ or $Z_{3}$), the second of those three on either of the second three qubits (i.e. $Z_{4}$, $Z_{5}$ or $Z_{6}$), and the third of those three is on either of the last three qubits (i.e. $Z_{7}$, $Z_{8}$ or $Z_{9}$).

Or, for something more symmetric again, just apply a $Z$ flip to all qubits: $$ X_{L} \hat{=} Z_{1}Z_{2}Z_{3}Z_{4}Z_{5}Z_{6}Z_{7}Z_{8}Z_{9} $$

  • $\begingroup$ What is the meaning of the second and third = in your answer? $\endgroup$
    – M. Stern
    Apr 1 '21 at 18:56
  • $\begingroup$ @M.Stern I've changed the text a bit, hope it is clarified now $\endgroup$
    – JSdJ
    Apr 1 '21 at 19:02
  • $\begingroup$ Maybe one should say that each of these operators is a valid choice for $Z_L$, but they are not equal. $\endgroup$
    – M. Stern
    Apr 2 '21 at 8:46
  • $\begingroup$ Well, I've stated that they can be performed like this, not that they_must_ be performed like this. Furthermore, they differ only up to a stabilizer, so they're equal for the codespace of the 9-qubit code. $\endgroup$
    – JSdJ
    Apr 2 '21 at 11:37
  • 1
    $\begingroup$ Ok but the equality sign is wrong then. $\endgroup$
    – M. Stern
    Apr 2 '21 at 13:11

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