0
$\begingroup$

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?

$\endgroup$

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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