REVIEW 3 major objections 4 minor 93 references
One dimensional Bose-Hubbard model with long range hopping
T0 review · 3 major / 4 minor · reviewed 2026-08-07 · deepseek-v4-flash
Pith's one-line read The paper establishes a phase diagram for one-dimensional bosons with power-law hopping: a Tomonaga-Luttinger liquid for α ≥ 3, broken U(1) symmetry for α < 3 when repulsion is weak, and at positive temperature broken symmetry only for α…
desk verdict A solid SCHA/RG treatment of 1D bosons with long-range hopping; the qualitative phase diagram is old, the quantitative finite-T exponent claim is new but overclaimed in the abstract and rests on an uncontrolled approximation. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The load-bearing object is the bosonized long-range hopping term $\int dx \sum_l \cos[\theta(x+la)-\theta(x)]/l^\alpha$, written in terms of the phase field $\theta$ dual to the density. Its scaling dimension $(4K)^{-1}$ enters the renormalization-group relevance condition $\alpha+1/(2K)<3$, and the flow equations for $K$ and the dimensionless hopping strength $g_{LR}$ determine whether the flow ends on the Tomonaga-Luttinger fixed line or runs to strong coupling. For the ordered and finite-temperature regimes, the self-consistent harmonic approximation replaces each cosine by a normal-ordered quadratic term whose coefficient $g(l)=\exp(-\langle(\theta(x+la)-\theta(x))^2\rangle/2)$ is fixed self-consistently; the resulting dispersion $\omega(k)$ decides between the Tomonaga-Luttinger phase with $\omega\sim k$ and the broken-symmetry phase with $\omega\sim k^{(\alpha-1)/2}$.
What would settle it
A numerical study (for example quantum Monte Carlo or DMRG) of a one-dimensional Bose-Hubbard chain with hopping $1/r^\alpha$ at $\alpha=2.5$ and weak repulsion could measure the single-particle correlation function as a function of temperature: the paper predicts its power-law exponent grows without bound as $T\to 0$, while any finite limiting exponent, or an exponential decay, would rule out the central finite-temperature claim.
Extended reading notes
Core claim
The paper's central claim is that the fate of a one-dimensional Bose-Hubbard chain with power-law hopping $r^{-\alpha}$ is decided by the competition between $\alpha$ and the Luttinger exponent $K$. For $\alpha\ge 3$ the long-range hopping is irrelevant in the renormalization-group sense and the ground state is always a Tomonaga-Luttinger liquid. For $\alpha<3$, the cosine operator that couples sites at distance $l$ with amplitude $l^{-\alpha}$ becomes relevant once $K>1/(6-2\alpha)$, and for sufficiently weak repulsion the ground state breaks the global $U(1)$ gauge symmetry, so that $\langle e^{i\theta}\rangle\neq 0$, with gapless excitations dispersing as $k^{(\alpha-1)/2}$. At positive temperature the same self-consistent calculation shows the broken-symmetry phase disappears for $\alpha\ge 2$, while for $2<\alpha<3$ the Tomonaga-Luttinger liquid survives but with an exponent that diverges as a power law when $T\to 0$, indicating increasingly fragile quasi-long-range order.
Load-bearing premise
The finite-temperature conclusions, including the diverging Luttinger exponent for $2<\alpha<3$, rest on the self-consistent harmonic approximation, which replaces the true non-quadratic cosine interaction by a quadratic one and then lets that replacement fix its own strength; the paper itself shows this approximation produces a spurious first-order transition and misses the likely phase at $\alpha=2$.
Editorial extensions
If this is right
- For $\alpha\ge 3$, long-range hopping is irrelevant: the Tomonaga-Luttinger liquid ground state remains stable for any repulsion, with only renormalized velocity and Luttinger exponent.
- For $1<\alpha<3$ and weak enough repulsion, the ground state develops a superfluid order parameter $\langle e^{i\theta}\rangle\neq 0$ that breaks the global $U(1)$ symmetry; the gapless excitations disperse as $k^{(\alpha-1)/2}$ and the momentum distribution acquires a delta peak.
- At any $T>0$, broken symmetry survives only for $\alpha<2$; in the window $2<\alpha<3$ the Tomonaga-Luttinger liquid persists but its exponent diverges as a power law as $T\to 0$.
