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AN INVESTIGATION ON THE SOLUTION OF SIMPLE HARMONIC
OCILLATOR problem USING SCHRODINGER METHOD
A PROJECT REPORT
Submitted in the partial fulfillment of the requirement for the degree of
BACHELOR OF SCIENCE
In
PHYSICS
DEPARTMENT OF PHYSICS
CHAPTER-1
Schrodinger Equation and Wave function
Time dependent Schrödingerequation:
Thenon-dissipation of the wave packet of the material particle has been explained by
assuming the necessity of the guiding wave obeying Schrödinger wave equation which we shall
derive here.
Consider a system of stationary wave to be associated with the particle. Let  (r , t ) be the wave



displacement for the de Broglie wave at any location r  i x  j y  kz at time t. Then the
differential equation of the wave motion in three dimension in accordance with wax well’s wave
equation can be written as
1  2  2  2  2 1  2
I.e.    2 2 or 2  2  2  2 2
u t
x
y
z
u t
2
Where u is thewave velocity.
The solution of equation (1) gives  as a periodic displacement in terms of time i.e.
 ( r , t ) =  0 (r )e  it .
(2)
Where  0 is the amplitude at the point considered. Itis a function of position r i.e. of coordinate
(x y z) and not of time t.
The equation (2) May be expressed as
 (r , t )   0 (r )e  it
(3)
Differentiating equation (3) twice with respect to t we get
 2
  2  0 (r )e  it
2
t
  2  (r , t )
Substituting this in (1) we get
 2  2  2
2


 2 
x 2 y 2 z 2
u
(4)
  2 v
But

I.e.
Also
2 u

 2

u 
(5)
 2  2  2
 2  2  2
2
x
y
z
(6)
Using (5) and, (6) equation (5) becomes
2 
4 2
2
0
(7)
So far we have not introduced wave mechanical concept and so the treatment is general. For
introducing the concept of wave mechanics we must put from de Broglie equation.

h
mv
(8)
Substituting this equation (7), we get
2 
4 2 m2 v 2
0
h2
(9)
If E and V are the total and potential energies of the particle respectively then its kinetic energy
1 2
mv is given by
2
1 2
mv  E  V
2
This gives m2u 2  2m( E  V )
Substituting this equation (9), we get
2 
4 2 m
( E  V )  0
h2
(10)
The above equation is called Schrödinger time independent wave equation .The quantity
 is usually referred as wave function.
Let us now substitute, inequation (10)

h
2
(11)
Then the Schrodinger time independent wave equation, in usually used from, and may be written
as
2 
2m
2
( E  V )  0
(12)
2) Schrödinger equation for a free particle
For the free particle V=0; therefore if we put V=0 in equation (12) it will become the
Schrödinger equation for a free particle
i.e.
2 
2mE
2
0
3) Time dependent Schrodinger equation
Time dependent Schrodinger equation
Differentiating equation (3) with respect to t we get

 i 0 (r )e  it
t
 i (2 v) 0 (r )e it
 2 iv (r )


iE

i
i
Which gives
2 iE

h
(13)

t
E  i
(14)
Substituting value of E from above equation in (12), we get
2 
2m  

i
V   0
2 
 t

Or
2  
I.e.

2
2m
2m  

i
V 
2 
 t

2  V   i

t
(15)
This equation contains the time and hence is called time independent Schrodinger
equation
Equation (15) may written as
(
The operator (
2m
2m
 2  V )  i

