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The Shor’s 9-qubit code has the following stabilizers

$\hat{S}_1= \hat{Z}_1\hat{Z}_2$ , $\hat{S}_2= \hat{Z}_2\hat{Z}_3$, $\hat{S}_3= \hat{Z}_4\hat{Z}_5$

$\hat{S}_4= \hat{Z}_5\hat{Z}_6$, $\hat{S}_5= \hat{Z}_7\hat{Z}_8$ , $\hat{S}_6= \hat{Z}_8\hat{Z}_9$

$\hat{S}_7= \hat{X}_1\hat{X}_2\hat{X}_3\hat{X}_4\hat{X}_5\hat{X}_6$ , $\hat{S}_8= \hat{X}_4\hat{X}_5\hat{X}_6\hat{X}_7\hat{X}_8\hat{X}_9$

Is it true that error syndrome for qubit 5 phase-flip is the same as that for qubit 6 phase-flip?

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    $\begingroup$ Do you want a simple yes or no answer (the answer is in fact yes) or do you want to see the mathematics? $\endgroup$
    – JSdJ
    Apr 2 at 13:49
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    $\begingroup$ Also, small nitpick regarding terminology (knowing which might benefit you): these are the generators of the stabilizer group - the wording stabilizers itself is a bit ambiguous. $\endgroup$
    – JSdJ
    Apr 2 at 13:50
  • $\begingroup$ @JSdJ I need to see the mathematics to get better understanding of it. $\endgroup$
    – heromano
    Apr 2 at 21:18
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    $\begingroup$ This comment doesn't answer your question, but you can play around with this code and check step by step what is being done in the Shor's 9-qubit algorithm for correcting quantum errors: github.com/sebastianvromero/qecc_shor9q . Just run it locally downloading it or launch it in Binder. Cheers! $\endgroup$ Apr 5 at 23:59
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TL;DR Yes, the two error syndromes are identical, because the two errors trip the same set of stabilizers.


Let $|\psi\rangle$ denote a state in the code subspace of the Shor's 9-qubit code. Every operator $\hat{S}$ in the stabilizer group $S$ of the code fixes $|\psi\rangle$, i.e. $\hat{S}|\psi\rangle = |\psi\rangle$. In particular, generators $\hat{S}_1, \dots \hat{S}_8$ fix $|\psi\rangle$, i.e. $$\hat{S}_i|\psi\rangle = |\psi\rangle\tag1$$ for $i=1,\dots,8$. Let $|\psi_5\rangle$ denote the result of a phase-flip error on qubit 5, i.e. $|\psi_5\rangle = \hat{Z}_5|\psi\rangle$ and similarly $|\psi_6\rangle = \hat{Z}_6|\psi\rangle$.

We begin computing the errors syndrome for $|\psi_5\rangle$ with the $Z$ type stabilizer generators, i.e. $\hat{S}_j$ for $j=1,\dots,6$. Note that $\hat{S}_j=\hat{Z}_a\hat{Z}_b$ for some qubits $a$ and $b$. We calculate

$$ \hat{S}_j|\psi_5\rangle = \hat{Z}_a\hat{Z}_b\hat{Z}_5|\psi\rangle = \hat{Z}_5\hat{Z}_a\hat{Z}_b|\psi\rangle = \hat{Z}_5\hat{S}_j|\psi\rangle = \hat{Z}_5|\psi\rangle = |\psi_5\rangle\tag2 $$

where we first used the definitions of $\hat{S}_j$ and $|\psi_5\rangle$, then the fact that $\hat{Z}_q$ and $\hat{Z}_r$ commute for any qubits $q$ and $r$, then the definition of $\hat{S}_j$ again, then $(1)$ with $i=j$ and finally the definition of $|\psi_5\rangle$. Thus, we see that $|\psi_5\rangle$ is an eigenvector of $\hat{S}_j$ for $j=1,\dots,6$ associated with eigenvalue $+1$ and therefore a measurement of $\hat{S}_j$ on this state yields $+1$ with probability $1$. Note that analogous calculation yields the same result on $|\psi_6\rangle$

$$ \hat{S}_j|\psi_6\rangle = \hat{Z}_a\hat{Z}_b\hat{Z}_6|\psi\rangle = \hat{Z}_6\hat{Z}_a\hat{Z}_b|\psi\rangle = \hat{Z}_6\hat{S}_j|\psi\rangle = \hat{Z}_6|\psi\rangle = |\psi_6\rangle\tag3 $$

for $j=1,\dots,6$ as before.

Now, let us turn to the $X$ type stabilizer generators $\hat{S}_7$ and $\hat{S}_8$. For the state $|\psi_5\rangle$, we find

$$ \begin{align} \hat{S}_7|\psi_5\rangle &= \hat{X}_1\hat{X}_2\hat{X}_3\hat{X}_4\hat{X}_5\hat{X}_6\hat{Z}_5|\psi\rangle \\ &= -\hat{Z}_5\hat{X}_1\hat{X}_2\hat{X}_3\hat{X}_4\hat{X}_5\hat{X}_6|\psi\rangle \\ &= -\hat{Z}_5\hat{S}_7|\psi\rangle \\ &= -\hat{Z}_5|\psi\rangle \\ &= -|\psi_5\rangle \end{align}\tag4 $$

where first we used the definitions of $\hat{S}_7$ and $|\psi_5\rangle$, then the facts that $\hat{X}_i$ and $\hat{Z}_j$ commute when $i\ne j$ and anticommute when $i=j$, then once again the definition of $\hat{S}_7$, then $(1)$ with $i=7$ and finally the definition of $|\psi_5\rangle$. Thus, we see that $|\psi_5\rangle$ is an eigenvector of $\hat{S}_7$ associated with eigenvalue $-1$ and therefore a measurement of $\hat{S}_7$ on $|\psi_5\rangle$ yields $-1$ with probability $1$. Analogous calculation for $\hat{S}_8$ shows that $|\psi_5\rangle$ is an eigenvector of $\hat{S}_8$ associated with eigenvalue $-1$. Finally, the same calculation with $|\psi_6\rangle$ in place of $|\psi_5\rangle$ shows that $|\psi_6\rangle$ is an eigenvector of $\hat{S}_7$ and $\hat{S}_8$ associated with eigenvalue $-1$.

We conclude that measuring $(\hat{S}_1, \hat{S}_2, \hat{S}_3, \hat{S}_4, \hat{S}_5, \hat{S}_6, \hat{S}_7, \hat{S}_8)$ on states $|\psi_5\rangle$ and $|\psi_6\rangle$ yields the same syndrome $(+1, +1, +1, +1, +1, +1, -1, -1)$.


Remark: The calculation above shows a useful fact about CSS codes: phase flips are detected by the $X$ type stabilizers and leave the $Z$ type stabilizers unaffected. Similarly, bit flips are detected by the $Z$ type stabilizers and leave the $X$ type stabilizers unaffected.

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