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I'm new to Quantum Error Correction, and I have a question on Shor's Code.

If we have a protected subspace, $V \subset \mathbf{C}^2\otimes \cdots \otimes \mathbf{C}^2$

$V=\operatorname{span}\{|0_{l}\rangle, |1_{L}\rangle.$ We also consider Pauli basis of $\mathbf{C}^2\otimes \cdots \otimes \mathbf{C}^2$ of 9 copies, and constructed as follows: Take the basis of $M_2$ consisting of: \begin{eqnarray} \nonumber X=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, Y= \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}, Z=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} & \text{and} & 1_2. \end{eqnarray} We list the 1-Paulis as $U_1,\cdots ,U_{28}.$ Define the error map as $\mathscr{E}:M_{2^9}\rightarrow M_{2^9}$ by $\mathscr{E}(X)=\frac{1}{28}\sum_{i=1}^{28}U_iXU_i^*$. $\mathscr{E}$ is completely positive and trace preserving. How do we say that it satisfies the Knill Laflamme Theorem and thus ensure the existence of a recovery operator?


Cross-posted on math.SE

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Strictly, what you have to calculate is that for all $i$ and $j$ $$ \langle 0_L|U_iU_j|1_L\rangle=0 $$ and $$ \langle 0_L|U_iU_j|0_L\rangle=\langle 1_L|U_iU_j|1_L\rangle. $$ (I've ignored the Hermitian conjugate because all the single-qubit errors are Hermitian.)

Obviously there's a lot of work involved in calculating all $28^2$ cases of $i,j$. You can at least simplify this by using symmetry - there's permutation invariance within blocks of three qubits and between blocks of three qubits. This means that you can reduce your work to two sets of cases: (i) two Pauli errors on the same block of 3 qubits (of which there are $9^2$ cases, but we can take the two errors to be on the first two qubits, or two on the first, reducing to $3+3^2$ cases) and (ii) one Pauli error on each of 3 qubits (of which there are $9^2$ cases, but we can assume the errors are on the first qubit of each block, reducing to $3^2$ cases). 21 error combinations is a much more tolerable calculation.

Could I also recommend finding out about the stabilizer formalism as applied to error correcting codes? This places the calculation in a completely different light which is much more convenient.

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  • $\begingroup$ I actually got this from Ved Guptas Book on Functional Analysis of Quantum Information Theory it was on Page 60, right after the proposition. Since, the book was primarily math oriented and cause they just introduced this, as an application of Choi's theorem I don't think they talked about Stabilizer Formalism. How would you explicitly show this though? $\endgroup$ – Anon Oct 21 at 7:22
  • $\begingroup$ @Anon That's far too broad to address here. Go away and read a bit about the stabilizer formalism, check any relevant questions that you might find on this site, and then ask something more specific. Even if it's more or less this question again, you'd have done some of the correct notational setup and help us to know where to pitch the answer. $\endgroup$ – DaftWullie Oct 21 at 7:49

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