Chapter 2 Quantum chemistry using auxiliary field Monte Carlo

Transcription

1 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

2 Acknowledgment Shlomit Jacoby Prof. Daniel Neuhauser Prof. Anna Krylov Prof. Jan Martin Support: Israel Science Foundation Lise Meitner Center 2

3 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

4 Many-Fermion ground state (2 nd quantization) Hˆ = h T ˆ + ˆT V ˆ 1 2 ρ ρ ρ Electron density (matrix) 4

5 Goals Ground state energy Low lying excited states properties (correlation functions) How to do that rigorously??? 5

6 e β Ĥ φ Boltzmann is Hot! φ = a ψ + a ψ + a ψ = E0ψ 0 0 ae β E ψ 2 ae β β βe ψ 0 0 ae 0 E ψ ae β + βh 0 βh φ He φ E φ e φ 6

7 Groundstate is Cool Energy (Eh) 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 β (au) 7 gs HF Determinant Variational!

8 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

9 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

10 2-body = very hard ˆ T V ˆ e βρ ρ φ φ N = huge # of det. 1 10

11 Slicing time Divide to N slices: Use Thus: βhˆ βhˆ βhˆ βhˆ e = e e e N β = 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

12 Hubbard-Stratonovich Strategy Fourier transform of a Gaussian is a Gaussian: 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 e e e d σ 12

13 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

14 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

15 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

16 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: 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

17 Results for H Correlation energy (ev) β =4 au β =1 au β =2 au Sylvestrelli, Baroni & Car PRL 71, Iterations 17

18 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 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 = Rom, Charutz & Neuhauser, CPL 270, 382 (1997)

20 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

21 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)

22 Successful Application to H 2 Correlation energy (ev) β= 1au β= 2au β= 4au β= 8au E E ( SC AFMC ) corr ( cc pvqz ) corr = 099. ( 2) ev = 110. ev 1.4 au Iterations 22

23 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

24 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

25 Inversion Barrier of Water H H O H H 25

26 Inversion Barrier of Water Barrier Correlation Energy (ev) MP5 L=12 au L=8 au β (au) MP2 Tarczay et al JCP 110, (1999) Baer, CPL 324, 101 (1999) 26

27 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

28 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

29 Force variance is finite! St. Dev. Force (au) dr=1e-3 au dr=1e-4 au dr=1e-5 au β (au) 29

30 N 2 : Bond length R e (A) Experiment L=8 au L=12 au β (au) 30

31 N 2 : Harmonic frequency ω e (cm -1 ) L=12 au Experiment L=8 au β (au) 31

32 N 2 Heat of Formation D(kcal/mol) Experiment L=8 au L=12 au β (au) 32

33 The N 2 Potential Curve Potential Energy (ev) R (au) Hartree Fock β = 0.7 au β = 1.3 au β = 2.5 au Gdanitz CPL (1998) Baer, JCP 113, 473 (2000) 33

34 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

35 Singlet-Triplet Splitting of CH 2 30 J (kcal/mole) singlet states 1-singlet state β (au) 35

36 Sources of Error in AFMC E(β) - E FCI (E h ) 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 β (au) 36

37 Well-tempered AFMC E(β) - E FCI (E h ) K = 6 K = 1 H 2 ccp-vdz I = 5x10 5 R HH = 1.2 A Using several determinants considerably improves the performance β (au) 37

38 Simple example: excited states of H 2 Lines: FullCI #Dets: 6 o β = 0 au o β = 5 au 38

39 H 2 states: close up 39

40 Torsional Ethylene bond rapture Potential energy (E h ) 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 G Θ (deg) 40

41 Singlet-Triplet states of Ethylene Potential energy (ev) B 1u 6-31G β = 4.5 Eh A g Torsional angle Θ (deg) 41

42 Single bond-breaking 7 HF H+F Energy above min. (ev) 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) R (Angstrom) 42

43 H 2 O PES for double bond breaking E-EFCI-min (ev) 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 R/Req 43

44 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