1
Operator based solution to harmonic oscillator
Consider 1d harmonic oscillator
h2 d 2 1 2
Ĥ 
 kx
2m dx 2 2
h2 d 2 1

 m 2 x 2
2m dx 2 2
k
k  m 2

m
Write in to a standard form using dimensionless coordinates
Q  mx
d
1 d

dQ
m dx
Ĥ 
Next define: Q 
h

h2 d 2
1
  2Q 2
2
2 dQ
2
q where q is dimensionless.
 1 
Unit analysis: Q  kg 2 m 


kgm 2 s 1
q ; q is indeed dimensionless.
s 1
d
 d

dQ
h dq
1
d2 1
Ĥ   h 2  h q 2
2
dq 2
 1 d2 1 2 
 h  
 q
 2 dq 2 2 
→ For any mass, any force constant, H.O. is always the same; up to a scale 
Solve Ĥ q   E q 
Let us try to factorize the Hamiltonian
 1  d

1  d
H  h
 q 
 q


 2  dq

2  dq

 d 2
 d
1
d 
h  2  q 2   q  q  
 dq dq  
2
 dq
1
 H  h
2
Define operators:
p̂  i

1  d
1
 q 
q̂  ip̂ 


2  dq
2
b̂ † 
b̂ 
d
dq

1  d
1
 q 
q̂  ip̂ 


2  dq
2
p̂, q̂   i
1
1
then  b̂ †b̂  Ĥ  h or Ĥ  h b̂ †b̂  h
2
2
Commutation Relation?
1
1
b̂ † , b̂   q  ip,q  ip   q,ip  i  p,q  1

 2
2
†
b̂, b̂   1


Use commutation relations to solve for the spectrum of H.O. If a is an
eigenfunction of Ĥ with
eigenvalues Ea then b̂† a is also an eigenstate

1
2
  b̂ †b̂   b̂ † a


1

 h b̂ †  b̂, b̂ †   b̂ †b̂   a

2
h  Ea b̂† a
Acting with b̂ † creates a new eigenfunction with eigenvalue raised by  .
Likewise b̂ a is an eigenstate

1
2
  b̂ †b̂   b̂ a


1

 h  b̂ † , b̂   b̂b̂ †   b̂ a

2
1

 h  1 b̂b̂ †   b̂ | a

2
 Ea  h b̂ a
Acting with b̂ creates a new eigenfunction with eigenvalue lowered by h
There is a lowest state 0 where b̂ 0  0 terminates
1

Ĥ | 0   b̂ †b̂   h 0

2
1
h 0
2
1
 Eo  h
2
1


En   n  h 


2

n  0,1,2
What is the ground state?
b̂ | 0  0
 d

 dq  q  (q)  0
(q)  e
(q  q)e
1
 q2
2
1
 q2
2
0
 d

All other eigenstates can be obtained by repeated action of b̂ †     q 
 dq

0  e
1
 q2
2
1  2qe
 2  e
1
 q2
2
1
 q2
2
~ qe
1
 q2
2
 2q 2 e
1
 q2
2
~ 2q 2  1 0
  4q 0  2q 2q 2  1 0  4q 2  6q  0
...etc.
All polynomials  0 , as you can verify.
Normalization:

n
1
b̂† |  0 
n!
|  n  | n 
Proof:
n|n 1
  | 0 
b̂ b̂ | n  n | n
| n ~ b̂ †
n
  | 0
1
b̂ †
n!
n
†
b̂ b̂  is the number operator it counts the # of quanta
†

1
2

1
2
  b̂ †b̂   | n   n   h | n




n | b̂ b̂ | n n | nn  n
†
b̂ | n | n  1  C | n  1  C 2
2
b̂ | n  n
C n
b̂ | n  n | n  1 even reasonable for n=0
b̂ † | n  D | n  1
n | b̂ b̂
†

| n  D 2
n | b̂b̂ † | n  n | b̂, b̂ †   b̂ †b̂ | n
 1 n n | n
D 2  1 n 
D  1 n
b̂ † | n  n  1 | n  1
1
| n  1 
b̂ | n
n
1
| n  1 
b̂ † | n
n 1
1 †
| n 
b̂ | n  1
n
| 0
|1  b̂ † | 0
1 †
1 † 2
| 2 
b̂ |1 
b̂ | 0
2
2
M

| n 

n
1
b̂ † | 0
n!

k
Therefore, all states can be expressed in terms of b̂† | 0
Moreover:

†
2 b̂  (q̂  ip̂)†  (q̂†  ip̂† )  (q̂  ip̂)  2b̂†
b̂ † and b̂ are indeed Hermitian conjugates

1
q̂  ip̂ 
 b̂ 
2


 b̂ †  1 (q̂  ip̂)

2

1
(b̂  b̂ † )
 q̂ 
2


 ip̂  1 (b̂  b̂ † )

2
2
p̂ 
i
(b̂ †  b̂)
2
Evaluation of general operators over harmonic oscillator basis states
Every classical function (in 1d) can be expressed as f(p,q), most often we would
have h(p) + g(q). This can be immediately expressed in terms of b̂ † and b̂
 i(b̂   b̂) 
 b̂   b̂ 
h

g
 2 
2 



If we know a lower order polynomial expansion then we can evaluate matrix
elements.
Procedure:
In general we would get terms like b̂ †b̂b̂, b̂b̂ † , b̂ †b̂b̂ †  the order of the terms
matters. We can always write a term b̂†b̂b̂†b̂b̂b̂† in so called normal order in which
  b̂
each term has the form b̂†
k
m
b̂ † to the left
b̂ to the right
Examples:
Ĥ 
1 2 1 2
p  q
2
2
2
1 i
1  b̂ †  b̂ 

