If electron-correlation is added to the wavefunction, all properties converge to experimentally observed values. B3LYP is the most popular DFT method;

Transcription

1 Lec Mon 6mar17 If electron-correlation is added to the wavefunction, all properties converge to experimentally observed values. B3LYP is the most popular DFT method; B3LYP/cc-pVTZ 99.87% of experiment; Takes 5 sec on a PC

2 b3lyp/cc-pv6z %nproc=16 %mem=1000mw %chk=jean-h0-b3lypcc-pv6z-opt.chk #p opt b3lyp/cc-pv6z pop=full water molecule from Jean Standard pdf

3 Molecular Orbital Coefficients: A1--O A1--O B--O A1--O B1--O Eigenvalues O 1S S S S S S S PX PY PZ PX PY PZ PX PY PZ PX PY PZ PX PY PZ PX PY PZ D D D D D D D D D D D D D D D D D D D D D D D D D F

4 109 6H H H H H H H H H H H H H H H H H H H I I I I I I I I I I I I I Last 3 CGTF coefficients for the O atom

5 31 B--V A1--V Total kinetic energy from orbitals= D01 No NMR shielding tensors so no spin-rotation constants. Leave Link 601 at Sat Mar 4 11:53:51 017, MaxMem= cpu: 18.8 Enter /g09/g09/l9999.exe 1\1\GINC-MFRUITS\FOpt\RB3LYP\CC-pV6Z\HO1\CALLIS\04-Mar-017\0\\#p opt b3lyp/cc-pv6z pop=full\\water molecule from Jean Standard pdf\\0,1\o, 0.,0., \H,0., , \H,0., , \\Version=AS64L-G09RevD.01\State=1-A1\ HF= \RMSD =1.093e-09\RMSF=7.6e-05\Dipole=0.,0., \Quadrupole= ,1.3066, ,0.,0.,0.\PG=C0V [CO1,SGVH]\\@ THE POLHOLDE ROLLS WITHOUT SLIPPING ON THE HERPOLHOLDE LYING IN THE INVARIABLE PLANE. H.GOLDSTEIN, "CLASSICAL MECHANICS", PG 161 Job cpu time: 0 days 3 hours 36 minutes 53.4 seconds. Clock time = 3.6 hrs/16 cpus = 13.5 minutes with 16 cpus. 3.6 hours at least on personal computer File lengths MBytes: RWF= 67 Int= 0 DE= 0 Chk= 8 Scr= 1 Normal termination of Gaussian 09 at Sat Mar 4 11:53: out file was lines

6 How can CI mixing higher energy state with ground state cause correlation??? Higher states have more nodes. Squaring the linear combination gives cross terms interference terms, which are negative when electrons are on different sides of a nodal surface. Ψ*Ψ is thereby reduced. Example: The H molecule ionic and covalent terms, difference between the MO and Valence Bond VB methods x x σ u * _ Both electrons in ANTI-BONDING MO σ u *1 σ u * σ g σ u * σ g 1 σ g x x σ g Both electrons in BONDING MO

7 H Molecule Configuration State Functions CSFs A1 = 1s1 on Nucleus A B1 = 1s1 on Nucleus B Doubly excited CSF σ u Ground CSF σ g Energy ionic covalent σ u = [A1 - B1] [A - B] = AA BB - AB BA σ g = [A1 B1] [A B] = AA BB AB BA not normalized not normalized ionic covalent = molecular orbital method covalent* = AB BA = valence bond method *Heitler-London 197 the first wavefunction proposed for H at the dawn of quantum mechanics.

8 Do a linear variation calculation: H = <σ g Hσ g > <σ g H σ u > <σ u Hσ g > <σ u Hσ u > σ u c Hg σ g c Hu σ u σ g c Lg σ g c Lu σ u Ground = c Lg σ g > c Lu σ u > Will σ u > be added or subtracted from σ g >? Depends on the sign of <σ g Hσ u >

9 B A 1s B, 1s A where = = = > = > >= > >= >= B A u B A g uu gg u u g g g σ σ σ σ σ ] 4[ 1/ 4[ 1/ 4 1/ 1/ 1 0 1/ BA BA AB BA BB BA AA BA BA AB AB AB BB AB AA AB BA BB AB BB BB BB AA BB BA AA AB AA BB AA AA AA BA AB BB AA BA AB BB AA B A B A B A B A uu gg uu r gg H u g u h g uu r h h gg H = = = >= >=< < >= >< < = > >=< <

