Chapter 2 Quantum chemistry using auxiliary field Monte Carlo 1. The Hubbard-Stratonovich Transformation 2. Neuhauser s shifted contour 3. Calculation of forces and PESs 4. Multireference AFMC 5. Examples of applications 1
Acknowledgment Shlomit Jacoby Prof. Daniel Neuhauser Prof. Anna Krylov Prof. Jan Martin Support: Israel Science Foundation Lise Meitner Center 2
Overview New quantum chemistry method: Works in terms of basis sets A continuous range of approximations from Hartree-Fock to full CI Based on Monte Carlo, but - No fixed nodes! Discuss the formalism Discuss application issues Show few results 3
Many-Fermion ground state (2 nd quantization) Hˆ = h T ˆ + ˆT V ˆ 1 2 ρ ρ ρ Electron density (matrix) 4
Goals Ground state energy Low lying excited states properties (correlation functions) How to do that rigorously??? 5
e β Ĥ φ Boltzmann is Hot! φ = a ψ + a ψ + a ψ + 0 0 1 1 2 2 = E0ψ 0 0 ae β E 1 1 + 1ψ 2 ae β β βe ψ 0 0 ae 0 E ψ 2 2 + ae β + βh 0 βh φ He φ E φ e φ 6
Groundstate is Cool Energy (Eh) -108.41-108.43-108.45-108.47-108.49-108.51-108.53-108.55 STO-3G Full CI ˆ ˆ βh Φ He Φ E( ) ˆ β β = Hˆ = H E β β Φ e Φ d E ( β) = ( H ˆ E ( β) ) 2 < 0 dβ β 0 1 2 3 4 5 β (au) 7 gs HF Determinant Variational!
How to do it??? If electrons were non-interacting easy: h T ˆ 1 N 1 Slater Det. ( ) ( ) e β ρ φ φ = φ β φ β N New Slater Det. 8
Proof (BCH) T T βh ˆ ρ βh ˆ ρ e φφ ˆˆ ˆ 1 2 φn e c1c2 cn 0 Slater Det. T T T T T T βh ˆ ρ βh ˆ ρ βh ˆ ρ βh ˆ ρ βh ˆ ρ βh ˆ ρ = e cˆ ˆ ˆ 1e e c2e e cne 0 dˆ dˆ dˆ 1 2 N ( ) ( ) ( ) φ β φ β φ β 1 2 N Slater Det. N T T βh ρˆ βh ρˆ βh βh ˆi = ( ) ˆj φi ( β) = ( ) φ ij ij j j j= 1 e c e e c e 9
2-body = very hard ˆ T V ˆ e βρ ρ φ φ N = huge # of det. 1 10
Slicing time Divide to N slices: Use Thus: βhˆ βhˆ βhˆ βhˆ e = e e e N β = 1 2 1 ˆ T ˆ T β ρ V ρ ˆ βh βt ρˆ 2 2 e = e e + O ( β ) N β 1 T 1 T 1 T ˆ β ρˆ Vρˆ β ρˆ Vρˆ β ρˆ Vρˆ ˆ ˆ ˆ T T T βh βt ρ 2 βt ρ 2 βt ρ 2 e = e e e e e e + O( β) 1 2 Goal: Represent 2-body exponential as sum of 1-body exponentials N 11
Hubbard-Stratonovich Strategy Fourier transform of a Gaussian is a Gaussian: 1 2 1 2 aρ 2 2 a x iρx e e e dx An operator! a = V β x = σv β V σ V σ β iσ V ρ β 1ρ ρ β 1 2 2 e e e d σ 12
Problem in evaluation of integrals Consider a multidimensional integral f W 1 T σ W σ 2 (... ) e f d 1 T σ W σ N σ1 σn σ 2 N e d σ Can t do on a grid! Why? Most points are on grid boundaries 1D 2:1 2D 8:1 3D 26:1 1D 2:2 2D 12:4 3D 56:8 13
Solution: Monte Carlo sampling The integral f W 1 T σ W σ 2 (... ) e f d 1 T σ W σ N σ1 σn σ 2 N e d σ Is calculated by producing random numbers ( σ1... σ N ) that are Gaussian distributed and summing f W ( σ... σ ) 1 N f ( σ... σ ) 1 N 14
