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I'm following the construction from Section 3.5 of Daniel Gottesman's thesis (third paragraph), which is supposed to allow us to construct an $\left[ \hspace{-1pt} \left[ n-1, \, k + 1, \, d - 1 \right] \hspace{-1pt} \right]$ stabilizer code $C'$ from an $\left[ \hspace{-1pt} \left[ n, \, k, \, d \right] \hspace{-1pt} \right]$ stabilizer code $C$, by:

  1. Choosing a generating set $\left \lbrace M_j \right \rbrace$ for the stabilizers of $C$ such that there exists a qubit $q$ where $M_1$ has an $X$ on $q$, $M_2$ has a $Z$ on $q$, and the remaining $M_j$'s are not supported on $q$.
  2. Removing $M_1$ and $M_2$ from $\left \lbrace M_j \right \rbrace$, and removing the now-unnecessary $I$ from each remaining generator to obtain $\left \lbrace M'_j \right \rbrace$.

For example, we can begin with the $\left[ \hspace{-1pt} \left[ 5,\,1,\,3 \right] \hspace{-1pt} \right]$ code: \begin{equation} S = \begin{array}{ccccc} X & Z & Z & X & I \\ I & X & Z & Z & X \\ X & I & X & Z & Z \\ Z & X & I & X & Z \end{array}, \end{equation} we can choose a new generating set: \begin{equation} S = \begin{array}{ccccc} X & Z & Z & X & I \\ I & X & Z & Z & X \\ X & I & X & Z & Z \\ Y & X & X & Y & I \end{array}, \end{equation} choose $q$ to be the rightmost qubit, $IXZZX = M_1$, $XIXZZ = M_2$. After removing these two generators and the rightmost qubit, we're left with \begin{equation} S' = \begin{array}{cccc} X & Z & Z & X \\ Y & X & X & Y \end{array}, \end{equation} which is a $\left[ \hspace{-1pt} \left[ 4,\, 2,\, 2 \right] \hspace{-1pt} \right]$ code, as intended.

However, if we instead begin with the $\left[ \hspace{-1pt} \left[9,\, 1,\, 3 \right] \hspace{-1pt} \right]$ rotated surface code: \begin{equation} S = \begin{array}{ccccccccc} Z & Z & I & I & I & I & I & I & I \\ X & X & I & X & X & I & I & I & I \\ I & Z & Z & I & Z & Z & I & I & I \\ I & I & X & I & I & X & I & I & I \\ I & I & I & X & I & I & X & I & I \\ I & I & I & Z & Z & I & Z & Z & I \\ I & I & I & I & X & X & I & X & X \\ I & I & I & I & I & I & I & Z & Z \end{array}. \end{equation} I can take the leftmost qubit to be $q$, and the top two stabilizers to be $M_2$ and $M_1$ respectively. No other generator has support on $q$, so: \begin{equation} S' = \begin{array}{ccccccccc} Z & Z & I & Z & Z & I & I & I \\ I & X & I & I & X & I & I & I \\ I & I & X & I & I & X & I & I \\ I & I & Z & Z & I & Z & Z & I \\ I & I & I & X & X & I & X & X \\ I & I & I & I & I & I & Z & Z \end{array}. \end{equation} This code is intended to have distance 2, but it instead has distance 1. Am I doing something wrong?

glS
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Ben Criger
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1 Answers1

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I think that Gottesman's proof implicitly assumes that the stabilisers of the code all have weight which is at least that of the distance of the code: his construction of the distance says that assume you have a weight $w$ operator on the new set of qubits that commutes with them. Then there is an operator of weight $w+1$ that commutes with the original set of stabilizers. But he does not distinguish between whether that operator is a stabilizer or a logical operator, he just relates that to the original distance of the code.

I believe that your calculation is correct.

DaftWullie
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