Ch 10. Many-Electron Atoms MS310 Quantum Physical Chemistry - Schrödinger equation cannot be solved...

Ch 10. Many-Electron AtomsCh 10. Many-Electron Atoms

MS310 Quantum Physical Chemistry

- Schrödinger equation cannot be solved analytically - Schrödinger equation cannot be solved analytically when an atom has more than one electron.when an atom has more than one electron.

- Approximate numerical methods can be used to obtain- Approximate numerical methods can be used to obtain the eigenfunctions and eigenvalues of the Schrödinger the eigenfunctions and eigenvalues of the Schrödinger equation for many-electron atoms.equation for many-electron atoms.

- New issues such as indistinguishability of eNew issues such as indistinguishability of e --s, es, e-- spin, spin, interaction between and need to be considered. interaction between and need to be considered.

The Aim of the ChapterThe Aim of the Chapter

→

lμ→

sμ


10.1 He : the smallest many-electron atom10.1 He : the smallest many-electron atomWe can solve analytically the hydrogen atom, but cannot solve analytically the helium atom.Separate the total motion and internal motion, Schrödinger equation is given by

(Laplacian of e2 is similar to e1)

The wavefunction of S.E depends on the all cocordinates of electrons cannot be obtainedHow one can solve S.E? Orbital Approximation on many-electron eigen functions

|r-r||,r||,r|

),()r,r()44

2

4

2

22(

21122211

2121120

2

20

2

10

22

22

22

11

2

rrr

rrEr

e

r

e

r

e

mm ee

ee

)(sinsin

1

sin

1)(

1

11

1122

12

2

122

111

12

1

21

rrr

rrre


)r()...r()r,...r( n11n1 n

Orbital approximation : wavefunction of many-electron atoms is given by the product of one-electron orbitals

φn(rn) : similar to hydrogen atom orbital and associated with one-electron orbital energy εn

when using the orbital approximation : big problem is Potential of He atom is given by

120

2

20

2

10

2

21 44

2

4

2)r,r(

r

e

r

e

r

eV

Orbital : eigenfunction of H atomIt doesn’t think about the last term : electron-electron repulsionWhen dealing with Coulomb potential, electron correlation is neglected.

120

2

4 r

e


What does neglect of the electron correlation in He atom lead to?We know both electrons of He atom are in the 1s orbital→ the wavefunctions similar to 1s orbital wavefunction

0

2

3

0

)(1 a

r

ea

Zeta(ζ) : ‘effective nuclear charge’ : felt by the electronNo electron correlation : electron 1 interact with nucleus and feels averaged charge distribution of electron 2.Spatially averaged charge distribution of electron 2 : φ*(r2)φ(r2)Negative charge in the volume dτ : -eφ*(r2)φ(r2)dτ

Under this approximation* n-electron Schrödinger E → n·one-electron


10.2 Introducing electron spin10.2 Introducing electron spin

In Stern-Gerlach experiment, there are 2 deflected beams.Case of l>0 : nonzero magnetic moment → splittingValence electron of Ag is only 1, 5s → cannot split(l=0)However, splitting is observed! → solved by the ‘spin’, the intrinsic angular momentum sIn this case, it splits into 2s+1 beams : s=1/2z-component element of angular momentum : sz = ms =ℏ ± /2ℏIntrinsic : spin independent to environment of electron(r,θ,φ)

What about the many-electron case? : ‘doubly occupied’ in the orbitals of many-electron atom one electron has ms = +1/2, the other electron has ms = -1/2 → added to 4th quantum number to the H atom eigenfunction, ψnlmlms

(r,θ,φ)


Define the 2 spin wavefunction as α and β, and spin angular momentum operator and .They satisfy the following properties :

2s zs

1,0

2ˆ,

2ˆ

)12

1(

2)1(ˆ,)1

2

1(

2)1(ˆ

****

222

222

dddd

msms

ssssss

szsz

σ : spin variable (not a spatial variable). It is orthogonal to spatial variables.


