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The Harmonic Oscillator
Turning Points, A, -A
m
Stretch spring, let go.
Mass, m, oscillates back and forth.
Hooke's Law
F  k x
force
no friction
linear restoring force
spring constant
F  ma
mx   kx
Harmonic oscillator - oscillates sinusoidally.
d 2 x(t )
k


 m  x(t )
dt 2
 
A is how far the spring is stretched initially.
k
x ( t )  A sin  
m
amplitude
1/ 2
t
mass
At the turning points, A, -A, motion stops.
All energy is potential energy.
Copyright – Michael D. Fayer, 2007
Potential is Parabolic
F 
A
 V ( x)
x
V ( x )   k x dx 
V
1
k x2
2
k  4 2 m 2  m 2
x
oscillator
frequency, Hz
oscillator
frequency, rad/s
Energy of oscillator is
E  1 / 2kA2
A - classical turning point.
A can take on any value. Energy is continuous, continuous range of values.
Copyright – Michael D. Fayer, 2007
Quantum Harmonic Oscillator
Simplest model of molecular vibrations
Bond dissociation energy
E
Molecular potential energy
as a function of atomic separation.
x
Bonds between atoms act as "springs".
Near bottom of molecular potential well,
Molecular potential approximately parabolic
Harmonic Oscillator.
Copyright – Michael D. Fayer, 2007
Potential V ( x ) 
1
k x2
2
Classical particle can never
be past turning point.
V
Turning point
Kinetic energy
zero; potential
energy max.
Turning
point
x
Turning
point
Particle can be stationary at bottom of well,
know position, x = 0; know momentum, p = 0.
 x p  0
This can't happen for Q.M. harmonic oscillator.
Uncertainty Principle indicates that minimum Q.M. H.O. energy  0
Copyright – Michael D. Fayer, 2007
One Dimensional Quantum Harmonic Oscillator
in the Schrödinger Representation
H E
(H  E)   0
2
d2
1
2
H 

k
x
2m d x 2 2
kinetic energy
Schrödinger Representation
potential energy
d 2 ( x ) 2m
2
2 2


E

2

m

x   ( x )  0.
2
2 
dx
  2 m /

Substitute H and
definition of k.
Mult. by -2m/2.
Define
2mE
2
d 2 ( x )
2 2




x  ( x)  0
2
dx


Copyright – Michael D. Fayer, 2007
Find
 ( x)
d 2 ( x )
2 2




x  ( x)  0
2
dx


Good from -  .
Must obey Born Conditions
1. finite everywhere
2. single valued
3. continuous
4. first derivative continuous
Use polynomial method
1. Determine
 ( x ) for x  
2. Introduce power series to make the large x solution correct for all x.
Copyright – Michael D. Fayer, 2007
d 2 ( x )
2 2




x  ( x)  0
d x2


For very large x
 2 x 2  
Therefore,  can be dropped.
d 2
2
2


x

2
dx
Try
 e


2
x2
Then,


 x2
 x2
d 2
2
2
  x e 2  e 2
2
dx
This is negligible compared to the first term
as x goes to infinity.
Copyright – Michael D. Fayer, 2007
Two solutions
e


2
x2
e
This is O.K. at


x2
2
This blows up at
x  
x  
Not finite everywhere.
Therefore, large x solution is
 ( x)  e


2
x2
For all x
 ( x)  e


2
x2
f ( x)
Must find this.
Copyright – Michael D. Fayer, 2007
 ( x)  e


2
x2
f ( x)
Need second derivative in Schrödinger equation

 x2
d 2  ( x)
 e 2 ( 2 x 2 f   f  2 x f ' f '')
2
dx
With
df
f '
dx
and
d 2  ( x)
d x2
Substitute
d2 f
f '' 
d x2
into the original equation
d 2 ( x )
2 2




x  ( x)  0
2
dx

and divide by

e


2
x2
gives
f   2 x f      f  0
Equation only in f.
Solve for f and have  ( x ) .
Copyright – Michael D. Fayer, 2007
divide by 


f   2 x f    1  f  0.



