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Jack Simons Henry Eyring Scientist and ProfessorChemistry Department

University of Utah

Electronic Structure Theory Electronic Structure Theory Session 1Session 1

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The Schrödinger equation for N electrons and M nuclei of a molecule:

H(r,R) (r,R,t) = i ∂(r,R,t)/∂t, or if H is t-independent,

H(r,R) (r,R) = E (r,R)

|(r,R)|2 gives probability density for finding electrons at

r = r1r2 r3 ... rN and nuclei at R1 R2 R3 ...RM.

H contains electronic kinetic energy: Te = -2/2 j=1,N me-1 j

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nuclear kinetic energy: TM= -2/2 j=1,M mj-1 j

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electron-nuclei Coulomb potentials: – j=1,MZj k=1,N e2/|rk-Rj|

VeM nuclear-nuclear Coulomb repulsions: + j<k=1,M ZjZke2/|Rk-Rj|

and electron-electron Coulomb repulsions: + Vee= j<k=1,Ne2/rj,k

It can contain more terms if, for example, external electric or magnetic fields

are present (e.g., k=1,N erkE).

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In the Born-Oppenheimer approximation/separation, we (first)

ignore the TM motions of the nuclei (pretend the nuclei are fixed at specified locations R) and solve

H0 (r|R) =EK(R) (r|R),

the so-called electronic Schrödinger equation.

H0 contains all of H except TM (Te plus all of the potential energy terms).

Because H0 is Hermitian, its eigenfunctions form a complete and

orthonormal set of functions of r. So, can be expanded in the K:

(r,R) = K K(r,R) K(R).

The K(r,R) and the K(R) depend on R because H0 does through

j=1,MZj k=1,N e2/|rk-Rj| Z+ j<k=1,M ZjZke2/|Rk-Rj|

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This expansion can then be used in

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) = 0to produce equations for the K(R):

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.

If we ignore all of the non-adiabatic terms, 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)

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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

The non-adiabatic couplings can induce transitions among these states (radiationless transitions).

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There are major difficulties in solving the electronic SE:

Vee= j<k=1,Ne2/rj,k

makes the equation not separable- this means is not rigorously a product of

functions of individual electron coordinates.

e.g., 1s(1) 1s(2) 2s(3) 2s(4) 2p1(5))

Also, has cusps

The factors (-2/me1/rk /rk –Ze2/rk)

and (-22/me1/rk,l /rk,l +e2/rk,l)

will blow up unless so-called cusp conditions are obeyed by :

/rk = -meZe2/ 2 as rk0);

/rk,l = 1/2 mee2/ 2as rk,l0)

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Cusp near nucleus Cusp as two electrons approach

This means when we try to approximately solve the electronic SE, we should use

“trial functions” that have such cusps. Slater-type orbitals (exp(-rk)) have cusps

at nuclei, but Gaussians (exp(-rk2)) do not.

We rarely use functions with e-e cusps, but we should (this is called using explicit e-e correlation).

/rk = Ze2 as rk0) and /rk,l = - e2 as rk,l0).

The electrons want to pile up near nuclei and they want to avoid one another.

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Often, we use so-called atomic units.

We let each coordinate be represented in terms of a parameter a0 having units of length and a dimensionless quantity: rj a0 rj.

The kinetic and potential energies thus contain

T = -(2/2m)(1/a0)2 j2 and Ven= -Ze2(1/a0) 1/rj, Vee= e2(1/a0) 1/rj,i

respectively, where the variables now are the dimensionless ones.

Factoring e2/a0 out from both the kinetic and potential energies as

T = e2/a0{-(2/2m)(1/e2a0) j2} and V = e2/a0 {-Z/rj +1/rj,i}

Choosing a0 = 2/(e2m) = 0.529 Å,

when m is the electron mass, allows T and V to be written in terms of a common factor:

e2/a0 = 1 Hartree = 27.21 eV:

T = -1/2 j2 while V = - Z/rj.

Digression into atomic units.

