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 $?

  • $\begingroup$ 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


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

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.

  • 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

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