Sandwich test for Quantum Phase Estimation

TL;DR

This paper introduces the SANDWICH test, a novel quantum phase estimation algorithm that improves efficiency by addressing limitations of the Sequential Hadamard test, enabling faster and more practical quantum computations.

Contribution

The paper presents the SANDWICH test, a new quantum phase estimation method that reduces runtime complexity by using a sandwiching technique with the SPROTIS operator.

Findings

The SANDWICH test has a runtime of O(k^2 ln k / ε^2 s_min^6).

It effectively estimates ⟨ψ|U^k|ψ⟩ with improved efficiency over previous methods.

The algorithm's performance depends on the parameter s_min, which is typically not very small in practice.

Abstract

Quantum Phase Estimation (QPE) has potential for a scientific revolution through numerous practical applications like finding better medicines, batteries, materials, catalysts etc. Many QPE algorithms use the Hadamard test to estimate $\langle \psi|U^{k}|\psi\rangle$ for a large integer $k$ for an efficiently preparable initial state $|\psi\rangle$ and an efficiently implementable unitary operator $U$. The Hadamard test is hard to implement because it requires controlled applications of $U^{k}$. Recently, a Sequential Hadamard test (SHT) was proposed (arXiv:2506.18765) which requires controlled application of $U$ only but its total run time $T_{\rm tot}$ scales as $\mathcal{O}(k^{3}/\epsilon^{2}r_{\rm min}^{2})$ where $r_{\rm min}$ is the minimum value of $|\langle \psi|U^{k'}|\psi\rangle|$ among all integers $k' \leq k$. Typically $r_{\rm min}$ is exponentially low and SHT becomes too slow. We present a new algorithm, the SANDWICH test to address this bottleneck. Our algorithm uses efficient preparation of the initial state $|\psi\rangle$ to efficiently implement the SPROTIS operator $R_{\psi}^{\phi}$ where SPROTIS stands for the Selective Phase Rotation of the Initial State. It sandwiches the SPROTIS operator between $U^{a}$ and $U^{b}$ for integers $\{a,b\} \leq k$ to estimate $\langle \psi|U^{k}|\psi\rangle$. The total run time $T_{\rm tot}$ is $\mathcal{O}(k^{2}\ln k/ \epsilon^{2} s_{\rm min}^{6})$. Here $s_{\rm min}$ is the minimum value of $|\langle \psi|U^{\hat{k}}|\psi\rangle$ among all integers $\hat{k}$ which are values of the nodes of a random binary sum tree whose root node value is $k$ and leaf nodes' values are $1$ or $0$. It can be reasonably expected that $s_{\rm min} \not\ll 1$ in typical cases because there is wide freedom in choosing the random binary sum tree.