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REVIEW 3 major objections 5 minor 79 references

Sandwich test for Quantum Phase Estimation

T0 review · 3 major / 5 minor · reviewed 2026-08-06 · deepseek-v4-flash

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Pith's one-line read The Sandwich test estimates $\langle\psi|U^{k}|\psi\rangle$ in time $\mathcal{O}(k^{2}\ln k / \epsilon^{2} s_{\min}^{6})$, improving on the Sequential Hadamard test's $\mathcal{O}(k^{3}/\epsilon^{2}r_{\min}^{2})$ by replacing the…

desk verdict A genuinely new control-free phase estimation idea, but the central runtime proof in Sec. II is internally inconsistent and the headline scaling claim is not established. read the letter →

arxiv 2507.23716 v2 pith:OPIU7VZ6 submitted 2025-07-31 quant-ph

Avatar Tulsi This is my paper
classification quant-ph MSC 81P6868Q12 PACS 03.67.Ac
keywords quantumphaseestimationHadamardtestSequentialSPROTISoperatorrandombinarysumtreesandwichcontrolledunitarysamplecomplexity
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

The paper presents the Sandwich test, a quantum algorithm for estimating the complex expectation value $\langle\psi|U^{k}|\psi\rangle$ that sits at the heart of quantum phase estimation. Instead of requiring controlled applications of $U^{k}$ as the Hadamard test does, it needs only controlled applications of $U$ itself, and it improves on the recently proposed Sequential Hadamard test by replacing the worst-case factor $1/r_{\min}^{2}$, where $r_{\min}$ is typically exponentially small, with $1/s_{\min}^{6}$, where $s_{\min}$ is the smallest overlap over the nodes of a random binary sum tree chosen by the algorithm. Because the tree gives wide freedom to skip over steps with tiny overlap, the paper argues $s_{\min}$ can reasonably be expected not to be small in typical cases. If that conjecture holds, estimating $\langle\psi|U^{k}|\psi\rangle$ to accuracy $\epsilon$ takes time $\mathcal{O}(k^{2}\ln k / \epsilon^{2} s_{\min}^{6})$, a qualitative improvement over the Sequential Hadamard test's $\mathcal{O}(k^{3} / \epsilon^{2} r_{\min}^{2})$. The same scheme generalizes to sandwiching the SPROTIS operator between arbitrary unitaries, not just powers of one $U$.

What carries the argument

The load-bearing object is the Sandwich operator $S = U^{a} R_{\psi}^{\phi} U^{b}$, built from the SPROTIS operator $R_{\psi}^{\phi} = \mathbb{1}_{N} + (e^{i2\phi}-1)|\psi\rangle\langle\psi|$, which selectively rotates the phase of the initial state. Inserted between two powers of $U$, it converts the phase-addition problem into a modulus-measurement problem: measuring $|\langle\psi|S|\psi\rangle|$ together with the three moduli $r_a, r_b, r_{a+b}$ determines the phase defect $\omega_{a+b}$ through Eq. (11), and the recursion $\theta_{a+b} = \theta_a + \theta_b - \omega_{a+b} + \phi$ propagates estimates up a random binary sum tree of height $\Theta(\ln k)$. The tree is the second mechanism: its random splits, with child values drawn as ceiling/floor of fractions of the parent, give exponentially many choices of which intermediate integers appear, allowing the algorithm to avoid integers $k'$ where $|\langle\psi|U^{k'}|\psi\rangle|$ is tiny.

What would settle it

Take a specific family of instances, e.g., $U$ a random $n$-qubit diagonal unitary with independently uniformly random phases and $|\psi\rangle$ the uniform superposition, fix $k$ around $10^3$, and sample random binary sum trees with the paper's splitting rule. If for typical instances $s_{\min}$ decays exponentially in $n$ (or in $k$), the claimed run-time improvement over the Sequential Hadamard test collapses; conversely, verifying $s_{\min} = \Omega(1)$ for such random instances would support the conjecture.

