I'm trying to understand the method from Grover and Rudolph to initialise state based on probability distribution. There is a example described in This post, however I don't understand why we need to have to do the rotation conditionally to a $\theta_0$ register. Since we add this register specifically already in this value, why it wouldn't work if we just perform the rotation in the additional qubits without it being controlled? Like: $$|0\rangle|0\rangle \rightarrow |0\rangle R_{\theta} |0\rangle = |0\rangle (\cos(\theta)|0\rangle + \sin(\theta)|1\rangle)$$ If you could show a quick example of why it doesn't work it'd be very nice.
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1$\begingroup$ Because you want to tie the probability term $\sqrt{p_i}$ to the value of $|i\rangle$. If you don't have controlled rotations then you're constantly rotating the amplitudes on the ancilla register independent of what you have in register $|i\rangle$. If you've tried working out the maths maybe shared what you have so far to see where things went wrong. $\endgroup$– diemilioCommented Oct 1 at 12:25
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$\begingroup$ I see, so like the ancilla register is eq. to $\theta_i$ only when "tensored" to $\ket{i}$. But why does it needs to have this specific value? It could work with any other value as long as $\ket{i}$ is the only value it is "tensored" with. $\endgroup$– L.DZCommented Oct 1 at 14:07
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$\begingroup$ to which step are you referring to exactly? The overall goal of the scheme is to find a sequence of operations (in a qubit system) to obtain a final target state $\sum_i \sqrt{p_i}|i\rangle$. Are you asking about the step $\sqrt{p_i^{(m)}}|i\rangle\to |i\rangle(\sqrt{\alpha_i} |0\rangle+\sqrt{\beta_i}|1\rangle)$, in the notation of the linked question? $\endgroup$– glS ♦Commented Oct 2 at 11:10
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$\begingroup$ eg having in mind the steps as I describe them in quantumcomputing.stackexchange.com/q/33739/55, I'd say introducing the ancillary states $|\theta_i\rangle$ before the actual rotation isn't strictly necessary, at a theoretical level. They use them to provide an explicit recipe to use those states to implement the rotation in terms of qubits, and they are in fact removed in the uncomputation step afterwards. In the overall action $U^\dagger RU$ (again using the notation in the linked post) there is indeed no $\theta_i$ showing up at all $\endgroup$– glS ♦Commented Oct 2 at 11:18
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$\newcommand{\ket}[1]{|#1\rangle}$The ancilla qubits is necessary and it takes a value $\theta_i$ that is computed in the quantum circuit for the whole superposition. Then the rotation is done with a gate of the form $$U\ket{anc}\ket{reg} = \ket{anc} * Ry(\theta_i)*\ket{reg}$$. by doing this, we only need 1 rotation gate to compute the conditional amplitude of all the value in the superposition.
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