(1)

Pauli-principle Hund’s rule 1 slide + 3 slidesAntisymmetric functions for n-particles - Slater Determinants 3 slidesFilling Shells Figure 2 slidesIonization energies - Figure 2 slidesHartree - Selfconsistent field - Iterations 8 slides

only Iterations part 4 slidesSelfconsistent field Result - Figures 3 slidesConfigurations - Ionization potentials of atoms - Tables 3 tablesScreened potential and Centrifugal barrier - Explains Tables 2 slidesEnergy for N-particles - Using Slater Determinants 5 slides

Helium example 1 slideLithium example Counting nonzero terms 2 slides

N particles - Energy Summary 1 slideSchrodinger equation from variational method 1 slideVariational method - deriving Hartree-Fock Equations 2 slidesHartree-Fock Equations 1 slideTotal energy and the selfconsistent orbital energies 3 slidesEvaluation of electron repulsion Configuration mixing 2 slides + 2slides

2

1 Filling up the shells with electrons

In order to know and to, more or less, understand in which state an electronis bound we can use some basic rules. The generel idea is that the lowestenergy-state is the most stable one. The exited states can in most cases fall inthe lower state or ground state by emission of light.The principles one need to know how to fill up the shells are the Pauli-principleand the Hund’s rule.

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3

2 Pauli-principle

The filling of the shells observed was explained by Pauli by formulating thePauli exclusion principle:

Two electrons can not be in the same state - defined here by possible quantumnumbers n, l, m and ms

n, the principal quantum number of the shell,

l, the angular momentum quantum number of the subshell,

m, the magnetic quantum number

ms denotes spin ”up” and spin ”down”, ms = 1/2 or ms = −1/2

The Pauli principle mathematical: require the multiparticle wavefunc-tion antisymmetric with respect to exchange of two particles.

If two particles coordinates are exchanged, the function must change sign.

If two particles are in the same state (function),

this requirement leads to zero wavefunction, thus impossible.

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4

Figure 1: Atomic levels (shells)List of topics

5

Figure 2: Atomic levels (shells) filled up to the element ....List of topics

6

3 Antisymmetric product function for n-particles

Ψ(x1, x2, ...xn) =∑

perm(α,β,...ν)

(−1)P (perm(α,β,...ν))perm (φαφβ....φν) (x1)(x2)....(xn)

where each term in the sum looks as φβ(x1)...φν(x2)...φα..., summing over allpermutations, and P (perm(α, β, ...ν)) is the number of swaps of the givenpermutation perm(α, β, ...ν)

This is very close to the definition of the determinant

det(A) =∑

σ∈Sn

(sgn(σ)

n∏

i=1

Ai,σ(i)

)

The above in this notation

Ψ(x1, x2, ...xn) =∑

σ∈Sn

sgn(σ)n∏

i=1

φασ(i)(xi)

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7

Slater determinant

The antisymmetric combination for n-particles can be written as a determinantin this way: ∣∣∣∣∣∣∣∣

φα(x1) φα(x2) ... φα(xn)φβ(x1) φβ(x2) ... φβ(xn)... ... ... ...

φν(x1) φν(x2) ... φν(xn)

∣∣∣∣∣∣∣∣

For 3 particles:

3 particle Slater determinant

∣∣∣∣∣∣

φα(x1) φα(x2) φα(x3)φβ(x1) φβ(x2) φβ(x3)φγ(x1) φγ(x2) φγ(x3)

∣∣∣∣∣∣List of topics

8

3 × 3 determinant

∣∣∣∣∣∣∣

φα(1) φα(2) φα(3)φβ(1) φβ(2) φβ(3)φγ(1) φγ(2) φγ(3)

∣∣∣∣∣∣∣

φα(1) φα(2) φα(3)

φβ(1) φβ(2) φβ(3)

φγ(1) φγ(2) φγ(3)

φα(1) φα(2) φα(3)

φβ(1) φβ(2) φβ(3)

φα(1) φβ(2) φγ(3) + φβ(1) φγ(2) φα(3) + φγ(1) φα(2) φβ(3)

− φγ(1) φβ(2) φα(3) − φα(1) φγ(2) φβ(3) − φβ(1) φα(2) φγ(3)

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9

4 Hund’s rule

Hund’s rule is the manifestation of the same effect as we have seen in theparahelium - orthohelium effect ( the ”second” rule )

Hund’s first rule

Full shells and subshells do have a total circular momentum of zero.