t
(16)
 2  V ) is called Hamilton and is represented by H

operated on. Gives E which may be seen from (14).This equation (16)
t
may be written as
While operator i
H   E
(17)
The above forms of Schrodinger equation describe the motion of a non-relativistic material
particle.
PHYSICAL INTERPRETATION OF WAVEFUNCTION
It would be now interesting to interpret wave-function  appearing in Schrodinger equation,
physically in terms of the observable properties of the system. In the beginning it was
considered that the wave-function  is merely an auxiliary mathematical quantity employed to
facilitate computations relative to experimental results. This is true to a certain extent in
direction of the detector; but it does not seem reasonable to introduce an isolated mathematical
function without inquiring into its physical significance. Schrodinger himself attempted the
physical interpretation of  in terms of charged density. As the beams of electrons exhibit
diffraction phenomenon like X-ray, one might use the optical analogy to arrive at the physical
significance of  . In any electromagnetic wave system if A is the amplitude of the wave, them
the energy density, i.e., energy per unit volume is equal to A2 , so that the number of photons per
unit volume, i.e. Photon density is equal to A2 / hv where hv is the energy of a photon. This
means that the photon density is proportional to A2 . As hv is constant. By analogy, if  is the
amplitude of matter wave at any point in space, then the number of material particles per unit
volume i.e., the particle density must be proportional to  2 Thus the square of absolute value
2
of  ,i.e.,  , is a measure of the particle density. If q is the electric charge on a particle, the
charge density will be equal to the product of the charge q and the particle density. Thus the
2
quantity.  Is measure of charge density as Schrodinger considered. Usually  *  is written
instead of  , where  * .is the complex conjugate of 
2
This interpretation was found to lead to very satisfactory results when wave mechanics
was applied to the directional distribution of photo-electrons, intensity distribution in Compton
scattering, the stable states of Bohr atom, the emission of spectral lines etc., but the difficulty
arises when we wish to follow the flight of a single electron or any other material particle. The
wave packet associated with material particle dissipates in a course of time, so that it cannot
represent for long the material particle which maintains its identity. Moreover in such a case it is
natural to ask. “Where is the particle in relation to the wave packet?”.
“To remove the above discrepancy another physical interpretation of the wave function,
generally accepted at present, was suggested by Max - Bron in 1926 and then developed by
2
Bohr,Dirac, Heisenberg and others. According to this view.   * =  represents the
probability density of the particle in the state  Then the probability of finding the particle in a
volume element   dx dy dz about any point r at time t is expressed as
P(r )   (r , t ) 
2
The function  is sometimes called probability amplitude for the position of the particle.
The postulate suggested by Born shows that the
quantum mechanical laws and the results of
their measurements can be interpreted on the basis of probability considerations
NORMALISED WAVE FUNCTIONS.
We have seen that * represents the probability density and so * or.
2
  .represents the probability of finding the particle in the volume element 
In physical problems we come across situations where the particle is bound by forces to a
limited region. The examples of this kind are: - the electron in an atom, the particle in a box with
impenetrable walls. In such cases the total probability of finding the particle in the entire space
is of course, unity, i.e.
P(r )   (r , t )   1
2
Where the integration extends over all space.
  (r , t )
*
( r , t )  1
v
A wave function which satisfies the above equation is said to be normalized to unity (or
briefly, normalised).
When the particle is bound to the limited regions the probability of finding the particle at
infinite distance is zero, i.e., [psi-psi star] at x=plus or minus infinity or zero.
Any solution of the wave equation may be normalised by multiplying or dividing by a
constant and it can be readily seen that the result is also a solution of the wave function.
ORTHOGONALITY OF WAVE FUNCTIONS
An orthogonal wave function is of paramount importance in quantum mechanics. When two
wave functions are orthogonal, it means that they represent two mutually exclusive physical
states. We state this in quantum mechanics by taking the inner product, or the integral of the two
wave quality means that the two "vector spaces" which are being described by the wave
functions are not linear combination of each other.
The problem of liner harmonic oscillator is of importance since many systems of interest can be
1
approximated to it. It potential energy V  kx 2 , k is the force constant, is a continuous function
2
of the coordinate x and therefore is completely different from systems we considered so far
where the potential is constant over a region.
WAVE EQUATION
In SHM restoring force is proportional to displacement
F  kx
According to Newton's second law
F

md 2 x
dt 2
md 2 x
  kx
dt 2
md 2 x
 kx  0
dt 2
d2x k
 x0
dt 2 m
This equation represents a periodic motion of angular frequency


k
m , or
frequency

1 k

 
2 2  m 
The potential energy of oscillator is
x
V    kxdx
0

1 2
kx
2
With the force constant expression in equation (1.40), the potential V is given by
V
1
1
4 Vo2 x 2  m 2 x 2
2
2
(4.72)
The time-independent Schrodinger equation of the linear harmonic oscillator is then
d 2 2m 
1

 2  E  m 2 x 2   0
2
2
dx


Or
d 2
m
2 2
2
dx

mk  E 1

kx 2   0
2 
 k 2

1 d 2 
 2
mk dx 2 
m
mk 2 
E

x   0
2
2
k

Or
2
Let
mk
2
2E
m

2
k
Then
1 d 2
     2 x 2   0
2
2
 dx
Let
y x
d d dy
d


dx
dy dx
dy
2
d 2
d  d 



2
dx  dx 
dx
d  d  dy



dy  dx  dx
d  d 


dy 
dy 
d 2
2
dy 2

d 2
    y 2   0
2
 dy
I.e.equation (4.73) reduces to
 2
    y 2   0
x 2
(4.75)
ASYMPTOTIC SOLUTION
We shall investigate first the solution of equation (4.75), when y   when y is very large,
  y 2  y 2 And equation (4.75) becomes
 2
 y 2  0
2
x
Its asymptotic solutions are
 ( x)  e  y
2
/2
2
2
d 2
 y 2 e  y /2  e  y /2
2
dy
  y 2  1 e  y
2
/2
As y is very large
y

