Lecture 4, September 10, 2009

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EEWS-90.502-Goddard-L04 1

Nature of the Chemical Bond

with applications to catalysis, materials science, nanotechnology, surface science,

bioinorganic chemistry, and energy

William A. Goddard, III, [email protected] WCU 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 E11-101 Hours: 0900-1030 Tuesday and Thursday

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

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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

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4

^th

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(ia) = e^ia

eîa HΦ = H (eîa Φ) = E (eîa Φ)

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

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The Hamiltonian for H

₂⁺

For H atom the Hamiltonian is Ĥ = - (Ћ²/2m_e)² – e²/r or

Ĥ = - ½ ² – 1/r (in atomic units)

r

Coordinates of H atom

Coordinates of H₂⁺

For H₂⁺ molecule the Hamiltonian (in atomic units) is Ĥ = - ½ ² + V(r) where

1 1 1

We will rewrite this as

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The Schrödinger Equation for H

₂⁺

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 e is the electronic energy

Then the total energy E becomes

E= e + 1/R R

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

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

E(R)

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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 e^

= ( )² =

Thus The inversion operator is of order 2

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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) = eφ(r) I^

I^

φ_g(r) = + φ_g(r) φ_u(r) = - φ_u(r)

g for gerade or even u for ungerade or odd

φ_g(r) φ_u(r)

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Now consider symmetry for H

₂

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

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inversion symmetry for H

₂

wavefunctions

g

u u

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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 t where tΦ(1,2) = Φ(2,1) Applying t twice leads to the identity t² = e,

t²Φ(1,2) = Φ(1,2)

Thus the previous arguments on inversion apply equally to transposition

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permutational symmetry for H

₂

wavefunctions

symmetric

antisymmetric

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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 5^th 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 a for spin up and b for spin down. Thus the ground states of H atom are φ(xyz)a(spin) and φ(xyz)b(spin)

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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)a, with up-spin, m_s = +1/2 φ(r)b, with down-spin, m_s = -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 field

Similarly for H₂⁺ the ground state becomes φ_g(r)a and φ_g(r)b

While the excited state becomes φ_u(r)a and φ_u(r)b

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Electron spin

+½ or a or up-spin and -½ or b 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

a b

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

where r refers to the vector of 3 spatial coordinates, x,y,z while s to the internal spin coordinates

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

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

DE = -gB_zs_z

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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(s), called an orbital, and a spin function, х(s) = a or b.

ψ(r,s) = φ(s) х(s)

where r refers to the vector of 3 spatial coordinates, x,y,z while s to the internal spin coordinates.

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spinorbitals for two-electron systems

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

Ψ(1,2) = Ψ(r₁,s₁,r₂,s₂) = ψ_a(r₁,s₁) ψ_b(r₂,s₂)

= φ_a(r₁) х_a(s1) φ_b(r₂) х_b(s₂)

=[φ_a(r₁) φ_b(r₂)][х_a(s1) х_b(s₂)]

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

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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) a(1) a(2) Φ(1,2) a(1) b(2) Φ(1,2) b(1) a(2) Φ(1,2) b(1) b(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, s₁ and s₂

Neither symmetric nor antisymmetric Symmetric spin

Symmetric spin

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Spin states for 2 electron systems

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

Φ(1,2) a(1) a(2)

Φ(1,2) [a(1) b(2) + b(1) a(2)]

Φ(1,2) b(1) b(2)

Φ(1,2) [a(1) b(2) - b(1) a(2)]

Adding the spin quantum numbers, m_s, to obtain the total spin projection, M_S = m_s1 + m_s2 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

M_S +1 0 -1 0

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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: r₁ ➔ r₂

Spin only: s₁ ➔ s₂

Space and spin simultaneously: (r₁,s₁) ➔ (r₂,s₂)

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

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Permutational symmetries for H

₂

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 6^th postulate of

QM

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The 6

^th

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

t

_ij

Ψ = - Ψ for all I and j

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Implications of the Pauli Principle

Consider two independent electrons, 1 on the earth described by ψ_e(1)

and 2 on the moon described by ψ_m(2)