For example, eigenfunctions of H atom ψnlmlms (r,θ,φ) are

00 2

3

02

1100

2

3

02

1100

1111 a

r

a

r

e)a()r(,e)

a()r(

--

==+

They also satisfy the orthonormal relation.

1)()(),(),(

0)()(),(),(

*100

*100

2

1100

*

2

1100

*100

*100

2

1100

*

2

1100

ddrrddrr

ddrrddrr

Hamiltonian doesn’t depend on the spin coordinate. → these two wavefunctions have the same energy.

10.3 Wave functions must reflect the 10.3 Wave functions must reflect the indistinguishability of electronsindistinguishability of electrons


Discuss the He : electrons are numbered 1 and 2Macroscopic object can be distinguished from one anotherHowever, no way to distinguish between any 2 electrons!How can indistinguishability be introduced by the orbital approximation?

Consider the n-electron wavefunction ψ(1,2,…,n)=ψ(r1,θ1,φ1,σ1,r2,θ2,φ2,σ2,…,rn,θn,φn,σn)

See the case of He.Wavefunction is not observable, but square of wavefunction is proportional to electron density and observable.2 electrons of He are indistinguishable → if we change the electron 1 and 2, there are no change of observable! → ψ2(1,2) = ψ2(2,1)


ψ2(1,2) = ψ2(2,1) means ψ(1,2) = ψ(2,1) or ψ(1,2) = - ψ(2,1)ψ(1,2) = ψ(2,1) : symmetric wave functionψ(1,2) = - ψ(2,1) : antisymmetric wave function

For a ground-state He atom, symmetric and antisymmetric wavefunctions are given by

)r()(where

)()()()()()()()(),(

)()()()()()()()(),(

ssssricantisymmet

sssssymmetric

1

1111

1111

1

1122221121

1122221121

=

=

+=

-

Postulate 6 : Wave functions describing a many-electron system must change sign(be antisymmetric) under the exchange of any two electrons.

W. Pauli showed only the antisymmetric wave function is allowed for electrons.

This postulate is also known as the Pauli exclusion principle.


Meaning of this postulate→ single-product wave function cannot be antisymmetric.

Ex)

How can solve it? → Slater determinant

)1()1()2()2()2()2()1()1( 1111 ssss

)()(...)()()()(

............

)2()2(...)2()2()2()2(

)1()1(...)1()1()1()1(

!

1),...,2,1(

11

11

11

nnnnnn

nn

m

m

m

m=n/2(n:even) or m=(n+1)/2(n:odd)Why determinant can construct antisymmetric wave function? → value of determinant automatically change the sign when 2 rows are interchanged.


Ground state of He : represented by 2x2 determinant

)2()2(1)2()2(1

)1()1(1)1()1(1

2

1)2,1(

ss

ss

Ground state of He : two electrons have the same quantum number n, l, and ml and only the difference is ms (+1/2 for one e- & -1/2 for the other).

Electrons : assigned to orbitals by a configurationConfiguration : specifies the values of n and l for each electron ex : ground state of He : 1s2 F : 1s22s2p5

In a configuration ml and ms are not specified : total energy doesn’t depend on .


Consider the ground state of Li.If third electron is in the 1s orbital, wavefunction is given by

)3()3(1)3()3(1)3()3(1

)2()2(1)2()2(1)2()2(1

)1()1(1)1()1(1)1()1(1

!3

1)3,2,1(

sss

sss

sss

However, third column is same to first column → ψ(1,2,3)=0!Therefore, third electron must in the next level, 2s (L1 : 1s22s1) and real wavefunction is given by

It shows Pauli exclusion principle(at most 2 electron can occupy in the one orbital)

Orbitals with the same n value : shellOrbitals with the same n and l values : subshell

)3()3(2)3()3(1)3()3(1

)2()2(2)2()2(1)2()2(1

)1()1(2)1()1(1)1()1(1

!3

1)3,2,1(

sss

sss

sss

10.4 Using the variational method to solve the 10.4 Using the variational method to solve the Schrödinger equationSchrödinger equation


Many-electron atom : only can approximateOne of the method : ‘Hartree-Fock self-consistent field method’ combined with the variational method