Monte Carlo scheme Φ e Hˆβ Φ = Φ U σ I ( β ) Φ W T T T N ( t iv ) ( t iv ) ( t iv ) 1 2 U σ ( β) Φ = e e e I β + σ ρˆ β + σ ρˆ β + σ ρˆ Φ W N 1 2 σ e n = 1 { } = T n β σ V σ n G. Sugiyama and S. E. Koonin, Ann. Phys. N.Y. 168, 1 (1986). 15
Exercise A 1D particle is in a potential well V(x) ˆ 1 ˆ 2 H = p + V ( x) 2 Based on the Hubbard-Stratonovich transformation derive a Monte Carlo method to calculate its GS energy Hint: 1 2 1 2 pˆ β σ β 2 ˆ ( ) 2 iσp β = ( ) e ψ x A e e ψ x dσ 1 σ 2 β 2 = A e ψ( x σ β) dσ 16
Results for H 2 1.00 Correlation energy (ev) 0.50 0.00-0.50-1.00-1.50-2.00-2.50 β =4 au β =1 au β =2 au Sylvestrelli, Baroni & Car PRL 71, 0 2000 4000 6000 1148 1993 Iterations 17
What s causing the trouble? e Hˆ β = e ( h+ ivσ ) T ˆ ρ β W { σ } = 1 i σ T V ˆ ρ β O W ( β ) σ T Vσ β 1 Noise ( β ½ ) Signal ( β) Signal/Noise β ½ 0 Baer, Head-Gordon & Neuhauser, JCP 109, 6219 (1998) 18
19 Solution: contour-shift invariance ( ) W H U e β σ β = ( ) ( ) ( ) τ α τ σ τ σ i ( ) W ia d V i d V H U e e e T T β σ τ α σ τ α α β β τ τ β τ τ = 0 0 2 1 Rom, Charutz & Neuhauser, CPL 270, 382 (1997)
τ τ τ Time slice stabilization Φ e βh Φ = Φ β τ e Φ τ τ ( ) H τ ( ) { } ( ) is τ = D W e σ τ σ τ S τ = i ln ( K iv( i )) ( ) T + σ α ρ τ β τ e ( τ τ) τ τ Φ Φ τ T σ Vα τ τ δ S τ = 0 δσ τ 20
The stabilizing shift α ( τ ) = Φ ( β τ ) ˆ ρ Φ( τ ) Φ ( β ) Φ( 0) when β, τ are very large: α( τ ) = Φ ˆ ρ Φ gs gs Need to know exact GS density! Use HF density as partial stabilizer 21 Baer, Head-Gordon & Neuhauser, JCP 109, 6219 (1998)
Successful Application to H 2 Correlation energy (ev) -0.85-0.90-0.95-1.00-1.05 β= 1au β= 2au β= 4au β= 8au E E ( SC AFMC ) corr ( cc pvqz ) corr = 099. ( 2) ev = 110. ev 1.4 au 0 2000 4000 6000 Iterations 22
Things to note No uncontrolled approximation is made (no fixed nodes etc.) The auxiliary fields are sampled from a universal distribution W ( ) ( ) = 1 β T σ τ Vσ τ σ e 2 d 0 { } τ 23
Some application issues Two codes: Plane waves/pseudopotentials based code Gaussian based code (Shlomit Jacoby): Reads RHF/UHF GAMESS output Can get data from CASSCF/GVB or spin flip methods to perform multireference calculation 24
Inversion Barrier of Water H H O H H 25
Inversion Barrier of Water Barrier Correlation Energy (ev) 0.01 0-0.01-0.02-0.03-0.04-0.05-0.06-0.07 MP5 L=12 au L=8 au 0 0.5 1 1.5 2 β (au) MP2 Tarczay et al JCP 110, 11971 (1999) Baer, CPL 324, 101 (1999) 26
Electronic force on nuclei de ( + δ ) ( ) E R R E R F = lim dr δ R 0 R δ Force variance (statistical error) is infinite: σ ( F ) ( E) 2σ = lim = δ R 0 δ R 27
Solution: correlated sampling Auxiliary fields are sampled from a universal Coulomb-Gauss distribution: W 1 β T σ ( τ ) Vσ ( τ ) σ 2 d 0 { } = e τ Straightforward correlated sampling: Accurate potential surface Geometries Vibrational frequencies 28
