Skip to main content
added 55 characters in body
Source Link
kong
  • 59
  • 2

Suppose we have a $[[n,2,d]]$ stabilizer code whose one choice of basis of logical operators is $\{\bar{X}_1,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_2\}$. We can also choose different basis $\{\bar{X}_1\bar{X}_2,\bar{Z}_2\}$$\{\bar{X}_1\bar{X}_2,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_1\bar{Z}_2\}$. The question is, if the distanceweight of the lowest-weight logical operators of the former case is $d_1$, and the latter case is $d_2$ ($d_1>d_2$$d_1<d_2$), what is the distance of this code? One way of thinking this is that the distance should be determined by the lowest-weight logical operator among all possible basis, so $d_2$$d_1$ is the distance of the code. Another way of thinking is that once we decide the basis of the logical operators, distance should be the lowest-weight logical operator of that basis, so we can use this code as $[[n,2,d_1]]$$[[n,2,d_2]]$ code by choosing $\{\bar{X}_1,\bar{Z}_1\}$$\{\bar{X}_1\bar{X}_2,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_2\}$$\{\bar{X}_2,\bar{Z}_1\bar{Z}_2\}$ as the basis of the two logical qubits. I am confused which way of thinking is true.

Suppose we have a $[[n,2,d]]$ stabilizer code whose one choice of basis of logical operators is $\{\bar{X}_1,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_2\}$. We can also choose different basis $\{\bar{X}_1\bar{X}_2,\bar{Z}_2\}$ and $\{\bar{X}_2,\bar{Z}_1\bar{Z}_2\}$. The question is, if the distance of the former case is $d_1$, and the latter case is $d_2$ ($d_1>d_2$), what is the distance of this code? One way of thinking this is that the distance should be determined by the lowest-weight logical operator among all possible basis, so $d_2$ is the distance of the code. Another way of thinking is that once we decide the basis of the logical operators, distance should be the lowest-weight logical operator of that basis, so we can use this code as $[[n,2,d_1]]$ code by choosing $\{\bar{X}_1,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_2\}$ as the basis of the two logical qubits. I am confused which way of thinking is true.

Suppose we have a $[[n,2,d]]$ stabilizer code whose one choice of basis of logical operators is $\{\bar{X}_1,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_2\}$. We can also choose different basis $\{\bar{X}_1\bar{X}_2,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_1\bar{Z}_2\}$. The question is, if the weight of the lowest-weight logical operators of the former case is $d_1$, and the latter case is $d_2$ ($d_1<d_2$), what is the distance of this code? One way of thinking this is that the distance should be determined by the lowest-weight logical operator among all possible basis, so $d_1$ is the distance of the code. Another way of thinking is that once we decide the basis of the logical operators, distance should be the lowest-weight logical operator of that basis, so we can use this code as $[[n,2,d_2]]$ code by choosing $\{\bar{X}_1\bar{X}_2,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_1\bar{Z}_2\}$ as the basis of the two logical qubits. I am confused which way of thinking is true.

Source Link
kong
  • 59
  • 2

Does the distance of a stabilizer code depend on the logical basis?

Suppose we have a $[[n,2,d]]$ stabilizer code whose one choice of basis of logical operators is $\{\bar{X}_1,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_2\}$. We can also choose different basis $\{\bar{X}_1\bar{X}_2,\bar{Z}_2\}$ and $\{\bar{X}_2,\bar{Z}_1\bar{Z}_2\}$. The question is, if the distance of the former case is $d_1$, and the latter case is $d_2$ ($d_1>d_2$), what is the distance of this code? One way of thinking this is that the distance should be determined by the lowest-weight logical operator among all possible basis, so $d_2$ is the distance of the code. Another way of thinking is that once we decide the basis of the logical operators, distance should be the lowest-weight logical operator of that basis, so we can use this code as $[[n,2,d_1]]$ code by choosing $\{\bar{X}_1,\bar{Z}_1\}$ and $\{\bar{X}_2,\bar{Z}_2\}$ as the basis of the two logical qubits. I am confused which way of thinking is true.