Harmonic Oscillators
• F=-kx or V=cx2. Arises often as first approximation
for the minimum of a potential well
• Solve directly through “calculus” (analytical)
• Solve using group-theory like methods from
relationship between x and p (algebraic)
V
Classically F=ma and
d 2x
dt 2
 Cm x  0 
x  sin 2t  
1
2
C
m
Sch.Eq.
  2 d 2
2 mdx


2

C
2
x 2  E
 
2m

d 2
du 2


(
2 mE
2
define
u
x
 u )  0
2
P460 - harmonic oscialltor
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Harmonic Oscillators
• Solve by first looking at large |u|
d 2
du 2
 u   0    Ae
2
 u2
2
 Be
u2
but B  0 as  ()  0
• for smaller |u| assume
 (u )  Ae
d 2H
du 2
 2u
 u2
dH
du
2
H (u )


 (  1) H  0
• Hermite differential equation. Solved in 18th/19th
century. Its constraints lead to energy eigenvalues
• Solve with Series Solution technique
P460 - harmonic oscialltor
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2
Hermite Equation
d 2H
du 2
 2u
dH
du


 (  1) H  0

H (u )  a0  a1u  a2u ...   al u
2
l
l 0

dH
du
  lal u
l 1
l 1
d 2H
du 2

  (l  1)lal u
l 2
l 2
• put H, dH/du, d2H/du2 into Hermite Eq.
• Rearrange so that you have
2
3
A  Bu  Cu  Du ....  0
• for this to hold requires that A=B=C=….=0 gives


A term : 2a2  (  1)a0  0


B term : 6a3  2a1  (  1)a1  0
P460 - harmonic oscialltor
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Hermite Equation
• Get recursion relationship
• if any al=0 then the series can end (higher terms are
0). Gives eigenvalues
al  2 
 (   1 2 l )
( l 1)( l  2 )
al
• easiest if define odd or even types
a1  0 a0 , a2 ....non  zero (even)
a0  0 a1 , a3 ....non  zero (odd )
ends when
2 mE


 1  2l  0 gives
2E

 2l  1 (  class . freq.)
2m
h

El  h ( 12  l ) l  0,1,2...
2
Wave functions from recursion relations
 0 (u )  A0 e
u 2 / 2
 1 (u )  A1ue
 ( x)   ( x)
u 2 / 2
P460 - harmonic oscialltor
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H.O. - algebraic/group theory
• See Griffiths Sect. 2.3.1
• write down the Hamiltonian in terms of p,x
operators
H  E let  
1
2m
p
2

c
m
 2
 (mx) 2   E
• try to factor but p and x do not commute. Explore
some relationships, Define step-up and step-down
operators
a 
1
2 m
a a 
(  ip  mx)
1
2 m
( p  (mx)  im2  x, p )
2
2
commutator  [ A, B ]  AB  BA
a a 
1

H
i
2
x, p 
P460 - harmonic oscialltor
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H.O.- algebraic/group theory
• Look at [x,p]
x, p  f ( x)  
 if ( x)
or just x, p   i  Group A lg ebra
x df
i dx
d ( xf )
idx
• with this by substitution get
a a 
H


1
2
a a 
H


1
2
[ a , a  ]  1
• and Sch. Eq. Can be rewritten in one of two ways
 (a a  12 )  E
 (a a  12 )  E
• Look for different wave functions which satisfy
S.E.
P460 - harmonic oscialltor
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H.O.- algebraic/group theory
• Start with eigenfunction with eigenvalue E
H  E can then show
H '  ( E   ) ' with  '  a
H ' '  ( E   ) ' '
with  ' '  a
• so these two new functions are also eigenfunctions
of H with different energy eigenvalues
• a+ is step-up operator: moves up to next level
• a- is step-down operator: goes to lower energy
.a+(a+)
.
a+ 
a- 