1
substitute
  x
f ( x )  H ( )
Gives
d 2 H ( )
d H ( )  


2

   1  H ( )  0.
2
d
d


Hermite's equation
Substitute series expansion for H()
H ( )    av    a0  a1   a2  2  a3  3 

dH ( )
  a   1  a1  2 a2   3 a3  2 
d

d 2H
    1 a    2  2a2  6a3  
2
d

Copyright – Michael D. Fayer, 2007
d 2 H ( )
d H ( )  


2

   1  H ( )  0.
2
d
d


2a2  6a3   12a4  2  20a5  3 
2a1   4a2  2  6a3  3 
substitute in series
The sum of these infinite
number of terms in all powers
of  equals 0.







3
2  


1
a


   1 ao    1 a1     1  a2  
3









 0.
In order for the sum of all the terms in this expression to vanish identically
for any ,
the coefficients of the individual powers of  must vanish separately.
To see this consider an unrelated simpler equation.
a5 x 5  a4 x 4  a3 x 3  a2 x 2  a1 x  a0  0
Fifth degree equation. For a given set of the ai, there will be 5 values of x
for which this is true. However, if you know this is true for any value of x,
then the ai all must be zero.
Copyright – Michael D. Fayer, 2007


2a 2    1  a 0  0


 0 


6a3    3  a1  0


 1 


12a4    5  a2  0


 2 


20a5    7  a3  0


 3 
Coefficients of like powers of .
In general


( +1)( +2)a  2    1  2  a  0


a  2



2


1



a

  1 (  2) 
Even and odd series.
Pick a0 (a1 = 0), get all even coefficients.
Pick a1 (a0 = 0), get all odd coefficients.
Recursion Formula
Copyright – Michael D. Fayer, 2007
Have expression in terms of series that satisfy the diff. eq.
But not good wavefunction.
Blows up for large |x| if infinite number of terms. (See book for proof.)
 ( )  e
e

2
2
H ( )
 2 / 2
2
e e
 2 /2
For infinite number of terms and large |x|.
(   x )
blows up
Unacceptable as a wavefunction.
Copyright – Michael D. Fayer, 2007
Quantization of Energy
If there are a finite number of terms in the series for H(),
wavefunction does not blow up. Goes to zero at infinity.
e
 2 /2
n
The exponential goes to zero faster than n blows up.
To make series finite, truncate by choice of .
= (2n + 1)
n is an integer.
Then, because
a  2



2


1



a

  1 (  2) 
if a0 or a1 is set equal to zero (odd or even series)
series terminates after
=n
a finite number of terms.
Copyright – Michael D. Fayer, 2007
Any value of  with
 = (2n + 1)
is O.K. Any other value of  is no good.
Therefore,

2mE
2
 (2n  1) 2 m /
definition of 
definition of 
Solving for E
1

En   n   h
2

n0
E0  1/2 h
n is the quantum number
Lowest energy, not zero.
Energy levels equally spaced by h.
Copyright – Michael D. Fayer, 2007
Hermite Polynomials
Energy Levels
1

En   n   h
2

Wavefunctions
 n ( x )  N ne
H1    2 
2
2
H n  
  2 m /
  x
1


   2 1 
N n    n 
   2 n ! 

H 0    1
H 2    4  2  2
H3    8  3  12 
H4    16  4  48  2  12
H5    32  5  160  3  120 
1
2
normalization constant
H6    64  6  480  4  720 2  120
Copyright – Michael D. Fayer, 2007
Lowest state
n = 0
1
4
 
 0 ( x)    e
 

 x
2
2
1
4
 
  e
 

0(x)
2
()2
-4 -3
-2
-1
0

1
2
3
Classical turning points
 x2 
2
h
k
x   h / k  
-4 -3
4
-2
-1
0

1
2
3
4
1/2 kx 2  1/2 h
potential
energy
total
energy
classical turning points - wavefunction extends into
classically forbidden region.
Copyright – Michael D. Fayer, 2007
More wavefunctions - larger n, more nodes
n=1
n=2
n=3
-6 -4 -2 0 2 4
n=4
-6 -4 -2 0 2 4
n=5
-6 -4 -2 0 2 4 6
n=6
-6 -4 -2 0 2 4
-6 -4 -2 0 2 4
-6 -4 -2 0 2 4 6
Copyright – Michael D. Fayer, 2007
Probability for n = 10
Classical turning points
~ =  4.6
 *
Time oscillator spends
as a function of position.
-6
-4
-2
0