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Extended reading notes

Core claim

The central claim is that the phase $\theta_k = \arg\langle\psi|U^{k}|\psi\rangle$ can be recovered by decomposing $k$ into a random binary sum tree whose leaves are $0$ and $1$, estimating the leaf phases with a Hadamard test, and then estimating each internal node's phase with a Sandwich test. The sandwich operator is $S = U^{a} R_{\psi}^{\phi} U^{b}$, where $R_{\psi}^{\phi} = \mathbb{1}_{N} + \Phi|\psi\rangle\langle\psi|$ is the SPROTIS (Selective Phase Rotation of the Initial State) operator. A short calculation gives $\theta_{a+b} = \theta_a + \theta_b - \omega_{a+b} + \phi$, with $\sin\omega_{a+b}$ expressed through the moduli $r_a, r_b, r_{a+b}$ and the measured $|\langle\psi|S|\psi\rangle|$, so each internal node's phase is obtained from four projectively measured quantities. The paper shows that the total number of applications of $U$ needed to reach variance below one is $\Theta(k^{2})s_{\min}^{-6}$, and hence estimating $\theta_k$ to accuracy $\epsilon$ costs $\Theta(k^{2}\epsilon^{-2}s_{\min}^{-6})$; the abstract reports this as $\mathcal{O}(k^{2}\ln k / \epsilon^{2}s_{\min}^{6})$, with the $\ln k$ reflecting the tree height. This beats the Sequential Hadamard test's $\mathcal{O}(k^{3}\epsilon^{-2}r_{\min}^{-2})$ chiefly because $r_{\min}$ ranges over all integers $k' \le k$ and is typically exponentially small, whereas $s_{\min}$ ranges only over the tree nodes, which the algorithm is free to choose by randomization.

Load-bearing premise

The speedup stands or falls on the conjecture that the smallest overlap $s_{\min}$ among the nodes of the random binary sum tree is not exponentially small; the paper gives a freedom-of-choice argument for why this should typically hold, but no analytic proof or numerical data.

Editorial extensions

If this is right

  • Quantum phase estimation can be run with controlled applications of $U$ alone, cutting the circuit depth overhead of the Hadamard test by a factor of $O(n)$.
  • For typical instances where $r_{\min}$ is exponentially small but $s_{\min}$ is not, total run time improves from $O(k^{3}/\epsilon^{2}r_{\min}^{2})$ to $O(k^{2}\ln k / \epsilon^{2}s_{\min}^{6})$.
  • The SPROTIS operator becomes a recognized resource for estimating matrix elements of unitary powers, alongside its earlier use in phase estimation with approximate eigenstates.
  • The same sandwich construction applies to any pair of efficiently implementable unitaries, not only powers of a single $U$, and to continuous-time settings.
  • Multi-layer sandwich operators (two or more SPROTIS insertions) offer additional freedom to shunt small-overlap terms and can effectively raise $s_{\min}$, at the cost of more complex estimation.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • The $s_{\min}$ conjecture is the point most worth testing numerically; until representative random instances (random unitaries, local Hamiltonian evolutions, molecular ground-state overlaps) are checked, the practical advantage over the Sequential Hadamard test remains conditional rather than established.
  • The abstract's $\ln k$ factor does not appear in the in-text derivation's $\Theta(k^{2}\epsilon^{-2}s_{\min}^{-6})$ bound; if real, it likely originates from the $\Theta(\ln k)$ tree height, and the discrepancy between the two statements may deserve clarification.
  • Multi-layer sandwiches could turn the tree-skipping heuristic into a more controllable worst-case parameter, since shunting a small modulus term in Eq. (24) removes it from the measurement burden, effectively increasing the achievable $s_{\min}$.
  • The method may transfer to other overlap-estimation tasks, such as power iteration or overlap estimation between two states, wherever the bottleneck is controlled application of a large power of a unitary.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