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Addressing the non-separability problem:

If Vee could be replaced (or approximated) by a one-electron additive potential

VMF = j=1,N VMF(rj)

each of the solutions L would be a product (an antisymmetrized product called a

Slater determinant) of functions of individual electron coordinates (spin-orbitals) j(r):

| r1)(r2) (r3) (r4) (r5) |

= (N!)-1/2 P=1,N! SP P r1)(r2) (r3) (r4) (r5)

where SP is the sign of the permutation P.

For example:

| r1)(r2) |

= (2!)-1/2 [r1)(r2) - 2(r1) 1(r2)]

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Before considering how we might find a good VMF, let’s examine how important permutational antisymmetry is by considering some spin-orbital product (Slater determinant) wave functions for two electrons in and *

orbitals.

The purposes of this exercise are to remind one how to form singlet and triplet functions, to see how the different spin states have different physical content (i.e., charge distribution) even when the orbital occupancy is the same, and to introduce the idea that it is simply not possible, in certain circumstances,

to use a single Slater determinant function as an approximation to K.

So, let’s think of how the low-energy states of an olefin with two electrons in its orbital framework should behave as we twist and break the

bond.

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SingletSinglet 22 || = 2-1/2

TripletTriplet ** ||= 2-1/2 [

||= 2-1/2 [

2-1/2 [||+|

SingletSinglet 2-1/2 [|| – |

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SingleSinglet t **22 ||= 2-1/2

Note: single determinant not possible here

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Now think of 2-1/2(L + R) and2-1/2 (-L + R)

to consider the behavior when -bond cleavage occurs upon rotation.

Singlet Singlet 22

|| = 2-1[|RR|+ |LL|

+ |RL|+ |LR|]ionic + diradical

Singlet Singlet 22

|| = 2-1[|RR|+ |LL|-|RL|- |LR|]

ionic - diradical

Triplet Triplet

|| = 2-1[|LR|- |RL|]

= |LR|diradical

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Singlet Singlet **

2-1/2 [|| - |-|RR|+

|LR|- |RL|

– |LL|]

– 2-3/2[|RR|+|LR|

– |RL|- |LL|]

-|RR|+|LR|] ionic

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So, the spin state and orbital occupancy plus the antisymmetry have effects on the ionic/radical character of the function.

To adequately describe the (singlet) bond breaking, we need to mix the and configuration state functions (CSF). This shows how single configuration functions and single determinants may not be adequate.

*2

π2

Triplet ππ*

Singlet ππ*

Twist Angle0 90

Diradical

Ionic

Ionic +/-Diradical

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To adequately describe the (singlet) bond breaking to give the singlet

diradical, we need to mix the and configuration state functions (CSF).

|| = 2-1[|RR|+ |LL|

+ |RL|+ |LR|]ionic + diradical

|| = 2-1[|RR|+ |LL|

-|RL|- |LR|]ionic - diradical

So, one must combine 2-1/2{|| - ||}

to obtain a diradical state and 2-1/2{|| + ||}

to obtain an ionic state.

A single determinant function won’t work!

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Analogous “trouble” occurs if one uses a single determinant wave function to describe a bond that one wants to break:

H2 (2) H(1sA) + H(1sB)

H3C-CH3 (2) H3C + CH3

The |(1)(2)| determinant has diradical and ionic terms at large-R.

It is possible to use a trial function of the form |(1)’(2)| and to allow and ’ to evolve from being and near the equilibrium bond length

into L and R at large-R.

However, the function |R(1)L(2)| is not a singlet; it is a mixture of singlet and triplet, and, how do you let and ’ evolve?

Nevertheless, these are the approaches used in so-called unrestricted and

restricted Hartree-Fock theory (RHF, UHF)

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|(1) ’(2)| is neither a singlet nor a triplet, but a mixture:

2-1/2[ |(1) ’ (2)| - |(1) ’ (2)|] + 2-1/2[ |(1) ’ (2)| + |(1) ’ (2)|]

|(1) (2)| is a singlet, but dissociates incorrectly

|(1) ’(2)| dissociates to diradicals, but is neither singlet nor triplet

One experiences jerks (not in the energy, but in its derivatives as well as spin impurity when the RHF and UHF curves connect.