2
 1  y 2
2
d 2
 y 2 e  y /2  y 2
2
dy
y 2 /2
y 
Out of the two asymptotic solutions, e .is not acceptable as it diverges when
the exact
solution of equation (4.75) may be written as
  e y /2 and   e  y
2
2
/2
  e y /2 Is not acceptable since it increases with x
2
While
  e  y /2 Satisfies the condition
2
 The exact solution =(complimentary function +particular integral)
  H ( y) e y
  e y
2
/2
2
/2
H ( y)
Since substitution of equation (4.77) in equation (4.76) gives
2
 2
  y 2  1 e  y /2  y 2
2
x
  e  y /2 H ( y )
Where H(y) is a function of y and the product e
y 
 y /2
H ( y ) tends to zero as
SERIES SOLUTION
Substitution of equation (4.78) in equation (4.75) gives


2
d 2  y 2 /2
e
H     y 2  e  y /2 H  0
2
dy
2
2

d   y 2 /2 dH
e
 He  y /2 y      y 2  e  y /2 H  0

dy 
dy

e y
e
2
/2
 y 2 /2
2
2
 2 dH

d 2 H dH  y 2 /2

e
y  e  y /2
 He  y /2 y  y     y 2  e  y /2 H  0
2
dy
dy
dy


2
2
2
2
2
d2H
dH
dH
 ye  y /2
 e  y /2
 y 2 e  y /2 H   e  y /2 H  y 2 e  y /2 H  0
2
dy
dy
dy
d2H
dH
dH
y
y
 y2 H   H  y2 H  0
2
dy
dy
dy
d2H
dH
 2y
  H  2 y2 H  0
2
dy
dy
d2H
dH
 2y
 H  H  0
2
dy
dy
d2H
dH
 2y
    1 H  0
2
dy
dy
Hermite Equation
H "  2 yH ' 2kH  0
Solution

H ( y )   an y n
n 0

H '( y )   nan y n 1
n 1

H "( y )   n(n  1)an y n  2
n2
We shall look for H(y), a series solution of the type

H ( y )   an y n
n 0
Equation (4.80) when substituted in equation (4.79), we have

  k  1 k  2  a
k 0
k 2
  2k  1    ak  y 2  0




 n(n  1)an y n2  2 y  nan y n1  (  1) an y n  0
n2
n 1
n 0


n 0
n 0
ie  (n  2)(n  1)an  2 y 2  2 y  an y n  0

ie   (n  2)(n  1)an  2  2nan  (  1)an  y n  0
n 0
ie (n  2)(n  1)an  2  2nan  (  1)an  0
an  2 
2n  1  
a
 n  1 n  2  n
For the validity of this equation, coefficient of each power of y must vanish separately.
k
Coefficient of y when equated to zero gives the recurrence relation
ak  2 
2k  1  
a
 k  1 k  2  k
a
This formula allows the calculation of all even coefficients in terms of 0 and the odd
a
a 0
coefficients in terms of. 1 Equation (4.80) will have only odd coefficients if 0
and odd
a 0
coefficients if 1
Thus we have two independent solutions for equation and a linear
combination of the two will be the most general solution.
The solutions are:
H e ( y )  a0  a2 y 2  a4 y 4  ...
And
H 0 ( y )  y  a1  a3 y 2  a5 y 4  ...
ENERGY EIGEN-VALUES
equation (4.82),we get
when k  
ak  2 
2k  1  
a
 k  1 k  2  k
e y  1  y2 
2
y4 y4

