# Pauli Matrix tolerant and encoding

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

• 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? Nov 25 '20 at 16:41
• 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?
– JSdJ
Nov 25 '20 at 23:32
• 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.
– Haha
Nov 26 '20 at 2:21
• 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)
– Haha
Nov 26 '20 at 2:21

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?

• 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
– Haha
Nov 26 '20 at 2:28