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					Electronic Structure Theory
      TSTC Session 2
                  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




Jack Simons Henry Eyring Scientist and Professor
            Chemistry Department
              University of Utah                              0
                       Digression into atomic units.
  Often, we use so-called atomic units.
  We let each (e and nuc.) coordinate be represented in terms of a parameter a0
  having units of length and a dimensionless quantity: rj  a0 rj; Rj  a0 Rj

The kinetic and potential energies in terms of the new dimensionless variables:
T = -(2/2me)(1/a0)2 j2 , Ven= - ZKe2(1/a0) 1/rj,K, Vee= e2(1/a0) 1/ri,j

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

T = e2/a0{-(2/2me)(1/e2a0) j2} and V = e2/a0 {-ZK/rj,K +1/rj,i}

Choosing a0 = 2/(e2me) = 0.529 Å = 1 Bohr,

where me is the electron mass, allows T and V to be written in terms of

e2/a0 = 1 Hartree = 27.21 eV in a simple manner:
T = -1/2 j2 while Ven = - ZK/rj,K and Vee = 1/ri,j                          1
Let’s now look into how we go about solving the electronic SE

                H0 (r|R) =EK(R) (r|R)

  for one electronic state (K) at some specified geometry R.




                                                               2
There are major difficulties in solving the electronic SE. The potential terms

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


make the SE equation not separable- this means (r|R) is not a product of
functions of individual electron coordinates.


                      (r|R)(r1)(r2)N(rN)
             e.g., 1s(1) 1s(2) 2s(3) 2s(4) 2p1(5) for Boron).


This means that our desire to use spin-orbitals to describe (r|R) is not really
correct. We need to make further progress before we can think in terms of spin-
orbitals, orbital energies, orbital symmetries, and the like.

                                                                                    3
The correct (r|R) have certain properties that we need to know
about so that, when we try to create good approximations to (r|R),
we can build these properties into our approximate functions.


 1. Because Pi,j H0 = H0 Pi,j , the (r|R) must obey Pi,j (r|R)
 = ± (r|R) [-1 for electrons].

 2. Usually (r|R) is an eigenfunction of S2 (spin) and Sz

 3. (r|R) has cusps near nuclei and when two electrons get close

 a. Near nuclei, the factors (-2/me1/rk /rk –ZAe2/|rk-RA|) (r|R)

 will blow up unless /rk  = -meZAe2/ 2 as rkRA).

 b. As electrons k and l approach, (-22/me1/rk,l /rk,l +e2/rk,l) (r|R)

 will blow up unless /rk,l  = 1/2 mee2/ 2as rk,l0)
                                                                        4
Cusps            /rk  = -meZAe2/2 as rkRA) and

                   /rk,l  = 1/2mee2/ 2 as rk,l0).

  Cusp near nucleus
                                                    Cusp as two electrons
                                                    approach




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




                                                                            5
In the electronic kinetic energy, in addition to the terms like

                         (-2/me1/rk /rk –ZAe2/rk) (r|R)

there are terms involving angular derivatives

                                  L2/2mer2 (r|R)

         = -2/(2mer2) {(1/sinq)∂/∂q(sinq∂/∂q+(1/sinq)2∂2/∂2}(r|R)

These terms will also blow up (for any state with L > 0) unless (r|R)

                            (r|R)  0 (as rk RA).

So, for L = 0 states, one has /rk  = -meZAe2/2 as rkRA),

and for L > 0 states, both (r|R)  0 (as rk RA) and /rk  = -meZAe2/2
as rkRA) hold, but the latter is “useless” because (r|R)  0 anyway.

This is why the cusp condition /rk  = -meZAe2/2 as rkRA) is useful only
for ground states.                                                           6
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.


                                               d/dr(exp(-rk2) = -2rk(exp(-rk2) =0
                                               at rk = 0,
        tight Gaussian

                  orbital with cusp at r = 0

                                               d/dr(exp(-rk)= -(exp(-rk) = -
                         loose Gaussian        at rk = 0.

                     medium Gaussian
                                               So, sometimes we try to fit STOs by
                                               a linear combination of GTOs, but this
                                               can not fix the nuclear cusp problem
                                               of GTOs
                                   r




                                                                                       7
It is very difficult to describe the ee cusp (Coulomb hole) accurately.
Doing so is important because electrons avoid one another. We call
this dynamical correlation. We rarely use functions with e-e cusps, but
we should (this is called using explicit e-e correlation).