- For hard-core bosons at half-filling, the critical decay exponent between superfluid order and Luttinger liquid lies between $2$ and $3$ and is shifted by the hopping strength.
- Frustrated long-range hopping with alternating sign is always irrelevant and cannot destroy the Tomonaga-Luttinger liquid, and the same holds for long-range easy-axis exchange at incommensurate filling.
Reading between the lines
- Going beyond the paper: if the quadratic truncation is relaxed, the finite-temperature transition for $2<\alpha<3$ may be continuous rather than first-order, with the diverging-exponent law possibly surviving with modified coefficients.
- The special case $\alpha=2$, which the approximation cannot resolve, is a plausible host of a continuous transition to a quasi-long-range ordered phase at finite temperature, paralleling the classical long-range XY model.
- The same bosonized treatment could be pushed to commensurate filling, where long-range hopping would compete with density-wave order and could shift the critical $\alpha$ away from the incommensurate values.
- Trapped-ion or Rydberg-atom arrays that tune the hopping decay exponent could test the predicted divergence of the Luttinger exponent by tracking the temperature dependence of single-particle correlations.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript studies the one-dimensional Bose-Hubbard model with power-law hopping ~1/r^α using bosonization, an operator-product-expansion renormalization group, and the self-consistent harmonic approximation (SCHA). The main claims are: for α ≥ 3 the ground state is always a Tomonaga-Luttinger liquid; for α < 3 a state with continuous U(1) symmetry breaking is possible when the repulsion is sufficiently weak; at positive temperature symmetry breaking survives only for α < 2; and for 2 < α < 3 the TLL exponent diverges as T→0 according to Eq. (109). The paper also treats frustrated hopping, long-range spin interactions, and time-dependent correlations, and it compares RG and SCHA predictions throughout.
Significance. The ground-state phase boundary and the mapping to easy-plane XXZ spin chains are of clear interest for trapped-ion and Rydberg-atom experiments. The RG treatment is standard and carefully presented, with explicit flow equations, invariants, and asymptotic correlation functions, and the paper is unusually candid about the limitations of the SCHA, including the spurious first-order transition at T>0 and the failure at α=2. If the finite-temperature exponent divergence were confirmed, it would be a distinctive signature; at present that particular claim rests on an uncontrolled approximation, so the significance is conditional rather than established.
major comments (3)
- [Sec. V.D.2, Eq. (109); Abstract] The abstract's statement that for 2<α<3 a TLL with a diverging Luttinger exponent is found omits a necessary condition. Equation (107) only has a diverging solution when the zero-temperature system is already in the JLR-relevant regime, i.e. when the bare K satisfies K > 1/(6−2α). For K below that threshold, the T→0 limit of the sum in Eq. (107) is finite and K̃ tends to the finite solution of Eq. (51). The divergence in Eq. (109) is therefore a low-temperature crossover near the zero-temperature critical/CSB boundary, not a property of the TLL phase for all K in the window 2<α<3. Please state the K condition explicitly and adjust the abstract accordingly.
- [Sec. V.D.1 and V.D.2] The derivation of Eq. (109) uses the same thermal self-consistency, Eq. (95), that the paper itself shows produces a spurious first-order CSB-TLL transition at T>0 and fails to capture the expected BKT-like quasi-long-range order at α=2 (Sec. V.D.1 and V.D.2). The approximation is uncontrolled precisely in the regime 2<α<3, where the zero-temperature fixed point is at strong coupling or in the ordered phase. I therefore do not regard Eq. (109) as an established prediction. Please either label it explicitly as a SCHA-based conjecture and soften the abstract/conclusion, or add an independent check such as quantum Monte Carlo or DMRG on finite chains.
- [Sec. V.D.2, Eqs. (107)-(109)] The asymptotic estimate in Eq. (C3) retains only the k=0 term of the expansion (C2) and assumes K̃→∞. The paper should state more clearly that this estimate is not valid for finite K̃, where the polylogarithm term in Eq. (C2) remains finite and the sum saturates. Without this caveat, Eqs. (108)-(109) give the impression of a universal power law when the actual behavior is parameter-dependent.
minor comments (4)
- [Eqs. (19)-(20), App. A] The flow equations are mutually consistent and consistent with Appendix A when Eq. (19) is read as du/dℓ = πJLR A0^2 a/K. Please ensure the typesetting makes the denominator K unambiguous, since the current line break could be misread as du/dℓ = πJLR A0^2 a K.
