Sign-Free Quantum Monte Carlo and Hubbard-Phase Inference

Auxiliary-field quantum Monte Carlo is sign-free only in specific Hubbard parameter domains where fermion determinants pair to a nonnegative weight. Inside those domains it removes exponential sign cancellations, not Trotter, autocorrelation, finite-size, projection, estimator, or analytic-continuation errors.

Required background. Use Hubbard symmetries, Markov-chain sampling, and sign-problem anatomy. Helpful background. Covariance and resampling governs uncertainty propagation.

Two standard nonnegative domains

After a Trotter decomposition and Hubbard–Stratonovich transformation, the partition function becomes

$$Z_{\Delta\tau}=\sum_{\{s\}}p(s) \det M_\uparrow[s]\det M_\downarrow[s] +O(\Delta\tau^2)$$

for a symmetric second-order breakup. For the attractive Hubbard model with balanced populations, the two real spin matrices are identical, so the product is $(\det M)^2\ge0$. For the repulsive model at half filling on a bipartite lattice with the particle–hole-symmetric chemical potential, transforming one spin species pairs the determinants and again makes their product nonnegative. Doping, frustrating hopping, spin imbalance, or many additional interactions generally destroy these proofs.

The determinant formulation originates with Blankenbecler, Scalapino, and Sugar 1981, pp. 2278–2286; the half-filled Hubbard application and scaling checks are exemplified by White et al. 1989, pp. 506–516.

From an estimator to a phase claim

For antiferromagnetism define

$$S(\mathbf Q)=\frac1{L^d}\sum_{ij}e^{i\mathbf Q\cdot(\mathbf r_i-\mathbf r_j)} \langle\mathbf S_i\cdot\mathbf S_j\rangle, \qquad m_L^2=\frac{S(\mathbf Q)}{L^d}.$$

Long-range order requires $m_L^2$ to approach a nonzero limit along sizes and aspect ratios compatible with the expected scaling. A correlation ratio formed from $S(\mathbf Q+\delta\mathbf q)/S(\mathbf Q)$ can locate a transition with less amplitude dependence, but crossing drift and covariance remain essential.

Report warmup, integrated autocorrelation times, binning or resampling, $\Delta\tau$ extrapolation, boundary conditions, and the full covariance among sizes and observables. In projector QMC, also vary projection length and trial state. Analytic continuation should not be used when an imaginary-time or equal-time diagnostic answers the question.

Evidence ceiling

Within a proved sign-free domain and converged finite-size analysis, QMC can give unbiased statistical estimates for the specified lattice model. That does not prove the model realizes a material, nor does absence of conventional order establish a different phase. Competing channels must be measured rather than inferred from one structure factor.

Exercises

Why does $(\det M)^2\ge0$ not guarantee an exact numerical answer?

Solution

It guarantees a nonnegative sampling weight in that formulation. Finite $\Delta\tau$, incomplete equilibration, autocorrelation, finite volume, projection error, unstable linear algebra, and biased model or estimator choices remain and require independent convergence tests.

References

• Richard Blankenbecler, Douglas J. Scalapino, and Robert L. Sugar, “Monte Carlo Calculations of Coupled Boson–Fermion Systems. I,” Physical Review D 24 (1981) 2278–2286, doi:10.1103/PhysRevD.24.2278.
• Steven R. White, Douglas J. Scalapino, Robert L. Sugar, E. Y. Loh Jr., James E. Gubernatis, and Richard T. Scalettar, “Numerical Study of the Two-Dimensional Hubbard Model,” Physical Review B 40 (1989) 506–516, doi:10.1103/PhysRevB.40.506.