For a generic stabilizer code, there's a standard form for $H,X,Z$; see for example previuos post.

For CSS codes, there are at least these simplifications : $B=0$; $C_2=0$; I think $A_1$ and $D$ can also be transformed to $0$ but I'm not sure. There could also be other simplifications. Has anyone worked out the details or knows of a reference for such a standard form?


1 Answer 1


Certainly $B=C_2=0$ for CSS codes. But for the other two.

Argument reductio ad absurdum

A CSS code is made out of two classical codes, with parity check matrices $H_X, H_Z$, and these matrices are the submatrices of the quantum code's stabilizer generator matrix.

Classical codes can also be transformed into equivalent codes by row operations on the parity-check matrices.

Let's imagine if $H_i$ is a $k_i\times n$ matrix for $i=X,Z$, and represents a $n$ bit code. Now, if $H_i$ could be simplified further so that entire columns could be made zero, then it is actually a code on fewer than $n$ bits! That is clearly not possible.

Argument impossible algorithm

We have the matrix $$ H = \left(\begin{array}{ccc|ccc} I & A_1 & A_2 & 0 & 0 & 0 \\ 0 & 0 & 0 & D & I & E_2 \end{array}\right). $$

How do we intend to turn any of the submatrices $0$ with row operations? Take $D$ for instance. There are no $1$ in any row above it. So we can't do any operations that remove all $1$ from every column of $D$. We will always be left with at least one $1$ in each column no matter what row operations we do.

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    $\begingroup$ I came to the same conclusion last night after I posted the question. In standard form the destabilizers have form : $((0,0,0|I,0,0),(0,I,0|0,0,0))$ and that lead me to believe they can be used to eliminate $A_1$ and $D$ but that screws up the commutation relations so it's a no go. There is one restriction on the matrices that follows from $H_x H_z^T=0$ : $D^T + A_1 I^T + A_2 E_2^T = 0$ $\endgroup$
    – unknown
    Jun 14 at 17:59

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