- [Sec. VI and throughout] There are several typos: 'self-consistent Hamonic approximation' in the Conclusion, 'atiferromagnetic' and 'echange' in Sec. VI, and 'No Title' in Ref. [53]. A careful proofreading pass is needed.
- [Eq. (16)] The replacement of the discrete lattice sum by a continuum integral is stated to be justified when the characteristic length scale is much larger than the lattice spacing; this criterion should be stated explicitly before Eq. (16) and recalled when the RG result is used for α close to 3, where the correlation length can be of order a for moderate JLR.
- [Sec. V.D.2, Eq. (111)] The disagreement between the transition temperature obtained from the TLL instability at α<2 and the SCHA result in Eq. (103) is noted in the text, but a short discussion of which estimate is more trustworthy would help the reader. As written, the two estimates are left side by side without guidance.
Circularity Check
No significant circularity: derived quantities are self-consistent solutions, with acknowledged approximation artifacts.
full rationale
The claimed results (TLL/CSB phase boundary, Ktilde, g(infinity), xi, Tc, and the diverging low-temperature exponent) are outputs of explicit bosonization, RG, and SCHA equations, not fitted inputs. The bosonized long-range hopping, Eq. (9), uses standard operator coefficients A_m from prior independent work, and the RG equations (17)-(22) are derived via OPE in Appendix A; the paper explicitly notes that the only difference from Ref. [33] is a regularization-dependent constant. The SCHA is a self-consistent approximation: Eqs. (34)-(37) define g(l) through the quadratic Hamiltonian, and subsequent equations solve that self-consistency, so target quantities are solutions rather than prescribed values. In particular, the finite-temperature exponent divergence, Eq. (109), follows by solving Eq. (107) with the Appendix C asymptotic estimate; it is not a parameter tuned to reproduce the divergence. The paper itself flags the approximation's failures (first-order SCHA artifact at T>0, Sec. V.D.1; inability to capture BKT-like quasi-long-range order for alpha=2, Sec. V.D.2), which are correctness or robustness limitations, not circularity. The only self-citations (Refs. [69], [72]) supply standard constants or general SCHA references and are not load-bearing: no 'uniqueness' or ansatz is imported from them. No step reduces a prediction to an input by construction.
Assumptions & free parameters
assumptions (8)
- domain assumption The low-energy sector of the Bose-Hubbard model away from commensurate filling is a Tomonaga-Luttinger liquid described by Eq. (5) with velocity u and Luttinger parameter K.
- domain assumption The long-range hopping (2) bosonizes to the leading cos[θ(x+la)-θ(x)] terms of Eq. (9); higher harmonics in Eq. (10) are irrelevant.
- domain assumption The discrete sum over lattice sites in Eq. (9) can be replaced by the continuum double integral of Eq. (16), valid when θ varies on scales much larger than the lattice spacing.
- standard math The operator product expansion of cosθ is dominated by the identity and (∂xθ)^2 terms (Eq. A2), with higher-dimension operators neglected.
- ad hoc to paper The self-consistent harmonic approximation truncates the normal-ordered cosine at second order (Eqs. 34-36) and uses the resulting quadratic Hamiltonian for self-consistency.
- ad hoc to paper The momentum cutoff can be imposed by the factor e^{-ka} in Eq. (46) and later integrals.
- standard math For the CSB phase, the low-momentum dispersion is determined by the asymptotic identity (60) for ∑(1-cos kla)/l^α, with the O(k^2) term subleading for 1<α<3.
- ad hoc to paper The finite-temperature self-consistency (Eq. 95) assumes a free-boson thermal occupation factor and the validity of the SCHA at nonzero temperature.