 ...
2! 3!
ak  2 2k 2
 2 
ak
k
k

ak  2 2

ak
k
2
Consider the Taylor series expansion of ( y )
exp( y 2 )  1  y 2 

1
  k / 2  !y
y4 y6
 ...
2! 3!
k
0,2,4
(In general for even solutions)
The ratio of the coefficients of the successive terms in equation (4.86) is
 k / 2 !
ak  2
1
2



ak
 k / 2   1 !  k / 2   1 k
2
Where k is large. Therefore, for large values of k,   exp( y / 2) H ( y ) tends to behave like
exp( y 2 / 2) , if the series is even, and y exp( y 2 / 2) if the series is odd; which is not
acceptable. This unrealistic solution can be avoided if the series in equation (4.80) terminates
after a finite number of terms. In such a situation ( y )
2
Will tend to zero as y   because of the factor exp( y / 2) .The series can be terminated by
selecting  in such a way that 2k  1   vanishes for k=n. Thus one of the series becomes a
polynomial and the other can be eliminated by setting the first coefficient to zero. Substitution
of the value of  gives
consider k th
term
 2k  1    ak
2n  1 
2E

0
The series can be terminated by selecting  in such a way that 2k+1vanishes for k=n
i.e.
2n  1 
2E

0
i.e.
1

En   n    , n  0,1, 2,...
2

The energy value of an oscillator based on quantum theory is
En  n  , n  0,1, 2,...
From the two expressions, it is evident that the quantum mechanical energy value is higher than
1

the quantum theory value by 2
, which is the energy possessed by the lowest state n = 0.
1

This energy of 2
is called the zero-point energy which is a manifestation of uncertainty
principle. E = 0 means that both position and momentum are well defined oscillator without
1

violating uncertainty principle is 2
Figure 4.11(a) illustrates the energy values as given by
equation (4.88).
ENERGY EIGENFUNCTIONS
When   2n  1 , equation (4.79) reduces to
d 2 H n ( y)
dH n ( y )
 2y
 2nH n ( y )  0
2
dx
dy
The solutions of equation (4.89) are Hermite polynomials........... of degree n. The energy Eigen
functions can now be expressed as
  y 2  2 m 2
x
, y 
2


 n ( y )  N n H n ( y ) exp 
y2 
m
x2
The first-four Hermite polynomials are
y 2  H 0 ( y)  1
A Hermite Polynomials of degree n is defined by
d n e y
Hn( y )  (1) .e .
dy n
n
2
y2
 H 0 ( y)  1
H1 ( y )  2 y
H 2 ( y)
d  y2
( e H  2 y )
dx
2
2
2
 e y  2  y.e  y .  2 y  e  y 


 1 e y
2
 e y  4 y 2 .e  y  2
2
 4 y2  2
2
d 3  y2
H 3 ( y )  (1) e . 3 (e )
dy
y2
3.
 1  e
 e y .2
2
y2

d 2  y2
e  2 y
dy 2

2
d   y2
ye  2 y  e  y  1

dy 


2
2
2
2
 2e y e  y  2 y  2 y 2  e  y  2 y  e  y  4 y 


 2e y e  y  2 y  2 y 2  e  y  2 y  e  y  4 y 


3
 4 y  8 y  8 y
2
2
2
2
 8 y 3  12 y
The normalization condition leads to
1/2 
Nn
2




 m 
H
2
n
( y ) exp( y 2 )dy  1

Or
1/2
Nn
2


1/2 n

  2 (n !)  1
m



Or
1/2
1/2

1 

N n  


n
 m  2  n ! 
The normalized Eigen functions and the ground state Eigen function are given by
1/2
1/2

1 

 n ( y )  


n
 m  2  n ! 
  y2 
H n ( y ) exp 

 2 
 m 

  
1/4
 0 ( x)  
 m x 2 
exp  

2 

The wave functions................. and the probability density ................. of the lowest four states are
also illustrated in figure 4.11.
It may also be noted from figure that oscillator Eigen functions are even for even values of n
2
and odd for odd values of this is understandable as exp( y / 2) is always even and. H n ( y ) is
even or odd according as n is even or odd. It may also be noted that the quantum oscillator can
be found outside the parabolic potential barrier since  does not vanish at the classical turning
points. In other words, the particles can penetrate the barrier to some extent. This barrier
penetration is an important feature of quantum mechanics.
Another interesting point to be noted in connection with the results is the nature of
probability distribution of classical and quantum oscillators. Classically, the probability of
finding the oscillator at a given point is inversely proportional to its velocity at that point. The
total energy
E
1 2 1 2
mv  kx
2
2
Or

2E  kx
m
Therefore the classical probability
Pc 
m
2 E  kx 2
GRAPHICAL REPRESENTATION OF RESULT.
REFERENCE:
1.
2.
3.
4.
Aruldhas
- Quantum Mechanics
NouredineZettili- Quantum Mechanics-Concepts and Applications
Townsend- A Textbook of Quantum Mechanics
Arthur Beiser - Concept of Modern Physics, Sixth Edition.