†
 
(b̂  b̂)  

2 2
2  2 


 
1 †2
1
b̂  b̂ 2  b̂ †b̂  b̂b̂ †  b̂ †2  b̂ 2  b̂ †b̂  b̂b̂ †
4
4




1 †
b̂ b̂  b̂b̂ †
2
1
 b̂ †b̂  1 b̂ †b̂
2
1

  b̂ †b̂  

2
Consider anharmonic contribution
3
 1  †
3
q :  
b̂  b̂
 2 

2


3


3

 1  †
 
b̂  b̂ 2b̂ †b̂  1 b̂ †2  b̂ 2
 2 

 2b̂ †2b̂  b̂ †  b̂ †3  b̂ †b̂ 2  b̂b̂ †b̂  b̂  b̂b̂ †2  b̂ 3
 3b̂ †2b̂  3b̂ †  b̂ †3  3b̂ †b̂ 2  3b̂  b̂ 3
This is the normal form. It is easily seen that this can always be done.
3 Evaluating matrix elements
  b̂
Ž |m where Ô  b̂†
We want to evaluate n | O
k
l
b̂ | m  m | m  1
b̂ † | n  n  1 | n  1

  m   m n | b̂
2
n | b̂ † b̂ | m  m n | b̂ †2 | m 1
†
 m m 1n | m 1
| m
 0 nm
n | m   n,m  
 1 n  m
†2
Therefore m  1| b̂ b̂ | m  m m  1 and all other matrix elements are zero.
Let us try another one:
n | b̂ †b̂ 3 | m  n | b̂ † | m  3 m(m 1)(m  2)
 n | m  2(m  2) m(m 1)
m  2 | b̂ †b̂ 3 | m  (m  2) m(m 1)
m=0,1,2 elements are all 0 (do not exist)
 It is very easy to evaluate matrix elements, no need to do integrals! You will
use this in the exercises.
Recap: we can evaluate matrix elements of operators of the type
b̂  b̂ c
†
k
l
kl
over
Harmonic oscillator basis states.
What can we do with this tool?
a) evaluate corrections to harmonic oscillator due to anharmonic corrections.
1

V (q)  h  q 2   q 3   q 4 
2

1

Ĥ  h  b̂ †b̂    h  q 3   q 4 

2
Evaluate Hamiltonian matrix elements:
m | H | n m,n  0M
• diagonalize matrix
• monitor convergence with # of basis functions. Quite easy to do in Mathcad or
Matlab
The energy E obtained by diagonalizing the Hamiltonian will always decrease or
stay the same as the size of basis functions M is increased. This is known as the
“Interleaving Theorem”. Consider a basis in which H M  M is diagonal. Consider
extension of the basis set by one function, or H M 1M 1 .
 D11

1

M 
M 

M 
M 
M 

M 
M 
M 1 
 *
 V1
D22
D33
O
O
O
O
O
DMM
V2*

    

V1 

V2 
M 

M 
M 

M 
M 
M 

VM 

F 

 D V  X 
 X 
 

 E

F  Y 
 Y 
 V
DX  VY  EX(1)
V  X  FY  EY (2)
From (1):
VY  (E  D)X
X  (E  D)1VY
From (2):
V

E  D 1 V  F Y  EY


1
  Vi* E  Dii  Vi  F  Y  EY
 i

from which we get the equation:
1
V  E  D  V  E  F
Another way for representing the Interleaving Theorem is as follows:
Let Dii  Ei , eigenvalues of H M .
M
| Vi |2
EE
i1
EF
i
All of E that satisy this equation are eigenvalues of H M 1
 M |Vi |2 M -|V|2
 =
E i=1 E-E i i=1 (E-E i )2
The terms on the right-hand side of the equation are all negative. This means that
the function decreases everywhere (except at poles, E i ). Look at the sketch of this
function.
Note also that if M approaches  , E keeps decreasing till it reaches the exact
value→ approximate Ei is upperbound to exact E i .
Consider double well potential
V(q)  h   q 2  q 3   q 4 
 ,  0
A double well might describe vibrational levels in a chemical reaction such as:
The motion of the proton in H-C≡N  C≡N-H, or the NH3 tunneling process.
Something else we can easily address: Franck-Condon transition moments
 1 d2 1 2 
Ĥ groundstate  h  
 q 
2
 2 dq 2 
 1 d2 1

Ĥ x  h  
 (q  a)2   Ea
2
 2 dq 2

Simplifying assumptions:
• Frequency of excited state is the same as the ground state
• The potential for excited state is displaced harmonic oscillator.
• Ea is the difference in energy between the bottom of the wells. (adiabatic energy
difference)
1

Energy levels ground state: | n; En   n   h

2
1
1

Energy levels excited state: n ; En   n%  h  Ea  h

2
2
Absorption Spectra:
E•  E   n% 12  h  E
n
0
%
 Ea  nh
a
1
 h
2
% |2
Intensity is proportional to | n | 0
Emission Spectra:
% | n from excited | 0
% to ground state | n
| 0
1
1

