1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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

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.