CH.III
Chapter III
The Linear Harmonic Oscillator
The potential energy of the linear Harmonic Oscillator is given by
V x   12  2 x 2 .

2 2
 x   12  2 x 2  E  x 
2

Let  
 The Schrodinger equation becomes


d 

x

(1)
dx 
d
 d
d 2  d 2

 2 
dx
 d
 d 2
dx
d 2    2 E


  2    0
2
 

d
(2)
Eq.(2) doesn't lead to a two-terms recurrence relation. To solve it we first note that as
 
Eq.(2)
approximated to
d 2  
d
2
  2    0
    e 

2
2
 e
2
2


 0  we propose a solution for Eq.(2) of
But the second term violates the boundary condition    

the form
    e 
so
and
2
2
v 
2
d  
 e 
d
d 2  
d 2
 e 
2
2
(3)
v   e 
2
2
2
v    2e 
dv 
d
2
2
v   2e 
2
2
dv   2
e
d
2
d 2v 
d 2
 Equation (2) becomes
d 2 v 
d
2
 2
dv   2 E 

 1v   0
d
  
(4)
Equation is called Hermit's differential equation where its solution is represented by infinite series of the form

v    ak  k
k 0
Substituting back into equation (4) we get


k 0
k 0
 2E  
 1  ak  k  0



k  0
 k k  1ak  k  2  kak  k  
1
CH.III
Equating the coefficient of 
k

to zero
k  2k  1ak  2  2kak   2 E  1ak
 
ak 2


0
 2E

   1  2k  
 
a k
 k  2 k  1 


(5)
If we set a1  0 and choosing a0 arbitrary we generate the even solution. In the other hand, choosing a1
arbitrary and letting a0  0 we generate the odd solution 
   e 
2
2
veven    vodd  
It is clear from Eq.(5) that
ak  2
2



k


ak
k
Now let us examine the series e
2

 
j 0
a k 1
ak
(6)
2j
j!

 
k
k 0  k

 !
2

k
 !
1
2
2
   
k




k
k  k 
  1!   1
2  2 
2
 v  behaves like e as     both the even and the odd series violate the boundary conditions
   0.
 
To solve this dilemma we have to terminate the series after a finite number of terms, say n, 
an  2  0
 from Eq.(5) we get
2E
 1  2n 

(7)
If n is even  a n  2 and all higher terms vanish and since the odd series get unacceptable solution we let
a1  0 . If n is odd we retain only the odd series by letting a0  0 .  the solution exits for integer n  from
Eq.(7) we get


E  n  12 
(8)
And the eigen functions are
    Ne 
2
2
v  or
2
CH.III
 x   Ne


2
x2
  
H n 
x 



(9)
Where H n are the Hermit's polynomials with the following properties:
g  x, t   e  t
2

  H n x 
 2tx
n 0
H n x    1n e x
dn
2
dx n
e x
tn
n!
generating function
2
Rodrigues formula
H n  x    1n H n x 
Parity
H n x   2 xH n x   2nH n x   0

e
x2

Differential equation
H n  x H m  x dx  2 n  n! nm
Orthogonality
H n 1 x   2 xHn x   2nH n 1 x   0
H n x   2nH n 1 x   0
H n x   2 xH n x   2nH n x   0





Recurrence relations
Now using the Rodrigues formula one can prove that
H 0 x   1
H1 x   2 x
H 2 x   4 x 2  2
H 3 x   8 x 3  12 x
Now since the wave function given by Eq.(9) must be normalized we have





2

  x  x dx  N  e
2 2
x
H n x H n x dx  1
with  


Using the orthogonality property we get
N2

2
n

  
 n! 1  N  n


2  n!   
  
n x   

  
1
4

1
2 n n!

1
   4  2 x
0 x   
 e
  
 4 3 3 

1 x   
  3 


1
4
e
2
x2
4
1
2 n n!
  
H n 
x 





2

xe

1
Eo  12 

2
x2
E1  32 
3
CH.III

1
   4   2   2 x
2 x   
x  1e
 
 4   4

2
E2  52 
 n ( )
Exercise:
1- Prove that
And
p
kn
x
i
kn


2 

2
2- Prove that x 2 


|  n ( ) |2
n k , n 1  n  1 k , n 1
n  1 k , n 1  n k , n 1
n  12 


&
4



p 2   n  12

CH.III
The Dirac Notation Method
Let n be the eigen ket of the system

H n  En n
(1)
With
H
p2 1
  2 x 2
2 2
(2)
Let us introduce the dimensionless coordinate and momenta operators

x 
2
x
p 
and

1
2 
p


H   p2  x2  x  ipx  ip  ix, p
But x, p  
1
x, p   i
2
2




H   p2  x2   x  ipx  ip  12

and x  ip  a †
x   ip   a
Now let

H   a †a  12

(3)

(4)
Substituting back in Eq. (1) 


 a †a  12 n  En n  a †a n  12  n  En n
(5)
Letting a † a n  n n

(6)
Eq.(5) becomes


 n  12 n  En n or

E n   n  12

(7)
To find the values of n we operate on Eq.(6) by a we get
aa † a n  n a n
(8)
From the definition of Eq.(3) it is easy to prove that
a, a  1 or aa
†
†
 1 a † a
(9)
Substituting Eq.(9) into Eq.(8) we get
5
CH.III
1  aa a n    a n
aa a n     1a n 
†
n

