1 Jack Simons Henry Eyring Scientist and Professor Chemistry Department University of Utah Electronic Structure Electronic Structure Theory Theory TSTC Session 1 TSTC Session 1 1. Born-Oppenheimer approx.- energy surfaces 2. Mean-field (Hartree-Fock) theory- orbitals 3. Pros and cons of HF- RHF, UHF 4. Beyond HF- why? 5. First, one usually does HF-how? 6. Basis sets and notations 7. MPn, MCSCF, CI, CC, DFT 8. Gradients and Hessians 9. Special topics: accuracy, metastable states
16
Embed
Jack Simons Henry Eyring Scientist and Professor Chemistry Department University of Utah
Electronic Structure Theory TSTC Session 1. 1. Born-Oppenheimer approx.- energy surfaces 2. Mean-field (Hartree-Fock) theory- orbitals 3. Pros and cons of HF- RHF, UHF 4. Beyond HF- why? 5. First, one usually does HF-how? 6. Basis sets and notations 7. MPn, MCSCF, CI, CC, DFT - PowerPoint PPT Presentation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
1
Jack Simons Henry Eyring Scientist and ProfessorChemistry Department
University of Utah
Electronic Structure Theory Electronic Structure Theory TSTC Session 1TSTC Session 1
1. Born-Oppenheimer approx.- energy surfaces2. Mean-field (Hartree-Fock) theory- orbitals3. Pros and cons of HF- RHF, UHF4. Beyond HF- why?5. First, one usually does HF-how? 6. Basis sets and notations7. MPn, MCSCF, CI, CC, DFT8. Gradients and Hessians9. Special topics: accuracy, metastable states
This expansion ((r,R) = K K(r|R) K(R)) can then be substituted into
H(r,R) (r,R) = E (r,R)
[H0 –2/2 j=1,M mj-1 j
2 -E] K K(r|R) K(R) = 0
to produce equations for the K(R) by multiplying by < L(r,R)| and integrating over dr1dr2…drN :
0 = [EL(R) -2/2 j=1,M mj-1 j
2 -E] L(R)
+ K< L(r|R)| -2/2 j=1,M mj-1 j
2 K(r|R)> K(R)
+ K< L(r|R)| -2j=1,M mj-1 j
K(r|R)> j K(R)These are called the coupled-channel equations.
7
If we ignore all of the non-BO terms
K< L(r|R)| -2/2 j=1,M mj-1 j
2 K(r|R)> K(R)
+ K< L(r|R)| -2j=1,M mj-1 j
K(r|R)> j K(R)
we obtain a SE for the vib./rot./trans. motion on the Lth energy surface EL(R)
0 = [EL(R) -2/2 j=1,M mj-1 j
2 -E] L(R)
The translational part of L(R) separates out
(e.g., exp(iPR/)) and won’t be discussed further.
8
Each electronic state L has its own set of rot./vib. wave functions and energies:
[EL(R) -2/2 j=1,M mj-1 j
2 -EL,J,M,] L,J,M, (R) = 0
This is the electronic-vibrational-rotational separation one sees in textbooks.
9
The non-BO couplings
K< L(r|R)| -2/2 j=1,M mj-1 j
2 K(r|R)> K(R)
+ K< L(r|R)| -2j=1,M mj-1 j
K(r|R)> j K(R)
can induce transitions among the BO states (radiationless transitions).
When the nuclear motions are treated classically, these wave functions are replaced by trajectories on the two surfaces.
10
The surfaces drawn below are eigenvalues of the electronic SE
H0 (r|R) =EK(R) (r|R)
where H0 contains all but the nuclear kinetic energy. Such surfaces
are called adiabatic. Each surface adiabatically evolves as the geometry is changed and T2 is always above T1.
11
Sometimes, one leaves out of H0 some small terms V (e.g., spin-orbit coupling A k Sk Lk ) in defining the BO states. The
resulting BO states are called diabatic. One then includes the non-BO couplings
K< L(r|R)| -2/2 j=1,M mj-1 j
2 K(r|R)> K(R)
+ K< L(r|R)| -2j=1,M mj-1 j
K(r|R)> j K(R)
as well as couplings
K< L(r|R)|V|K(r|R)> K(R)
due to the “ignored terms”.
Singlet-triplet diabatic states’ curve crossing
12
Sometimes, one leaves out of H0 some small terms that couple different electronic configurations (e.g., n* or *) in defining the BO states. The resulting BO states are also called diabatic.
At geometries where these diabatic states cross, the couplings K< L(r|R)|V|K(r|R)> are especially important to consider.
13
Again, one includes the non-BO couplings
K< L(r|R)| -2/2 j=1,M mj-1 j
2 K(r|R)> K(R)
+ K< L(r|R)| -2j=1,M mj-1 j
K(r|R)> j K(R)
as well as couplings
K< L(r|R)|V|K(r|R)> K(R)
due to the “ignored terms”.
14
Can adiabatic BO surfaces cross? Suppose that all but two exact BOstates have been found and consider two orthogonal functions (r|R) and L(r|R) that span the space of the two “missing” exactBO states. Form a 2x2 matrix representation of H0 within the space spanned by these two functions:
<|H0|> -E <|H0|L>
<L|H0|> <L|H0|L> -E= 0det
E2 –E(HK,K+HL,L) +HK,K HL,L –HK,L2 = 0
E = ½{(HK,K+HL,L) ±[(HK,K-HL,L)2 +4HK,L]1/2}The two energies can be equal only if both
HK,K(R) = HL,L(R) andHK,L(R) = 0 at some geometry R.
R is a 3N-6 dim. space; so the “seam” of intersection is a space of 3N-8 dimensions.
15
BO energy surfaces have certain critical points to be aware of
Minima (all gradients vanish and all curvatures are positivecharacteristic of stable geometries
Transition states (all gradients vanish and all but one curvature are positive; one is negative) characteristic of transition states.
16
Summary: Basic ingredients in BO theory are:
Solve H0 (r|R) =EK(R) (r|R) at specified R for states K and L “of interest”.Keep an eye out for geometries R* where EK and EL intersect or come close.
Solve [EL(R) -2/2 j=1,M mj-1 j
2 -EL,J,M,] L,J,M, (R) = 0or compute classical trajectories on the EL and EK surfaces.
Stoppng here {(r,R) = K(r|R) K(R) or (r,R) = L(r|R) L(R)} = pure BO
To go beyond the BO approximation, compute all of the couplings:
K< L(r|R)| -2/2 j=1,M mj-1 j
2 K(r|R)> K(R)
+ K< L(r|R)| -2j=1,M mj-1 j
K(r|R)> j K(R)
K< L(r|R)|V|K(r|R)> K(R) and
Evaluate the effects of the couplings on the nuclear-motion state L(R) or on the
classical trajectory coupling surface EL(R) to EK(R). How is this done?