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Quantum Error Correction
Problem: the quantum computer cannot be isolated from the environment, therefore, operations can be faulty.
This leads to errors in quantum gates, initial state preparation, measurements.
classical: redundancy (by copying)
Quantum challenges:
Effective error model: Stochastic independently distributed \(X,Y,Z\) errors with probability \(p\)
\(3\)-qubit code
\[\begin{aligned}|\psi\rangle_L&=\alpha|0\rangle+\beta|1\rangle\\ &=\alpha|000\rangle+\beta|111\rangle\end{aligned}\]
\(\alpha|000\rangle+\beta|111\rangle\xrightarrow[]{X_1}\alpha|100\rangle+\beta|011\rangle\)
Apply \(C_1NOT_3, C_2NOT_3\) on state \(|a\rangle|b\rangle|0_C\rangle\) and measuring \(|0_C\rangle\)
where \(|0_C\rangle\) is an ancilla state.
If \(|ab\rangle=|00\rangle\) or \(|ab\rangle=|11\rangle\), \(|0_C\rangle\) will not change. \(\Rightarrow\) even parity
If \(|ab\rangle=|01\rangle\) or \(|ab\rangle=|10\rangle\), \(|0_C\rangle\) will change to \(|1_C\rangle\). \(\Rightarrow\) odd parity
consider \(n\)-qubit operation \(U\) with eigenvalues \(\pm 1\)
\[\begin{aligned}|0\rangle|\psi\rangle&\xrightarrow[]{H_1}&1/\sqrt{2}(|0\rangle+|1\rangle)|\psi\rangle\\ &\xrightarrow[]{Controlled-U}&1/\sqrt{2}|0\rangle|\psi\rangle+1/\sqrt{2}|1\rangle U|\psi\rangle\\ &\xrightarrow[]{H_1} & 1/2(|0\rangle+|1\rangle)|\psi\rangle+1/2(|0\rangle-|1\rangle) U|\psi\rangle \\ & &=|0\rangle (1/2)(\mathbb{I}+U)|\psi\rangle+|1\rangle (1/2)(\mathbb{I}-U)|\psi\rangle\end{aligned}\]
where \((1/2)(\mathbb{I}\pm U)\) is \(P_\pm:\) projector operator on the \(\pm 1\) eigenspace of \(U\)
\(\Rightarrow\) realize a measuremnt of \(U\)
For \(U=Z_1Z_2\)
\(\Rightarrow\) This is a contrast to measure \(Z_1\) and \(Z_2\) individually.
\[\begin{aligned}|\psi\rangle|00\rangle\xrightarrow[]{error}&|\psi'\rangle|00'\rangle|00\rangle_{ancilla} \\ \xrightarrow[]{C_1NOT_4, C_2NOT_4, C_2NOT_5, C_3NOT5}&|\psi'\rangle|00'\rangle Z_1Z_2|0\rangle Z_2Z_3|0\rangle \\ \xrightarrow[]{correction}& |\psi''\rangle|00''\rangle\end{aligned}\]
\[p_{correct}=(1-p)^3+3p(1-p)^2\]
no error on \(3\) qubits + the probability of one \(X\) error
\[p_{fail}=3p^2(1-p)+p^3\]
where \((1-p)\) is the probability of single qubit which successes.
\(\Rightarrow\) \(3\)-qubit is better than single qubit.
\(p_{correct}^{(5)}\geq p_{correct}^{(3)}\) for small \(p\)
Encode codes in codes in codes…
\[p_{fail}^{(n)}=3(p_{fail}^{(n-1)})^2-2(p_{fail}^{(n-1)})^3\leq 3 (p_{fail}^{(n-1)})^2\]
For \(n\) layers of concatenation:
\[p_{fail}^{(n)}\leq\frac{1}{3}(3p)^{2^n}\]
\(\Rightarrow p_{fail}\rightarrow 0\text{ for }p\leq 1/3\)
Stabilize Formalism
i.e. \([S_1,S_2]=0\)
\(S_1|\psi\rangle=|\psi\rangle\\S_2|\psi\rangle=|\psi\rangle\)
\[\{\mathbb{I},Z_1Z_2,X_1X_2,-Y_1Y_2\}\]
\[\sum_{j=0}^t{n\choose j}3^j\cdot 2^k\leq 2^n\\\text{ called Quantum Hamming Bound}\]
\({n\choose j}:j\) errors occur in \(n\)-qubits
\(3^j:\) \(X,Y,Z\) errors
\(2^k:\) error codespace to an \(2^k\) dimension orthogonal subspace
\[2(1+3n)\leq2^n\]
\(\Rightarrow n\geq 5\) qubits needed for smallest complete code.
can be detected by \(\{Z_1Z_2, Z_2Z_3,Z_4Z_5,Z_5Z_6,Z_7Z_8,Z_8Z_9\}\)
can be detected by \(\{X_1X_2X_3X_4X_5X_6,X_4X_5X_6X_7X_8X_9\}\)
detect the sign of any of three qubits.
but we can applying \(Z_1\)-correction
we get the same correct state.
Quantum Fault Tolerance
\(X\) error propagates from a controlled qubit through \(CNOT\) gate onto a target qubit
( propagate errors from \(X_1\) to \(X_2\))
Topological QEC
Kitaev's toric code
\[\prod_i S_Z^{(i)}=\mathbb{I}\Rightarrow\prod_i^{L^2-1} S_Z^{(i)}=S_Z^{(L^2)}\]
\[H^{\otimes n}=H_1\otimes H_2\otimes...\otimes H_n\]
if one \(H_i\) is wrong, only effect one qubit \(\Rightarrow\) Fault Tolerant