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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?

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  • $\begingroup$ Are you still wondering about this? $\endgroup$
    – kηives
    Nov 16, 2022 at 21:15

1 Answer 1

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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.

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