 
E0  En   Ea  h    n   h

 
2
2
 Ea  nh

In ;

n | 0%
2
The mirroring property is true under the assumption of a displaced harmonic
oscillator with the same frequency.
In exercises you are asked to calculate the spectrum analytically.
% and 0%| n (they are the same up to a sign).
 evaluate n | 0
It is even easier to do it numerically:
Set up the Hamiltonian for the excited state with the H.O. basis of the ground state.
 1 d2 1
2
n | h 
 q  a    Ea | m
2
 2 dq 2

1
 n | h b̂ †b̂  Ea  q̂a  a 2 | m
2
1 †
q̂ 
b̂  b̂
2
 diagonalizing H yields the energies (should be equidistant) giving the eigenvector
%: the first component of eigenvector | n
%, energy En.
we can calculate 0 | n
%|2
 absorption intensity: | 0 | n


% relative intensity | n | 0
% |2 i.e.
If we are interested in the emission calculate n | 0
%.
the nth component of excited state eigenvector | 0
What is the advantage of the numerical approach?
a) I think it is conceptually easier, more transparent
b) We can easily generalize to more complicated potentials
 anharmonic
 excited state frequency is different.
Only minor changes to procedure adjust calculating H-matrix.
c) Later on we will generalize and consider higher dimensions.
4 Another Technique to solve vibrational eigenvalue problems: The
discrete variable representation
Up to now we did not consider very realistic forms of the potential, e.g.
cubic/quartic terms yield potentials that go to infinity.
More realistic:
Such a potential is well represented by a Morse potential
V  D 1 exp  x  xe 
2
2
1
2


 D  x  xe   2 x  xe  
2


1
2
3
4


 D  2 x  xe    3 x  xe    4 x  xe  
4


1
1
 D 2  k  m 2
2
2
Question: Can we use our current techniques to obtain the vibrational spectrum?
1 d2
Let us consider a more general problem: diagonalize H  
 V(q) . Where V(q)
2 dq 2
is a general potential.
Can we evaluate n | V (q) | m ?
Let us make the problem even more general. Consider general operator  with
known matrix elements n | Â | m .

We can evaluate matrix elements of a function f  .
It would be easy in the eigenvector basis:
 | ai   ai | ai 

f   f (0) f (0)  
1
1
f (0) Â 2  f (0) Â 3
2
3!
Taylor Series

1


f  | ai    f (0)  f (0)ai  f (0)ai2  | ai 
2


 f (ai ) | ai 

or a j | f  | ai   f (ai ) ij

f  is diagonal in the  basis.

What is a recipe to calculate n | f  | m for a general basis?
Ž |m and diagonalize
a) Construct n | A
→ eigenvalues ai eigenvectors
 n  m
m ai  n  ai
m

n  ai  ai n ai
v
v
A ai  ai ai
v
ai : components n ai in the basis.
b) Construct f  in the  -basis

a j | f  | ai   f (ai ) ij
 f (a1 )
0
0
0 


f (a2 ) 0
0 
 0
 0
0
O
0 


0
0 f (an) 
 0

c) Transform the matrix from the  -basis to the m -basis
 n | a a
i
i, j
j

| f  | ai ai | m

 n | f  | m
ai | m  Uim
*

m | ai   Uim
 U mi
m | f (A) | n  U  f (a)U

If we can diagonalize operator  , we can construct any matrix f  Is this an exact
approach? No, an error is introduced because the basis is not complete.
m | Â 2 | n   m | Â | kk | Â | n
k,complete


% k%| Â | n
m | Â | k
k%finitebasis
If we enlarge the basis we generally get closer to the exact results.
Application to potentials:
To evaluate n | V (q) | m in harmonic oscillator basis:
a) evaluate:
1
n | q | m 
n | (b̂  b̂ † ) | m
2
1

m n | m  1  m  1n | m  1
2
n, m  0M
b) diagonalize n | q | m
 eigenvalues qi, eigenvectors n | qi  (matrix)
c) construct q j | V(qi ) | qi | (diagonal)


d) transform to original basis
 m | qi qi | V(qi ) | qi qi | n  m | V | n
i, j
2
 b̂ †  b̂ 
1
e) add kinetic energy T̂  h m | 
 | n
2
 2 
f) diagonalize Ĥ  Tˆ  V̂
Check: evaluate simple harmonic oscillator.
Will you get the right result?
No not quite: In the finite basis  n | q̂ | kk | q̂ | m  n | q̂ 2 | m We can get exact
k
results if we make consistent approximations for m | p̂ 2 | n   m | p̂ | kk | p̂ | n
k
 Tˆ  V̂ : equidistant H.O. levels, correct eigenfunctions. You will investigate
these in the exercises.
Applications
a) evaluate energy spectrum in the ground state
b) evaluate Franck-Condon eigenvalue spectrum using realistic Morse potentials.
Apply the procedure to both ground and excited states.
|  0    |nn |  0    |nCn0
n
n
•n
|      |nC
n
Franck-Condon overlap: 0 | %   Cn0  C%n
n
Intensity is | 0 | % |2 for absorption E0  E%
n
%   Ckn  C%k 0 intensity: n | 0
%2
Likewise for emission: n | 0
k
 Important: sum the amplitude first, then square the result.
I will assemble an exercise to let this sink in.
5 Generalization of formalism to multiple oscillators
If we have multiple coordinates in their reduced dimensional form:

qi ,
,  i ,i  1N then for each coordinate or “mode” we can define
qi
b̂i 
1 
 
qi 

qi 
2
b̂i† 
1 
 
q

i
qi 
2 
Derivatives are now partials. As before b̂i , b̂i†   1 but we also have b̂i , b̂ †j 