Ψ(1,2)=

ψ_e(1) ψ_m(2)

And test whether this satisfies the Pauli Principle

Ψ(2,1)=

^ψ_m^{(1) ψ}_e^{(2) ≠ - ψ}_e^{(1) ψ}_m⁽²⁾

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

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Quick fix to satisfy the Pauli Principle

Combine the product wavefunctions to form a symmetric combination

Ψ

_s

(1,2)=

^ψ_e^{(1) ψ}_m^{(2) + ψ}_m^{(1) ψ}_e⁽²⁾

And an antisymmetric combination

Ψ

_a

(1,2)=

ψ_e(1) ψ_m(2) - ψ_m(1) ψ_e(2) We see that

t₁₂

Ψ

_s

(1,2) = Ψ

_s

(2,1) = Ψ

_s

(1,2) (boson symmetry)

t₁₂

Ψ

_a

(1,2) = Ψ

_a

(2,1) = -Ψ

_a

(1,2) (Fermion symmetry)

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

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Consider some simple cases: identical spinorbitals Ψ(1,2)=

ψ_e(1) ψ_m(2) - ψ_m(1) ψ_e(2)

Identical spinorbitals: assume that ψ_m = ψ_e Then

Ψ(1,2)=

^ψ_e^{(1) ψ}_e^{(2) -} ^ψ_e^{(1) ψ}_e^{(2) = 0}

Thus two electrons cannot be in identical spinorbitals Note that if ψ_m = e^iaψ_e where a is a constant phase factor, we still get zero

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Consider some simple cases: orthogonality

Consider the wavefuntion

Ψ

_old

(1,2)=

^ψ_e^{(1) ψ}_m^{(2) -} ^ψ_m^{(1) ψ}_e⁽²⁾

where the spinorbitals ψ_m and ψ_e are orthogonal hence <ψ_m|ψ_e> = 0

Define a new spinorbital θ_m = ψ_m + l ψ_e (ignore normalization) That is not orthogonal to ψ_e. Then

Ψ

_new

(1,2)=

^ψ_e^{(1) θ}_m^{(2) -} ^θ_m^{(1) ψ}_e^{(2) =}

ψ_e(1) θ_m(2) + l ψ_e(1) ψ_e(2) - θ_m(1) ψ_e(2) - l ψ_e(1) ψ_e(2)

= ψ_e(1) ψ_m(2) - ψ_m(1) ψ_e(2)

=Ψ

_old

(1,2)

Thus the Pauli Principle leads to orthogonality of

spinorbitals for different electrons, <ψ_i|ψ_j> = d_ij = 1 if i=j

=0 if i≠j

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Consider some simple cases: nonuniqueness

Starting with the wavefuntion

Ψ

_old

(1,2)=

^ψ_e^{(1) ψ}_m^{(2) -} ^ψ_m^{(1) ψ}_e⁽²⁾

Consider the new spinorbitals θ_m and θ_e where θ_m = (cosa) ψ_m + (sina) ψ_e

θ_e= (cosa) ψ_e - (sina) ψ_m Note that <θ_i|θ_j> = d_ij Then

Ψ

_new

(1,2)=

^θ_e^{(1) θ}_m^{(2) -} ^θ_m^{(1) θ}_e^{(2) =}

+(cosa)² ψ_e(1)ψ_m(2) +(cosa)(sina) ψ_e(1)ψ_e(2) -(sina)(cosa) ψ_m(1) ψ_m(2) - (sina)² ψ_m(1) ψ_e(2) -(cosa)² ψ_m(1) ψ_e(2) +(cosa)(sina) ψ_m(1) ψ_m(2) -(sina)(cosa) ψ_e(1) ψ_e(2) +(sina)² ψ_e(1) ψ_m(2)

[(cosa)²+(sina)²] [ψ_e(1)ψ_m(2) - ψ_m(1) ψ_e(2)]

=Ψ

_old

(1,2)

Thus linear combinations of the spinorbitals do not change Ψ(1,2)

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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 Principle Thus we will take determinants of our wavefunctions.

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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

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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