chem%356:%introductory%quantum%mechanicsscienide2.uwaterloo.ca/~nooijen/chem356/chem+356+pdf/ch_8.pdf ·...

Winter 2013 Chem 356: Introductory Quantum Mechanics

Chapter 8 – Approximation Methods, Hueckel Theory 108

Chapter 8 – Approximation Methods, Hueckel Theory ........................................................................... 108

Approximation Methods ...................................................................................................................... 108

The Linear Variational Principle ........................................................................................................... 111

Chapter 8 – Approximation Methods, Hueckel Theory

Approximation Methods

A) The variational principle

For any normalized wave function ψ , the expectation value of 0ˆ ˆ,H H E≥ , the exact

groundstate energy.

Proof: ψ = cnφn

n∑ with Hφn = Enφn

If we would measure the energy we would find nE with probability 2n nP c=

H = ψ *(τ )Hψ (τ )dτ ≡ PnEnn∑∫

0 0n n nn nP E E P E≥ =∑ ∑ ( 0nE E≥ )

This argument is a bit shaky when H has degenerate eigenvalues. You will do a correct proof in the assignments. If ψ would not be normalized we can calculate

2 * ( ) ( )N dψ τ ψ τ τ= ∫

and then 02

1 H EN

ψ ψ ≥

or in the final form: The variational principle

!E0 =ψ *(τ )Hψ (τ )dτ

Domain∫ψ *(τ )ψ (τ )dτ

Domain∫≥ E0

where 0E is the exact ground state energy

The variational energy !E0 is exact when 0( ) ( )ψ τ φ τ= , the exact ground state wavefunction.



Examples: use a trial wavefunction that depends on one or more parameters ,α β … Then minimize the

trial energy E α ,β( ) Some simple (trivial) examples:

Hh.o. = !ω − 1

2d 2

dx2 +12

x2⎛⎝⎜

⎞⎠⎟

Take trial wavefunction of the type

2 /2( ) xx e αψ −=

Or 2 /2xNe α−

Q: what is α ? What is 0E ?

A: The exact wavefunction has the form 2 /2xNe α− ,

E0 =

12!ω

→by minimizing the energy we should get 1α = , !E0 =

12"ω

Q: Take the Hamiltonian for the Hydrogen atom, and an 0l = state

H = !2

2µr 2

ddr

r 2 ddr

⎛⎝⎜

⎞⎠⎟− e2

4πε0r

Take the trial wavefunction re α−

-‐ What are the integrals to evaluate? -‐ What is the optimal value for α ?

-‐ What is the value for !E0 ?

A:

!E0(α ) =4π r 2e−αr He−αr dr

0

∞

∫4π r 2e−αre−αr dr

0

∞

∫

Minimizing α :

∂ !E0(α )∂α

= 0

0

1opt a

α→ = !E0 = E0 = − 1

2e2

4πε0a0

again: exact a0 =

4πε0!2

µe2

You can do those problems yourself and see if you get the correct answer

Nontrivial example:

Take the Hamiltonian for H-‐atom, s-‐orbital, and use the trial wavefunction 2re α−



→

!E0(α ) =4π r 2e−αr2

He−αr2

dr0

∞

∫4π r 2e−αr2

e−αr2

dr0

∞

∫ ...=

= 3!2α

2me

− 2e2α12

ε0(2π )3/2

→

∂E0

∂α= 3!2

2me

− e2

(2π )3/2ε0α1/2 = 0

α

12 =

2mee2

3(2π )3/2ε0!2

α opt =

me2e4

18π 3ε02!4

!E0(α opt ) = − 43π

mee4

16π 2ε0"2

⎛

⎝⎜⎞

⎠⎟

!E0(α opt ) ≈ −0.424 e2

4πε0a0

2

00 0

12 4

eEaπε

= −

!E0 > E0 now, since the trial wavefunction cannot be exact for any α

Note: Gaussian trial orbitals (basis sets) are widely used in electronic structure programs. This is because integrals are easily evaluated over Gaussians. This is the origin of the name for the Gaussian Program: It uses Gaussian basis functions!