2
4
6
Looks increasingly classical.
For large object, nodes so closely spaced because n very large that
can't detect nodes.
Copyright – Michael D. Fayer, 2007
Dirac Approach to Q.M. Harmonic Oscillator
Very important in theories of
2
p
1 2
H
 kx
vibrations, solids, radiation
2m
2
Want to solve
eigenkets, normalized
H E E E
We know commutator relation
 x, P   i
1
To save a lot of writing, pick units such that
m 1
k 1
1
In terms of these units
1 2
2
H  (P  x )
2
 x, P   i 1
identity operator
Copyright – Michael D. Fayer, 2007
Define operators
a 

a a
i
2
( P  i x)

a 
1
i 2
( P  i x)

a is the complex conjugate (adjoint) of a since P and x are Hermitian.
Then
1

a a  [( P  i x )( P  i x )]
2
1 2
2
 [P  i x P  i P x  x ]
2
1 2
2
 [ P  i( x P  P x)  x ]
2
1 2
i
2
 [ P  x ]  [ x, P ]
2
2

1

aa  H  1
2
Hamiltonian
commutator
Copyright – Michael D. Fayer, 2007
1
aa  H  1
2

Similarly
1 2

2
a a  [ P  i( x P  P x)  x ]
2
1

a aH 1
2
Therefore
1


H  (a a  a a )
2
and
a, a    1


Very different looking from Schrödinger Hamiltonian.
Can also show
[a , H ]  a

[a , H ]  a

Copyright – Michael D. Fayer, 2007
Consider E; eigenket of H.
a E  Q
Q  Ea Ea

Q Q 0
scalar product of vector with itself
Q Q 0
only if
Q 0
We have
Q Q  E a  a E  0.
Then
1
1

E a a E  E H  1 E  (E  ) E E  0
2
2
Therefore,
1
E
2
normalized, equals 1
Copyright – Michael D. Fayer, 2007
Now consider
aH E  Ea E
eigenket of H
commutator
a, H   a H  H a  a
H a E   ( E  1) a E 
these are same
rearrange
aH  Haa
Operate H on ket, get same ket back
times number.
 a E  is eigenket with
eigenvalue, E - 1.
Then,
( H a  a) E  a H E  E a E
H a E   ( E  1) a E 
eigenvalue
Ha E  a E  E a E
transpose
Ha E  E a E  a E
factor
eigenket
a E  E 1
Maybe number multiplying.
Direction defines state, not length.
Copyright – Michael D. Fayer, 2007
H a E   ( E  1) a E 
a is a lowering operator.
It gives a new eigenvector of H with one unit lower energy.
a E  E 1
a E  E 2
2
Each application gives new ket one with one unit
lower energy.
a E  E 3
3
Could keep doing this indefinitely,
1
but
E
2
Therefore, at some point we have a value of
E, call it E0,
such that if we subtract 1 from it
1
E0  1 
2
But E0 - 1 can't be < 1/2. Therefore a E0  0
For eigenvector E0
1

a  a E0   ( H  1) E0
2
1
 ( E0  ) E0  0
2
1
2
1
E0  h
2
E0 
not zero
in conventional
units
Copyright – Michael D. Fayer, 2007
Raising Operator


a  H E   E a E



a  H E   ( H a  a ) E
using the commutator
rearranging, operating, and factoring as before


H  a E   ( E  1) a E 
These are the same.
Therefore,  a  E  is an eigenket of H with eigenvalue E + 1.