 
b̂i , b̂ †j    qi 
,q j 


qi
qi 

     
 
 
  qi ,q j   
,
,q j    qi ,


 qi q j   qi
  qi 
0
In particular : qi


(qi ,q j ,) 
qi (qi ,q j ,)
q j
q j
0
because :

qi  0
q j
(i  j)
The commutation relations can be summarized as
b̂i , b̂ †j    ij


b̂i , b̂ j   0


b̂i† , b̂ †j   0


Using the operators b̂i we can define the orthonormal basis
b̂  b̂  b̂  | 0
†
1
n1
†
2
n1 !
n2
n2 !
nk
†
k
nk !
here|0 is the totalgroundstate
| 0 | 01  | 0 2  | 0 3 
b̂1 | 01   0 | 01   e
1
 q12
2
b̂k | 0 k   0 | 0 k   e
| 0  e


1 2 2
q1 q2 qk2
2
1
 qk2
2

We can successfully act with b̂i† on the wave function and create all powers.
We can create these basis states as | n1,n2 ,n3 nk  specifying the quantum numbers
in each state.
Then in analog to before:
b̂i | n1 ,n2 ,ni ,nk   ni | n1 ,n2 ,ni  1,nk 
lower n i by1,plus factor ni
b̂i† | n1 ,n2 ,ni ,nk   ni  1 | n1 ,n2 ,ni  1,nk 
raise n i by one,plus factor ni  1
Using these very simple rules we can evaluate matrix elements of the type

k
m1, m2 ,mk | b̂1†b̂ j b̂i† | n1,n2, nk  . The matrix element factorizes for each normal
mode. How many states? Assume maximum M i  1 quanta.
M 1  M 2  M 3 M k
For a general molecule I would have 3N-6 vibrations, e.g. C2H4  12 modes. If I use a
basis with maximum 10 quanta (M i  9) 1012 basis states.
 The number of quantum states grows astronomically with the number of
normal modes.
The Hamiltonian for a multimode harmonic oscillator can be written
k
1

Ĥ   h i  b̂i†b̂i  

2
i1
The states | n1 ,n2 nk  are eigenstates with energy

i

1
2
 i  ni  


ni  0,1,2
The state of lowest energy has ni  0 i
1
Ground state energy:  h i
i 2
 called the zero point energy.
6 Vibrations in polyatomic molecules in harmonic approximation
Using electronic structure methods we can solve the electronic Schrodinger
r
r
r
r
equation at a particular nuclear configuration Ĥ el R  r; R   E R  (r, R) in which


r indicate electronic coordinates (collectively) and R indicates nuclear coordinates.
This way we can calculate points on electronic surface. The nuclear coordinates are
assumed fixed.
e.g. for diatomic, but in principle we can calculate potential energy surface on a grid
r
E Ri where R  x1, y1, z1, x2 , y2 , z2 ,x N , yN , zN  3N coordinates. Interpolating
 
between the points we get the surface.
For polyatomic molecules this is rarely done (only for high accuracy spectroscopy)
We wish to locate minima on the surface associated with molecular structure. To do
E
i  13N and Hessian
this we must calculate gradients of energy
 Ri Rr
Kij 
uru
E
1  2E
 0 Ri
at the minimum, where
 Ri
2  Ri R j r
R
This is used to find equilibrium geometry Re (or R0 ). Then we can approximate the
potential energy by second order Taylor series:
r
r r
r r
1
E R  E0 Re   R  Re i K ij R  Re j
2 i, j
Pictorially in one dimension:


 

You can imagine what it looks like in many dimensions.
In the harmonic approximation we replace the potential energy surface, which is a
very complicated 3N or 3N-6 dimentional function, by a quadratic surface.
 1  2 1 2
 qi
In the notes I show how one can derive the Hamiltonian Ĥ    i  
 2  qi2 2 
i
for the vibrational part. We know how to solve for all the eigenfunctions of this
Hamiltonian. We can add in the rotational part of Ĥ (complicated to derive) and
obtain all ro-vibrational levels  Statistical mechanics.
7 Derivation of rigid rotor, harmonic oscillator form of Hamiltonian for
polyatomic molecules.
a) Rotational part of the problem
The derivation of the rotational part of the Hamiltonian is complicated, and I will not
attempt to give a derivation but only list the pertinent points. Even classically,
rotational motion is most easily described in a molecule-fixed (or body-fixed) frame,
rather than a laboratory-fixed frame. If we assume a molecule having 3N atoms, we
can list the coordinates as Rj , j  1N,  x, y, z , and each of the nuclei having
mass m j . The center of mass is defined as
m R 

m
j
cm
R
j
j
j
j
and the moment of inertia tensor is defined as


I  m j R2jx  R2jy  R2jz   m j R j R j
j
j
This symmetric matrix can be diagonalized, yielding eigenvalues I a , I b , I c , and a set
of so-called principal axis a, b, c. These axes, the eigenvectors of the moment of
inertia tensor, rotate with the molecule. In terms of these (instantaneous) axes the
J2
J2
J2
kinetic energy operator due to rotations can be written as T̂R  a  b  c ,
2I a 2I b 2I c
where J is the angular momentum operator J  r  p , with respect to the moleculefixed frame. To describe the eigenfunctions of the angular momentum operator, one
uses both the space fixed angular momentum operators J x , J y , J z , J   J x  iJ y , and
the molecule fixed angular momentum operators J a , Jb , J c , J   J a miJb . Please note
that the ladder operators for the molecule fixed frame differ in sign from the spacefixed operators, and is distinguished by the superscript  . The total angular
momentum operator is independent of of the axis of rotation and is given by
J 2  J x2  J y2  J z2  Ja2  Jb2  Jc2 . The basis functions used to solve the rigid rotor
problem are defined as
 j,k,m  ,  ,  | j, k, m  Pjkm  eik eim
and the angles  , ,  are the so-called Euler angles that describe the general
rotation of a point (or axis system) in space. The following relations are useful to
solve the problem.
J 2 | j, k, m  j  j  1 | j, k, m
J z | j, k, m  m | j, k, m
J c | j, k, m  k | j, k, m
J  | j, k, m  J x  iJ y | j, k, m | j, k, m  1


