William A. Goddard, III, [email protected]

EEWS-90.502-Goddard-L04 1© copyright 2009 William A. Goddard III, all rights reserved

Nature of the Chemical Bond with applications to catalysis, materials

science, nanotechnology, surface science, bioinorganic chemistry, and energy

William A. Goddard, III, [email protected] Professor at EEWS-KAIST and

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

Course number: KAIST EEWS 80.502 Room E11-101Hours: 0900-1030 Tuesday and Thursday

Senior Assistant: Dr. Hyungjun Kim: [email protected] Assistant: Ms. Ga In Lee: [email protected]

Lecture 4, September 10, 2009


Last time: The nodal Theorem

E00 < E10 < E20< E21 and E00 < E01 < E11 < E21

but nodal argument does not indicate the relative energies of E10 and E20 versus E01

+

Φ00 Φ10

Φ01 Φ20+-

+-

+-+

- Φ11+-

++-+

+- -Φ21

The nodal Theorem sometimes orders excited states in 2D, 3D

The ground state of a system is nodeless (more properly, the ground state never changes sign). Useful in reasoning about wavefunctions. Implies that ground state wavefunctions for H2

+ are g not u)

For one dimensional finite systems, we can order all eigenstates by the number of nodes E0 < E1 < E2 .... En < En+1


4th postulate of QM

Consider the exact eigenstate of a system

HΦ = EΦ

and multiply the Schrödinger equation by some CONSTANT phase factor (independent of position and time)

exp(i) = ei

ei HΦ = H (ei Φ) = E (ei Φ)

Thus Φ and (ei Φ) lead to identical properties and we consider them to describe exactly the same state.


The Hamiltonian for H2+

For H atom the Hamiltonian is

Ĥ = - (Ћ2/2me)– e2/r or

Ĥ = - ½ – 1/r (in atomic units)

r

Coordinates of H atom

Coordinates of H2

+

For H2+ molecule the Hamiltonian (in atomic units) is

Ĥ = - ½ + V(r) where 1 1 1

We will rewrite this as


The Schrödinger Equation for H2+

The exact (electronic wavefunction of H2+ is obtained by solving

Here we can ignore the 1/R term (not depend on electron coordinates) to write

where is the electronic energy Then the total energy E becomes

E= + 1/RR

Since v(r) depends on R, the wavefunction φ depends on R.

Thus for each R we solve for φ and and add to 1/R to get the total energy E(R)

E(R)


Inversion Symmetry

The operation of inversion (denoted as ) through the origin of a coordinate system changes the coordinates as

x -x

y -y

z -z

Taking the origin of the coordinate system as the bond midpoint, inversion changes the electronic coordinates as illustrated.

I

The identity or do-nothing operator, is called einheit. e

Applying the inversion twice, leads to the identity

x -x x; y -y y z -z z Nothing changes!

I I I e= ( )2 = Thus The inversion operator is of order 2


Symmetry Theorem

If h(r) is invariant under inversion: h(-r) = h(r)

Then for all eigenfunctions φ(r) of h

h(r)φ(r) = φ(r)

I

I

φg(r) = + φg(r)

φu(r) = - φu(r)

g for gerade or even

u for ungerade or odd

φg(r) φu(r)


Now consider symmetry for H2 molecule

For H2 the Hamiltonian is

1/r12 interaction between 2 electrons

all terms depending only on electron i

For multielectron systems, inversion, , inverts all electron coordinates simultaneously

I

H(1,2)Φ(1,2) = E Φ(1,2) If H(1,2) is invariant under inversion: H(-r1,-r2) = H(r1,r2) Then for all eigenfunctionsΦ(1,2) of H


inversion symmetry for H2 wavefunctions

g

g

u

u


Permutation Symmetry

Transposing the two electrons in H(1,2) must leave the Hamiltonian invariant since the electrons are identical

H(2,1) = h(2) + h(1) + 1/r12 + 1/R = H(1,2)

We denote transposition as where Φ(1,2) = Φ(2,1)

Applying twice leads to the identity e,

Φ(1,2) = Φ(1,2)

Thus the previous arguments on inversion apply equally to transposition


permutational symmetry for H2 wavefunctions

symmetric

antisymmetric


Electron spin

Consider application of a magnetic field

Our hamiltonian has not no terms dependent on the magnetic field.