The Linear Variational Principle Consider a trial wave function ( ) ( )n n

nc fψ τ τ=∑

Let us assume for simplicity -‐ Real coefficients, functions, (nf τ ) -‐ Orthonormal expansion functions:

*( ) ( )n m nmDomainf f dτ τ τ δ=∫

Then we can try to optimize the coefficients

!E0("c) =ψ *(τ )Hψ (τ )dτ∫ψ *(τ )ψ (τ )dτ∫

≡ ND

N = cn fn(τ )Hcm fm(τ )dτ∫

= cncm fn(τ )Hfm(τ )dτ∫

n,m∑

,

n nm mn mc H c≡∑

,

( ) ( )n n m mn m

D c f c f dτ τ τ=∑∫

,

( ) ( )n m n mn mc c f f dτ τ τ=∑ ∫

2,

,n m n m n

n m nc c cδ= =∑ ∑

Then 0k

Ec∂ =∂

k∀

→ 2 0k k

k

N DD Nc cN

c D D

∂ ∂−∂ ∂∂ ⎛ ⎞ = =⎜ ⎟∂ ⎝ ⎠

Or k k

N N Dc D c∂ ∂⎛ ⎞= ⋅⎜ ⎟∂ ∂⎝ ⎠

k k

N DEc c∂ ∂=∂ ∂

km m n nkm nk

N H c c Hc∂ = +∂ ∑ ∑

2 2n knk k

D c cc c∂ ∂= =∂ ∂ ∑



→ 0km m k n nk km nH c Ec c H Ec⎛ ⎞ ⎛ ⎞− + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑

Since nk knH H= , this is twice the same equation.

→ 0km m kH c Ec− =∑

This has the form of a matrix eigenvalue equation!

H = k Hkm

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

m

and

H

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

!c⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ =

!c⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ E

Hkmcm = ck Em∑

We can also write [ ]( ) 0kmE cδ− =H

This has the form of a linear equation.

A!c = 0 , with A = H − E1

This type of equation only has a solution if ( )det 0A = .

Hence ( )det 0E− =H 1 → equation for E “secular determinant”

Let us discuss examples later. For now I want to draw the analogy:

Schrödinger equation H Eψ ψ=

If we make a basis expansion

n nnc fψ =∑ |n m nmf f δ=

Then we get a matrix type Schrodinger equation H

!c = E!c

With H → Hnm = φn *∫ (τ )Hφm(τ )dτ

Such an eigenvalue equation has M solutions for an M M× matrix. They represent approximations to the ground and excited states.

If the basis is not orthonormal the define Snm = φn *(τ )Hφm(τ )dτ∫ and H

!c = S!cE (see MQ)

Or det H − SE = 0 → eigenvalues E .



Example Linear Variations: Consider particle in the box

H = −!2

2md 2

dx2

φn(x) = 2

nsin nπ x

a⎛⎝⎜

⎞⎠⎟

Now add to H a linear potential

V0

ax

Use as a trial wave function 1 22 2 2sin sinx xc ca a a a

π π+

, 1,2i j = Hij =

2a

sin iπ xa∫ − !

2

2md 2

dx2 +V0

ax

⎛⎝⎜

⎞⎠⎟

sin jπ xa

0 01 2

0 022

162 9

1629

V VE

V VE

π

π

⎛ ⎞+ −⎜ ⎟⎜ ⎟=⎜ ⎟− +⎜ ⎟⎝ ⎠

En =

n2π 2!2

2ma2 υ0 =

2ma2

!2π 2 V0

H = !2π 2

2ma2 ⋅1+

υ0

2−

16υ0

9π 2

−16υ0

9π 2 4+υ0

2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

The eigenvalues of this Hamiltonian are !2π 2

2ma2 εn



εn =5+υ0

2± 1

29+

32υ0

9π 2

⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

1/2

≈5+υ0

2± 3

2≈

1+υ0

2

4+υ0

2

This happens to be pretty good solution, especially if 0V is small

Other instructive example: consider particle on the ring

− !2

2mR2

∂2

∂ϕ 2 , with the degenerate 1m = solutions

1 cos

1 sin

ϕπ

ϕπ

2

1 22E

mR= h

Now apply a magnetic field, which adds − e

2me

BeLz = γ Lz to the Hamiltonian Lz = −i! ∂

∂ϕ

Under the influence of the perturbation the levels split. Calculate the energy splitting.

