# Qutrit Steane Code

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 $$Z^\dagger$$?

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

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