3 major / 5 minor

Summary. The manuscript introduces a quantum algorithm, the "Sandwich test," to estimate the expectation value ⟨ψ|U^k|ψ⟩ for a unitary U and an efficiently preparable state |ψ⟩. The algorithm avoids controlled applications of U^k by using the SPROTIS operator Rψ^φ sandwiched between U^a and U^b, and it recursively combines estimates along a random binary sum tree with root value k. The author claims a total runtime of Θ(k^2 ε^{-2} s_min^{-6}) (Eq. (23)), with the abstract stating O(k^2 ln k / ε^2 s_min^6), where s_min is the minimum overlap |⟨ψ|U^{k̂}|ψ⟩| over node values k̂ of the tree. This is claimed to improve on the Sequential Hadamard test's O(k^3 ε^{-2} r_min^{-2}) if s_min is not exponentially small. The paper also sketches multi-layer generalizations and concludes that numerical experiments are needed to support the s_min assumption.

Significance. If the runtime claim were valid and s_min were typically not too small, the Sandwich test would provide a meaningful improvement for low-depth quantum phase estimation by replacing controlled U^k with controlled U and a SPROTIS operator. The core idea of recursively combining phase estimates over a sum tree is interesting. However, the paper provides no machine-checked proofs or numerical data, and the central complexity derivation contains several errors. The runtime claim is therefore not established, and the advertised advantage over the Sequential Hadamard test is conditional on an unproven conjecture.

major comments (3)
  1. [Section II, Eq. (22)] The assertion "N_{h'} = Θ(k)" is not justified. From the fact that the values of all nodes at height h' sum to k and each non-trivial node has value at least 2, one obtains only the upper bound N_{h'} ≤ k/2. In particular, at the root (h'=0) we have N_0 = 1, so Eq. (22) is invalid at the root: its left side is Θ(k s_min^{-6}), while the right side is Θ(k^2 s_min^{-6}). Since the per-height cost is used to obtain the total runtime, this is a load-bearing error.
  2. [Section II, Eqs. (21)-(23)] The variance argument mixes two incompatible regimes. The bound "the total variance is upper bounded by (1 − γ_max^{q−1})^{-1} ≤ 1" requires q>1. For q=1 the geometric sum diverges and the total variance is Θ(ln k), not O(1). For q>1, the per-height cost contains the factor Σ γ^{1-q} ≤ N_{h'} x_min^{h'(1-q)}, and summing over h' yields a growing factor (for a balanced tree it is Θ(k^{q-1})), so the total runtime is Θ(k^{q+1} s_min^{-6}), not Θ(k^2 s_min^{-6}). The abstract's O(k^2 ln k) is also inconsistent with Eq. (23)'s Θ(k^2). The derivation as written cannot produce the claimed Θ(k^2) scaling.
  3. [Section II, after Eq. (13); Section III] The central advantage over the Sequential Hadamard test depends on the conjecture that s_min is not very small. The author states "It is difficult to analytically prove that s_min is not very small" and "Numerical experiments are needed to confirm this." Since s_min enters the runtime as s_min^{-6}, even a modest exponential suppression would destroy the claimed improvement. No proof or numerical evidence is provided, so the main result is conditional on an unverified assumption.
minor comments (5)
  1. [Abstract and Eq. (23)] The abstract states O(k^2 ln k / ε^2 s_min^6) while Eq. (23) states Θ(k^2 ε^{-2} s_min^{-6}); the discrepancy concerning the ln k factor should be resolved.
  2. [Eq. (22)] The equality "Θ(k)s_min^{-6} Σ γ^{1-q} = Θ(k^2)s_min^{-6} x_min^{h'(1-q)}" is not an equality but, even under the N_{h'}=Θ(k) assumption, an upper bound; the notation should be corrected accordingly.
  3. [Eq. (13) and subsequent text] The definition of s_min in Eq. (13) is local to a single Sandwich test (min{r_a, r_b, r_{a+b}}), while the global s_min used in the runtime claims is the minimum over all tree nodes; these two notions should be distinguished explicitly.
  4. [Section II, after Eq. (20)] The sentence "But it can also be easily estimated by estimating θ1 to an accuracy of ∞/∥" appears to contain a typo; the required accuracy should be specified (presumably O(1/k)).
  5. [Section II, protocol for SPROTIS] The claim that the SPROTIS operator can be implemented with circuit depth Θ(n) using Refs. [72,73] would benefit from a more precise resource statement, including the constant factors and the locality-constrained overhead.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: runtime bound derives from explicit identities plus an acknowledged s_min conjecture; self-citation is non-load-bearing.

