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I've been searching about quantum amplitude estimation techniques and read that it is customary to use an extra qubit to separate good and bad states using algorithm $A$ as

$$ A|{0} \rangle^{\otimes^{n+1}} = |0\rangle |\psi_0\rangle + |1\rangle |\psi_1\rangle $$

I was wondering the following:

a) If we don't have a guarantee that a solution $|\psi_1\rangle$ exists, what happens if we perform amplitude estimation/amplification?.

b) Is quantum amplitude estimation bounded by the initial probability of $|{\psi_1}\rangle$ or is it bounded with the size of the hillbert space? With the assumption that all $2^n$ possible states are initialized with the same amplitude, we would need $\sqrt{2^n}$ steps to boost the amplitude of $|{\psi_1}\rangle$, but wouldnt it be possible to perform quantum amplitude amplification if the inversion about the mean would be performed directly on the additional qubit? I was wondering this because if we look at the n qubits from our problem, initially, we would see the same amplitude for every state, but if we only focused on the additional qubit we only have two states and the inversion about the mean would be around $\sqrt{\frac{1}{2}}$ rather than $\sqrt{\frac{1}{2^n}}$ (using $\mathbf{Q} = -AS_oA^{-1}S_x$ but $S_o$ and $S_1$ only in the additional qubit). If this is not possible, could you give me some insight as to why it cant be done? I would really appreciate it.

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