                   D
                   T
            5




The coulomb hole for He in cc-pVXZ (X=D,T,Q,5) basis
sets with one electron fixed at 0.5 a0
                                                                  8
The nuclear cusps /rj  = -meZAe2/2 as rjRA) depend on Z.

Given a wave function (r|R), one can compute the electron density
r(r) = N  *K (r,r2,r3 ...rN | R)K (r,r2,r3 ...rN | R)dr2dr3 ...drN

Using the cusp condition that obeys, one can show that
/rr(r) = -2meZAe2/2 r(r) as rRA) .



So, if we knew the ground-state r(r) and could find the locations of
its cusps, we would know where the nuclei are located. If we also
could measure the “strength” -2meZe2/2 of the cusps, we would know
the nuclear charges. If were to integrate r(r) over all values of r,
we could compute N, the number of electrons.
This observation that the exact ground-state r(r) can be used to
find R, N, and the {ZK} and thus the Hamiltonian H0 shows the
origins of densityfunctional theory.                               9
Addressing the non-separability problem and the permutational and spin
symmetries:

 If Vee could be replaced (or approximated) by a one-electron additive potential
                                     VMF = j=1,N VMF(rj|R)

each of the solutions (r|R) would be a product (an antisymmetrized product
called a Slater determinant) of functions of individual electron coordinates (spin-
orbitals) j(r|R):
                       (r1 , r2 ,..., rn ) 

                                       1 P 1 (1)2 (2)...n (n)  O 12 ...n 
                                 1
                                   
                                           p(P)

                                 n! P

    A spinorbital is product of spin-and spatial function
     k (rj )   k (rj ) j
     k  (rj )   k (rj ) j
      k  k    dv k (r) k (r)  
                        *

                                                                                                10
Is there any optimal way to define VMF = j=1,N VMF(rj|R) ?

Does such a definition then lead to equations to determine the
optimal spin-orbitals?

Yes! It is the Hartree-Fock definition of VMF .

Before we can find potential VMF and the Hartree-Fock (HF)
spin-orbitals j(r)( or ), we need to review some background
about spin and permutational symmetry.




                                                                 11
   A brief refresher on spin
                                                |   1
    SZ 1/2                                   |   0
    SZ   1/2                                |   1
S 2        1/2(1/2 +1)  3/4 2
             2
                                                                1 1      1 1
                                                    S         ( + 1)  ( 1)  
                                                                2 2      2 2
 S 2    1/2(1/2 +1)  3/4 2
             2

                                                                       S  0
  Special case of                                            Special case of
J 2 | j,m      2
                     j( j +1) | j,m     
                                              J  | j,m      j( j + 1)  m(m  1) | j,m  1 
                                                              
    For acting on a product of spin-orbitals, one uses
 SZ   SZ ( j)        S   S ( j)
                                       S 2  S S+ + SZ + SZ
                                                        2

         j                    j

Examples:            SZ(1)(2) 1/2 (1)(2) +1/2 (1)(2)  (1)(2)
                     S(1)(2)  (1)(2) + (1)(2)
                                                                                   12
       
 Let’s practice forming triplet and singlet spin functions for 2 e’s.
 We always begin with the highest MS function because it is “pure”.
      (1)(2)                                              So, MS =1; has to
SZ(1)(2) 1/2 (1)(2) +1/2 (1)(2)  (1)(2)           be triplet
SZ (1)(2)  1/2 (1)(2) 1/2 (1)(2)   (1)(2) So, MS =-1; has to
                                                            be triplet


                                                        1
 S (1) (2)   (1) (2) +  (1) (2)   So, |1,0       [ (1) (2) +  (1) (2)]
                                                         2
     1(1+1) 1(11) | S 1, MS  0         This is the MS =0 triplet

                                   
 How do we get the singlet? It has to have MS = 0 and be orthogonal
 to the MS = 0 triplet. So, the singlet is          1
                                           | 0,0    [ (1) (2)   (1) (2)]
                                                        2
                                                                             13
Slater determinants (Pi,j) in several notations. First, for two electrons.
               1  (r1 )   (r1 )
  (r1 ,r2 ) 
                2  (r2 )   (r2 )                                  Shorthand
               1  (r1 ) (1)  (r1 ) (1)                             (r1 ,r2 )   
             
                2  (r2 ) (2)  (r2 ) (2)


