How does applying Z gate to $|0\rangle$ change the phase of other states during reflection about $|s\rangle$ in Grover algorithm in Qiskit textbook

Question

I am trying to understand the Reflection Gate - Us explained for 2 qubits in the qiskit textbook. In the explanation it is mentioned that first Hadamard gate is applied to change the state $|s\rangle$ to $|0\rangle$ then a circuit adds negative phase to all the states orthogonal to $|s\rangle$ and this is done by applying 2 $Z$ (one each on both the qubits) gates followed by a controlled $Z$. I am confused about this because once $|s\rangle$ goes to $|0\rangle$ there won't be any superposition so $Z$ gates won't do anything at all because the effect of $Z$ gates on computational basis is to change $|1\rangle$ to $-|1\rangle$. Can someone please explain. This question is different from other questions asked so please do not link it to other questions.

Answer 1

In a nutshell: The key fact to remember is that what you called the reflection gate is only applied after the oracle. Therefore, the state is not $|s\rangle$ anymore.

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Detailed answer:

To show in more detail what's going on we can calculate a concrete example. In general, Grover's algorithm apply the Grover operator $Q$ defined as $$ Q = H^{\otimes n}S_0 H^{\otimes n} S_f $$ where $H$ is the Hadamard gate, $S_0 = 2|0\rangle^{\otimes n}\langle 0|^{\otimes n} - \mathbb I$ is a reflection about the $|0\rangle^{\otimes n}$ state and $S_f$ is the oracle. Sometimes we group the first three operations and call it "diffusion" because it equals a reflection about the maximally superposed state: $$ Q = S_{+} S_f $$ with the diffusion operator $S_+ = 2|+\rangle^{\otimes n}\langle +|^{\otimes n} - \mathbb I$.

In the standard formulation of Grover's algorithm, we initialize a maximally superposed state and apply $Q$ a specific number of times $p$ to amplify the amplitude of the solution bitstring: $$ Q^p H^{\otimes n} |0\rangle^{\otimes n}. $$ If you know the number of solutions you can calculate $p$, otherwise we usually apply different powers of $Q$, measure, and check if the output is a solution.

Let's look at your particular example to see that the diffusion operator really amplifies the solution state. I've attached that as notes, since the math takes a bit long to TeX

Sidenote: you can also use different operations than $H^{\otimes n}$ to for the general amplitude amplification algorithm.