Transcript No Slide Title

Lecture 11: Harmonic oscillator-I.
Vibrational motion
(This lecture introduces the classical harmonic oscillator
as an introduction to section 12.4 .
Lecture on-line
Classical harmonic oscillator (PowerPoint)
Classical harmonic osciillator (PDF format)
Quantum mechanical harmonic oscillator (derivation)
(PowerPoint)
Quantum mechanical harmonic oscillator (derivation)
(PDF format)
Handout for this lecture (Classical Harmonic Oscillator)(PDF)
Handout for this lecture (Derivation of QM Harmonic Oscillator
(PDF))
Tutorials on-line
Basic concepts
Observables are Operators Postulates of Quantum Mechanics
Expectation Values More Postulates
Forming Operators
Hermitian Operators
Dirac Notation
Use of Matricies
Basic math background
Differential Equations
Operator Algebra
Eigenvalue Equations
Extensive account of Operators
Audio-visuals on-line
Overview of the harmonic oscillator (PDF)
(Good overview from the Wilson
group,****)
Overview of the harmonic oscillator (Powerpoint)
(Good overview from the Wilson
group,****)
Vibrating molecule-I (Quick Time movie 1.4 MB)
(From the CD included in Atkins
,***)
Vibrating molecule-II (Quick Time movie 1.4 MB)
(From the CD included in Atkins ,***)
Slides from the text book (From the CD included in Atkins ,**)
The material in this lecture covers the following in Atkins.
Harmonic oscillator...classical
Let us consider a particle of mass m attached to a spring
x
xo
o
xo
Equilibrium
x=0,t=0
Stretch
x=xo
compress
x=-xo
QuickTime™ and a
Video decompressor
are needed to see this picture.
At the beginning at t = o the particle is at equilibrium,
that is no particle is working at it , F = 0,
Harmonic oscillator...classical
In general F = -kxo . The force propotional to
displacement and pointing in opposite direction
x
xo
o
xo
xo
F=-kxo
Equilibrium
F= kxo
x=0,F=0
xo
k is the force constant
of the spring
QuickTime™ and a
Video decompressor
are needed to see this picture.
Harmonic oscillator...classical
We might consider as an other example two particles
attached to each side of a spring
Case I: Equilibrium
re
Equilibrium
F= 0
r = re
B
A
QuickTime™ and a
Video decompressor
are needed to see this picture.
Harmonic oscillator...classical
Case II: Stretch
r = re+x
F= -kx
B
A
Stretch
r = re+x
Again we have that
the force F is proportional
to the displacement
x and pointing in the
opposite direction
F = - k x
QuickTime™ and a
Video decompressor
are needed to see this picture.
Case III: Compress
re-x
F= -k(-x)
A
Equilibrium
B
r = re x
Harmonic oscillator...classical
QuickTime™ and a
Video decompressor
are needed to see this picture.
Harmonic oscillator...classical
We have from Newtons equation
{mass}x{accelaration} = force
m
d2xo
dt 2
 kxo or
d2 x o
dt 2
k
  x o
m
The general solution to this is
 k  
 k  
xo  A sin   t  Bcos  t '
 m  
 m  
however if xo  o at t = o we must
have B = o
 k  
xo  A sin   t 
 m  
Harmonic oscillator...classical
k
position x = A sin (
t)
m
Let us look at this solution
k 
(a) for  t  0 or t = o we have xo  0
m
(b) for
(c) for
(d) for
(e) for
k  
 m 
t  or t =

  xo  A
m 
 2
2 k 
x

or t =
k
A
m
k
t
-A
k 
m 
t   or t =     xo  o

m 

k 
k  3
t

m 

2
2
m
3 m 
 xo  A




2
k
k 
m 
t

2

or
t
=
2

 x o  o




m 
k 

m
2
k
3
m
2
k
Harmonic oscillator...classical
x
k
x = A sin (
t)
A
m
dx
velocity v =
=
dt
k
k
A
cos (
t)
m
m

k
-A
2
m

m
2
k
m
k
k
t
k
-A
m
 m
2
k
m
k
t
v
k
A
m

2
m
3 m
2
k
3
m
2
k
Harmonic oscillator...classical
x

k
A
k
k
Force  - kx = -A sin (
t)
m
m
v

k
A
m
m
2
k
3 m
 m
2
k
2
k
m
k
t
-A

m
2
k
m
k
t
k
-A
m
2
m
3
m
2
k
Harmonic oscillator...classical
x
It follows that the time to complete
A
m
one cycle is t cycle  2 
k
as a consequence one can

-A
1
1 k
complete  =

2
t cycle 2  m
cycles per time unit
The frequency is often written

k
m
k
t
m
k
3
m
2
k

as  =
or   2
2
where  is referred to as the circular frequency.
We clearly have
k
=
m
2
m
Harmonic oscillator...classical
x
We might also look at the
A
kinetic energy
 k 2
1 2 1  k
T = mv  m A
cos
t
 m 
2
2  m

-A


1 2
k
2
T  A k cos2 
t
 m 
2

T
m
2
k
1 2
A k
2
m
k
t

m
3
m
2
k
2
k

2
m
k
m
k
t
m
k
3
m
2
k
What about potential energy Harmonic oscillator...classical
V(x) ??
x
m
m
2

k
We always have
k
A
dV( x)
F= t
dx
Thus
 m
-A
3 m
dV( x)
2
k
2
k
F = -kx = dx
or


1 2
2  k 
x dV

V
A k sin
t
 m 
2
dx = V( x) - V(o)

m
m
d

x
V


2
0
k
k
1 2
x
Ak
1
2
2
=  kx dx = kx  o
2
0
t
Thus
1
V( x) = kx2
2
 m
3 m
2
2
k
k
Harmonic oscillator...classical
 k  1 2
 k 
1 2
2
2
E  T  V  A kcos 
t  A ksin 
t
 m  2
 m 
2
1 2
E A k
2
We note total energy independent of t
V(x) = 1/2k1x2 V V(x) = 1/2k2x2
E
-A2 -A1 k1 > k2 A1 A2
x
k1  k 2
From the relation
1 k
=
 k = 4 2 m
2 m
Harmonic oscillator...classical
V(x) = 1/2k1x2 V V(x) = 1/2k2x2
Thus
E
E = 2 2  m A2
Note that the amplitude A
2E
A=
k
depends on E and k. For a
given E the smaller k the
larger A.
Note that the frequency  is
independent of A
-A2 -A1 k1 > k2 A1 A2
x