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Lecture 22 – Harmonic oscillator
Ch 9
pages 465-469
Summary of lecture 21
 A quantum mechanical particle is described through its wave
function Y(x,t)
 The square of the wave unction Y(x,t)2 characterizes the particle
distribution in space and is a measure of the probability of finding
a particle at a certain time and place
For example, the probability of finding a particle within a certain
volume V is:
2
Y
 ( x, t )dx
V
Summary of lecture 21
 The duality of matter introduces a probabilistic nature to
measurements, so that we can calculate observables only in a
probabilistic sense. This is done through the expectation value of a
variable O, which can be calculated for a quantum system using
the expression:
 O   *Odx
For example, the average position of an electron in a molecule is
given by:
 x   * xdx
Summary of lecture 21
 The wave function can be calculated by solving the Schroedinger
Equation:
 2 Y ( x, t )
ih Y( x, t )

V
(
x
,
t
)
Y
(
x
,
t
)

2
t
8 2 m x 2
h2
 The time-independent component of the wave function (x) is
obtained by solving the time-independent Schroedinger equation
(when the potential is independent on time):
 2 ( x )
 V ( x ) ( x )  E ( x )
8 2 m x 2
h2
Summary of lecture 21
 We have calculated the wave function for a particle in a box:
 ( x) 
2
 nx 
sin 

L  L 
And the corresponding energy levels:
n2h2
En 
8mL2
Mathematically, quantization results when particle waves are
constrained in space by boundary conditions.
The Classical Linear Harmonic Oscillator
Classical mechanical problems are very often solved by
introducing the so-called Hamiltonian, which is defined as:
p2
H ( x, p ) 
 V ( x)
2m
The classical linear harmonic oscillator has the following
Hamiltonian:
2
2
H ( x, p ) 
p
x

2m
2
Where the frequency of the oscillator is:
  2 

m
The Classical Linear Harmonic Oscillator
The trajectory x(t), p(t) of the oscillator is obtained by solving
Hamilton’s equations of motion:
H ( x, p )
dp(t )

x
dt
H ( x, p ) dx(t )

p
dt
The solution is (homework) (A is the amplitude of the motion)
x (t )  A cos t
p(t )  mA sin t
We can introduce a quantum mechanical Hamiltonian as:
h2 d 2
H 
 V ( x)
2
2
8m dx
The Time-Independent Schroedinger Equation can be written:
H  ( x )  E ( x )
Quantum Mechanical Linear Harmonic Oscillator
The quantum mechanical treatment of the linear harmonic
oscillator (LHO) is one of the most important applications of
quantum mechanics
The LHO is used as a simple approximation to molecular
bond vibrations and rotations, for example, and forms the
basis of much spectroscopy.
The time-independent Schroedinger equation for the LHO is:
 2 x 2
 2

 ( x )  E ( x )
2
8 m x 2
h2
 2 m 2 x 2
 2

 ( x )  E ( x )
2
8 m x 2
h2
This equation may be rewritten as:

2 m
h
qx 

4E
h
 2 ( q)
 q 2 ( q)   ( x )
2
q
Quantum Mechanical Linear Harmonic Oscillator
 2 ( q)
 q 2 ( q)   ( x )
2
q
Although there are no solid boundary conditions as there was
with the particle in the box, the wave function is localized in
the sense that is must approach zero as x increases toward
infinity. This just means that the probability of finding the
particle must decrease as we move toward very large
extensions.
The solution to Schroedinger’s equation for the LHO is
 n ( q)  An e
 q2 / 2
H n ( q)
H n ( q)  ( 1) e
n
q2
 
 n q2
e
q n
An is a constant and Hn(q) is called a Hermite polynomial of the
nth order
Quantum Mechanical Linear Harmonic Oscillator
 2 ( q)
 q 2 ( q)   ( x )
2
q
H 0 ( q)  1
 n ( q)  An e
H 1 (q)  2q
 q2 / 2
H n ( q)  ( 1) n e q
H n ( q)
2
qx 
H 2 ( q)  4q 2  2
As with the Particle-in-a-Box, the probability of finding a
particle at q  x 
P   (q)
2


