Suppose that we wish to find $x$ s.t. $f(x) = 1$. Instead of having access to an oracle like $U_f: |i\rangle \mapsto (-1)^{f(i)}|i\rangle$ or $U_f: |i\rangle|z\rangle \mapsto |i\rangle|z\oplus f(i)\rangle$, suppose we have access to copies of a state prepared using the oracle in a simple way, e.g. $\sum_i (-1)^{f(i)}|i\rangle$ or $\sum_i |i\rangle|f(i)\rangle$.

Is there a way to use these states to perform a search algorithm that uses $O(\sqrt{N})$ copies of the state, where $N$ is the size of the search space (the range of values $i$ can take)?


1 Answer 1



Intuitively, we don't use properties of the oracle $U_f$, we're just making a quantum tomography of a state which equals $|\psi_f\rangle = \frac{1}{\sqrt{N}}\sum_i (-1)^{f(i)}|i\rangle$ for some $f$ (and there are $N$ possible $f$).

Let's also denote $|\psi_0\rangle = \frac{1}{\sqrt{N}}\sum_i |i\rangle$.

One can use the proof of Grover's algorithm optimality to derive the bound on the number of copies needed for tomography.

There is an inequality which gives

$$ 2N-2\sqrt{N}\sqrt{p} - 2\sqrt{N}\sqrt{N-1}\sqrt{1-p} \le $$ $$ \le \sum_f ||\psi_{f}\rangle^{\otimes K} - |\psi_{0}\rangle^{\otimes K}|^2, $$

for the average success probability $p$ of determining $f$ from $K$ copies of $|\psi_{f}\rangle$.

But $$ \langle \psi_0 | \psi_f \rangle = \frac{N-2}{N}, $$ and $$ ||\psi_{f}\rangle^{\otimes K} - |\psi_{0}\rangle^{\otimes K}|^2 = 2-2{\rm Re}\langle \psi_0 | \psi_f \rangle^K, $$ so that $$ 2N - 2\sqrt{N} \le N(2 - 2 (\frac{N-2}{N})^K), $$ if we want $p$ to be close to 1.

Thus $$ \frac{1}{\sqrt{N}} \ge (\frac{N-2}{N})^K \ge 1-\frac{2K}{N} , $$ hence $$ K \ge (N-\sqrt{N})/2, $$ which means the complexity is $O(N)$ at least.

  • $\begingroup$ How does one prove that the tomography you describe is the optimal method for finding $i$ such that $\langle i| \psi_f\rangle = 1$? $\endgroup$
    – forky40
    Commented Apr 2, 2023 at 21:58
  • $\begingroup$ The proof of the inequality is in the appendix of that paper. Essentially, for a tomography we can't do anything better than a von Neumann measurement in some enlarged Hilbert space. $\endgroup$
    – Danylo Y
    Commented Apr 3, 2023 at 7:02

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