γ = −e

2me

H = H0 + γ Lz

1 21 1cos sinc cψ ϕ ϕπ π

= +

γ z Lz

1

πcosϕ

⎛⎝⎜

⎞⎠⎟= γ !i 1

πsinϕ

γ Lz

1

πsinϕ

⎛⎝⎜

⎞⎠⎟= −γ !i 1

πcosϕ

→ H − E1 =

!2

2mR2 − E −iγ !

+iγ ! !2

2mR2 − E⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

det H − E1( ) = !2

2mR2 − E⎛⎝⎜

⎞⎠⎟

2

+ γ 2!2 = 0

!2

2mR2 − E⎛⎝⎜

⎞⎠⎟= ±γ !



E± =

!2

2mR2 ± γ !

Can I find eigenfunctions?

!2

2mR2 −iγ !

+iγ ! !2

2mR2

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

1i

⎛

⎝⎜⎞

⎠⎟= !2

2mR2 + γ !⎛⎝⎜

⎞⎠⎟

1i

⎛

⎝⎜⎞

⎠⎟

!2

2mR2 −iγ !

+iγ ! !2

2mR2

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

1−i

⎛

⎝⎜⎞

⎠⎟= !2

2mR2 −γ !⎛⎝⎜

⎞⎠⎟

1−i

⎛

⎝⎜⎞

⎠⎟

What are the eigenfunctions then?

1

cos sin ~ ii ei

ϕϕ ϕ⎛ ⎞→ +⎜ ⎟

⎝ ⎠

1

cos sin ~ ii ei

ϕϕ ϕ −⎛ ⎞→ −⎜ ⎟−⎝ ⎠

(we normalize, 12π

factor)

− !2

2mR2

∂2

∂ϕ 2 − iγ ! ∂∂ϕ

⎡

⎣⎢

⎤

⎦⎥eiκ → !2

2mR2 + γ !⎛⎝⎜

⎞⎠⎟

eiϕ

− !2

2mR2

∂2

∂ϕ 2 − iγ ! ∂∂ϕ

⎡

⎣⎢

⎤

⎦⎥e− iκ → !2

2mR2 −γ !⎛⎝⎜

⎞⎠⎟

e− iϕ

Bottom line: We can indicate perturbation 0H H V= +

Diagonalize H over degenerate states.

Examples: H-‐atom: H = H0

(ne) + g(v)!L ⋅!S( )

Diagonalize H over 2p orbitals → … 32

2 p , 2 p1

2

eigenfunctions from diagonalizing 6 6× Hamiltonian.

Everything comes out by brute force. Example 2: add in additional magnetic interaction

H = H0

(ne) + g(v)!L ⋅!S + e

2me

Bz Lz +2e

2me

BzSz

→ diagonalize H over 2p and 2s orbitals



→ all the splitting from diagonalization The linear variational principle is a very powerful tool to calculate approximate eigenfunctions It is widely used to calculate the splitting of energies in a degenerate manifold, when adding a perturbation. When the energies of a Hamiltonian 0H are not degenerate, one can get a good estimate of the energy

correction due to a perturbation V , by calculating V .

Hence if (0)0 n n nH Eφ φ= , then eigenvalues of ˆ ˆ ˆ

eH H V= + are given to first approximation by

φn H φn = φn *(τ ) H0 +V⎡⎣ ⎤⎦∫ φn(τ )dτ

= En

(0) + φn *(τ )∫ Vφn(τ )dτ

These are just the diagonal elements of the Hamiltonian matrix. = First order Perturbation Theory: → If we go back to box + linear field

H = − !