†
(10)
n
From Eq.(10) we conclude that a n
 is also eigenket for aa†  with the eigen value n  1 ,i.e.,
a n  C n n  1
(11)
By operating on Eq.(6) by a † one can prove that
a † n  C n n  1
(12)
Now multiplying Eq.( 11) by its bra form
n a † a n  Cn
2

n 1 n 1
(13)
Using Eq.(6), Eq.(13) becomes
2
n  Cn
(14)
Similarly, multiplying Eq.(12) by its bra form
n aa † n  Cn
2

n 1 n 1
Using Eqs.(9) and (6) we get
1  n  Cn
2
(15)
Now from Eq.(14) we conclude n  0
 there must be a nonnegative minimum value  min such that
,using Eq.(11)

a nmin  C min
nmin  1


Cmin
0

From Eq.(14) we conclude that min  0 .
By applying the operators a and a † repeatedly on Eq.(6) we can generate from any given eigenket n new
eigenkets with different eigen values that is integrally spaced
and zero. Setting

 n composed of a set of positive integers
n  n and using Eq.(7) we obtain
E n   n  12

(16)
Now from Eqs.(14 & 15) we have
Cn  n and Cn  n  1
 Eqs. (11 & 12) now read
a n  n n 1
(17)
6
CH.III
a† n  n 1 n 1
(18)
†
Equations (17) and (18) specify completely the operators a and a as the lowering (annihilation) and raising
(creation) operators, respectively.
Now from Eq.(3) we have
and x  ip  a †
x   ip   a

2
1
xi
2
pa


&
2
1
x i
2 
p  a†
Adding and subtracting the above two equations we get, respectively
2
x  a  a†

x
and
2
i

p  a  a†

a  a 
2 

†

pi
2
a
†
a
(19)

(20)
Let is now evaluate the matrix elements of some operators
X mn  m x n 
X mn 

2
a  a  m a  a  n
†
 n m n 1 
2 

†
n 1 m n 1

X mn 

x



 
2  



0
1
0
1
0
2
0
0
2
0
0
3





0 

3 
0 



0
amn  m a n  n m n  1  n m, n 1
0

0
a  0

0


1
0
0
2
0
0
0
0





0 

3 
0 



0
And
7

2 
 n m, n 1 
n  1 m, n 1

CH.III
a 
 m a † n  n  1 m n  1  n  1 m, n 1



† 
a 




0
1
0
0

†
mn
0
0
2
0

Using Eq.(4) we get
0
0
0
3








 
0
0
0
0

 


H mn  m H n   m a † a n  12  m n   n  12  mn
1

0
 
H
0
2 
0


0
3
0
0

0
0
5
0

0
0
0
7








 
Let us generate some eigen functions:
Using Eq.(17) we have
a0 0
Expressing a in terms of x & p
 

1
 
xi
p  0  0
2  
 2
 
 d 


 2 x  2 dx  0  0



 
 d 


 2 x  2  dx 0  0 


1 d 

 x 
0  0 
 dx 

d
0   2 x0  0 
dx
Now from Eq.(18) we have
Again expressing

1
d0
  2 x 
0
  4 
0    e
 
 2 x2
2
a† 0  1
a† in terms of x & p
 

1
xi
p o  1
 
2  
 2
 
1 d 

o  1
x

 2
2 dx 

Substituting for o it is easy to prove that
8

CH.III
 
1   
 
1
4
2 x e

x 2
2
The Motion of the Wave Packets:
It is known that  x, t 
i
 Et

  Cn e
n
n 0

x 
(21)


  x,0    C n n  x 
n0
If   x,0  is normalized then

2
 Cn  1
(22)
n0
Substituting for En from Eq.(16) into Eq.(21) we get
  x, t 
i
 t 
e 2
 Cn e  int n x 
(23)

n0

H      x, t H  x, t dx



 2 2 1

H      x, t  
  2  2 x 2   x, t dx
 2




Substituting for x, t  from Eq.(23)

H   Cn
2
n0

 2 2 1

2 2



x
 n   2   2  x n x dx




Using the Schrödinger equation we have
 2 2 1


  2  2 x 2 n  x   En n  x 
 2







n0

H   C n En  n   x n  x dx   C n En
2
n0
2
(24)
From Eq. (22) and (24) we conclude the two facts that
(1) The only possible results of measuring the energy of the system are the energy values En.
(2) If the system is in the state x, t  , the probability of obtaining the result En is equal to Cn
postulate number (IV).
Let us now find the expectation value of the position operator with respect to the state x, t  .
9
2
as stated in
CH.III




x   n  x, t xk x, t dx    C nC k e i n  k t  n   x xk x dx
n0 k 0


But

 n x xk x dx  2  n k , n 1  n  1 k , n 1 






x 
 n CnCn 1e it  Cn1Cn e  it
2  n  0


Setting Cn  Cn e i n 


n 
i t  n  n 1 
 Cn Cn 1 e  i t  n1  n  
 Cn Cn 1 e
2 n  0
x 

2n
x 

 C n Cn1 cost   n1   n 
(25)
n 0
Let us test the last result at the limit n → ∞. At this limit we can assume  n1   n   &
Cn1  Cn
Again for n → ∞ we can write for the energy En  n .Then Eq.(25) now reads

2
x 

2
 Cn
2
n 0
E n cost   
Knowing that

2

2
 Cn
n 0
2
En  E

x 
In classical mechanics we have
x  A cost   
x 
2E
 2
and E  12  2 A 2 , then
cost   
(27)
Which is similar to Eq.(26).
10
2E
 2
cost   
(26)
CH.III
In chapter (I) we proved that
d
1
A, H   A
A 
dt
i
t
but x, H  
ip

d
1
x , H 
x 
dt
i


p
d
x 
dt

(28)
Similarly
d
1
 p, H  , and  p, H    2i  2
p 
dt
i
d
p    2 x
dt

(29)
Differentiate Eq.(28) with respect to t we get
d2
dt 2
x 
1d p
 dt
Using the result of Eq.(29) we get
d2
dt
2
x   2 x
x  x
o
cost   

(30)
Which is similar to Eq.(27).
11