0

The following is taken from "Quantum Error Correction Via Codes Over GF(4)" Calderbank, Rains, Shor, Sloane.

We are told that the group $E$ of tensor products $\pm w_{1} \otimes \dots \otimes w_{n}$ and $\pm iw_{1} \otimes \dots \otimes w_{n}$, where each $w_{j}$ is one of $I, \sigma_{x}, \sigma_{y}, \sigma_{z}$, describes the possible errors in $n$ qubits.

So $E$ is a subgroup of the unitary group $U(2^{n})$

We want to construct a quantum code from a pair of subgroups of the quantum error group $E$. The group $E$ has order $2^{2n + 2}$.

Would someone be able to explain to me why $E$ has order $2^{2n + 2}$?

am567
  • 781
  • 3
  • 8

1 Answers1

2

Note that $2^{2n+2} = 4^{n+1}$. Now it is easy to solve the combinatorial problem.

There are $n$ elements in the tensor product. Each element can be one of four operators $I, X, Y, Z$. That gives us $4^n$ possibilities.

Finally, there are four phases $\pm 1, \pm i$.

Hence $4^{n+1}$ operators in the group.

Abdullah Khalid
  • 1,993
  • 7
  • 17