By requiring that:
We find:
( q) dq  1
2
n

1/ 2


1
An   n

 2 n!  
1/ 2
Finally:

 
 n  x    n

 2 n!  
e x
2
/2

Hn x 

 
 n q2
e
q n
Quantum Mechanical Linear Harmonic Oscillator
 2 ( q)
 q 2 ( q)   ( x )
2
q
The energy has the form:
1/ 2

 
 n  x    n

2
n
!




e x
2
/2

Hn x 

4E
 2n  1
h
1
E n  hv ( n  )
2
Note this is shifted by h/2 from Planck’s energy. This is called
the zero point energy, the existence of which is required by the
Heisenberg Uncertainty Principle.
Quantum Mechanical Linear Harmonic Oscillator
1/ 2

 
 n  x    n

 2 n!  
e x
2
/2


Hn x 
It is interesting to calculate probabilities Pn(x) for finding a
harmonically oscillating particle with energy En at x; it is easier
to work with the coordinate q; for n=0 we have:
1/ 2
 1 
 0  q   A0 

  
e q
2
/2
1/ 2
 2 
 1  q   A1 

  
qe
1/ 2
 1 
 2  q   A2 

2  
1/ 2
 1 
 3  q   A3 

3  
 q2 / 2
 2q
 2q
2
3
 P0  q    0  q  
1
 P1  q    1  q  
2q 2
2
2
 1 e  q
2
/2
 3q  e  q
2


e q
eq
 P2  q    2  q 
/2
2
2
 P3  q    3  q 
2
 2q

 1
2
2 
2
 2q

3
2
e q
 3q 
3 
2
2
eq
2
Quantum Mechanical Linear Harmonic Oscillator
Probabilities:
0.6
0.5
0.4
0.3
0.2
0.1
0
n=0
n=1
n=2
3
2.
3
1.
6
0.
9
0.
2
-0
.5
-1
.2
-1
.9
-2
.6
P(x)
P(q) for n=0,1,2
q
Quantum Mechanical Linear Harmonic Oscillator
Potential well, wave functions and probabilities:
Wavefunctions for particle in the box
2 2
nh
E
8mL2
n = 1, 2, …
E

*
Quantum Mechanical Linear Harmonic Oscillator
Potential well, wave functions and probabilities:
As the quantum number gets larger, the probability increases
towards larger displacement values. This corresponds to a
classical phenomenon, as the energy of an oscillator increases,
motion becomes more extended away from the status of lowest
energy. The fundamental frequency of the oscillator is also the
same both classically and in quantum mechanics:
Quantum Mechanical Linear Harmonic Oscillator
Potential well, wave functions and probabilities:
The fundamental frequency of the oscillator is also the same
both classically and in quantum mechanics:
0 


There are however several differences between classical and
quantum mechanics
Quantum Mechanical Linear Harmonic Oscillator
Potential well, wave functions and probabilities:
the first is that there is motion even in the lowest energy state
(see the shape of the probability for n=0); the second is that the
wave function extend beyond the classical limits for the motion
in a region of space where the potential is very large and that
are not expected to be observed classically. The term 1/2hv is
called zero point energy. This states that an oscillator cannot be
at complete rest; if it was at rest, we would know the
momentum (p=0) and position precisely; the zero point energy
allows Heisenberg’s principle not to be violated. What that
means is that molecules or solids vibrate even at 0 degree K.
Expectation values
The expectation value of a variable O can be calculated for a
quantum system using the expression
 O   *Odx
We can calculate the expectation value for variables such as
position and momentum; as in the classical oscillator
x  pX  0
However, if we calculate the mean-square position and the
mean-square momentum we will find some differences
Expectation values
Let us calculate the mean-square position and the mean-square
momentum for a linear harmonic oscillator in the nth energy
level as follows (notice that the wave functions are real and so
complex conjugate is the same as the function itself)
 x 
2
 q2 

In general
 x 2 

1


 0 (q)q  (q)dq 
2

1
1
n  

2
1


q

2
1
e

2
q2
dq 
 h  
 p      n 
 2  
2
x
2
 

2 q
 q e dq 
2

1
2
1

2
Thus, the squares are not zero although the expectation value
for position and momentum are; in fact, they are related to the
energy of the oscillator in the nth energy level.