 j mm  j  m  1
J  | j, k, m  J a miJ b | j, k, m | j, k  1, m
 j mk  j  k  1
In rotational spectra the eigenvalues are the same always for all m   j j , and the
rotational levels are hence 2 j 1-fold degenerate (the same as for a diatomic
molecule). The eigenvalues of the rotational Hamiltonian can be obtained by
diagonalizing the matrix over the basis functions | j, k, m, k   j j , for a fixed j and
m. The rotational Hamiltonian can be written as

H r   J 2   J c2   J   J 
2
2

  J a2  J b2  J c2   J c2  2 J a2  J b2 
   2 J a2    2 J b2     J c2

We hence choose
J a2
J2
J2
 b  c
2I a 2I b 2I c
1 1
1
1 1
1
    ,     ,
4  I a Ib 
8  Ia Ib 
1 1
1 1
1 
      
2  I c 2  Ia Ib  
The most complicated term in the Hamiltonian is the third one, involving , and if
two moments of inertia are equal (symmetric top molecules), we would choose
them as the a and b axis, such that  = 0. In that case the functions | j, k, m are
eigenfunctions themselves with eigenvalues  j 1   k 2 . If the moments of inertia
are all equal (spherical top), also =0 , and the eigenvalues are simply  j  j 1 . For
the general asymmetric top the Hamiltonian has to be explicitly diagonalized. Using
ladder operators it is easy to construct the Hamiltonian, for a particular j and m, but
over all k   j j , and perform the diagonalization. This would allow one to
explicitly calculate the partition functions in thermodynamics without having to use
the high temperature limit. For the general problem, there is no closed form solution
and one needs to diagonalize (small) matrices of dimension 2 j 1 to calculate the
rotational eigenvalues and wave functions.
b) Vibrational part of the problem
If we denote the coordinates of all the nuclei as the 3N components of a vector
R  x1, y1, z1,, xN , yN , zN 
we can expand the electronic energy (including nuclear repulsion) around the
equilibrium conformation as (Taylor series expansion):
r
r
r r
E r r
1
2 E r r
E R  E Re  
R  Re i  
R  Re i R  Re j 
2 i, j Ri R j
i Ri
  





and note that at Re where the energy is minimal, the first derivative vanishes. The
second derivative is called the Hessian and is a generalization of the force-constant
2 E
of the one-dimensional Harmonic oscillator. Let us denote it K ij 
, evaluated
Ri R j
at Re . The Hamiltonian for the nuclear problem then takes the form
r
r r
r r
h2
1 2 1
Ĥ  E Re  
  R  Re i K ij R  R e j
2
2 i M i Ri 2 i, j
re
The energy E R is a constant and does not play a role, unless one considers
 

 

 
electronic transitions. The aim of this write up is to show that by a suitable choice of
coordinates the Hamiltonian reduces to
 h2 2 1

Ĥ    
  q2 
2
2 q 2

 
6 (5 for a linear molecule) of the  will be zero, corresponding to coordinates q
that describe the center of mass motion and the overall rotation of the molecule. The
underlying assumption of the Taylor series expansion is that one examines only
small amplitude motion. The above Hamiltonian does not provide a proper
treatment of rotations, which are large amplitude motions. In particular the
“rotational normal modes” only describe infinitesimal rotations, but would distort
the molecule under larger displacements. Hence the potential energy function,
which is independent of the rotational normal modes (   0 ) is not accurate at all
in this limit. A fully proper derivation of the molecular Hamiltonian is really quite
involved, and I will not pursue it. The remaining 3N-6 (3N-5) coordinates are socalled normal modes corresponding to nuclear vibrations. These normal modes can
be pictured classically and correspond to stretches, bending, twisting and so forth.
All of these motions are occurring simultaneously in a molecule but they can be
decomposed into elementary harmonic motions that all have their own frequency.
This underlying simplicity allows us to understand vibrational spectra fairly easily.
Because in the new coordinate system the Hamiltonian separates completely the
construction of nuclear eigenfunctions and eigenvalues is easy (although the
rotational part is actually fairly complicated).
r r
r
 R1, R2 ,, RN  1 q1 2 q2 3N 6 q3N 6  R  ,  ,   X T X cm ,Y cm , Z cm




where the  v qi  are eigenfunctions of the one-dimensional H.O. The total energy
would be
1
1
1



E  E Trans  E Rot   v1   h 1   v2   h 2  v3N 6   h 3N 6



2
2
2
Translational motion is little discussed in spectroscopy (but probed for example in
scattering experiments, and it plays a vital role in statistical mechanics) while
rotational motion has been treated above. Here we are concerned with the
derivation of Eqn. 4 starting from Eqn. 3. Define mass weighted coordinates
r r
r
r
1
yi  M i R  Re i  Ri 
yi  Rie
Mi
and, using the chain rule,

