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?
I would really appreciate the help! Thank you!
Best Answer
It may help to define the following operators for $k\in\{1,\dots,9\}$ \begin{align*} X_k &= I\otimes \cdots \otimes I \otimes X\otimes I \otimes \cdots \otimes I\\ Y_k &= I\otimes \cdots \otimes I \otimes Y\otimes I \otimes \cdots \otimes I\\ Z_k &= \underbrace{I\otimes \cdots \otimes I}_{k-1\text{ times}} \otimes Z\otimes \underbrace{I \otimes \cdots \otimes I}_{9-k\text{ times}}. \end{align*} We may now explicitly define the operators $U_1,\dots, U_{27}$ as $$ U_{9a+k} = \left\{\begin{array}{ll} X_{k}& \text{ }a =0\\ Y_{k}& \text{ }a =1\\ Z_{k}& \text{ }a =2\\ \end{array}\right. $$ for $a\in\{0,1,2\}$ and $k\in\{1,2,\dots,9\}$ and $U_{28} = I\otimes I \otimes I\otimes I\otimes I \otimes I\otimes I\otimes I \otimes I$.
To show that these operators satisfy the conditions of the Knill-Laflamme Theorem, you must show that $$ \langle 0_L|U_i^*U_j|1_L\rangle=0 $$ and $$ \langle 0_L|U_i^*U_j|0_L\rangle=\langle 1_L|U_i^*U_j|1_L\rangle $$ for all $i,j\in\{1,\dots,28\}$.
We can cluster the nine qubits into three groups of three by writing the logical states as: $$ |0_L\rangle=\left(\frac{|000\rangle + |111\rangle}{\sqrt{2}}\right)^{\otimes 3}\qquad\text{and}\qquad |1_L\rangle=\left(\frac{|000\rangle - |111\rangle}{\sqrt{2}}\right)^{\otimes 3}. $$ We see that each of the operators $U_1,\dots,U_{28}$ acts on at most one of the clusters of three qubits. It should be obvious that $\langle 0_L|U_i^*U_j|1_L\rangle = 0$ for all $i,j\in\{1,\dots,28\}$. (Can you see why?) It remains to check that $\langle 0_L|U_i^*U_j|0_L\rangle=\langle 1_L|U_i^*U_j|1_L\rangle$.
To see how each of the above operators acts on $|0_L\rangle $ and $|1_L\rangle$, it will help to see how they act on each cluster of three qubits. It may easily be verified that each of the following is orthogonal to each other and orthogonal to $\lvert 000\rangle + \lvert 111\rangle$: \begin{align*} X\otimes I\otimes I \left(\lvert 000\rangle + \lvert 111\rangle\right) &= \lvert 100\rangle + \lvert 011\rangle\\ I\otimes X\otimes I \left(\lvert 000\rangle + \lvert 111\rangle\right) &= \lvert 010\rangle + \lvert 101\rangle\\ I\otimes I\otimes X \left(\lvert 000\rangle + \lvert 111\rangle\right) &= \lvert 001\rangle + \lvert 110\rangle\\ Y\otimes I\otimes I \left(\lvert 000\rangle + \lvert 111\rangle\right) &= i(\lvert 100\rangle -\lvert 011\rangle)\\ I\otimes Y\otimes I \left(\lvert 000\rangle + \lvert 111\rangle\right) &= i(\lvert 010\rangle - \lvert 101\rangle)\\ I\otimes I\otimes Y \left(\lvert 000\rangle + \lvert 111\rangle\right) &= i(\lvert 001\rangle - \lvert 110\rangle) \end{align*} From this, it is clear that the vectors $$ \lvert 0_L\rangle,X_1\lvert 0_L\rangle, \cdots X_9\lvert 0_L,\rangle,Y_1\lvert 0_L\rangle, \cdots Y_9\lvert 0_L\rangle $$ are pairwise orthogonal. It follows that $$ \langle 0_L|X_kY_l|0_L\rangle = \langle 0_L|X_k|0_L\rangle= \langle 0_L|Y_l|0_L\rangle=0 $$ for all $k,l\in\{1,\dots 9\}$.
Analogously, it may easily be verified that each of the following is orthogonal to each other and orthogonal to $\lvert 000\rangle - \lvert 111\rangle$: \begin{align*} X\otimes I\otimes I \left(\lvert 000\rangle - \lvert 111\rangle\right) &= \lvert 100\rangle - \lvert 011\rangle\\ I\otimes X\otimes I \left(\lvert 000\rangle - \lvert 111\rangle\right) &= \lvert 010\rangle - \lvert 101\rangle\\ I\otimes I\otimes X \left(\lvert 000\rangle - \lvert 111\rangle\right) &= \lvert 001\rangle - \lvert 110\rangle\\ Y\otimes I\otimes I \left(\lvert 000\rangle - \lvert 111\rangle\right) &= i(\lvert 100\rangle +\lvert 011\rangle)\\ I\otimes Y\otimes I \left(\lvert 000\rangle - \lvert 111\rangle\right) &= i(\lvert 010\rangle + \lvert 101\rangle)\\ I\otimes I\otimes Y \left(\lvert 000\rangle - \lvert 111\rangle\right) &= i(\lvert 001\rangle + \lvert 110\rangle) \end{align*} From this, it is clear that the vectors $$ \lvert 1_L\rangle,X_1\lvert 1_L\rangle, \cdots X_9\lvert 1_L,\rangle,Y_1\lvert 1_L\rangle, \cdots Y_9\lvert 1_L\rangle $$ are pairwise orthogonal. It follows that $$ \langle 1_L|X_kY_l|1_L\rangle = \langle 1_L|X_k|1_L\rangle= \langle 1_L|Y_l|1_L\rangle=0 $$ for all $k,l\in\{1,\dots 9\}$.
It now follows that $$ \langle 0_L|U_i^*U_j|0_L\rangle=\langle 1_L|U_i^*U_j|1_L\rangle =0 $$ in the case when $i,j\in\{1,2,\dots,18,28\}$. It therefore remains to check the case when at least one of $i$ or $j$ is in $\{19,20,\dots,27\}$ (i.e. when either $U_i=Z_k$ or $U_j = Z_l$ for some $k\in\{1,\dots,9\}$ or $l\in\{1,\dots,9\}$).
Note that $$ Z\otimes I \otimes I(\lvert 000\rangle+\lvert 111\rangle) = I\otimes Z \otimes I(\lvert 000\rangle+\lvert 111\rangle) = I\otimes I \otimes Z(\lvert 000\rangle+\lvert 111\rangle) = \lvert 000\rangle-\lvert 111\rangle $$ and that $$ Z\otimes I \otimes I(\lvert 000\rangle-\lvert 111\rangle) = I\otimes Z \otimes I(\lvert 000\rangle-\lvert 111\rangle) = I\otimes I \otimes Z(\lvert 000\rangle-\lvert 111\rangle) = \lvert 000\rangle+\lvert 111\rangle. $$
Note that if $U_i$ and $U_j$ do not act on the same cluster of 3 qubits, then it must be the case that $$ \langle 0_L|U_i^*U_j|0_L\rangle=\langle 1_L|U_i^*U_j|1_L\rangle=0. $$ It remains to check the cases when $U_i$ and $U_j$ do act on the same cluster of 3 qubits.
I'll leave the rest to you.