a E  E 1
a

number, but direction defines state
takes state into new state, one unit higher in energy.
It is a raising operator.
Copyright – Michael D. Fayer, 2007
E0
is the state of lowest energy with eigenvalue (energy) 1/2.
Apply raising operator repeatedly. Each application gives state
higher in energy by one unit.
H E0
1
 E0
2
eigenvalue, one unit higher in energy
3

H  a E0   E0  1
2
H  a
H  a
2
3
5
E0   E0  2
2
7
E0   E0  3
2
1 3 5 7
E , , , ,
2 2 2 2
1

En   n  
2

1

With normal units En   n   h
2

Same result as with Schrödinger Eq.
Copyright – Michael D. Fayer, 2007
E0
is the state of lowest energy with eigenvalue (energy) 1/2.
Apply raising operator repeatedly. Each application gives state
higher in energy by one unit.
H E0
1
 E0
2
eigenvalue, one unit higher in energy
3

H  a E0   E0  1
2
H  a
H  a
2
3
5
E0   E0  2
2
7
E0   E0  3
2
1 3 5 7
E , , , ,
2 2 2 2
1

En   n  
2

1

With normal units En   n   h
2

Same result as with Schrödinger Eq.
Copyright – Michael D. Fayer, 2007
Eigenkets
E  n
labeled with energy
1
2
Can relabel kets with quantum number
E  n
1
 n
2
Take n to be normalized.
Raising and Lowering operators

a n  n n  1
a n  n n  1
n  n  1

a n  (n  1) n  1
n  n
numbers multiply ket when raise
or lower
a n  n n 1
Will derive these below.
Copyright – Michael D. Fayer, 2007
Consider operator

a a operating on n

a a n a

n n 1
n n
Therefore

a a n n n

n
is an eigenket of operator a a with eigenvalue n.

a a
number operator. Eigenvalue – quantum number
Important in Quantum Theory of Radiation and Solids
a

and a
called creation and annihilation operators.
Number operator gives number of photons in radiation field
or number of phonons (quantized vibrations of solids) in crystal.
Copyright – Michael D. Fayer, 2007
a n  n n  1

a n  n n  1
To find  n and  n
n  1 a n  n

n  1 a n  n
Take complex conjugate
n a n  1   n   n1
Now

n a a n  ( n  1)

because a a  H  1/2
( H  1/2) n  ( n  1/2  1/2) n
Work out

n a a n  n a n  1 n
 n n  n 1  n

n a a n   n 1  n  n  1
from here
Copyright – Michael D. Fayer, 2007
 n1  n  n  1
But
 n   n1
Then  n  n   n  n  1
2

and  n1 n1   n1  n  1
2
Therefore,
 n 1   n  n  1
2
2
True if
n  n  1
n  n
Copyright – Michael D. Fayer, 2007
Using the occupation number representation with normal units
H
1


 (a a  a a )
2
  2  ( k / m )1/ 2
Consider H n
1


H n 
 aa n  a a n
2
1


 a( n  1)1/ 2 n  1  a n1/ 2 n  1
2





1
 ( n  1)1/ 2 ( n  1)1/ 2 n  n1/ 2 n1/ 2 n
2
1

 (2n  1) n
2


1
1


   n   n   n   h n
2
2


1

Therefore, n are eigenkets of H with eigenvalues  n    .
2

Copyright – Michael D. Fayer, 2007
Units in the raising and lowering operators
a 

a 

 1
1/ 2 
P

i
k
x
1/ 2 
1/ 2
(2  )  m

i
 1
1/ 2 
P

i
k
x
1/ 2 
1/ 2
i (2  )  m

aa
1



1
2

 2k 


 
 
x 

2
k


1/ 2

1/ 2
1/ 2
x

Add operators, P cancels.
1/ 2
x
aa
 m 
P  i 

 2 
 2k
Many constants. This is the reason
why derivation was done in units
1 .
such that m  1 k  1
Need constants and units to work
problems.
1/ 2


x in terms of raising and lowering operators.
a  a 

Subtract operators, get P in terms of
raising and lowering operators.
Copyright – Michael D. Fayer, 2007
Can use the raising and lowering operator representation to calculate
any Q.M. properties of the H. O.
Example
x 4  for ground state, average value of x4
4
0x 0
In Schrödinger Representation



