Winter 2013 Chem 356: Introductory Quantum Mechanics
108
Chapter 8 – Approximation Methods, Hueckel Theory ............................................................................ 108
Approximation Methods ....................................................................................................................... 108
The Linear Variational Principle ............................................................................................................ 111
Example Linear Variations..................................................................................................................... 113
Chapter 8 – Approximation Methods, Hueckel Theory
Approximation Methods
A) The variational principle
For any normalized wave function , the expectation value of , the exact
groundstate energy.
Proof: with
If we would measure the energy we would find with probability
( )
This argument is a bit shaky when has degenerate eigenvalues. You will do a correct proof in the
assignments.
If would not be normalized we can calculate
and then
or in the final form: The variational principle
where is the exact ground state energy
0ˆ ˆ,H H E
n n
n
c ˆn n nH E
nE2
n nP c
ˆ ˆ*( ) n n
n
H H d P E
0 0n n n
n n
P E E P E 0nE E
H
2 *( ) ( )N d
02
1H E
N
(1)
0 0
ˆ*( ) ( )
* ( ) ( )
Domain
Domain
H dE E
d
0E
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 109
The variational energy , is exact when , the exact ground state wavefunction.
Examples: use a trial wavefunction that depends on one or more parameters … Then minimize the
trial energy
Some simple (trivial) examples:
Take trial wavefunction of the type
Or
Q: what is ? What is ?
A: The exact wavefunction has the form ,
by minimizing the energy we should get ,
Q: Take the Hamiltonian for the Hydrogen atom, and an state
Take the trial wavefunction
- What are the integrals to evaluate?
- What is the optimal value for ?
- What is the value for ?
A:
Minimizing :
again: exact
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
0E 0( ) ( )
,
,E
22
. 2
1 1ˆ2 2
h u
dH x
dx
2 /2( ) xx e 2 /2xNe
0E
2 /2xNe 0
1
2E
1 0
1
2E
0l 2 2
2
2
0
ˆ42
d d eH r
dr dr rr
re
0E
2
0
02
0
ˆ4( )
4
r r
r r
r e He drE
r e e dr
0 ( )0
E
0
1opt
a
2
0 0
0 0
1
2 4
eE E
a
2
0
0 2
4a
me
2re
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 110
now, since the trial wavefunction cannot be exact for any
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 basic functions!
2 2
2 2
2
0
02
0
ˆ4( )
4
r r
r r
r e He drE
r e e dr
...
1
2 2 2
3/2
0
3 2
2 (2 )e
e
m
2 2
0
3/2 1/2
0
30
2 (2 )e
E e
m
1 2
23/2 2
0
2
3(2 )
em e
2 4
3 2 4
018
e
opt
m e
4
0 2 2
0
4( )
3 16
e
opt
m eE
2
0
0 0
( ) 0.4244
opt
eE
a
2
0
0 0
1
2 4
eE
a
0 0E E
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 111
The Linear Variational Principle
Consider a trial wave function
Let us assume for simplicity
- Real coefficients, functions,
- Orthonormal expansion functions:
Then we can try to optimize the coefficients
,
( ) ( )n n m m
n m
D c f c f d
Then
Or
( ) ( )n n
n
c f
nf
*( ) ( )n m nmDomain
f f d
0
*( ) ( )( )
* ( ) ( )
H d NE c
Dd
( ) ( )n n m mN c f Hc f d
,
ˆ( ) ( )n m n m
n m
c c f Hf d
,
n nm m
n m
c H c
,
( ) ( )n m n m
n m
c c f f d 2
,
,
n m n m n
n m n
c c c
0k
E
c
k
2
0k k
k
N DD N
c cN
c D D
k k
N N D
c D c
k k
N DE
c c
km m n nk
m nk
NH c c H
c
2 2n k
nk k
Dc c
c c
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 112
Since , this is twice the same equation.
This has the form of a matrix eigenvalue equation!
m
kmk H
H and c c E
H
We can also write 0kmE c H
This has the form of a linear equation.
, with A E H 1
This type of equation only has a solution which .
Hence det 0E H 1 equation for “secular determinant”
Let us discuss examples later. For now I want to draw the analogy:
Schrodinger equation
If we make a basis expansion
Then we get a matrix type Schrodinger equation
c EcH
With ˆ*( ) ( )nm n mH H d H
Such an eigenvalue equation has solutions for an matrix. They represent approximations
to the ground and excited states.
If the basis is not orthonormal the define and c cEH S (see MQ)
Or det 0SE H eigenvalues .
0km m k n nk k
m n
H c Ec c H Ec
nk knH H
0km m kH c Ec
km m k
m
H c c E
0Ac
det 0A
E
H E
n n
n
c f |n m nmf f
M M M
*( ) ( )nm n mS H d E
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 113
Example Linear Variations
Consider particle in the box
Now add to a linear potential
Use as a trial wave function
2 2 2
22n
nE
ma
The eigenvalues of this Hamiltonian are
This happens to be pretty good solution, especially as is small
2 2
22
dH
m dx
2( ) sinn
n xx
n a
H 0Vx
a
1 2
2 2 2sin sin
x xc c
a a a a
, 1,2i j 2 2
0
2
2sin sin
2ij
Vi x d j xH x
a a m a adx
0 0
1 2
0 0
22
16
2 9
16
29
V VE
V VE
2
0 02 2
2maV
0 02 2 2
2
0 0
2
161
2 9
1624
29
Hmn
2 2
22n
ma
1/22
0 0
2
5 3219
2 2 9n
0
0
0
15 3 2
2 2
42
V
V
0V
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 114
Other instructive example: consider particle on the ring
, with the degenerate solutions
Now apply a magnetic field, which adds to the Hamiltonian ˆzL i
Under the influence of the perturbation the levels split. Calculate the energy splitting.