Cite this review
Pith. "Pith review of One dimensional Bose-Hubbard model with long range hopping." pith.science (2026). https://pith.science/paper/TDE6267M
@misc{pith2026250603629,
author = {Pith},
title = {Pith review of: One dimensional Bose-Hubbard model with long range hopping},
year = {2026},
howpublished = {\url{https://pith.science/paper/TDE6267M}},
note = {Machine review of arXiv:2506.03629}
}
abstract
Interacting one-dimensional bosons with long range hopping decaying as a power law $r^{-\alpha}$ with distance $r$ are considered with the renormalization group and the self-consistent harmonic approximation. For $\alpha\ge 3$, the ground state is always a Tomonaga-Luttinger liquid, whereas for $\alpha <3$, a ground state with long range order breaking the continuous global gauge symmetry becomes possible for sufficiently weak repulsion. At positive temperature, continuous symmetry breaking becomes restricted to $\alpha<2$, and for $2<\alpha<3$, a Tomonaga-Luttinger liquid with the Tomonaga-Luttinger exponent diverging at low temperature is found.
Figures
Reference graph
Works this paper leans on
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[1]
(58) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 KSCHA KRG K K FIG
Dispersion relation and order parameter In the CSB phase, ⟨eiθ⟩ ̸= 0 and as a consequence, g(∞) = lim l→+∞ g(l) = |⟨eiθ⟩|2 > 0. (58) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 KSCHA KRG K K FIG. 3. Plot of the Luttinger exponent predicted by the SCHA and the renormalization group for α = 5 /2 for vari- able K and 2πJLRA2 0a/u = ...
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[2]
Correlation functions in the CSB phase a. equal time correlations In the superfluid phase, the momentum distribution is given by n(k) = A2 0 Z dxeikx⟨eiθ(x)e−iθ(x)⟩ = A2 0 2πg(∞)δ(k) + Z dxeikx(g(l) − g(∞)) = A2 0 2πg(∞)δ(k) + g(∞) 4Kω (k) e−|k|a + . . . , (68) yielding a momentum distribution with a power law di- vergence ∼ (|k|ξ)−(α−1)/2 for 0 < |k|ξ ≪ ...
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[3]
So the divergence might be an artefact of the SCHA
In the case of the quantum sine-Gordon model, it is known that the SCHA gives a similar divergence when calculating the correlation functions of the dual field[70] as it overestimates the energy cost of propagating solitons in time[86]. So the divergence might be an artefact of the SCHA. For α >2, no divergence exists, and 10 1 2 ⟨Tτ [ϕ(x, τ) − ϕ(0, 0)]2⟩...
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[4]
(46) is replaced by g(l) = exp − Z ∞ 0 d(uk)e−ka 2Kω (k) 1 + 2 e ω(k) T − 1 (1 − cos(kla))
Stability of the CSB phase When T >0, Eq. (46) is replaced by g(l) = exp − Z ∞ 0 d(uk)e−ka 2Kω (k) 1 + 2 e ω(k) T − 1 (1 − cos(kla)) . (95) In order to have a CSB solution with g(∞) > 0, we need to solve g(∞) = exp − Z ∞ 0 d(uk)e−ka 2Kω (k) 1 + 2 e ω(k) T − 1 . (96) with ω(k) given by Eq. (59). To find g(∞) > 0, the integral T Z 1/a 0 dk ω(k)2 ∼ T Z 1/a 0...
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[5]
Temperature dependence of the Tomonaga-Luttinger exponent In the Tomonaga-Luttinger liquid phase, the velocity and the Luttinger exponent are still given by Eqs. (49)– 13 (50), but with g(l) = πa β ˜u sinh πla β ˜u ! 1 2 ˜K , (106) where β = 1/T so that ˜u ˜K = uK + 2πJLRA2 0a +∞X l=2 l2−α πT a ˜u sinh πT la ˜u ! 1 2 ˜K , (107) and ˜u/ ˜K = u/K. At high t...
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[6]
Density-density correlations For T >0, we have ⟨ϕ(k)ϕ(−k)⟩ = πKω (k) 2uk2 1 + 2 eω(k)/T − 1 , (114) and 1 2 ⟨(ϕ(x) − ϕ(0))2⟩ = K 4 Z ∞ −∞ dkω(k)[1 − cos(kx)] 2uk2 × 1 + 2 eω(k)/T − 1 . (115) In the superfluid phase with α <2 and at long distance, this yields ⟨(ϕ(x)−ϕ(0))2⟩ = πK Γ 5−α 2 cos π 4 (1 − α) x ξ 3−α 2 + πT |x| 2u , (116) so that the zero-tempera...
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