Schrödinger equation for ground state is Ĥψ0=Eψ0

We can rewrite the equation by the integrating :

If we cannot know the eigenfunction, how can we approach to energy(also, eigenvalue)? → variational theorem

d

dHE

0*0

0*0

0

ˆ

0*

* ˆE

d

dHE

Φ : trial wave function(prove : end-of-chapter problem)Set Φ and optimization parameter α, we can find optimum value


Example : particle in a boxReal ground-state wavefunction is

Take the trial function

xaa

x sin

2)(1

))(2

1()()( 9

9

7

7

5

5

3

3

a

x

a

x

a

x

a

x

a

xx

2

2

03

3

3

320

3

3

2

2

3

32

133.0

))((2

)()(2

ma

h

dxax

ax

ax

ax

m

dxax

ax

dxd

ax

ax

mE a

a

i) α=0

Real minimum energy : 0.125h2/ma2

Trial energy is bigger than real energy and there are small difference between real wavefunction and trial function


a

a

dxax

ax

ax

ax

ax

ax

ax

ax

ax

ax

m

dxax

ax

ax

ax

ax

dxd

ax

ax

ax

ax

ax

mE

09

9

7

7

5

5

3

3

9

9

7

7

5

5

3

320

9

9

7

7

5

5

3

3

2

2

9

9

7

7

5

5

3

32

))](21

()))][((21

()[(2

))](21

()[())](21

()[(2

ii) α≠0

Integrate it, we can obtain

)2309451514

2738

1058

(

)109395

40247231

11654

(

2 2

2

2

2

ma

E

Calculate the dE/dα=0 and find extreme value of α → α = -5.74 and α = -0.345, minimum : α = -0.345In this case, energy of trial function is 0.127h2/ma2 and it is very close to real value.


See the graph of the optimized trial function

Contribution of term : blue line

We can see E→E0 and Φ→ψ0

Obtain the ‘best’ energy : depends on the choice of trial function

We can obtain the lower energy if we choose the trial function Φ(x)=xα(a-x)α

This example shows Schrödinger equation can be solved by the variational method.

))(2

1( 9

9

7

7

5

5

a

x

a

x

a

x


10.5 The Hartree-Fock self-consistent field method

Antisymmetric wavefunction is given by the slater determinant

)()(...)()()()(

............

)2()2(...)2()2()2()2(

)1()1(...)1()1()1()1(

!

1),...,2,1(

11

11

11

nnnnnn

nn

m

m

m

It is represented by the modified Hydrogen-like orbitals, φj(k)

Hartree-Fock method : single Slater determinant gives the lowest energy for ground-state, absence of electron correlation

We discussed in 10.1, HF method assumed the electrons are uncorrelated and that a particular electron feels the spatially averaged electron charges distribution of remaining n-1 electrons.


Use HF method : n-electron Schrödinger equation → n 1-electron Schrödinger equations

1-electron Schrödinger equation is given by

niVm iii

effii ,...,2,1),r()r())r(

2( iii

22

HF method allows the best one-electron orbitals and the corresponding orbital energies to be calculated.

Neglect of electron correlation → effective potential is spherically symmetric → angular part of the wave functions is identical to the solutions for the hydrogen atom


To optimize the radial part of the determinantal wave function → Variational method in 10.4

Individual entries in the determinant are expressed as a linear combination of suitable basis functions

The criterion for a “good” basis set → the number of terms in the sum representing φj(r) is as small as possible → basis functions enable the HF calculations to be carried out rapidly


Two examples of basis set expansions for atomic orbitals→ Figures 10.3 & 10.4

2p atomic orbital of Ne obtained in a HF calculation


H 1s AO and the contributions of each member of the m=3 basis set → Figure 10.4


To solve the Schrodinger equation for electron 1, V1

eff(r1) must be known → we must know the functional form of all other orbitals

For an initial set of φj(k) → effective potential is calculated→ energy and improved orbital functions, φ'j(k) , for each of the n electrons are calculated

φ'j(k) → φ''j(k)

Until the solutions for the energies and orbitals are self-consistent

Coupled with the variational method in optimizing the parameters in the orbitals→ effective in giving the best one-electron orbitals and energies available for a many-electron atom in the absence of electron correlation


Accuracy of a HF calculation depends primarily on the size of the basis set

He 1s orbital cannot be accurately represented by a single exponential functions as was the case for the hydrogen atom


HF radial functions→ To obtain the radial probability distribution for many-electron atoms


Important result of HF calculation : εi for many-electron atoms depends on the both n and l. εns < εnp < εnd < … : not same as the H atomSee the radial distribution of 3s, 3p, and 3d.