a-(a- )
P460 - harmonic oscialltor
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H.O.- algebraic/group theory
• Can prove
H '  Ha    (a a  12 )a
 a (a a  12 )
(use a , a   1)
 a [ (a a  1  12 ) ]
 a ( H   )  a ( E   )
 ( E   )( a )
• this uses
H   (a a  12 )
• a similar proof can be done for step-down
• can raise and lower wave functions. But there is a
lowest energy level…..
P460 - harmonic oscialltor
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H.O. eigenfunctions
• For lowest energy level can’t “step-down”
a 0  0
1
2 m
or
d
( dx
 mx) 0  0
• easy differential equation to solve
ln  0  
m
2
 0 ( x)  Ae

x  const
2
m 2
x
2
• can determine energy eigenvalue
 (a a  12 ) 0  E0 0  E0  2
• step-up operator then gives energy and wave
functions for states n=1,2,3…..
 n  An (a )  0 and En  (n  12 )
n
P460 - harmonic oscialltor
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H.O. Example
• Griffiths problem 2.11. Compute <x>, <p>, <x2>,
<p2> and uncertainty relationship for ground state.
Could do just by integrating. Instead using step-up
and step-down operator.
a 
x
1
2 m

2 m
(ip  mx) 
( a   a )
pi
m
2
( a   a )
• With this
 x   * xdx   * ( a  a )dx
  n* ( n 1   n 1 ) dx  0(orthogonal )
  p  0 ( same)
• But consider
 (a ) 2   0*a a 0 dx   0* 2 dx
 0(orthogonality ) also  (a ) 2  0
but  a a  0  x 2  0
P460 - harmonic oscialltor
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H.O. Example
• Griffiths problem 2.11.
a 
x
1
2 m

2 m
(ip  mx) 
( a   a )
pi
m
2
( a   a )
• Can show using normalization (see Griffiths)
a n 
n n 1 and a n 
 a ( a n )  a
n  1 n 1
n n 1  n n
a ( a n )  a n  1 n 1  ( n  1) n
• using “mixed” terms
 x 2 


2 m
m
2
*

 0 (a a  a a ) 0 dx
( n  1  n) 
 p 2 


2 m
m
2

2 m
*

 0 (a a  a a ) 0 dx
(drop a a , a a as  0)
P460 - harmonic oscialltor
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H.O. Example 2
• Griffiths problem 2.41. A particle starts in the state:
 ( x,0)  A(1  2u ) e
2
u 2 / 2
u
  A(1  4u  4u )e
2
0  Ae
u 2 / 2
1  Bue
m

x
u 2 / 2
e iE0t / 
u 2 / 2
e
 i 3 E0 t / 
2  C (1  2u )e
2
u 2 / 2
e  i 5 E0 t /  
  D (30  41  22 )
• what is the expectation value of the energy?
32 E0  4 2 E1  2 2 E2
 E 
 2.7 E0
9  16  4
• At some later time T the wave function has evolved
to a new function. determine T
( x, T )  B(1  4u  4u )e
2
P460 - harmonic oscialltor
u 2 / 2
12
H.O. Example 2
• Griffiths problem 2.41. A particle starts in the state:
 ( x,0)  A(1  4u  4u )e
 ( x, t ) 
2
D (3 0 e

iE0t

 4 1e

i 3 E0t

u 2 / 2
 2 2 e

i 5 E0t

)
• At some later time T the wave function has evolved
to a new function. determine T
 ( x, T )  A(1  4u  4u )e
 ( x, T ) 
2
D (3 0 e
 De

iE0T


iE0T

 4 1e
(3 0  4 1e


i 3 E0T

i 2 E0T

u 2 / 2
 2 2 e
 2 2 e


i 5 E0T

i 4 E0T

• solve by inspection. Need:
 i 2 E0T / 
e
 1 and
 2 E0T /   
e
 i 4 E0T / 
P460 - harmonic oscialltor
1
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