Hence no effect.

But experimentally there is a huge effect. Namely

The ground state of H atom splits into two states

This leads to the 5th postulate of QM

In addition to the 3 spatial coordinates x,y,z each electron has internal or spin coordinates which leads to a magnetic dipole that is either aligned with the external magnetic field or it is opposite.

We label these as for spin up and for spin down. Thus the ground states of H atom are φ(xyz)(spin) and φ(xyz)(spin)


Spin states for 1 electron systems

Our Hamiltonian does not involve any terms dependent on the spin, so without a magnetic field we have 2 degenerate states for H atom.

φ(r) with up-spin, ms = +1/2

φ(r)with down-spin, ms = -1/2

The electron is said to have a spin anglular momentum of S=1/2 with projections along a polar axis (say the external magnetic moment) of +1/2 (spin up) or -1/2 (down spin). This explains the observed splitting of the H atom into two states in a magnetic fieldSimilarly for H2

+ the ground state becomes φg(r) and φg(r)

While the excited state becomesφu(r) and φu(r)


Electron spin

+½ or or up-spin and -½ or or down-spin

But the only external manifestation is that this spin leads to a magnetic moment that interacts with an external magnetic field to splt into two states, one more stable and the other less stable by an equal amount.

B=0 Increasing B

Now the wavefunction of an atom is written as ψ(r,)

where r refers to the vector of 3 spatial coordinates, x,y,z

while to the internal spin coordinates

So far we have considered the electron as a point particle with mass, me, and charge, -e.

In fact the electron has internal coordinates, that we refer to as spin, with two possible angular momenta.

E = -Bzsz


Spinorbitals

The Hamiltonian does not depend on spin the spatial and spin coordinates are independent. Hence the wavefunction can be written as a product of

a spatial wavefunction, φ(called an orbital, and

a spin function, х( or .

ψ(r,) = φ(х(

where r refers to the vector of 3 spatial coordinates, x,y,z

while to the internal spin coordinates.


spinorbitals for two-electron systems

Thus for a two-electron system with independent electrons, the wavefunction becomes

Ψ(1,2) = Ψ) = ψa() ψb()

= φa(хa(φb(хb(

φa(φb(хa(хb(

Where the last term factors the total wavefunction into space and spin parts


Spin states for 2-electron systems

Since each electron can have up or down spin, any two-electron system, such as H2 molecule will lead to 4 possible spin states each with the same energy

Φ(1,2)

Φ(1,2)

Φ(1,2)

Φ(1,2)

This immediately raises an issue with permutational symmetry

Since the Hamiltonian is invariant under interchange of the spin for electron 1 and the spin for electron 2, the two-electron spin functions must be symmetric or antisymmetric with respect to interchange of the spin coordinates, 1 and 2

Neither symmetric nor antisymmetric

Symmetric spin

Symmetric spin


Spin states for 2 electron systems

Combining the two-electron spin functions to form symmetric and antisymmetric combinations leads to

Φ(1,2)

Φ(1,2) [ +

Φ(1,2)

Φ(1,2) [ -

Adding the spin quantum numbers, ms, to obtain the total spin projection, MS = ms1 + ms2 leads to the numbers above.

The three symmetric spin states are considered to have spin S=1 with components +1.0,-1, which are referred to as a triplet state (since it leads to 3 levels in a magnetic field)

The antisymmetric state is considered to have spin S=0 with just one component, 0. It is called a singlet state.

Antisymmetric spin

Symmetric spin

MS

+1

0

-1

0


Permutational symmetry

Our Hamiltonian for H2,

H(1,2) =h(1) + h(2) + 1/r12 + 1/R

Does not involve spin

This it is invariant under 3 kinds of permutaions

Space only:

Spin only:

Space and spin simultaneously: )

Since doing any of these interchanges twice leads to the identity, we know from previous arguments that Ψ(2,1) = Ψ(1,2) symmetry for transposing spin and space coord

Φ(2,1) = Φ(1,2) symmetry for transposing space coord

Χ(2,1) = Χ(1,2) symmetry for transposing spin coord


Permutational symmetries for H2 and He

H2

He

the only states observed are

those for which the

wavefunction changes sign

upon transposing all coordinates of electron 1 and

2

Leads to the 6th postulate of

QM


The 6th postulate of QM: the Pauli Principle

For every state of an electronic system

H(1,2,…i…j…N)Ψ(1,2,…i…j…N) = EΨ(1,2,…i…j…N)