R 
1 yi 
1

1

 i


 ik

yk
M i yk Ri
M i Ri
M k Rk
i yk Ri
i
i


(take the steps one by one, as it shows you how to manipulate summations and
indices). It follows that the kinetic energy operator in terms of the yk becomes
h2
1 2
h2
2
 
  2
2 k M k Rk2
2 k yk
which looks more like eqn. 4. The potential energy term in terms of the coordinates
becomes
1
1
1
1
yi
K ij
  yiGij y j

2 i, j
Mi
M j 2 i, j
such that the Hamiltonian expressed in the mass-weighted coordinates y j becomes
Ĥ  
h2
2
1
  yi Gij y j

2
2 j  y j 2 i, j
This is not quite Eqn. 4 yet because of the coupling terms for i  j . For this reason
we next obtain the eigenvectors U of the matrix G with elements Gij (this is done in
actual computations)
G U
ij
jk
 Uik k
j
where U jk , j  1,, 3N are the components of the k th eigenvector that has eigenvalue
 k . The eigenvectors of a symmetric matrix ( Gij  Gij ) can be chosen orthonormal
(proof is analogous to orthogonality of eigenfunctions of hermitean operators), and
this is expressed as
U kiU kj  UikU jk   ij
k
k
The normal modes in terms of the eigenvectors U are defined as
U jk qk  y j
k
It may appear that the definition of the normal coordinates in Eqn. (14) is
backwards. We will see that everything works out properly. Using relation (13), we
easily obtain the inverse relationship


U ji  U jk qk   U ji y j   ik qk  U ji y j
j
k
j
k
j
i.e.
qi  U ji y j
j
Using the above information Eqn. (10) becomes
1
1
1
1
yi Gij y j   yi GijU jk qk   yiUik k qk   qk k qk

2 i, j
2 i, j,k
2 i,k
2 k
Hence the potential energy term has become a sum of diagonal terms with 'force
constants'  k .
To complete the derivation we have to show that the kinetic energy operator
satisfies
h2
2
h2
2
  2   2
2 j y j
2 k qk
Using the definition (14) and the chain rule we see that
y 


 j
 U jk
qk
y j
j qk y j
j
Hence
2


2
2

U
U



 q2  jk y ik y  ij y y  y2
k
i, j,k
i, j
i
k
j
i
i
j
i
It then follows that the Hamiltonian takes the final form of eqn. 4. Some important
techniques involving summations are used here. The derivation is non-trivial and
going through it will give you excellent practice. For example: The names of
summation labels are irrelevant: i, j , it does not matter. Each independent
summation uses a different summation label. Moreover, the order in which you sum
terms in principle doesn't matter, but simplifications in derivations only occur if you
do it in the right order!
Finally the Hamiltonian can be expressed in terms of dimensionless reduced normal
h
;  i . The Hamiltonian then takes on a
mode coordinates by defining qi  q%
i
i i
sum of familiar harmonic oscillators
 1 2 1 2 
Ĥ   h i  
 q%
i
2
2 
 2 q%
i
i
which can be written immediately in second quantization
1

Ĥ   h  b̂i†b̂i  

2
i
and we can find the solutions following the procedure in the hand-written notes.
8 A discussion of the full coupled electrons and nuclei problem
Construction of electronic basis at particular nuclear configuration R .
r
r
r r
r r
Ĥ el R  r; R  E R  r; R
  
  
The full Hamiltonian is given by:
z z
h2
1 2 h2
1
1
z
Ĥ 
   i2          

2  M
2 i mi
ri
R ,
i, j rij
i,
In Ĥ el
 TˆN  T̂e  V̂ee  V̂Ne  V̂NN
r
R we include everything, except TˆN . This Hamiltonian depends on the

nuclear configuration R .
Before we discussed that we can calculate the ground state at point R , Now I
assume we can also calculate excited states, and their energies. These electronic
eigenstates will be referred to as the adiabatic states or Born-Oppenheimer states.
The exact eigenstates of the full Hamiltonian can be expanded as:
r
r r
r r
 r, R   r; R   R
 
  

r
This expansion is exact in principle . What are equations satisfied by   R ?
r
r r
Ĥ r, R  TˆN  Ĥ el R 
  

 


1 r2
  
  Ĥ el  
  2M 

substitute expansion for :
r


1 r r
r r
   2M     Ĥ el    r; R   R

 
 
r r
r
r
1
 
[  R 2  r,r R   r,r R

2M  

   


r
r
r
r
r
r r r
r r
2   R  2 r, R     R  E R  r, R   R
   

 

     ]
r r
Next we integrate this expression against   r, R and use that these states at a
 
single R are orthonormal. We integrate over all electronic coordinates, we can
write:
r


1
r r
   * r, R   2m 2  r, R dr   R  A
 



r


1
    
2    R 
B
2m


 




r
r r
r
1 r
   R 
   r, R  r, R dr   M
*

r
r
  E R   R 
r


C
D
r
  E    R


If we make the approximations to expand in terms of a single state and keep only
terms (B+D) and neglect the others (A+C) we obtain the famous Born-Oppenheiner
approximation:
r
r
r
T̂N  V R    ,n R  En   ,n R





now can take potential energy surface for electronic state , and solve for the
problem of nuclear motion on a single surface.
This is what we discussed up to now. A slight improvement (usually not very
important unless very accurate electronic wave function is available) is to keep term
A, while still using a single state.
The term A can make diagonal Born-Oppenheimer correction:
r
r
2
r r
* r

r,
R
 2m  r, R dr  A R correction to potential. This can be seen
 


r
sometimes if different isotopes are used (same E R , but different M  and
r
therefore different correction A R .
 