                                                                          Symmetric space;
  (r1 ,r2 ) 
                 1
                       (r ) (1) (r ) (2)   (r ) (1) (r ) (2)
                             1          2            1        2           antisymmetric spin
                  2
                                                                          (singlet)
               (r1 ) 2 (r2 )
                                  1
                                        (1) (2)   (1) (2)
                                        
                                   2
                 1                                      1
| 12 |         [1 (1)2 (2)  1 (2)2 (1)]     [1 (1)2 (2)  2 (1)1(2)] (1) (2)
                  2                                      2
                                  Antisymmetric space; symmetric spin (triplet)
 (r1 ,r2 )   (r2 ,r1 )
 1  (r1 )   (r1 )     1  (r2 )   (r2 )           Notice the Pi,j
                       
  2  (r2 )   (r2 )     2  (r1 )   (r1 )
                                                                                                    14
                                                         antisymmetry
  More practice with Slater determinants

  (r1 ,r2 ,...,rn )  1 , 2 ....., n        Shorthand notation for general case

                          1 (1)    2 (1)            n (1)
                                                                   Odd under interchange of
                          1 (2)  2 (2)              n (2)    any two rows or columns
  (r1 ,r2 ,...,rn ) 

                          1 (n)  2 (n)              n (n)

                                                                  The dfn. of the Slater determinant
 (r1 , r2 ,..., rn )                                            contains a N-1/2 normalization.
        1 P 1 (1)2 (2)...n (n)  O 12 ...n 
  1
    
            p(P)

  n! P

     P permutation operator                                    parity ( p(P) least number of
                                                1
                                                       p( P)

                                                               transpositions that brings the indices
                1                                              back to original order)
       O            1 P
                          p( P)

                n! P                  antisymmetrizer
                                                                                                15
Example : Determinant for 3-electron system


                   1                     
           
O 1  2  3                                
                       1   Pij +  Pijk  1  2  3   
                    6     i, j    i, j,k 
                   1 (1) 2 (2) 3 (3)   2 (1)1 (2) 3 (3) 
                 1                                               
                    3 (1) 2 (2)1 (3)  1 (1) 3 (2) 2 (3) 
                  6
                    + 2 (1) 3 (2)1 (3) +  3 (1)1 (2) 2 (3) 
                                                                  



permutations       1, P , P , P23 , P231 , P312
                       12  13

transpositions     0 1       3    1    2     2
parity             +                 +     +




                                                                      16
The good news is that one does not have to deal with most of these
complications. Consider two Slater determinants (SD).
                    1
                   A
                    N!
                       (1) P (1) (2) (3)... (N)
                                       pP
                                              1     2    3       N
                                P

                           1
                              (1)q Q '1 (1) '2 (2) '3 (3)... 'N (N)
                                    Q
                   B 
                           N! Q
          
Assume that you have taken t permutations1 to bring the two SDs into
        coincidence. Now, consider evaluating the integral
maximal
                d1d2d3...dNA [  f ( j) +
                             *
                                                                 g(i, j)]   B
                                            j1,N             jk1,N

where f(i) is any one-electron operator (e.g., -ZA/|rj-RA|) and g(i.j)
is any two-electron operator (e.g., 1/|rj-rk|). This looks like a
horrible task (N! x (N + N2) x N! terms).

                                                                                  17
1. A factor of   (-1)t   will then multiply the final integral I
                       I       d1d2d3...dN [       *
                                                       A            f ( j) +          g(i, j)]          B
                                                           j1,N                    jk1,N

                   1                                                                     1
                                                                                           (1)q Q '1 (1) '2 (2) '3 (3)... 'N (N)
                                                                                                  Q
                                                                                B 
                              P
            A                                      
                        (1) p P1(1) 2 (2) 3 (3)... N (N)
                   N! P                                                                  N! Q


     1. The permutation P commutes with the f + g sums, so 
      
                                                                 

   1
I
   N!
                d  (1)       pP
                                       *1 (1) *2 (2) *3 (3)... *N (N)[  f ( j) +                                  g(i, j)]P        B
                       P                                                                          j1,N             jk1,N


     2.  P B  (1)  B and (1) p
                                pP
                                                                                              so
                                                                   P
                                                                               pP
                                                                        (1)         N!
                                                       P
        N!
     I
        N!
                    d   * (1) *  1       2   (2) *3 (3)... *N (N)[                         f ( j) +       g(i, j)]          B
                           P                                                               j1,N               jk1,N

           d *1 (1) *2 (2) *3 (3)... *N (N)[  f ( j) +
                                                                          g(i, j)](1) q Q '1 (1) '2 (2) '3 (3)... 'N (N)
                                                                                              Q
                                                                                                                       
                                                   j1,N           jk1,N           Q




      Now what?

