A computational-basis vector $|x_1x_2x_3\rangle$ has eigenvalue $(-1)^{x_i+x_j}$ under $Z_iZ_j$. Simultaneous eigenvalue $+1$ for $Z_1Z_2$, $Z_2Z_3$, and $Z_1Z_3$ therefore requires
$$
x_1=x_2=x_3.
$$
The <stabilizer subspace> is
$$
\boxed{V_S=\operatorname{span}\{|000\rangle,|111\rangle\}}.
$$
It is two-dimensional because the three displayed nonidentity stabilizers contain only two independent generators; indeed $(Z_1Z_2)(Z_2Z_3)=Z_1Z_3$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 324 / 4 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-324/4/a/i/solution