 



Term C is zero for real wavefunctions.
In general:


r r
r r
r   r, R  r, R dr  r    0
 R
 R
r
r
r
r
r
r r r
r r
   r, R  r, R dr    r, R  r, R dr
   
   
 F  F  0

F  F  F
0
Summary:
 
 
r
T̂N  E R : B.O.
r
r
T̂N  E R  A R : include diagonal B.O. correction



Remaining terms involve off-diagonal contributions. This is referred to as vibronic
coupling or non-adiabatic effects.
When is this important:
a) when different electronic states are close in energy.
b) If adiabatic state strongly changes character as a function of geometry.
 Tˆ  E Rr
N.A.C. 
1


 N.A.C. Tˆ  E Rr 
2


Where N.A.C. is the non-adiabatic coupling. N.A.C. is larger when E1  E2 is small, or
when looking at high vibrational states.


At configuration (a) the states are far apart but they might cross at another
geometry. The character of states 1 and 2 might be well preserved. Often the true
surfaces might look like:
around point (b) the character of the state changes rapidly 1  2, 2  1. Energies
are close and N.A.C. is large, this is a quite typical situation. For many excited states
close to intersections or so-called avoided crossings the Born-Oppenheimer
approximation may be poor.
Can we understand N.A.C.?
r r
r r
r
r r
r r  r, R  dR   r, R
* r
]dr
  r, R  r, R dr    r, R [
dR
Matrix element is large if :
r r
a)  r, R changes rapidly
r r
b) The change is proportional to (or like)   r, R
 
  
 
 
  
 
This is precisely what happens near avoided crossings. Term C is usually most
important, Term A involving  * 2  dr less so (like Born-Oppenheiner
approximation).
r r
r r
From the analysis so far it appears we need to calculate  * r, R  r, R dr and
 
 
then incorporate this term into the solution for the nuclear motion equation.
However, in regions in which  changes rapidly it is difficult to calculate matrix
elements. Moreover, inclusion of coupling term in nuclear motion calculation is not
easy either.
Do we have to proceed like this?
NO!
r
r r
Let us return to our original expression    r, R   R . We assumed that we

r
r r
solved the electronic Schrodinger equation to obtain  r, R   R . However we
  
  
can in principle use any complete set of states. The expansion is exact. Let me write
the terms again for easy discussion:
r


1
r r
A
   * r, R  2M 2  r, R dr   R
 



r
r
r r r 1 r
   * r, R  r, R dr 
   R
C
M

 

 
 

r


1
 
2    R
B
  2M  
r
r
  E R   R
D

 
r
  E  R 




These equations are more complicated than before as we have off-diagonal energy
r
r r
terms E  R . Simplifications arise if we choose the basis functions  r, R such
r r
r r
that N.A.C.    r, R  r, R dr  0 for all  ,  ,  .
r r
What does this mean?  r, R changes little with geometry. In particular we can try

 
 
 
 
to make N.A.C. vanish for close lying states. The diagonal    term is always zero.
If we use the Born-Oppenheimer approximation, a single state it is very important to
r
use the adiabatic states: eigenstates of Ĥ el R . If we include more states this is less

important, and we can exploit freedom to simplify the calculation. This is what so
called diabatic states are all about.
The derivation we did earlier leading to the A, B, C, and D terms, would be applicable
r
r r
to a general expansion.    r, R   R ; not assuming
  

r
r r
r r
H el r, R  E R  r, R except we have to replace the term D with
r
r
r
r r
r r
E R   R ;
E R    r, R Ĥ el r, R
 
  
 

 
 
 potential matrix, not just diagonal element.
Term A we can neglect as usual  only terms B and D remain.
Diabatic states: electronic states that change little (preserve character) as a function
of nuclear geometry. Diabatic states are not uniquely defined, reflecting
r r
arbitrariness of basis  r, R .
r
r r
r r
What do we do in practice? Calculate adiabatic states Ĥ el r, R  E R  r, R
 
 
  
for just a few states of interest. Diabatic states are defined as linear combinations of
adiabatic states.
r
r r
r r
 a r, R   r, R U a R
 
r
U R is a unitary matrix at R 

 

U  U  1U aU b   ab
U  U 
a

b
  
a
preserves orthonormality
How is U determined? In the diabatic basis we want:
r r r
r r

r,
R


r,
R dr  0 (1)
a

b

 
 
N.A.C. in diabatic basis vanishes. This cannot be done exactly but in practise it can be
r
done to sufficient accuracy. Suppose we can choose/find U  a R such that (1)

holds, what does the problem look like? In the equation we can neglect terms A and
C. We are left with terms B and D.
r
r
r
r
1
  2m 2   R   E R   R  E   R



For example, 2*2 problem:

 



   


   
 Tˆ  E R%
E12 R%
11
 N

E21 R%
TˆN  E22 R%

  Rr 
1
  E
r 

 2 R  
 R 
 2


1 R  
r
Instead of a potential energy surface we have potential energy matrix Eab R and a
r
coupled set of differential equations for the vibrational wavefunctions  a R .