Force variance is finite! 0.035 St. Dev. Force (au) 0.03 0.025 0.02 0.015 0.01 0.005 dr=1e-3 au dr=1e-4 au dr=1e-5 au 0 0 0.5 1 1.5 2 β (au) 29
N 2 : Bond length R e (A) 1.11 1.1 1.09 1.08 1.07 1.06 Experiment L=8 au L=12 au 1.05 0 0.5 1 1.5 2 2.5 β (au) 30
N 2 : Harmonic frequency 2900 2800 2700 ω e (cm -1 ) 2600 2500 2400 L=12 au Experiment 2300 2200 L=8 au 2100 0 0.5 1 1.5 2 2.5 β (au) 31
N 2 Heat of Formation D(kcal/mol) 240 220 200 180 160 Experiment L=8 au L=12 au 140 120 0 0.5 1 1.5 2 2.5 β (au) 32
The N 2 Potential Curve Potential Energy (ev) 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 1.8 1.9 2 2.1 2.2 2.3 2.4 R (au) Hartree Fock β = 0.7 au β = 1.3 au β = 2.5 au Gdanitz CPL 283 253 (1998) Baer, JCP 113, 473 (2000) 33
Multireference AFMC ψ gs Ĥ e β n C n Φ n J ( C) = C HC ε { C SC 1} S mm ' β Hˆ β Hˆ = Φm' e Φm H m' m = Φm' He Φm ˆ HC = SCE 34
Singlet-Triplet Splitting of CH 2 30 J (kcal/mole) 25 20 15 10 2-singlet states 1-singlet state 5 0 0.5 1 1.5 2 2.5 3 β (au) 35
Sources of Error in AFMC E(β) - E FCI (E h ) 10-1 10-2 10-3 10-4 10-5 β = 1 au β = 0.5 au β = 0.2 au β = 0.1 au β = 0.01 au H 2 ccp-vdz I=10 5 β > 0 SE β < Also: Basis-set Frozen core Rid of small eigenvalues of V ijkl 10-6 0 1 2 3 4 5 6 β (au) 36
Well-tempered AFMC E(β) - E FCI (E h ) 10-1 10-2 10-3 K = 6 K = 1 H 2 ccp-vdz I = 5x10 5 R HH = 1.2 A Using several determinants considerably improves the performance 10-4 0 1 2 3 4 5 6 7 8 β (au) 37
Simple example: excited states of H 2 Lines: FullCI #Dets: 6 o β = 0 au o β = 5 au 38
H 2 states: close up 39
Torsional Ethylene bond rapture Potential energy (E h ) -77.8-77.9-78.0-78.1 β = 0 (RHF) β = 0.1 au β = 0.5 au β = 1.0 au β = 2.0 au β = 3.0 au β = 4.5 au C 2 H 4 6-31G -78.2 0 20 40 60 80 100 120 Θ (deg) 40
Singlet-Triplet states of Ethylene Potential energy (ev) 5 4 3 2 1 1 3 B 1u 6-31G β = 4.5 Eh -1 1 1 A g 0 0 20 40 60 80 100 120 Torsional angle Θ (deg) 41
Single bond-breaking 7 HF H+F Energy above min. (ev) 6 5 4 3 2 1 0 FCI b = 0.1 au b = 1 au b = 2 au b = 3 au b = 4 au Basis set: 6-31** Full CI taken from Dutta and Sherrill, J. Chem. Phys. 118, 1610 (2003). -1 0 1 2 3 4 R (Angstrom) 42
H 2 O PES for double bond breaking E-EFCI-min (ev) 14 12 10 8 6 4 Basis set: DVZ 8 Detereminants β = 0 β = 0.5 au β = 1.5 au β = 3 au Full-CI AFMC results at various values of β, vs. Full-CI R eq =0.96Å 2 0 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 R/Req 43
Summary New electronic structure method Formally exact, but has statistical error Can give PES s, break bonds, compute excited states No fixed nodes approximation Balance mutireference and MC: method to select a multi-reference space 44