1
Approximations in Quantum Mechanics
Variation Principle
Introduction
We often can’t solve for the exact wavefunction for a particular Hamiltonian.
- Even the exact basis set is often unknown.
To approximate wavefunction, we choose a basis set with arbitrary parameters to be
determined.
Unknown wavefunction is written as a sum of basis functions.   c11  c22  c33 
c1 ,c2 ,c3 ,
 arbitrary parameters
1 , 2 , 3 ,
 basis set functions
Recall that nature seeks the lowest possible energy. In the variation principle, an energy
eigenvalue is determined using a wavefunction with arbitrary parameters. Then a
minimum of the energy is found by varying the arbitrary parameters.
- In the above wavefunction, the ci coefficients are varied.
- Note: Sometimes parameters with the basis set functions are varied also.
Recall in calculus, that the minimum for a function, f(x), can be found by setting the first
derivative of the function to zero.
df  x 
0
dx
The variation principle creates a separate derivative equation for each parameter to be varied.
dE  c1 , c2 , c3 ,
dc1
Procedure
 0
dE  c1 , c 2 , c3 ,
dc2
 0
dE  c1 , c 2 , c3 ,
dc3
 0
etc
1. Construct trial function from a basis set with undetermined coefficients.
- Note: We will consider the simple example of a trial function with two undetermined
coefficients.
  c11  c22
2. Use the trial function to calculate the expectation value of the energy.
*H  d  H 
 c1 1  c2 2  H  c1 1  c2 2 

E 


*

 c1 1  c2 2  c1 1  c2 2 
   d

c12 1 H 1  c1c 2 1 H 2  c 2 c1 2 H 1  c22 2 H 2
c12 1 1  c1c 2 1 2  c 2 c1 2 1  c 22 2 2
3. Define quantities to simplify E
H11  1 H 1
H22  2 H 2
H12  1 H 2  2 H 1
S11  1 1
S22  2 2
S12  1 2  2 1 - overlap integrals
E 
c12H11  2c1c 2H12  c 22H22
c12S11  2c1c2S12  c 22S22
2
4. Multiply the energy expectation value by denominator of expression.
 c12S11  2c1c2S12  c22S22  E  c12H11  2c1c2H12  c22H22
5. Take derivative of expression with respect to coefficient using implicit differentiation and
set derivative equal to zero.
 E
- find
c1
 2c1S11  2c2S12 
- remember
E   c12S11  2c1c 2S12  c 22S22 
 E
 2c1H11  2c 2H12
c1
 E
0
c1
2c1H11  2c2H12   2c1S11  2c 2S12  E  0
 c1  H11  ES11   c 2  H12  ES12   0
- for
 E
c 2
 E
 c1  H12  ES12   c2  H22  ES22   0
c2
6. Rearrange two equations in two unknown as a matrix equation.
 H11  ES11 H12  ES12  c1 

   0
 H12  ES12 H22  ES22  c2 
7. Solve secular determinant to find E eigenvalues.
H11  ES11
H12  ES12
H12  ES12 H22  ES22
0
2
 H11  ES11  H22  ES22     H12  ES12    0


2
2
2

S11S22 E 2   H11S22  H22S11  E  H11H22   S12
 E  2H12S12 E  H12    0
S S
11 22
E
2
 S12
 E2  H11S22  H22S11  2H12S12  E  H11H22  H122   0
H11S22  H22S11  2H12S12   H11S22  H22S11  2H12S12 
2
2 S11S22  S12

2
2
 4 S11S22  S12
H11H22  H122 
c 
8. Use E eigenvalues to find eigenvectors,  1  which are the coefficients of the optimized
 c2 
trial function. The optimized trial function is the approximate wave function for which
we have been searching.
9. The optimized trial function now may be used to calculate other properties.
- note: Energy is already calculated.
3
Perturbation Theory
Introduction/Motivation
In the beginning of studying quantum chemistry, the problems considered are relatively easy
to solve (believe it or not). The simple problems have closed, analytic solutions, that is, the
solution are expressed as relatively simple functions. However, not all problems have nice,
neat answers.
Consider hydrogen atom in an electric field
Without electric field, the ground state electronic orbital is a 1s, which is spherical. The
wavefunction and the energy of the system is known exactly.
Consider what happens when an electric field is applied to the hydrogen atom.
+
-
Now how does one describe the electronic orbital?
How does one determine the energy?
The problem can be solved using techniques from perturbation theory.
perturbation – disturbance, stress, shock, etc…
General Perturbation Theory
Assumptions of perturbation theory
1. The Hamiltonian can be written as a series in terms of an arbitrary parameter,  and
1
2
perturbations, H   , H   .
H  H   H   2H  
2. The wavefunction can be written as a series in terms of an arbitrary parameter,  and
1
2
corrections,    ,    .
0
1
2
           2   
3. The energy can be written as a series in terms of an arbitrary parameter,  and corrections,
1
2
E  , E  .
0
1
2
E  E   E   2E  
0
1
2
4
The  parameter indicates the strength of the perturbations, e.g., the strength of the electric
1
H 
field. The above assumptions are very good as long as  1 i.e.
1 In other
0
H 
words, the perturbations have to be relatively weak.
General method
To find the energy and wavefunction corrections, equations need to be developed by
H = E
substituting the series approximations into the Schrödinger equation.
H    H     H                
  E   E    E               
0
1
2
2
0
1
0