2

2md 2

dx2 + V0

ax

0H v

φn =

2a

sin nπ xa

En

(0) = !2π 2

2ma2

φn V φn =

V0

a2a

sin nπ xa

⎛⎝⎜

⎞⎠⎟

2

x dx0

a

∫

=

V0

a⋅ 2a⋅ a2

4⎛⎝⎜

⎞⎠⎟=

V0

2

→ En ≈ En

(0) +V0

2 all energies are shifted by

V0

2

If zero-‐order states are degenerate, first-‐order perturbation theory is useless. Instead use linear variational principle

Example ˆzL in cos(mx) , sin( )mx basis

→ choose other basis: → results ime ϕ, e

− imϕ

→ always diagonalize over zeroth-‐order states: degenerate first-‐order perturbation theory



Another example of H!c = !cE : Huckel π -‐electron theory

In organic chemistry, many molecules are essentially planar. The plane contains “ 2sp ” carbon, oxygen,

nitrogen atoms. The out of plane pz -‐orbitals constitute the π -‐orbitals. The molecule’s π -‐ orbitals are

linear combinations of the atomic pz -‐orbitals. One can parameterize a one-‐electron effective

Hamiltonian matrix as follows. Let us restrict ourselves to 2sp carbons.

Hα ββ α

⎛ ⎞= ⎜ ⎟⎝ ⎠

0 00

00 0

H

α ββ α β

β α ββ α

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

0 00 00 0

H

α β β ββ αβ αβ α

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

00

00

H

α β ββ α β

β α ββ β α

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

D Rule: α on diagonal β for any two adjacent atoms connected by a π -‐bond

β < 0 ~ pz

(1)V Ne pz(2) dτ∫

Following the variational principle we

a) Diagonalize the Hamiltonian → orbital energies kE , eigenvectors !ck

b) Fill up orbital levels from the bottom up putting an α and a β electron in each level.

Occupy as many levels as you have π -‐electrons c) If levels are degenerate, fill them up with α -‐electrons first, then add additional β -‐

electrons



d) Total energy: occupiedorbitals

E λε= ∑

e) Density Matrix kl k loccupied

D c cλ λ= ∑ kl klkl

E H D=∑ (see McQuarrie)

This procedure would work fine in MathCad

How do we do it on paper? Take ( )det E−H 1

→ ethylene 0E

Eα ββ α−

=−

( )2 2 0Eα β− − =

( )Eα β− = ±

E α β= ±

2π -‐ electrons Using α and β is a bit tedious for larger problems

→ divide each column by β and define Ex α

β−=

→ E xα β= −

211 0

1x

xx= − = 1x = ± E α β= ±

Or

1 11 1 01 1

xxx=

1 11 1xx

3 31 1 3 2 0x x x x x x+ + − − − = − + = 1x = 1 3 2 0− + = 2x = − 8 6 2 0− + + =



Always: 0i

ix =∑

1x = is double solution.

( ) ( ) ( )( )2 21 2 2 1 2x x x x x− + = − + +

3 3 2x x= − + ok k kE xα β= −

3π -‐electrons (4-‐fold degenerate) Triplet (3-‐fold degenerate) Singlet

What if we would look at the singlet state of the anion? This is not a stable structure, the molecule would distort



Jahn-‐Teller Distortion (the picture is not clear, my apologies. Follow notes in my lecture)

I might ask questions of 4*4 determinant → too hard to solve

→ I would give you the solution 1 2 3 4, , ,x x x x

You would show that 1 2 3 4( )( )( )( )x x x x x x x x− − − − is your secular determinant

You can guess the orbitals (phases) from symmetry arguments: The orbitals are always symmetric or antisymmetric with respect to plane or axis of symmetry If you know value of x you can solve

1 11 1 01 1

x ax bx c

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

1x = : 0a b c+ + =

( )(1 -1 0)1 1 -2

orthogonal combination

2x = − :

2 1 1 1 01 2 1 1 01 1 2 1 0

−⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟− =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠

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