10 < 1 H >= 1/ [ AA AA AA BB AA AB AA BA BB AA BB BB BB AB BB BA AB AA AB BB AB AB AB BA BA AA BA BB BA AB BA BA] But r 1 is large, which means anything involving AB will be negligible relative to AAAA, BBBB, and AABB. Therefore, <1H> is clearly positive. Ψ= gg> - cuu>, where c is positive

11 Ψ = Ψ is gg > c uu proportional > = gg to [ > gg c > c uu uu > > ] c gg > uu > uu> can be positive or negative, depend on where electrons 1 and are located. If e1 and e are in same lobe -c uu> is neg. and Ψ is reduced If e1 and e are in different lobes, -c uu> is positive, and Ψ is increased -uu> = -uu> = 1 1 = _ Lower probability = _ Lower probability -uu> = -uu> = 1 1 = = Higher probability Higher probability

12 CI-singles CIS important for understanding spectra 1/ N! A A B B C C If ground state is v Ψ = = D G v Φ CIS o= A, C o G virtual F E D N! 1/ A A B B F C = Φ F C C occupied B A N! 1/ A A B B C C

13 Example of Full CI FCI High energy CSF for six electrons and 7 basis functions G F E D C B A N! N! 1/ 1/ EE FF GG A A B B C C

14 If the basis set is complete, and if we include all possible configurations, the eigenvalues and eigenstates will be exact.

15 Allowed Forbidden! Intensity comes perturbations from vibrations that mix in the allowed transition Just look at spectra on LOG SCALE cm -1 = 50 nm 00 nm 50 nm

16 Example of G09 CIS calculation for benzene %chk=c:\564-17\benzene1.chk # cis=nstates=5/3-1g pop=reg density=current Title Card Required 0 1 C C C C C C H H H H H H

17 π MOs of Benzene f 3 xy LUMOs x -y 1 y HOMOs x 0 s

18 After CIS y xy x x y x xy y x y -1/ y xyx x y -1/ x xyy x y -1/ y xy-x x y 1 B a,b, 1 E 1u 1 L a, 1 B 1u -1/ x xy-y x y 1 L b, 1 B u Ground

19 Reduced 1 st order Transition Density is Product of Ground and Excited state wavefunctions integrated over all but one electron

20

21 Valence Bond L a, B 1u ionic Part of Ground = - L b, B u covalent - = Part ofground =

22 *********************************************************************** Excited states from <AA,BB:AA,BB> singles matrix: *********************************************************************** Ground to excited state transition electric dipole moments Au: state X Y Z Dip. S. Osc Ground to excited state transition velocity dipole moments Au: state X Y Z Dip. S. Osc Ground to excited state transition magnetic dipole moments Au: state X Y Z

23 Excited state symmetry could not be determined. Excited State : Singlet-?Sym ev nm f= <S**>= > > Excited state symmetry could not be determined. Excited State 3: Singlet-?Sym ev nm f= <S**>= > > Excited state symmetry could not be determined. Excited State 4: Singlet-?Sym ev nm f= <S**>= > > Excited state symmetry could not be determined. Excited State 5: Singlet-?Sym ev nm f= <S**>= > > >

24 Ground to excited state transition densities written to RWF 633 Excitation energies and oscillator strengths: Excited state symmetry could not be determined. Excited State 1: Singlet-?Sym ev nm f= <S**>=0.000 exp 40,000 cm -1 = 65 nm 0 -> > Excited State : Singlet-?Sym ev nm f= <S**>= > > Excited State 3: Singlet-?Sym ev nm f= <S**>= > > exp =50,000 cm -1 = 00 nm exp 54,000 cm -1 = 185 Excited State 4: Singlet-?Sym ev nm f= <S**>=0.000 exp 54,000 cm -1 = 185 nm 0 -> > nm Excited state symmetry could not be determined.

25 π MOs of Benzene f 3 xy LUMOs x -y 1 y HOMOs x 0 s

26 How to get Acurrate excited state energies CASPT B Roos et al: gets within 0.1 ev = 1000 cm -1 First CASSCF example of MCSCF 1. Choose "active filled A,B,C and virtual orbitals D,E,F. Do Full CI FCI on the active set F E D C B A N! 1/ A A B B C C

27 CASPT B Roos et al: gets within 0.1 ev = 1000 cm -1 Effectively do an MP nd order perturbation calc. using all single and double excitations out of the CASSCF slater determinants