r
r
What is Eab R ? If we know U  a R then:
r
r
r
r
Eab R  Ua R E R U b R




  
 
r
 U E R U
r
r
Transform diagonal surfaces E R to diabatic matrix Eab R .
r
In practise we would make Taylor series expansion of Eab R along a set of normal



modes.
1
r
i
ij
Eab q   Eab (0)   Eab
qi   Eab
qi q j 
2 i, j
i
i
ij
, Eab
Where Eab
are constants (expansion coefficients).
In the following I want to address two questions:
i
ij
, Eab
, Eab how can we calculate:
1) Suppose we know Eab
• vibronic energy levels.
• absorption spectra
• time-dependent wave functions
• adiabatic potentials
 I will develop a procedure that can be used in Mathcad.
i
ij
, Eab
2) What is a basic algorithmic procedure to obtain coefficients Eab
in practice?
I will first address question 2). The basic situation is:
Assume we know normal modes of ground state qi , R0  qi  0 , and harmonic
vibrational frequencies.
1) Calculate excited states at R0 : |  q  0  | 0 
2) Make a small displacement along normal mode qi . Calculate adiabatic excited
states at geometry qi ,  qi  also calculate overlap matrix
r r
r
S   0 r, R0  r,q0  dqi dr overlap at different geometries.


Next find unitary transformation
 SU

a
 S%
 a  diagonal.
r r
r r
Hence we find diabatic states a r, R0  dqi   r, R0  dqi U a that have





maximum overlap with adiabatic states at R0 .
From U q0  dqi  we can obtain
U  E q0  dqi U  Eab q  dqi 
Take numerical derivatives
Eab q0  dqi  Eab q0  dqi 
i
 Eab
:the first order Taylor coefficient
2dqi
A similar procedure can be used to get second and higher order coefficients.
 Calculate: E q ,  q , S q 
 diabatize: Ua , Eab q 
 finite difference numerical differentiation.
r
 Assume we can obtain potential matrix Eab q  as Taylor Series expansion.
This procedure is fully automated in the Aces2 program.
Next consider our first question: How can we use this in actual calculations?
Let us focus on 2 X 2 problem.
 T  V1 (q)
 1 (q) 
V12 (q)   1 (q) 


  E

 V21 (q) T  V22 (q)    2 (q) 
  2 (q) 
Use harmonic oscillator expansion for 1 (q),  2 (q) .
 a   |nCna
n
or in long hand:
| 1    |nCn1
n
|  2    |mCm2
m
 Form 2M X 2M matrix
| m,2
| n,1
Tˆ  V11 (q)
n,1 |
 m,2 |
V12 q 
V12 q 
Tˆ  V22 (q)
each block is an M X M matrix.
Procedure: Construct matrix of Ĥ , diagonalize to get eigenvalues and eigenvectors.

General form of eigenvectors  : |n,aCna
, E
n,a
Eigenvectors are linear combinations of basis vectors that have both vibrational and
electronic label.
We can also calculate transition moments to vibrational ground states.



n  0,a  0 | X̂ | n,aCn,a
 X1C0,1
 X2C0,2
Xa  0 | X̂ | a :transition moment
0,0 | X̂ | n,a  Xa 0 | n  Xa n0
Only n=0 components survive. (compare F.C. spectrum for one electronic state)
Example of Calculation
(one of the problems)
 E  w b b   (b   b)
k(b   b)
1
1
 1

k(b   b)
E2  w2b b  2 (b   b)





 E
1
0

1

  E w
2 1
0
1
1
1

 0
2 1 E1  2w1
31

 M
0
31
O


0
k
0
0


k
0
2k
0

0
2k
0
3k


0
0
3k O


2M X 2M matrix
• diagonalize
• calculate spectrum
• how does this depend on 1 , 2 , k ?
0
k
0
0
k
0
2k
0
0
2k
0
3k
0
0
3k
O
E2
2
0

2
E2  w2
2 2
0
0
2 2
M
0
E2  2w2
32
32
O
















Illustrative Example
(see also notes non-adiabatic)
O=C=O ionize 1s electron on oxygen. I can ionize either the right or left atom. If I
ionize the bond C=O* gets shorter (electrons contract, less screening of nucleus. It is
like O turns into F). O=C=O* or *O=C=O If I think either oxygen ionizes I can draw
diabatic curves as functions of R(CO)1-R(CO)2
Asymmetric Stretch
 1
2
 2 w qn  a 


0

0
1
2
w qn  a 
2





This is not the full story. There is a coupling and the true curves look more like the
dashed curves (adiabatic).
The full potential energy matrix Hamiltonian is like
 1

2

 w q  a 

 2


1
2 

w q  a  

2


at q=0 the energies are separated by 2  . If q is large, the coupling  does not
matter anymore.

1
1SR  1SL 
 g 
2

at q=0 
   1 1S  1S 
R
L
 u
2
  g  1SR
at large q 
  u  1SL
The adiabatic state changes rapidly as a function of q
Hamiltonian in S R , S L basis:
 1

2
2

 wq  wa  waq

 2



1

wq 2  wa 2  waq 


2


1  1 1 
If you transform by
you obtain the Hamiltonian in  g ,  u basis
2  1 1 
 1 2
 1
  wa     wq 2
 2
 2


awq

awq
1 2
 1 2
 wa     wq
2
2
 
1
  E1 (q  0)  wq 2
2

 
(aw)q
 
 
(aw)q
1
E2 (q  0)  wq 2
2
Diagonalizing these two Hamiltonians would give exactly the same eigenvalues, and
same interpretation of states. The pictures of the potentials involved are very
different.