This can be calculated and is allways valid.

Hund’s second rule

The total spin S should allways have the highest possible value. So as manyof the single electron spins as possible should be parallel.

The second rule appears more empirical and applies to a different magneticquantum number of the electrons with parallel spins. If is of course not allowedto break the Pauli-principle.

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10

5 Example for the Hund’s rule

Figure 3: Electron spins of the first elements.

The elements carbon C, nitrogen N and oxygen O are those where the Hund’srule have the biggest influence. We see electrons with parallel spins in thestates m = −1, 0,+1, depending on the element. The magnetic quantumnumber m gives more or less the “direction” of the circular moment l. Thes-states are the states where l = 0 and the p-states those where l = 1. Thenames for l = (2, 3, 4, ...) are (d, e, f, ...).The shell-name of the shell n = (1, 2, 3, 4, ...) is (K,L,M,N, ...).

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11

In cases with more electrons do we get some exceptions. For example is the4s subshell earlier filled than the 3d. This is caused by the smaller distanceof the 4s to the core and thus by the lower energy, which is more stable. Thementioned rules work well for general considerations and for atoms with notto many electrons.

6 Number of states

Usefull to know is the largest possible number of states for a given n whichmeans until a special shell is filled.

Nmax = 2 ·n−1∑

l=0

(2l + 1) = 2n2. (2)

We get this formula by adding all the possible quantum number configurations:We get for each n every l in the range (0, 1, ..., n− 1),for each l every m in the range (−l, ...,−1, 0, 1, ..., l)and for each of these states 2 spin possibilities.

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12

7 Ionization energies

Figure 4: Ionization energies. The shell properties would explain the structurein general - Periodic table and the Selfconsistent field List of topics

13

Especially at the first elements, we see the minimums at the elements withjust one electron in the last shell and a maximum at the noble gas elements.The weak maxima are caused by filled subshells or by the maximum of parallelstanding spins. After argon, it is more difficult to observe general tendences.

The shell properties would explain the structure in general

But there should be no closed shell at argon

Details - Periodic table and the Selfconsistent field

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14

8 Hartree - Selfconsistent field

Interaction energy of two charges depends on their distance|~r − ~x|:

W (|~r − ~x|) =q1q2|~r − ~x|

The two charges are an electron and a little volume dV at ~x containing charge cloudof density ρ

q1 → (−e) q2 → ρ(~x)dV → ρ(~x)d3x

The interaction energy of these twocharges is

dW (|~r1 − ~r2|) =(−e)ρ(~r2)

|~r − ~x|d3r2

dV =

x − r

( )xρ

x

r

d x3

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15

Interaction with a cloud; summing over all the small volume elements - it meansintegrating over the whole volume of the cloud gives the potential energy

W (~r) =

∫(−e)ρ(~x)

|~r − ~x|d3x

dV =

x − r

( )xρ

x

r

d x3

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16

If the charge cloud represents one elec-tron in state ψi(~x)

ρ(~x) = (−e)|ψi(~x)|2

If we have N electrons, each in its state,the total density becomes

ρ(~x) = (−e)N∑

i=1

|ψi(~x)|2

and again integrating gives the poten-tial energy due to the interaction witha (probability based density) cloud ofelectrons

W (~r) =

∫ (−e)2N∑i=1|ψi(~x)|2

|~r − ~x|d3x

dV =

x − r

( )xρ

x

r

d x3

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17

Now solving the Schrodinger equation with W (~r),

(T + V +W )ψi(~x) = Eiψi(~x)

We first need to know the W (~r), but that depends on all the other N solutions

W (~r) =

∫ (−e)2N∑i=1|ψi(~x)|2

|~r − ~x|d3x

Approximation chain: First we choose some simple approximation, e.g. the hydrogen-like states, or we might know the states foranother atom. We call it

ψ(0)i (~x)

From the set of all N ψ(0)i we construct

W (1)(~r) =

∫ e2N∑i=1|ψ(0)i (~x)|2

|~r − ~x|d3x

In atomic units the whole Schrodinger equation is(−1

2∇2 − Z

r+W (1)(~r)

)ψ(1)i (~x) = E

(1)i ψ

(1)i (~x)

