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Consider a Boolean function with multiple outputs $f: \{0, 1\}^{n} \rightarrow \{0, 1\}^{m}$, and consider being given oracle access to the function $f$. Let us denote the oracle by $O_f$. For an $x \in \{0, 1\}^{n}$ and $b \in \{0, 1\}^{m}$,

\begin{equation} O_f |x\rangle |b\rangle = |x\rangle|b\oplus f(x)\rangle, \end{equation}

where $\oplus$ is the bitwise XOR operator. Now, let's say we have a quantum computer and we want to learn the value of $f$ in $k$ positions, $x_1, x_2, \ldots, x_k \in \{0, 1\}^{n}$. Let's also say that we know these positions beforehand.

I want to prove that for a worst case function $f$, it takes the quantum computer at least $k$ queries to $O_f$ to do this. If that is not true, what is the lower bound on the number of queries?

The problem is that the quantum computer can generate superpositions like \begin{equation} \sum_{x \in \{0,1\}^{n}} \alpha_{x} |x \rangle |f(x)\rangle, \end{equation} where $\alpha_{x} \in \mathbb{C}$, for each $x \in \{0, 1\}^{n}$. Do these reveal sufficient information about the function to solve the task I mentioned with less than $k$ queries?

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What you're asking to do is at least as hard as the following problem: Given oracle access to $f$ such that either (a): $f(x)=0 \, \forall \, x$ or (b): $\exists z: f(x) = \delta_x^z$ for $x,z \in \{0, \dots k\}$. Specifically, in your problem statement set $m=1$ and relax the requirement from using $n$ qubits to instead just using a $k$ dimensional space. Then, "learning the value of $f$ in $k$ positions" is at least as difficult as "learning the value of $f$ in $k$ positions given that $f$ satisfies either (a) or (b)". The latter task requires $\Omega(\sqrt{k})$ queries to $O_f$ - for example see Theorem 9.3.2 in (Kaye, Laflamme, Mosca) - so this certainly places a lower bound on your problem in a special case.

However if the task is to just evaluate $f(x)$ for $x\in\{0, \dots, k\}$ with no additional structure on $f$, then I think you can just apply Holevo's theorem: each use of $U_f$ encodes information about $f(x)$ in $m$ qubits, from which you cannot retrieve more than $m$ bits of classical information (which is the number of bits required to specify the value of $f$). So to learn $k$ values of $f$ would require $km$ classical bits of information, which requires at least $km$ qubits or $k$ uses of $U_f$ as its currently specified.

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