If $S|\alpha\rangle=|\alpha\rangle$ and $|\beta\rangle=U|\alpha\rangle$, then
$$
(USU^\dagger)|\beta\rangle
=US|\alpha\rangle
=U|\alpha\rangle
=|\beta\rangle.
$$
Thus conjugating the state conjugates its entire <stabilizer group>.

The triangle $G_B$ is obtained from the path $G_A$ by <local complementation of a graph state> at vertex $2$, which toggles the edge between its neighbors $1$ and $3$. The corresponding <Local Clifford operation> is
$$
U_2=exp\left(-\frac{i\pi}{4}X_2\right)
\exp\left(\frac{i\pi}{4}Z_1\right)
\exp\left(\frac{i\pi}{4}Z_3\right).
$$
Direct conjugation gives
$$
U_2S_1^AU_2^\dagger=Y_1Y_2=S_1^BS_2^B,
$$
$$
U_2S_2^AU_2^\dagger=S_2^B,
$$
$$
U_2S_3^AU_2^\dagger=Y_2Y_3=S_2^BS_3^B.
$$
These three commuting operators generate exactly the same <stabilizer group> as $S_1^B,S_2^B,S_3^B$. Therefore
$$
\boxed{|G_B\rangle=e^{i\phi}U_2|G_A\rangle}
$$
for an irrelevant global phase $e^{i\phi}$.

Solution

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

{c}

A local Clifford operation is a tensor product of one-qubit Clifford gates. On graph states, local Clifford operations implement graph transformations such as local complementation.


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