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18

1. step - choose arbitrary set of ψ(0)i

ψ(0)i (~x) → W (1)(~r) =

∫N∑i=1|ψ(0)i (~x)|2

|~r − ~x|d3x (3)

(−1

2∇2 − Z

r+W (1)(~r)

)ψ(1)i (~x) = E

(1)i ψ

(1)i (~x)

2. step: take the set of ψ(1)i from the 1. step

ψ(1)i (~x) → W (2)(~r) =

∫N∑i=1|ψ(1)i (~x)|2

|~r − ~x|d3x

(−1

2∇2 − Z

r+W (2)(~r)

)ψ(2)i (~x) = E

(2)i ψ

(2)i (~x)

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19


ψ(2)i (~x) → W (3)(~r) =

∫N∑i=1|ψ(2)i (~x)|2

|~r − ~x|d3x

(−1

2∇2 − Z

r+W (3)(~r)

)ψ(3)i (~x) = E

(3)i ψ

(3)i (~x)


ψ(3)i (~x) → W (4)(~r) =

∫N∑i=1|ψ(3)i (~x)|2

|~r − ~x|d3x

(−1

2∇2 − Z

r+W (4)(~r)

)ψ(4)i (~x) = E

(4)i ψ

(4)i (~x)

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20


ψ(4)i (~x) → W (5)(~r) =

∫N∑i=1|ψ(4)i (~x)|2

|~r − ~x|d3x

(−1

2∇2 − Z

r+W (5)(~r)

)ψ(5)i (~x) = E

(5)i ψ

(5)i (~x)


ψ(5)i (~x) → W (6)(~r) =

∫N∑i=1|ψ(5)i (~x)|2

|~r − ~x|d3x

(−1

2∇2 − Z

r+W (6)(~r)

)ψ(6)i (~x) = E

(6)i ψ

(6)i (~x)

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21

Iteration chain

ψ(n)i (~x) → W (n+1)(~r) =

∫N∑i=1|ψ(n)i (~x)|2

|~r − ~x|d3x → ψ

(n+1)i (~x)

This chain can continue, until the set of ψ(n)i produces a potential W (n+1) which

is the same as W (n), which was the one to determine ψ(n)i . The potentials and

functions become consistent, hence the name Selfconsistent field.

Criterium for self-consistency: the (n+1)-th solution does not differ from th n-thsolution ∫ N∑

i=1

∣∣∣ |ψ(n+1)i (~x)|2 − |ψ(n)

i (~x)|2∣∣∣ d3x < ε (4)

where ε ∝ 10−8 as a typical value.

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22

9 Ionization potentials of atoms

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O O O ) O ) )

O O O ) *%+,: O ) )

O O O O ) )

O O O *-+,: O ) )

O O O *-+,: O )

O O O *-+,: *-.,: O )

O O O *-+,: *-.,4 O )

O O O *-+,: *-.,< O )

O O O *-+,: *-.,= O )

O O O *-+,: *-.,6 O )

O O O O ) )

O O O */+,: O ) )

O O O */+,: O )

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O O O */+,: */.,4 O )

O O O */+,: */.,< O )

O O O */+,: */.,= O )

O O O */+,: */.,6 O )

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23

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Q Q Q *+0-= */,-= Q )

Q Q Q *+0-: */,-= Q )

Q Q Q *+0-( Q )

Q Q Q *+0-( */,-= Q )

Q Q Q *+0-+ */,-= Q )

Q Q Q *+0-* */,-= Q )

Q Q Q *+0-" */,-= Q )

Q Q Q *+0-!> Q )

Q Q Q *+0-!> */,-= Q )

Q Q Q *+0-!> */,-= Q

Q Q Q *+0-!> */,-= */.-= Q

Q Q Q *+0-!> */,-= */.-: Q

Q Q Q *+0-!> */,-= */.-9 Q

Q Q Q *+0-!> */,-= */.-( Q

Q Q Q *+0-!> */,-= */.-+ Q

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24

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W W W *+,-*? *./-C *.0-" W

W W W W ) )

W W W *1/-C W ) )

W W W *1/-C W )

W W W *.,-C *1/-C W )

W W W *.,-+ W )

W W W *.,-9 W )

W W W *.,-9 *1/-C W )

W W W *.,-/ W )

W W W *.,-7 W )

