I am interested in qubits which are biased in noise. For my purpose I assume that I have only bit-flip (i.e. Pauli $X$) noise in my algorithm.

In the presence of such noise, quantum gates can convert this bit-flip into phase flip (Pauli $Z$). For instance $HX=ZH$: a Hadamard will convert an $X$ error to a $Z$ error.

I am in particular interested in cNOT gates. By construction, they will satisfy:

$$cNOT X_1 = X_1 X_2 cNOT$$ $$cNOT X_2 = X_2 cNOT$$

Because of that we could believe that a cNOT does the job. However, problems come when we consider the continuous evolution for the gate. For instance, a way to implement a $cNOT$ is through the Hamiltonian:

$$H=\hbar g |1\rangle \langle 1| |-\rangle \langle - |$$

Because this Hamiltonian involves Pauli $Z$, if an $X$ error occurs during the continuous evolution it might be converted to a $Z$ error.

I am aware of this paper which ensure that cNOT preserve the $Z$ noise during the continuous evolution. However I would like to preserve the $X$ noise.

My question:

Is there any paper which shows that it is possible to create a cNOT that preserves bit-flip noise?


2 Answers 2


I don't know how it's done in physical qubits, but one way to create a CNOT that maintains bit flip bias is to use the surface code. In the surface code you can perform a CNOT by code deformation, such as with lattice surgery. Assuming the surface code qubits have X biased logical errors, the lattice surgery CNOT will preserve this X bias.

  • $\begingroup$ Hey. Thanks for your answer. Unfortunately, I forgot to add the important precision that I would like to avoid using QEC schemes. I am however ok to use qubits that are encoded in larger systems than a strictly two-level one. Some people might say it is somehow QEC but I do not consider cat qubits to be QEC for instance (this is more a smart way to encode data). Anyway I think I found a way to show what I need (see my answer). Thanks. $\endgroup$ Jul 13, 2022 at 14:05

Actually it is "somehow" possible to show that if it preserves the $Z$ bias, then, if we redefine the computational basis, by symmetry there should exist a cNOT gate preserving the $X$ bias.

Let me now be more precise in my handwavy statement.

A cNOT gate satisfies the following:

$$ cNOT=|0\rangle \langle 0| + |1\rangle \langle 1 | \otimes X_2 = |+\rangle \langle +| + Z_1|-\rangle \langle - |$$

It means that that the cNOT gate inverts the control and the target when rewritten in $+/-$ basis.

Then, if we decide to choose the computational basis to be $|+\rangle, |-\rangle$ instead of $|0\rangle, |1\rangle$, we can perform $c_X Z$ gates (which means controlled $Z$ where the control is in the $X$ basis). This gate will preserve the $Z$ bias because it simply comes from a way to rewrite the cNOT in another basis.

Hence, we can implement $c_X Z$ gates which preserve the $Z$ bias.

By "symmetry", if we exchange the role of $X$ and $Z$, it implies that it is conceptually possible to implement $c_Z X=cNOT$ gates which preserve the $X$ bias. It needs to choose a different basis to encode your data as I did here. Also, it does not imply that the original cNOT I started from preserves the $X$ noise as well. I just showed that it is conceptually possible to find another cNOT scheme that will preserve the $X$ noise.


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