I used wrong formula; sign is wrong
H E 1 =
2 2
2 22mR
1m
1cos
1sin
2
1 22E
mR
ˆ ˆ2
e z z
e
eB L L
m
2 e
e
m
0 zH H L
1 2
1 1cos sinc c
1 1ˆ cos sinz zL i
1 1ˆ sin coszL i
2
2
2
2
2
2
E imR
i EmR
2
22 2
2det 0
2H EZ E
mR
2
22E
mR
2
22E
mR
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 115
Can I find eigenfunctions?
What are the eigenfunctions then?
(we normalize, factor)
Bottom line: We can indicate perturbation
Diagonalize over degenerate states.
Examples: H-atom:
Diagonalize over 2p orbitals … , eigenfunctions from diagonalizing
Hamiltonian. Everything comes out by brute force.
Example 2: add in additional magnetic interaction
diagonalize over 2p and 2s orbitals
all the splitting from diagonalization
The linear variational principle is a very powerful tool to calculate approximate eigenfunctions
2
22
22
2
1 12
2
2
imR
i imRi
mR
2
22
22
2
1 12
2
2
imR
i imRi
mR
1cos sin ~ ii e
i
1cos sin ~ ii e
i
1
2
2 2 2
2 2 22 2
i ii e emR mR
2 2 2
2 2 22 2
i ii e emR mR
0H H V
H
( )
0 ( )neH H g v L S
H 3
2
2 p 1
2
2 p 6 6
( )
0
2 ˆ( )2 2
ne
z z z z
e e
e eH H g v L S B L B S
m m
H
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 116
It is widely used to calculate the splitting of energies in a degenerate manifold, when adding a
perturbation.
When the energies of a Hamiltonian are not degenerate, one can get a good estimate of the energy
correction due to a perturbation , by calculating .
Hence if , then eigenvalues of are given to first approximation by
These are just the diagonal elements of the Hamiltonian matrix = First order Perturbation Theory:
If we go back to box + linear field
all energies are shifted by
If zero-order states are degenerate, first-order perturbation theory is useless. Instead use linear
variational principle
Example in sin( )mx basis
choose other basis: results
always diagonalize over zeroth-order states: degenerate first-order perturbation theory
Another example of : Hückel -electron theory
In organic chemistry, many molecules are essentially planar. The plane contains “ ” carbon, oxygen,
nitrogen atoms. The out of plane -orbitals constitute the -orbitals. The molecule’s - orbitals are
0H
V V
(0)
0 n n nH E ˆ ˆ ˆeH H V
0ˆ *( ) ( )n n n nH H V d
(0) *( ) ( )n n nE V d
2 2
0
2
2
VdH x
m adx
0H v
2sinn
n x
a a
2 2(0)
22nE
ma
0
0
2ˆ sina
n n
V n xV xdx
a a a
0 02
2
V Va
a a a
(0) 0
n n
VE E
a 0V
a
ˆzL
ime ime
Hc cE
2sp
zP
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 117
linear combinations of the atomic -orbitals. One can parameterize a one-electron effective
Hamiltonian matrix as follows. Let us restrict ourselves to carbons.
D
Rule: on diagonal for any two adjacent atoms connected by a -bond
Following the variational principle we
a) Diagonalize the Hamiltonian orbital energies , eigenvectors
b) Fill up orbital levels from the bottom up putting an and a electron in each level.
Occupy as many levels as you have -electrons
c) If levels are degenerate, fill them up with -electrons first, then add additional -
electrons
d) Total energy:
e) Density Matrix ,
kl k l
occupied
D c c
(see McQuarrie)
zP
2sp
H
0 0
0
0
0 0
H
0 0
0 0
0 0
H
0
0
0
0
H
(1) (2)0 ~ Ne
z zP V P d
kE kc
occupiedorbitals
E
kl kl
kl
E H D
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 118
This procedure would work fine in MathCad
How do we do it on paper? Take det EH 1
ethylene
- electrons
Using and is a bit tedious for larger problems
divide each column by and define
Or
Always:
is double solution.
ok
0E
E
2 2 0E
E
E
2
E
x
E x
21
1 01
xx
x 1x E
1 1
1 1 0
1 1
x
x
x
1 1
1 1
x
x
3 31 1 3 2 0x x x x x x
1x 1 3 2 0
2x 8 6 2 0
0i
i
x TrN
1x
2 21 2 2 1 2x x x x x
3 3 2x x
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 119
3 -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!
I might ask questions of determinant too hard to solve
k kE x
4*4
Winter 2013 Chem 356: Introductory Quantum Mechanics
Chapter 8 – Approximation Methods, Hueckel Theory 120
I would give you the solution
You show that 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 you can solve
:
orthogonal combinations
:
1 2 3 4, , ,x x x x
1 2 3 4( )( )( )( )x x x x x x x x
x
1 1
1 1 0
1 1
x a
x b
x c
1x 0a b c
(1 -1 0)
1 1 -2
2x
2 1 1 1 0
1 2 1 1 0
1 1 2 1 0