How does error correction disentangle the environment from the encoded state after the quantum erasure channel? For example; the paper Codes for the Quantum Erasure Channel introduces a 4 qubit code for the erasure channel, with the first codeword: $$|\bar{0}\rangle=|0000\rangle +|1111\rangle,$$ If we assume that an erasure occurs on the fourth qubit, then applying the channel isometry, the state becomes $$|000e\rangle \otimes |0\rangle_E +|111e\rangle\otimes |1\rangle_E .$$ The reduced state accessible by the decoder is $$|000e\rangle\langle 000e| +|111e\rangle\langle 111e|.$$ If the decoder map preserves even parity of vectors, then it will map $$|000e\rangle \rightarrow |0000\rangle$$ and $$|111e\rangle \rightarrow |1111\rangle.$$ Then the reduced state is mapped to $$|0000\rangle\langle 0000| +|1111\rangle\langle 1111| \neq |\bar{0}\rangle\langle\bar{0}|,$$ which is not the original state? What am I missing about the decoder operation?
1 Answer
$\newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle #1|} $The erasure error is not a unitary transformation. You can't reverse it with a unitary mapping.
Instead, run the full gamut of error-detection and correction. You take your state post-erasure and add a blank qubit
$$ (\ket{000}\bra{000} + \ket{111}\bra{111}) \otimes \ket{0}\bra{0}. $$
Then do the syndrome measurements on this. In this case, the code is actually a stabilizer code with generators $XXXX, IZZI, ZIIZ$, so quite simple. The syndrome measurements will project the four-qubit state into a distorted code space $\tilde C$, i.e. the code space $C$ affected by a Pauli error. Here, because there is one erasure it will be a single-qubit Pauli error.
You will probabilistically get one of four random syndromes (corresponding to $I,X,Y,Z$), because the erasure error is not a Pauli error. Based on the syndrome you get, and knowing that the error is on the last qubit, you figure out the decoding operation to apply.