W W W *.,-*? W ) )

W W W *.,-*? W )

W W W *.,-*? *1/-C W )

W W W *.,-*? *1/-C W

W W W *.,-*? *1/-C *10-C W

W W W *.,-*? *1/-C *10-! W

W W W *.,-*? *1/-C *10-+ W

W W W *.,-*? *1/-C *10-9 W

W W W *.,-*? *1/-C *10-" W

W W W W ) )

W W W *2/-C W ) )

W W W *2/-C W )

W W W *2/-C W

W W W *.3-! *2/-C W )

W W W *.3-+ *2/-C W )

W W W *.3-9 *2/-C W )

W W W *.3-" *2/-C W )

W W W *.3-/ *2/-C W )

W W W *.3-/ *2/-C W

W W W *.3-= *2/-C W )

W W W *.3-*? *2/-C W )

W W W *.3-** *2/-C W )

W W W *.3-*C *2/-C W )

W W W *.3-*! *2/-C W )

W W W *.3-*+ *2/-C W )

W W W *.3-*+ *2/-C W

W W W *.3-*+ *1,-C *2/-C W

W W W *.3-*+ *1,-! *2/-C W

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25

Screened potential and Centrifugal barrier

0 1 2 3 4 5 6 7 8−0.5

0

0.5

L=0

L=1

L=2Coulomb potential

0 1 2 3 4 5 6 7 8−0.5

0

0.5

L=0

L=1L=2

Hartree potential

List of topics

(−Zr

)+L(L+ 1)

r2vs.

(−Z − 1

re−αr − 1

r

)+L(L+ 1)

r2(5)

26

0 1 2 3 4 5 6 7 8−0.5

0

0.5

L=0

L=1

L=2Coulomb potential

0 1 2 3 4 5 6 7 8−0.5

0

0.5

L=0

L=1L=2

Hartree potential

27

Figure 5: Selfconsistent field calculations compared to coulomb - real results

Here we note the position of the turning point - same for both Coulomb andSCF case. This is true only for the 1s states. List of topics

28

Figure 6: Hartree Atomic potential

This figure shows results for oxygen with the potential and energies obtainedfrom a calculation with SCF program

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29

results for oxygen with the potential and energies obtained from a calculationwith SCF program compared with hydrogen-like levels List of topics

30

10 Expectation value of energy for N-particles

Hamiltonian for N-particles - the repulsion - we sum over pairs of coordinates

H (r1, r2, ...rN) =

N∑

i=1

(−1

2∇2ri− Ze2

ri

)+

∑

(i,j)pairs

e2

|~ri − ~rj|

Without antisymmetrization - independent particles - product function -only 1 term -each particle has its orbital ψa,ψb,ψc.....ψn

Φ (r1, r2, ...rN)→ ψa(r1)ψb(r2)ψc(r3)......ψn(rN)

With antisymmetrization - the Slater determinant

ΦHFa,b,...n (r1, r2, ...rN)→ 1√

N !

∣∣∣∣∣∣∣∣

ψa(r1)ψb(r1)......ψn(r1)ψa(r2)ψb(r2)......ψn(r2).......ψa(rN)ψb(rN)......ψn(rN)

∣∣∣∣∣∣∣∣

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31

For N-particles - the repulsion - we sum over pairs of coordinates

N∑

i=1

(−1

2∇2ri− Ze2

ri

)+

∑

(i,j)pairs

e2

|~ri − ~rj|

Φ (r1, r2, ...rN) = EΦ (r1, r2, ...rN)

Expectation value with the independent particles - product function - only1 term

Φ (r1, r2, ...rN)→ ψa(r1)ψb(r2)ψc(r3)......ψn(rN)

becomes

E = 〈Φ|H |Φ〉 →n∑

α=a

〈ψα|T −Ze2

r|ψα〉+

∑

(α,β)pairs

〈ψβψα|e2

|~r − ~r ′ ||ψβ ψα 〉

The sum over coordinate pairs becomes sum over pairs of orbitalsThis remains true for Slater determinants - but it must be shown.There are N ! terms in the Slater determinant, the left and right give(N !)2 terms, and there are N(N − 1)/2 coordinate pairs.Thus (N !)2N(N − 1)/2 terms.Due to the normalization, this becomes only N(N − 1) terms,N(N − 1)/2 direct terms and N(N − 1)/2 exchange terms List of topics

32

It should be shown that when the Slater determinant is used

ΦHFa,b,...n (r1, r2, ...rN)→ 1√

N !

