5
$\begingroup$

There is a well known 5 qubit code $ [[5,1,3]] $ with stabilizer generators

$$ XZZXI \\ IXZZX \\ XIXZZ \\ ZXIXZ $$ There is a corresponding $ [[5,1,3]] $ code for qutrits given by

\begin{align*} & XZZ^\dagger X^\dagger I \\ & IXZZ^\dagger X^\dagger \\ & X^\dagger IXZZ^\dagger \\ & Z^\dagger X^\dagger IXZ \end{align*}

Another well known qubit code is the $ [[7,1,3]] $ Steane code with stabilizer generators \begin{align*} & XXXXIII\\ & XXIIXXI\\ & XIXIXIX\\ & ZZZZIII\\ & ZZIIZZI\\ & ZIZIZIZ \end{align*}

Is there a qutrit analogue of the Steane code that can be obtained in a similar way? In other words by replacing some $ X $ by $ X^\dagger $ and some $ Z $ by $ Z^\dagger $?

$\endgroup$
1
  • $\begingroup$ For the qutrit version of the [[5, 1, 3]] code, do you have a reference where I can learn more about its property? $\endgroup$
    – Yunzhe
    Jan 20 at 15:09

1 Answer 1

5
$\begingroup$

$$ XXXXIII \\ XXIIXXI \\ XIXIXIX \\ ZZ^\dagger Z^\dagger Z III \\ ZZ^\dagger II Z^\dagger Z I \\ Z I Z^\dagger I Z^\dagger I Z $$

After I asked the question I played around with the generators by hand and wasn't really getting anywhere, trying to use inverses for both the $ X $ and $ Z $ type stabilizers. But then I was like what if I just take all the $ X $ type stabilizers as is and then just use inverses for the $ Z $ type ones? And that seemed worth trying since treating $ X $ and $ Z $ symmetrically wasn't working out for me. So then I did that and just took all the $ X $ type to be regular with no inverses. Then I looked at $ Z $ and was like ok if $ X $ is all regular then to commute the $ Z $ needs to be half regular half inverses. And without a loss of generality I can probably take the $ Z $ on the first qubit to always be regular. From there it seems like the placement of the two $ Z^\dagger $ is uniquely determined. I don't know if this method would work for qudit CSS codes in general, but maybe?

I guess the idea of using this in a more general context would be if you have a qubit CSS code with $ H_X=H_Z $ then you can take all the qudit stabilizers of one type (say $ X $ type) to be regular. Then for the stabilizers of the other type (say $ Z $ type) you have to take half of the $ Z $ regular and half as $ Z^\dagger $. Then hopefully you can arrange it all in a way so the phases cancel out and all the $ X $ and $ Z $ qudit stabilizer generators commute?

It was definitely not what I expected with the asymmetry of treating $ X $ and $ Z $ differently, especially since the 5 qudit code treats $ X $ and $ Z $ so nice and symmetrically. So overall pretty weird and I guess I'm hoping for another answer that is bette/nicer/more symmetric/ more systematic/has cool theory behind it.

$\endgroup$
2
  • 1
    $\begingroup$ If this is the answer you were looking for, please add a few lines on the method of how you arrived at this so that this post will be helpful in future for other people. Also, accept your own answer. $\endgroup$
    – FDGod
    Nov 9, 2023 at 1:41
  • $\begingroup$ I just got it by playing around with the inverses for a while after I posted the question, nothing very systematic. I would definitely prefer a better answer that has more theory behind it. I just figured since I did get an answer I might as well post it? But I can definitely edit the answer and add a bit more about how I got it. $\endgroup$ Nov 9, 2023 at 14:38

Your Answer

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

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