                                                                                                                                 18
          d * (1) *       2 (2) *3 (3)... *N (N)[  f ( j) +                g(i, j)](1)q Q '1 (1) '2 (2) '3 (3)... 'N (N)
                                                                                                    Q
    I           1                                                                                                         
                                                       j1,N            jk1,N           Q



    Four cases: the Slater-Condon rules – you should memorize.
    Recall to multiply the final I by (-1)t
    A and B differ by three or more spin-orbitals: I = 0
    A and B differ by two spin-orbitals-AkAl;BkBl
            I         dkdl *       Ak   (k) *Al (l)g(k,l)[Bk (k)Bl (l)  Bl (k)Bk (l)]
    A and B differ by one spin-orbital-Ak;Bk


            I         dkdj * (k) * ( j)g(k, j)[ (k) ( j)   (k)
                                             Ak             j                        Bk         j           j        Bk   ( j)]
                     j A,B
                             +  dk * (k) f (k) (k)  Ak                     Bk

    A and B are identical
            I           dkdj *            k   (k) * j ( j)g(k, j)[ k (k) j ( j)   j (k) k ( j)]
                      
                     k j A

                                       +          dk *       k   (k) f (k) k (k)
                                           k A                                                                                   19
In HF theory, we approximate the true (r|R) in terms of a single
Slater determinant. We then use the variational method to minimize
the energy of this determinant with respect to the spin-orbitals
appearing in the determinant. Doing so, gives us equations for the
optimal spin-orbitals to use in this HF determinant. They are called
the HF equations.

A single Slater determinant
                   1
                     (1) p P1(1) 2 (2) 3 (3)... N (N) | 1(1) 2 (2) 3 (3)... N (N) |
                            P
            A 
                   N! P
can be shown to have a density r(r) equal to the sum of the densities of
the spin-orbitals in the determinant r(r)  | j (r) j (r) |
                                               *

                                                         j




                               
                                                                                                 20
The fourth of the Slater-Condon rules allows us to write the expectation
value of H0 for a single-determinant wave function
                      1
                        (1) p P1(1) 2 (2) 3 (3)... N (N) | 1(1) 2 (2) 3 (3)... N (N) |
                               P
             A 
                      N! P


                                                                      1
       *
      A H 0 A               dkdj *k (k) * j ( j)           rk, j
                                                                           [ k (k) j ( j)   j (k) k ( j)]
                        k j A

                                                                          ZM
            +         dk *k (k)[1/2 2 (k)                                 ] k (k)
              k A                                               M
                                                                     | rk  RM |
                                                ZM
                         *k |[1/2         2
                                                       ] | k 
                k A                      M
                                            | r  RM |

                                                 1
                +               *k  * j
                                                r1,2
                                                     [ k j   j k ] 
                     k j A
 

                                                                                                             21
     
                                                          ZM                                                1
       E     *k |[1/2 2                                  ] | k     +               *k  * j        [ k j   j k ] 
               k A                               M
                                                      | r  RM |                  k j A
                                                                                                           r1,2

       The integrals appearing here are often written in shorthand as
                                                               ZM
                             *k |[1/2 2                         ] |  k  k | f | k 
                                                       M
                                                           | r  RM |
                                    1
                 *k  * j             [ k j   j k ]  k, j | k, j    k, j | j,k  (Dirac)
                                   r1,2
                                   1
                    *k  * j           [ k j   j k ]  (k,k | j, j)  (k, j | j,k)(Mulliken)
                                    r1,2
     When we minimize E keeping the constraints <k|j>=dk,j,
       we obtain the Hartree-Fock equations
 
                                            1                                      1
     f k (1) +        [   *j (2)       r1,2
                                                 j (2)d2 k (1)      *j (2)   r1,2
                                                                                        k (2)d2 j (1)]  k  k (1)  h HF k (1)
                  j1, N



        fk (1) +         [J       j    K j ]k (1)  fk (1) + VHFk (1)
                           j1,N
                                                                                                                                   22
Physical meaning of Coulomb and exchange operators and
integrals:
J1,2=  *1(r)J21(r)dr=  |1(r)|2 e2/|r-r’|2(r’)|2 dr dr’


K1,2= *1(r)K21(r)dr =  1(r) 2(r’) e2/|r-r’|2(r) 1(r’)dr dr’




                                           2(r')
                                   Overlap region



                        1(r)                                         23
What is good about Hartee-Fock ?