The electronic wavefunction Ψ(1,2,…i…j…N) changes sign upon transposing the total (space and spin) coordinates of any two electrons

Ψ(1,2,…j…i…N) = - Ψ(1,2,…i…j…N)

We can write this as

ij Ψ = - Ψ for all I and j


Implications of the Pauli Principle

Consider two independent electrons,

1 on the earth described by ψe(1)

and 2 on the moon described by ψm(2)

Ψ(1,2)= ψe(1) ψm(2)

And test whether this satisfies the Pauli Principle

Ψ(2,1)= ψm(1) ψe(2) ≠ - ψe(1) ψm(2)

Thus the Pauli Principle does NOT allow the simple product wavefunction for two independent electrons


Quick fix to satisfy the Pauli Principle

Combine the product wavefunctions to form a symmetric combination

Ψs(1,2)= ψe(1) ψm(2) + ψm(1) ψe(2)

And an antisymmetric combination

Ψa(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)

We see that

12 Ψs(1,2) = Ψs(2,1) = Ψs(1,2) (boson symmetry)

12 Ψa(1,2) = Ψa(2,1) = -Ψa(1,2) (Fermion symmetry)

Thus the Pauli Principle only allows the antisymmetric combination for two independent electrons


Consider some simple cases: identical spinorbitals

Ψ(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)

Identical spinorbitals: assume that ψm = ψe

Then Ψ(1,2)= ψe(1) ψe(2) - ψe(1) ψe(2) = 0

Thus two electrons cannot be in identical spinorbitals

Note that if ψm = eiψe where is a constant phase factor, we still get zero


Consider some simple cases: orthogonality

Consider the wavefuntion

Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)

where the spinorbitals ψm and ψe are orthogonal

hence <ψm|ψe> = 0

Define a new spinorbital θm = ψm + ψe (ignore normalization)

That is not orthogonal to ψe. Then

Ψnew(1,2)= ψe(1) θm(2) - θm(1) ψe(2) =

ψe(1) θm(2) + ψe(1) ψe(2) - θm(1) ψe(2) - ψe(1) ψe(2)

= ψe(1) ψm(2) - ψm(1) ψe(2) =Ψold(1,2) Thus the Pauli Principle leads to orthogonality of spinorbitals for different electrons, <ψi|ψj> = ij = 1 if i=j

=0 if i≠j


Consider some simple cases: nonuniqueness

Starting with the wavefuntion

Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)

Consider the new spinorbitals θm and θe where

θm = (cos) ψm + (sin) ψe

θe = (cos) ψe - (sin) ψm Note that <θi|θj> = ij

Then Ψnew(1,2)= θe(1) θm(2) - θm(1) θe(2) =

+(cos)2 ψe(1)ψm(2) +(cos)(sin) ψe(1)ψe(2)

-(sin)(cos) ψm(1) ψm(2) - (sin)2 ψm(1) ψe(2)

-(cos)2 ψm(1) ψe(2) +(cos)(sin) ψm(1) ψm(2)

-(sin)(cos) ψe(1) ψe(2) +(sin)2 ψe(1) ψm(2)

[(cos)2+(sin)2] [ψe(1)ψm(2) - ψm(1) ψe(2)] =Ψold(1,2) Thus linear combinations of the spinorbitals do not change Ψ(1,2)


Determinants

The determinant of a matrix is defined as

The determinant is zero if any two columns (or rows) are identical

Adding some amount of any one column to any other column leaves the determinant unchanged.

Thus each column can be made orthogonal to all other columns.(and the same for rows)The above properities are just those of the Pauli PrincipleThus we will take determinants of our wavefunctions.


The antisymmetrized wavefunction

Where the antisymmetrizer can be thought of as the determinant operator.

Similarly starting with the 3!=6 product wavefunctions of the form

Now put the spinorbitals into the matrix and take the deteminant

The only combination satisfying the Pauil Principle is


Example:

From the properties of determinants we know that interchanging any two columns (or rows) that is interchanging any two spinorbitals, merely changes the sign of the wavefunction

Interchanging electrons 1 and 3 leads to

Guaranteeing that the Pauli Principle is always satisfied

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William A. Goddard, III, [email protected]