I'm stumped by these questions in Chuang's book.

If I have a state $|ψ\rangle=1/2\sqrt2(1+M_0)(1+M_1)(1+M_2)|0\rangle_7$, where $$M_0=X_0X_4X_5X_6; M_1=X_1X_3X_5X_6; M_2=X_2X_3X_4X_6$$

I rewrite its superposition expression without operators on the computational basis:

$$|ψ\rangle=1/2^5[(|0000000\rangle+|1111111\rangle)+(|1000111\rangle+|0111000\rangle)+(|0101011\rangle+|1010100\rangle)+(|0011101\rangle+|1100010\rangle)+(|1101100\rangle+|0010011\rangle)+(|1011010\rangle+|0100101\rangle)+(|0110110\rangle+|1001001\rangle)+(|1110001\rangle+|0001110\rangle)] $$

I am not sure if it is right. Hope you guys can give me some suggestions.

Also, I am not sure $X$ or $Z$ operations are fault-tolerant. I learned from Chuang's book that a complete set of gates consisting of $H$-gate, phase-gate, C-NOT, and $T$ gates can be constructed using fault-tolerant procedures. In my opinion, $X$ is not fault-tolerant and $Z$ is fault-tolerant, since $Z$ is one of the phase gates.

Am I wrong? Hope someone can help me with my problems!

Thank you

  • $\begingroup$ You can't really talk about "fault tolerant" with regards to a single state. Surely you need at least a pair of states to give you a logical qubit? $\endgroup$
    – DaftWullie
    Nov 25, 2020 at 16:41
  • $\begingroup$ Are you sure that decomposition is correct? Since there are $3$ operators, there should be $2^{3} = 8$ different elements in the superposition, but I count $16$. Or did you apply a fourth stabilizer? $\endgroup$
    – JSdJ
    Nov 25, 2020 at 23:32
  • $\begingroup$ Hi, Daft! Thank you for your notice. I am not sure about the results I got that the superposition states $|\varphi>$ in the computational basis are right. I meet this problem when I learning Steane Code (CSS). And the $Z= Z_7Z_6Z_5Z_4Z_3Z_2Z_1$ $X=X_7X_6X_5X_4X_3X_2X_1$. Sorry about the unclear expressions. $\endgroup$
    – Haha
    Nov 26, 2020 at 2:21
  • $\begingroup$ Hi JSdJ, I thought the $|0>$ should be changed to the $1/\sqrt2 |0>+|1>$. That's why I got 16 terms in this question. (Sorry I'm in a mess now) $\endgroup$
    – Haha
    Nov 26, 2020 at 2:21

1 Answer 1


The state that you should obtain is a superposition of all possible products of ($I$ and) $M_{0},M_{1}$ and $M_{2}$ acting on $|0\rangle_{7}$. That is:

\begin{equation} \begin{split} III |0\rangle_{7}&= |0\rangle_{7} \\ M_{0}II|0\rangle_{7} = X_{0}X_{4}X_{5}X_{6}|0\rangle_{7} &= |1000111\rangle \\ IM_{1}I|0\rangle_{7} = X_{1}X_{3}X_{5}X_{6}|0\rangle_{7} &= |0101011\rangle \\ IIM_{2}|0\rangle_{7} = X_{2}X_{3}X_{4}X_{6}|0\rangle_{7} &= |0011101\rangle \\ M_{0}M_{1}I|0\rangle_{7} = X_{0}X_{1}X_{3}X_{4}|0\rangle_{7} &= |1101100\rangle \\ M_{0}IM_{2}|0\rangle_{7} = X_{0}X_{2}X_{3}X_{5}|0\rangle_{7} &= |1011010\rangle \\ IM_{1}M_{2}|0\rangle_{7} = X_{1}X_{2}X_{4}X_{5}|0\rangle_{7} &= |0110110\rangle \\ M_{0}M_{1}M_{2}|0\rangle_{7} = X_{0}X_{1}X_{2}X_{6}|0\rangle_{7} &= |1110001\rangle \\ \end{split} \end{equation}

You just sum up these terms (including the proper normalization factor of $\sqrt{8}=2\sqrt{2}$ of course) and that's the state that you get.

Then, fault-tolerance only makes sense in the scope of QEC's and the logical operations you perform on encoded logical states. A single qubit-operation is not fault-tolerant nor 'not fault-tolerant' - the definition doesn't really apply to it (or, at least, it doesn't really make sense to think about it).

Saying that the $X$ gate is not and the $Z$ gate is fault-tolerant is a statement that therefore doesn't ring very nicely - maybe you can explain your thinking more, especially with what you think fault-tolerance is?

  • $\begingroup$ Thank you for your rich explanation! I seem to understand. So, if the conditions are $Z= Z_7Z_6Z_5Z_4Z_3Z_2Z_1$ $X=X_7X_6X_5X_4X_3X_2X_1$ $Z$ (here) is fault-tolerant and $X$ (here) is not. Does my idea Right? Actually I still confused why the logical operations will go wrong (if we do not consider the environment's noise). Thank you again for your kind explanation $\endgroup$
    – Haha
    Nov 26, 2020 at 2:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.