It is by making a mean-field model that our (chemists’) concepts of
orbitals and of electronic configurations (e.g., 1s 1s  2s  2s  2p1 )
arise.

Another good thing about HF orbitals is that their energies K give
approximate ionization potentials and electron affinities (Koopmans’
theorem). IP ≈ -occupied ; EA ≈ -unoccupied





                                                                             24
     Koopmans’ theorem- what orbital energies mean.
     N-electrons’ energy
                                                                                  1
      E HF               *k |[1/2  
                                        2       ZM
                                                       ] | k  +    *k  * j      [ k j   j k ] 
                                            | r  RM |           k j1,N
                                                                                 r1,2
                k1,N                     M

     N+1-electrons’ energy
                                                     Z M                              1
     E HF                *k |[1/2 2  
                                                 | r  RM |
                                                            ] |  k  +    *k  * j
                                                                                       r1,2
                                                                                            [ k j   j k ] 
              k1,N +1                      M                          k j1,N +1



     Energy difference (neutral minus anion) if spin-orbital m is the N+1st electron’s
                                                    Z M                             1
        E     *m |[1/2 2                         ] |  m      *m  * j      [ m j   j m ] 
                                            M
                                                | r  RM |           k1,N
                                                                                     r1,2

      But, this is just minus the expression for the HF orbital energy
              m  m | h HF | m  mk | f| m  +  m [J j  K j ]m 
                                                                   j1,N




                                                                                                            25

  The sum of the orbital energies is not equal to the HF energy:

                        hHFk (1)  fk (1) +    [J     j    K j ]k (1)  k k
    So                                           j1,N

               k  k | h HF | k  k | f | k  +  k [J j  K j ]k 
                                                                 j1,N

                  k | f | k  + [ k, j | k, j    k, j | j,k ]
                                       j1,N
   So
             k   { k | f | k  + [ k, j | k, j    k, j | j,k ]}
              k1,N  k1,N                j1,N
  But                                   ZM                                1
       E HF     *k |[1/2 2           ] |  k  +    *k  * j   [ k j   j k ] 
                k1,N                      M
                                               | r  RM |                k j1,N
                                                                                     r1,2

   
     So, the sum of orbital energies double counts the J-K terms. So, we can
   compute the HF energy by taking
                                            
                        E HF 1/2 k +1/2   k | f | k 
                                   k1,N            k1,N
                                                                                            26
 Orbital energies depend upon which state one is studying. So a p*
 orbital in the ground state is not the same as a p* in the pp* state.
   k  k | h HF | k  k | f | k  +  k [J j  K j ]k 
                                           j1,N
                                                                     Occupied orbitals
                     p “feels” 6 J and 3 K interactions             feel N-1 others;
                     a “feels” 5 J and 2 K interactions             virtual orbitals feel
                                                                     N others.
             p
                                        p “feels” 5 J and 2 K interactions
                                        p “feels” 6 J and 3 K interactions
                          p
                 a                      q “feels” 6 J and 3 K interactions
                           q
                                        a “feels” 5 J and 2 K interactions
                                        b “feels” 5 J and 2 K interactions
                          a
                                        b “feels” 6 J and 3 K interactions
                          b
                                          Occupied orbitals feel N-1 others; virtual
                                          orbitals feel N others.
This is why occupied orbitals (for the state of interest) relate to IPs and virtual
orbitals (for the state of interest) relate to EAs.                              27
In summary, the true electronic wave functions have Pi.j symmetry,
nuclear and Coulomb cusps, and are not spin-orbital products or Slater
determinants.

However, HF theory attempts to approximate (r|R) as a single
Slater determinant and, in so doing, to obtain a mean-field
approximation to j<k=1,N1/rj,k in the form VHF  [J j  K j ]
                                               j1,N



 To further progress, we need to study the good and bad of the HF
 approximation, learn in more detail how the HF equations are
                                  
 solved, and learn how one moves beyond HF to come closer and
 closer to the correct (r|R).


                                                                  28

				
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posted:6/18/2012
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