∣∣∣∣∣∣∣∣

ψa(r1)ψb(r1)......ψn(r1)ψa(r2)ψb(r2)......ψn(r2).......ψa(rN)ψb(rN)......ψn(rN)

∣∣∣∣∣∣∣∣the expectation value will give (final result shown here first)

⟨ΦHF

∣∣H∣∣ΦHF

⟩=

n∑

j=α

〈ψα|T −Ze2

r|ψα〉

+∑

(α,β)pairs

〈ψβψα|e2

|~r − ~r ′||ψβ ψα 〉

−∑

(α,β)pairs

〈ψβψα|e2

|~r − ~r ′||ψα ψβ 〉 (6)

The last term - exchange energy. On the previous page product function (1term) is shown - no exchange energy

Next 3 slides: 1. Helium, 2. Lithium examples, 3. Counting nonzero terms

List of topics

33

For Helium

2

r1r

2

r12

+Ze

r

−e−e

1

[− h2

2me

∇ 2r1− Z e2

r1− h2

2me

∇ 2r2− Z e2

r2+

e2

|r1 − r2|

]Ψ (r1, r2) = E Ψ (r1, r2)

ΦHFa,b (r1, r2) =

1√2

∣∣∣∣ψa(r1) ψb(r1)ψa(r2) ψb(r2)

∣∣∣∣ =1√2

[ψa(r1)ψb(r2)− ψb(r1)ψa(r2)]

and the energy becomes

⟨ΦHF

∣∣H∣∣ΦHF

⟩= 〈ψa|T −

Ze2

r|ψa〉+ 〈ψb|T −

Ze2

r|ψb〉

+ 〈ψaψb|e2

|~r − ~r′ ||ψa ψb 〉 − 〈ψaψb|

e2

|~r − ~r′ ||ψb ψa 〉 (7)

The last term - exchange energy - and there is only one pair.List of topics Lithium example

34

r1r

2

r12

rr3

23r

r13

−e3

−e1

−e2

+Ze

r

Lithium 6× 6× 3-pair terms

[− h2

2me∇ 2r1 −

Z e2

r1− h2

2me∇ 2r2 −

Z e2

r2− h2

2me∇ 2r3 −

Z e2

r3

+e2

|r1 − r2|+

e2

|r1 − r3|+

e2

|r2 − r3|

]Ψ (r1, r2, r3) = E Ψ (r1, r2, r3)

(8)

1√3! ψα(1) ψβ(2) ψγ(3) + ψβ(1) ψγ(2) ψα(3) + ψγ(1) ψα(2) ψβ(3)

− ψγ(1) ψβ(2) ψα(3) − ψα(1) ψγ(2) ψβ(3) − ψβ(1) ψα(2) ψγ(3)

List of topics Helium example Next slide: Counting nonzero terms

35

+c a b

+b c a

−−b a c

−−c b a

−a c b

+a b c

+a b c

+c a b

+a b c

r2 3

e2

r2 3

e2

r2 3

e2

−b a c

+a b c r2 3

e2−

+a b c r2 3

e2−

+a b c r2 3

e2−

−−c b a

−a c b

+a b c

r2 3

e2

r2 3

e2

r2 3

e2

r2 3

e2

r2 3

e2

r2 3

e2

b c

b c

b c

b c

b c

b c

+b c a

+a b c

+ + −

+

+

1

3!

b c

a b

c a

a c

b a

c b

a

a

a

a

a

c

b

b

c

a

a a

Evaluation of the first six terms of the total 36 terms. For each further groupof 6 a very similar procedure would follow. From each six, two remain, totaling12 nonzero terms List of topics

36

11 N particles in Slater determinant - Total

Energy Summary

〈Φ| H |Φ〉 =N∑

j=1

〈ψj|T −Ze2

r|ψj〉+

∑

(i,j)pairs

〈ψjψi|e2

|~r − ~r′||ψj ψi 〉

−∑

(i,j)pairs

〈ψjψi|e2

|~r − ~r′ ||ψi ψj 〉 (9)

We will use also a more compact notation, instead of orbitals ψi we use orbitals|α〉, i.e. the index becomes the name of the function