full rationale

The central derivation is self-contained. The Sandwich operator S = U^a R_psi^phi U^b (Eq. 8) is defined from the SPROTIS operator R_psi^phi = W R_{0^n}^phi W^dagger (Eq. 6), and s_ab is expanded in Eqs. (10)-(11) entirely in terms of r_a, r_b, r_{a+b}, phi, and the unknown omega_{a+b}; none of these quantities is defined in terms of the final estimate theta_k. The random-binary-sum-tree recursion gives the algebraic identity theta_k = k theta_1 + sum_h sum_p omega-hat_h^p (Eq. 20). The runtime bound (Eqs. 21-23) follows from a stated measurement budget M = Theta(k^q) and the lower bound s_min, which is a property of the chosen tree and the fixed state/unitary, not a parameter fitted to the target. The paper explicitly labels the guard s_min not much less than 1 as a conjecture needing numerical confirmation ('It is difficult to analytically prove that smin is not very small... Numerical experiments are needed to confirm this'), so it is an acknowledged assumption rather than a fitted input renamed as a prediction. The single self-citation [74] is historical motivation and is not used to justify any equation in the derivation. No uniqueness theorem is imported, no known result is renamed, and no ansatz is smuggled in through citation. Therefore no circular step is exhibited; the potential internal inconsistency of the k^2 scaling (e.g., N_h' = Theta(k) at h'=0, and the q=1 versus q>1 variance/cost tradeoff) is a correctness concern, not a circularity concern.

Assumptions & free parameters 3 free parameters · 3 assumptions · 0 invented entities

The central runtime claim depends on the size of s_min, which is not proven. The algorithm also assumes efficient state preparation and the ability to ignore small amplitudes in QPE. These are made explicit in the paper but are not derived.

free parameters (3)
  • s_min = assumed O(1) in typical cases
    The runtime depends inversely on s_min^6. The paper does not derive or measure s_min; its size is the central unverified assumption.
  • x_min and y_max (tree split bounds) = chosen to satisfy 0 < x_min ≤ 1/2? Actually Eq (16) says 0 < x_min ≤ x ≤ y ≤ y_max < 1.
    These bounds control the tree depth and the per-level cost; they are algorithmic design choices.
  • q (variance budget exponent) = q ≥ 1, effectively q=1
    M = Θ(k^q) sets the number of measurements per node. The abstract's O(k^2 ln k) corresponds to q=1; the body's final Θ(k^2) omits the sum over levels.
assumptions (3)
  • domain assumption The initial state |ψ> can be efficiently prepared, so SPROTIS R_ψ^φ can be implemented with small overhead.
    Used throughout Section II; the circuit depth of R_0^n is O(n) without locality and O(n^2) with locality, assumed negligible relative to c-U.
  • ad hoc to paper s_min, the minimum overlap over tree nodes, is typically not much smaller than 1.
    The entire speedup claim rests on this; explicitly admitted as difficult to prove and needing numerical experiments (Section II).
  • domain assumption The initial state has non-negligible overlap η with the target eigenstate, so small-amplitude k values can be ignored in QPE.
    Used when arguing bad k' values can be skipped; relies on references [26-30] and the condition η >> 1 (likely η not << 1).

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Cite this review

Pith. "Pith review of Sandwich test for Quantum Phase Estimation." pith.science (2026). https://pith.science/paper/OPIU7VZ6

@misc{pith2026250723716,
  author       = {Pith},
  title        = {Pith review of: Sandwich test for Quantum Phase Estimation},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/OPIU7VZ6}},
  note         = {Machine review of arXiv:2507.23716}
}
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.

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