Shorの9キュービットコードは、最初のキュービットの測定値を修正できますか?


2

Shorの $ | 0 \ rangle $ のコードワードから始める:

$ | \ psi \ rangle = \ frac {1} {\ sqrt {8}}(| 000 \ rangle + | 111 \ rangle)\ otimes(| 000 \ rangle+ | 111 \ rangle)\ otimes(| 000 \ rangle + | 111 \ rangle)$

ここで、 $ X $ フリップまたは $ Z $ フリップではなく、最初のキュビットが測定されます。Shorのエラー修正アルゴリズムはこのエラーを修正して、元の $ | 0 \ rangle $ を復元できますか?あなたの答えを正当化するために計算を表示します。

私の予想では、Deferred Measurement Principleのため、問題なく修正できると思います。

私の解決策の試み:最初のキュービットの測定結果が $ | 0 \ rangle $ であるとしましょう。したがって、状態全体は次のように折りたたまれます。

$ | \ psi_0 \ rangle = \ frac {1} {\ sqrt {4}}(| 000 \ rangle)\ otimes(| 000 \ rangle + | 111 \ rangle)\ otimes(| 000 \ rangle + | 111 \ rangle)$

ステップ1では、 $ X $ エラーを検出して修正します。ビットフリップがないので、このステップでは状態は変更されません。

ステップ2では、 $ Z $ エラーを探します。私の質問は、 $ | \ psi_0 \ rangle $ で位相反転チェックを計算する方法と、結果を解釈してエラーを修正する方法です。 $ | 1 \ rangle $ を測定した後の状態についても同じ質問です。

ありがとうございます。

2

You can think of this measurement as an 'error' on the (encoded) state that needs to be corrected. Quantum error correction is all about subspaces of the Hilbert space, and during QECC we are always trying achieve information in what subspace our state lies.

The state lies in some subspace, which is either the codespace or some orthogonal space. With every space we identify an error (with the codespace it is the trivial 'error' - $I$). There are many other errors that map to a specific codespace, but we cannot always* correct for these errors.

For the $9$-bit Shor code, the subspaces are those associated with all the single-qubit bit flips $X_{i}$, and furthermore with all the single qubit phase flips $Z_{i}$. There is some degeneracy in the code where there are cases that sometimes you can correct mulitple (correlated!) $Z$-flips, but we'll disregard that in our discussion.

Now we are ready to investigate our measurement: after measureing, the state $|\psi_{0}\rangle$ does not lie in any of the subspaces associated with the $9$-bit Shor code, but it is rather a specific superposition:

\begin{equation} \begin{split} |\psi_{0}\rangle = &\frac{1}{2}|000\rangle \otimes \big(|000\rangle+|111\rangle\big) \otimes \big(|000\rangle+|111\rangle\big) \\ = &\frac{1}{4}\big(|000\rangle+|111\rangle\big) \otimes \big(|000\rangle+|111\rangle\big) \otimes \big(|000\rangle+|111\rangle\big) \\ +& \frac{1}{4}\big(|000\rangle-|111\rangle\big) \otimes \big(|000\rangle|111\rangle\big) \otimes \big(|000\rangle+|111\rangle\big) \\ = & \frac{1}{\sqrt{2}}\big(|\psi\rangle + Z_{1}|\psi\rangle\big) = \frac{1}{\sqrt{2}}\big(|\psi\rangle + Z_{2}|\psi\rangle\big) = \frac{1}{\sqrt{2}}\big(|\psi\rangle + Z_{3}|\psi\rangle\big), \end{split} \end{equation}

We thus see that the state $|\psi_{0}\rangle$ is a superposition of no error having happened, and of a phase flip on either the first, second or third having happened.

Measuring out stabilizers will not only collapse this superposition to either of the two (so either no error or a phase flip), the measurement outcome (the error syndrome) will also indicate what subspace we have projected to. The correction is then straightforward.

If the measurement on qubit $1$ resulted in the state $|1\rangle$, we obtain a different 'superposition of subpsaces'. It is not that different though (the $+$ in the beginning of the third line changes to a $-$), and the code doesn't really 'care': it projects all the same, to the same kind of subspaces. The correction process is therefore (in this particular case) exactly the same.

*I say not always, because it depends if the error acts the same as the correctable error on the codespace or not - if it acts the same (then it only differs from the correctable error by a stabilizer) then it is correctable.