〈Φ| H |Φ〉 =∑

α

〈α|(T + V )|α〉+∑

pairs αβ

[〈αβ|Vee|αβ〉 − 〈αβ|Vee|βα〉] (10)

List of topics

37

Schrodinger equation from variational principle: minimize

〈Φ|H |Φ〉〈Φ | Φ〉

(≥ Eg.s.)

which can be written as minimalization with constraint 〈Φ | Φ〉 = 1 ; minimize

〈Φ|H |Φ〉 − λ 〈Φ | Φ〉〈Φ | Φ〉 − 1 = 0 λ Lagrange multiplier

Derivative - differential

f(x)→ df(x)

dx→ df =

df(x)

dxdx

Functional derivative - functional F(f); variation of the functions f(x), δf

F(f) → δF(f)

δf→ δF =

δF(f)

δfδf (11)

Variation of 〈Φ|H |Φ〉 is taken (considering the complex nature of |Φ〉 ) as

δ (〈Φ|H |Φ〉) = 〈δΦ|H |Φ〉

With arbitrary variations |δΦ〉

〈δΦ|H |Φ〉 − λ 〈δΦ | Φ〉 −→ H |Φ〉 − λ |Φ〉 = 0 −→ H |Φ〉 = λ |Φ〉

Thus the Lagrange multiplier plays the role of energy eigenvalue List of topics

38

Deriving Hartree-Fock: For N electrons this energy must be minimized

〈Φ| H |Φ〉 =∑

α

〈α|(T + V )|α〉+∑

pairs αβ


with respect to the N orbitals |α〉, |β〉, etc. ,

with N conditions 〈α|α〉 = 1, 〈β|β〉 = 1, i.e.

with N Lagrange multipliers εα, εβ .......

〈δΦ(α→ δα)| H |Φ〉 − εα 〈δα | α〉 = 0

Which results in N equations (one for each of the N orbitals)

〈δα|(T + V )|α〉+∑

β

[〈(δα)β|Vee|αβ〉 − 〈(δα)β|Vee|βα〉]− εα 〈δα | α〉 = 0 (13)

Leaving for a while out the exchange-related term, we can as in prev. slide reducethis by removing 〈δα|

(T + V )|α〉+∑

β

[〈β|Vee|β〉] |α〉 − εα | α〉 = 0

which is exactly the Hartree method. The exchange term leads to complications →List of topics

39

Reducing the full equation

〈δα|(T + V )|α〉+∑

β

[〈(δα)β|Vee|αβ〉 − 〈(δα)β|Vee|βα〉]− εα 〈δα | α〉 = 0 (14)

we get the Hartree Fock Equations (N orbitals)

(T + V )|α〉+∑

β

[〈β|Vee|β〉] |α〉 −∑

β

[〈β|Vee|α〉] |β〉 − εα | α〉 = 0

The complexity of the exchange term becomes clear when we write it explicitelywith coordinates and integrations.Hartree Fock Equations: The direct term and the exchange term:

WHF = Wd − Wex

WHFψa(r) =

[occ∑

b

∫ψ∗b (x)

e2

|r− x|ψb(x)d3x

]ψa(r)−

[occ∑

b

ψ∗b (r)

∫ψb(x)

e2

|r− x|ψa(x)d3x

]

where the exchange potential is nonlocal

Wdψa →Wd(r)ψa(r) Wexψa(r)→∫Wex(r,x)ψa(x)d3x

List of topics

40

Hartree-Fock total energy and sum of orbital energies

with a shorthand notation

ϕα(r) → |ϕα〉 → |α〉

Slater determinantΦα,β,....,ν → |α, β, ...., ν〉

We have started our work with the Hartree-Fock equations by evaluating

〈Φα,β,....,ν |

[T + V +

∑

pairs ij

e2

|ri − rj|

]|Φα,β,....,ν〉 (15)

This we have evaluated as

〈Φ| H |Φ〉 =∑

α

〈α|(T + V )|α〉+∑

pairs αβ


From this expression we have obtained Hartree-Fock equations by the varia-tional procedure.