Probability of finding the electron : 3s>3p>3dTherefore, degree of shielding is different and 3s electrons are more tight binding than 3p and 3d electrons.Similarly, energy of 3p orbital is lower than 3d orbital


Effective nuclear charge ζ is smaller than real nuclear charge.Why? ‘Shielding’Electron doesn’t feel only the charge of nucleus, but also feels ‘average charge distribution’ of another electrons!

In same period small n : a few electrons effects large n : a lot electrons effects → Z-ζ increases as the across the periodic table


How can use the orbital energy εi? ‘ionization energy’

Highest occupied orbital energy εi : first ionization energy(first ionization energy : energy of ground state to free) → Koopman’s theorem

Lowest unoccupied orbital energy εi : electron affinity

However, accuracy of electron affinity is much less than ionization energy.


See the case of Scandium(Sc) ε4s and ε3d : depends on the configuration of Sc(4s23d1, 4s13d2, 4s03d3) because of orbital energy of many-atom electron depends on ‘distribution’ of another electrons.

Known distribution of orbital energy is given by this pictureIt called as the ‘Aufbau principle’


Electron configuration of 4th period elements is given by

Configuration of Cr and Cu : ‘Hund’s rule’

In K and Ca, ε4s < ε3d , but after the Sc, ε4s > ε3d

Calculate the ∆E of 4s23dn → 4s13dn+1 is given by ∆E(4s→3d) ≈ (ε3d - ε4s) + [Erepulsive(3d,3d) – Erepulsive(3d,4s)](by Vanquickenbourne)


d electron : more locally distributed than s electron → Erepulsive(3d,3d) > Erepulsive(3d,4s) > Erepulsive(4s,4s)Therefore, even though ε4s > ε3d, magnitude of repulsion is greater than ε4s - ε3d → promotion 4s23d1 → 4s13d2 cannot be occurred.

HF calculation : neglecting the electron correlationTherefore, total energy of HF > true energyDifference : correlation energyIt increases roughly linearly with the number of electron pairs in the atom. However, it is small amount of total energy.For example, correlation energy of He : 110 kJ mol-1, 1.4% of total energy

However the correlation error exist in both reactant and product, therefore this term cancelled out in thermodynamic function

10.6 Understanding trends in the periodic table10.6 Understanding trends in the periodic table from HF calculationsfrom HF calculations


Result of HF calculation is 1) orbital energy depends on n and l, and εns < εnp < εnd < … 2) electrons in a many-atom are shielded from the full nuclear charge by other electrons. 3) ground-state configuration results from a balance between orbital energy and electron-electron repulsion

HF method : numerical calculation → no simple formula for ζ by the radius r → Ex) VDW radii VDW radii : radius of include the 98% of electron charge

Reaction : related to HOMO and LUMOIt predict the binding of NaCl i) Na+Cl- : ∆E = Eionization

Na - Eelectron affinityCl = 5.14eV - 3.61eV = 1.53eV

ii) Na-Cl+ : ∆E = Eelectron affinityNa - Eionization

Cl = 12.97eV – 0.55eV = 12.42eV → Na+Cl- is more favorable than Na-Cl+


Atomic radius decrease continuously in going across the period.It increases abruptly as n increases.

Electronegativity same pattern as the IE.


- Slater determinant can represent the orbital-type wavefunction satisfying the Pauli exclusion principle.

- Result of HF-SCF method for helium atom is approximately same.

- Study about the electron configuration, good quantum number.

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Ch 10. Many-Electron Atoms MS310 Quantum Physical Chemistry - Schrödinger equation cannot be solved...

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