# Reflection operator in amplitude amplification for block encoding

I am trying to figure out equation (2.33) here.

Given that

$$\begin{gather} U_{\psi_0}|{0^m}\rangle|{0^n}\rangle=\sqrt{p_0}|{0^m}\rangle|{\psi_0}\rangle+\sqrt{1-p_0}|{\perp}\rangle,\\ (\Pi\otimes I_n)|{\perp}\rangle=0, \quad \Pi=|{0^m}\rangle\langle{0^m}|, \end{gather}$$

we call $$|\psi_{\rm good}\rangle=|0^m\rangle|\psi_0\rangle$$ and would like to prepare $$|\psi_0\rangle$$ in the second register with $$O(1)$$ probability. The recipe is to define

$$\begin{gather} R_{\mathrm{good}}=(1-2|{0^m}\rangle\langle{0^m}|)\otimes I_n=(1-2\Pi)\otimes I_n,\\ R_{\psi_0}=U_{\psi_0}(2|{0^{m+n}}\rangle\langle{0^{m+n}}|-I)U^{\dagger}_{\psi_0}. \end{gather}$$

and to "apply $$G^k=(R_{\mathrm{good}}R_{\psi_0})^k$$ to $$|\psi_0\rangle$$ for $$k=O(1\sqrt{p_o})$$ times." (I think that there's a typo, and we should actually apply $$G^k$$ to the first equation above, correct?)

What I do not understand here is why the so-defined $$R_{\mathrm{good}}$$ is equivalent to $$(1-2|\psi_{\rm good}\rangle\langle \psi_{\rm good}|)= (1-\Pi \otimes\Pi_{\rm good})$$, which is the original definition requried for the amplitude amplification. The line after (2.33) says: "This is because $$|{\psi_{\mathrm{good}}}\rangle$$ can be entirely identified by measuring the ancilla qubits."

So at which point in this process do we measure ancillas? It seems that measuring ancillas each time after we apply $$R_{\mathrm{good}}$$ would ruin the whole procedure. If we don't measure anything until the very end, then, as I said above, it's unclear why $$R_{\mathrm{good}} = (1-2|\psi_{\rm good}\rangle\langle \psi_{\rm good}|)$$ and how this construction reduces to amplitude amplification.

UPDATE

OK, I still don't understand what the sentence "This is because $$|\psi_{\rm good}\rangle$$ can be entirely identified by measuring the ancilla qubits." after (2.33) means but I just checked the following:

$$\begin{gather} (1-2\Pi\otimes\Pi_{\rm good}) (\sqrt{p_0}|0^m\rangle |\psi_0\rangle + \sqrt{1-p_0}|\perp\rangle = (1-2\Pi) (\sqrt{p_0}|0^m\rangle |\psi_0\rangle + \sqrt{1-p_0}|\perp\rangle \,, \end{gather}$$ because in the first term $$\Pi_{\rm good}$$ acts trivially and in the second term $$\Pi$$ annihilates $$|\perp\rangle$$ anyways.

On later steps the same happens, as the only thing that's changing is the amplitudes.

So I'm assuming that no measurements of ancillas are required until the very end. Correct?

• Are you still wondering about this? Nov 16, 2022 at 21:15

This is because when you're in the subspace where the block-encoding was applied, the ancilla qubits are in the $$|0\rangle$$ state. So you can boost the success probability on the state $$|0\rangle|\psi\rangle_{\text{good}}$$, by using amplitude amplification with the Grover iterate providing a phase kickback. The reflector picks out the state in which to perform the sign flip on. It's basically a Pauli Z-gate in the $$\{|\psi\rangle_\text{good}, |\perp\rangle\}$$ basis.