List of topics

41

The Hartree-Fock equation can be then written as[T + V +

∑

β

〈β| e2

|r− r′||β〉

]|α〉−

∑

β

[〈β| e2

|r− r′||α〉]|β〉 = εα |α〉 (17)

We form the matrix element with the given 〈α|

〈α|

[T + V +

∑

β

〈β| e2

|r− r′||β〉

]|α〉 − 〈α|

[∑

β

〈β| e2

|r− r′||α〉

]|β〉 = εα

(18)since 〈α|α〉 = 1. This can be rewritten as

εα = 〈α|(T + V )|α〉+∑

β 6=α

[〈βα|Vee|βα〉 − 〈βα|Vee|αβ〉] (19)

And now we can explore what is the sum of all εα∑

α

εα =∑

α

〈α|(T + V )|α〉+∑

α

∑

β 6=α

[〈βα|Vee|βα〉 − 〈βα|Vee|αβ〉] (20)

which we can compare with the above 〈Φ| H |Φ〉

〈Φ| H |Φ〉 =∑

α

〈α|(T + V )|α〉+∑

pairs αβ


42

The two expressions are very similar, but they differ in fact by all the interac-tion term, since it is counted twice in the sum:

∑

α

∑

β 6=α

Fαβ = 2∑

pairs αβ

Fαβ

for any set of objects that are symmetric Fαβ = Fβα

Our objects are symmetric, because they are in fact of the type Hαβ,αβ. Thewavefunctions are antisymmetric, |αβ〉 = −|βα〉.Thus, surprisingly perhaps

〈Φ| H |Φ〉 6=∑

α

εα

but as we derived here List of topics

〈Φ| H |Φ〉 =∑

α

εα −∑

pairs αβ


43

Evaluation of the repulsion term using the multipole expansion

1

|r1 − r2|=∑

LM

4π

2L+ 1

rL<rL+1>

Y ?LM (r1)YLM (r2) (23)

wherer< = r1, r> = r2 for |r1| < |r2|

r< = r2, r> = r1 for |r1| > |r2|

Evaluation of the matrix element in general case∫d3r1

∫d3r2ψ

?n1l1m1

(r1)ψ?n2l2m2

(r2)1

|r1 − r2|ψn1l1m1 (r1)ψn2l2m2 (r2) (24)

is performed separately over the radial and angular parts∫r21dr1

∫dr1

∫r22dr2

∫dr2 R?

n1l1(r1)Y

?l1m1

(r1)R?n2l2

(r2)Y?l2m2

(r2)

1

|r1 − r2|Rn1l1(r1)Yl1m1 (r1)Rn2l2(r2)Yl2m2 (r2)(25)

where dri means the integration over dΩi = sin θidθidϕi. List of topics

44

The evaluation of general case - angular integrals of three Ylm’s

CL =

∫Y ?limi

(θ, ϕ)YLM(θ, ϕ)Ylimi(θ, ϕ)dΩ (26)

For the case of both s-states, li = 0 mi = 0 only L = 0 M = 0 are non-zero; The sum reduces to one term. The angular factors give value one, sincethe (YL=0M=0)

2 = (4π)−1 cancels the corresponding factor in the multipoleexpansion and due to the normalization.

List of topics

45

12 Configuration mixing

Consider the usual:Hx(x)ϕα(x) = Eαϕα(x)

Hy(y)χβ(y) = Eβχβ(y)

For any Φ(x)

Φ(x) =∑

cαϕα(x)

For any Ξ(x)

Ξ(y) =∑

dβχβ(y)

Take now a general Ψ(x, y). First look at y as a parameter, Ψ(x, y0)

Ψ(x, y0)→ Φ(x) =∑

cα(y0)ϕα(x)

for every y0 ; Thus we get a new function of y;

cα(y) =∑

dβ(α)χβ(y)

Inserting back:

Ψ(x, y) =∑

dβ(α)χβ(y)ϕα(x)

46

Or, with a simpler notation

Ψ(x, y) =∑

dβαχβ(y)ϕα(x)

In the case of Helium, for example, the H(x) and H(y) are identical

and so are the χβ(y) and ϕα(x). This becomes configuration mixing.

Ψ(x, y) =∑

dβαϕβ(y)ϕα(x)

The coefficients are found by diagonalization.

For three coordinate sets - e.g. for Lithium :

Ψ(x, y, z) =∑

Dγβαϕ(z)ϕβ(y)ϕα(x)

List of topics

47

END OF THE PRESENTATION

List of topics

48

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