0
 x  0dx
4

Copyright – Michael D. Fayer, 2007
 
x 

 2k 
1/ 2
a  a 

 
 4
0 x 0 
0
(
a

a
) 0

 2k 
2
4
constant - C
4
3 
2 
 C  0 a 0  0 a a 0  0 a a a 0 
 0 ( a  )4 0 
Many terms. Must keep order correct. Operators don’t commute.
Copyright – Michael D. Fayer, 2007
Could write out all of the terms, but easier way.
Any term that doesn’t have same number of a’s and a+ = 0
Example




0a a a a 0  04
orthogonal = 0
Any operator that starts with a is zero.
a 0 0
Can't lower past lowest state.


0 aa a a 0  0

Terms with 0 a are also zero
because
a 0  Q 0
0 Q  Q  0 a

Copyright – Michael D. Fayer, 2007
Only terms left are


0 aa aa 0


0 aa a a 0
  
4


 
0 x 0 
0
a
a
a
a
0

0
a
aa
a 0 


 2k 
2


a n  n n 1
a n  (n  1) n  1



0 aa aa 0  0 aa a 1




 0 aa 0
 0 aa 2
 0a1
 2 0a1
 0 0 1
2 00 2
3 2 2
0 x 0 
4 k2
4

0 a a a a 0  0 a aa 1
No integrals.
2
2
Must be able to count.
Copyright – Michael D. Fayer, 2007
Vibrational Wave Packet
A short optical pulse will excite many
vibrational levels of the excited state
potential surface.
pulse bandwidth
excited
electronic
state
vibrational levels
Launches
vibrational
wave packet
short pulse optical excitation
ground electronic state
Copyright – Michael D. Fayer, 2007
Model Excited State Vibrational Wave Packet with H. O. States
Time dependent H. O. ket
n( t )  n e  iEnt /
Superposition representing wave packet on excited surface
t   n n e  int
n
Calculate position expectation value - average position - center of packet.
t xt
 
x 

 2k 
1/ 2

aa


t x t    m* e i m t   n e  i nt m x n
m
n
   m*  n e  i ( n  m ) t
m ,n

2k

m aa n
Copyright – Michael D. Fayer, 2007

m aa n
if
only non-zero
m  n1
Then
t xt 
But


2k  n

 n*1 n e  i (
n  n 1 ) t
 n   n1   and  n   n1  
t xt 


2k  n

 

 

n   n*1 n e  i ( n  n1 ) t n  1 

E  


 n*1 n e  i t n   n*1 n e i t n  1 
This expression shows that x time dependent.
Time dependence is determined by superposition of vibrational states
produced by radiation field.
Copyright – Michael D. Fayer, 2007
Simplify
Take n large so
n >1
Also,
i = 
Otherwise
j = 0
Each state same amplitude in superposition
for some limited set of states.
Using these
t xt 

2k
t x t  2 2
 2  n  e  i t  e i t 
n


2k
n cos ( t )
n
Position oscillates as cos(t).
Copyright – Michael D. Fayer, 2007
Wave packet on harmonic
potential surface.
Packet moves back and forth.
I2 example
Ground state excited to B state
 ~ 565 nm
20 fs pulse
band width ~700 cm-1
Level spacing at this energy
~69 cm-1
Take pulse spectrum to be rectangle and all  excited same within bandwidth.
States n = 15 to n = 24 excited
(Could be rectangle)
Copyright – Michael D. Fayer, 2007
Cos
+1 to -1
distance traveled twice coefficient of Cos
4 2


2k
n
n
10 equal amplitude states.
 2  0.1
k   2
  1.05  1022 g
  1.3  1013 Hz
Distance traveled = 1.06 Å.
Comparable to bond length.
Copyright – Michael D. Fayer, 2007
Copyright – Michael D. Fayer, 2007
Copyright – Michael D. Fayer, 2007