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I am reading how Hadamard is performed in surface code at https://arxiv.org/pdf/1208.0928.pdf

At some point, they apply H on all data qubits, and they claim that it converts X to Z stabilizers and vice versa, and I still do not understand why this identity is true:

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Appendix J, page 49, bulletpoint 4:

We perform the key to this process, executing physical Hadamards on all the data qubits in the patch. As this changes the eigenbases from $\hat Z$ to $\hat X$ and vice versa, we change the identity of the measure qubits from $\hat X$ to $\hat Z$ stabilization and vice versa, as shown in Fig. 27e. Also, the $\hat X_L$ and $\hat Z_L$ logical operators swap their identities. This step, and the two that follow, are done in between two cycles of the surface code (so the measure-X and measure-Z qubits do not perform any stabilization on the isolated patch during these three steps).

Perhaps these is 2 different operations? 1) Apply H on data qubits, 2) change stabilizers

Question 2:

After this operation, we still have a stabilized logical qubit, so it can not be in +L state, but just in one of 0L or 1L. So it feels like the H did nothing. What am I missing?

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    $\begingroup$ I took the liberty of copypasting the paragraph, no need to use pictures when you can quote them directly. That helps to avoid broken image links in the future. $\endgroup$
    – Mauricio
    Commented Jan 31, 2023 at 11:07

2 Answers 2

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I think your confusion might come from the wording they use. They do apply the two different operations you mention.

To apply a logical Hadamard, they apply a physical Hadamard gate to each data qubit and each ancilla previously used for measuring an $X$ stabilizer now measures a $Z$ stabilizer (and vice versa).

The type of stabilizer an ancilla measures is switched by changing the readout circuit.

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  • $\begingroup$ Thank you! how it shows it created logical H? $\endgroup$
    – Ron Cohen
    Commented Jan 31, 2023 at 9:16
  • $\begingroup$ Because it has transformed logical X into logical Z and vice versa. This is the heisenberg/stabilizer formalism representation of a Hadamard gate. $\endgroup$
    – Peter-Jan
    Commented Jan 31, 2023 at 9:53
  • $\begingroup$ This paper explains the Heisenberg representation: arxiv.org/abs/quant-ph/9807006 $\endgroup$
    – Peter-Jan
    Commented Jan 31, 2023 at 9:54
  • $\begingroup$ But after this operation, we still have a stabilized logical qubit, so it can not be in +L state, but just in one of 0L or 1L. So it feels like the H did nothing. What am I missing? $\endgroup$
    – Ron Cohen
    Commented Feb 1, 2023 at 11:14
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The main point is that if you apply a unitary $U$ to a state described by stabilizers, the action $$ |\psi\rangle\rightarrow U|\psi\rangle $$ can be equivalently written as $$ g_i\rightarrow Ug_iU^\dagger $$ for all the stabilizers $g_i$. (For a code, the action also applies to the logical operators.)

Add that to the fact that $HXH=Z$ and $HZH=X$, you can see how applying $H$ on every qubit changes all the $X$s into $Z$s and vice versa.

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  • $\begingroup$ But after this operation, we still have a stabilized logical qubit, so it can not be in +L state, but just in one of 0L or 1L. So it feels like the H did nothing. What am I missing? $\endgroup$
    – Ron Cohen
    Commented Feb 1, 2023 at 11:12
  • $\begingroup$ If it was in the state $|0\rangle_L$ to start with, that means it was in the $+1$ eigenstate of $Z_L$. However, when we apply hadamards everywhere, $Z_L\rightarrow X_L$. Hence, the stabilized state is stabilized in the $|+\rangle_L$ state. $\endgroup$
    – DaftWullie
    Commented Feb 1, 2023 at 12:06

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