2
2
1
2
2
0
0
1
0
2
2



 E       E     E         E     E      E      
H       H     H      2 H     H      H   
0
1
0
1
0
2
0
0
1
1
0
0
1
0
1
2
2
2
0
1
1
0
2
For the equality to hold for any , i.e., any strength of perturbation, each “coefficient” of 
terms must be equal.
0
0
0
0
For 0  H       E    - zeroth order, unperturbed Schrödinger equation
1
0
0
1
1
0
0
1
For 1  H       H       E      E    
For 2  H       H      H       E     E     E   
2
0
1
1
0
2
2
0
1
1
0
2
First Order Perturbation Theory
 H    E       H    E    
The 1 equation rearranged becomes
0
0
1
1
1
0
1
1
1
This equation becomes useful, if we can solve for    and E   . Note that H  is given, it is
the perturbation.
1
To solve we need to assume a form for the corrected wavefunction,    .
     a n n
1
an – coefficients to be calculated
n
n – wavefunction from basis set of “zeroth-order” Hamiltonian.
- e.g., harmonic oscillator wavefunctions, hydrogenic atom
wavefunctions, spherical harmonics, etc…
First-order perturbation energy
Substitute assumed series form of corrected wavefunction into first-order perturbation
equation.
 H    E  a 


 a H    E   n  H    E   0
0
0
n
n

0
n
n
 H    E   0
1
n
0
1
1
1
5
Now allow the zeroth-order Hamiltonian to operate on n .
a
n

H    E   n  H    E   0
 a  E  E  n  H    E   0
0
0
1
1
n
1
n
n
1
0
n
Note that because of the basis set we have chosen, E 0  E0 and  0  0
(i.e., *0 ) and integrate.
Now multiply the equation by 0
0
 a E
n
n

n
 a E
n

 E0  n  0 H    E  0
1
1
n
 E0  0 n  0 H   0  0 E  0
n
 E 0 0n  0 H   0  E  
1
1
n
 a E
n
1
1
n
Now taking the sum on the left-hand side, note that 0n for all n  0.
1
1
a 0  E0  E0   0 H   0  E 
0  0 H   0  E 
1
1
The first-order correction to the energy is E   0 H   0
1
1
Thus to calculate the first-order correction to the energy, all that is needed is the zeroth-order
ground state wavefunction and the first-order perturbation. This is remarkably simple!
First-order perturbation wavefunction
To calculate the first-order correction to the wavefunction, we need to calculate the an
coefficients.
Return to
a E
n
n


 E 0  n  H    E  0
n
1
1
(i.e., *k ) and integrate. We will be
Now multiply the equation by an arbitrary state, k
careful not to allow the state to be the ground state, i.e., k0.
k
 a E
n
n

n
 a E
n

 E0  n  k H    E  0
1
1
n
 E0  k n  k H   0  k E  0
n
 E 0  kn  k H   0  E    k 0
1
1
n
 a E
n
n
1
1
6
Now taking the sum on the left-hand side, note that kn = 1 only when k=n.
Also note that k0 = 0
a k  Ek  E0   k H   0
1
k H  0
ak 
 Ek  E0 
1
Another relatively simple result!
Therefore the first-order correction to the wavefunction is
k H  0
  a k k  
k
k
k  Ek  E0 
Notation: The prime on the summation symbol indicates to take the sum over all
possible k except k = 0, i.e., don’t include the ground state.
1
1
The total wavefunction including the first-order correction is
k H  0
      0  a k k  0  
k
k
k  Ek  E0 
Without using bra-ket notation, the total wavefunction is
1
*kH  0d

 0
1
      0  a k k  0  

 Ek  E0  k
k
k
Interpretation of results
 0
1
1
Perturbation has the effect of adding to the zeroth-order wavefunction, excited state
wavefunction, i.e., the excited state wavefunctions “mix” with the ground state wavefunction.
Also, one can consider that the perturbation has induced “virtual” transitions to excited
states.
Second Order Perturbation Theory
2
0
1
1
0
2
2
0
1
1
0
2
The 2 equation is H       H      H       E     E     E    
Rearranged, the equation becomes
0
0
2
1
1
1
2
2
0
H    E     H    E     H    E    






To solve we need to assume a form for the corrected wavefunction,    .
2
     b n n
2
bn – coefficients to be calculated
n
Substitute into second-order perturbation equation.
H    E   b
0
0
n
b  E
n
n
n
n



n  H    E     H    E

1
1

1

2
 E0  n  H    E     H    E
1
1
1
2
2
2
0
0
Substitute first-order correction to the wavefunction, 
b  E
n
n

 E0  n  H    E 
n
1
1
 a
n

n  H    E
n
2
2
7
1
0
Multiply equation by 0 and integrate.
0
b  E
n
n

 E0  n  0 H    E 
n
b  E
n
b  E
n
n
1
 a
n

n  0 H    E
n
1
n
2
 E 0  0n  a n 0 H   n  a n E  0n  0 H 
1
n
 E 0  E 0   a n
0H
2
0
0  E
2
00
n
1
n
2
 E 0  0 n  a n 0 H   n  a n E   0 n  0 H 
1
n
n
1
2
0  E  00
2
n
1
1
n  a 0E  0 H 
2
0  E
2
n
E
 2
 a n 0 H   n  a 0 E    0 H 
1
1
2
0
n
At this point, substitute for an remembering that a0 = 0
Second-order perturbation energy
Often H  2 is zero, i.e., we don’t consider a second-order perturbation. Therefore the
second-order correction to the energy is
E
 2
0 H  n n H  0
2


 0 H  0
 En  E0 
n
1
1
Interpretation of results
Thus to calculate the second-order correction to the energy, all of the zeroth-order excited
state wavefunctions must be known.
As a practical matter, often only the first few excited states are needed.