Classical vs Quantum Error Correction

Classical Error Correction → Errors flip bits.

Quantum Error Correction:

→ No cloning theorem;

→ Errors are continuous;

→ Measurements destroy information.

Three Qubit flip code

The code:

• $\ket 0 \rarr \ket{000}$
• $\ket 1 \rarr \ket{111}$

Syndrome detection:

• $P_0 \equiv \ket{000} \bra{000} + \ket{111} \bra{111}$ : No error
• $P_1 \equiv \ket{100} \bra{100} + \ket{011} \bra{011}$ : Bit flip on qubit one.
• $P_2 \equiv \ket{010} \bra{010} + \ket{101} \bra{101}$ : Bit flip on qubit two
• $P_3 \equiv \ket{001} \bra{001} + \ket{110} \bra{110}$ : Bit flip on qubit three.

Example:

• Suppose there is an error at qubit one.

• Then:

  $$ \braket{\psi \ | \ P_1 \ | \ \psi}= \\(\overline a \ket{100} + \overline b \ket{011})(\ket{100}\bra{100}+\ket{011}\bra{011})(a\ket{100}+b\ket{011})= \\(\overline a \bra{100}+\overline b \bra{011})(a\ket{100}+b\ket{011})=\overline aa+ \overline bb = 1 $$

Checkpoint: Verify that $\braket{\psi | P_0|\psi} = \braket{\psi|P_2|\psi} =\braket{\psi|P_3|\psi} = 0$

After the measurement the state remains the same: $P_1\ket \psi = \ket \psi$.

Recovery: use the value of the error syndrom eto apply the appropriate recovering procedure:

• 0: Do nothing