Simple Solutions of the one-dimensional Schrodinger Equation

FAQs & their solutions for Module 2:

Simple Solutions of the one-dimensional Schrodinger Equation

Question1: Determine the energy levels and the corresponding Eigen functions of a particle of mass  in a one dimensional infinitely deep potential well characterized by the following potential energy variation

a x x

a x x

V















for and 0 for

0 for 0 ) (

(1) Solution 1: For 0 < x < a, the one dimensional Schrödinger equation becomes

0 )

2 (

2

2  k x 

dx

d  

(2) where

2

2 2

 k _ E

(3) The general solution of the Schrödinger equation is

 

x  AsinkxBcoskx

 (4)

Since the boundary condition at a surface at which there is an infinite potential step is that  is zero, we must have



x⁰







xa



⁰

 (5)

The above condition also follows from the fact that since the particle is inside an infinitely deep potential well, it is always confined in the region 0 < x < a and therefore  must vanish for x < 0 and x > L; and for  to be continuous, we must have



x0



B0

 (6)

and



xa



Asinka0

 (7)

Thus, either

0 A

or

 n

ka ; n1,2, (8)

The condition A = 0 leads to the trivial solution of  vanishing everywhere, the same is the case for n = 0. Thus the allowed energy levels are given by

, 3 , 2 , 1 2 ² ;

2 2

2 

 n

a E_n n



 

(9) The corresponding eigenfunctions are















 



 



 

a x x

a x a x

n

n a

and 0 0

0 2 sin 

 (10)

where the factor 2/a is such that the wave functions form an orthonormal set :



^{a m x n x dx  m n}

0

*( ) ( ) 

 (11)

It may be noted that whereas n(x) is continuous everywhere, dn(x)/dx is

discontinuous at x= 0 and at x = a. This is because of V(x) becoming infinite at x=

0 and at x = a .

Question2: Consider the potential energy variation given by



















b x V

b x x x

V

0

0 0

0 )

( (12)

Solution 2:

 

 x b

e kb A

kx A x



 







sin sin

Continuity of d /dx at x = b will give us

2

cot 2 

  

 (13) where

₂

2 0

2 2

 b

V

  (14)

and

2 2

2 2



Eb

  (15)

Question3: In continuation of the previous problem, assume

2 2

2

0

9

2   

 b V

(16)

Calculate the number of bound states and also the corresponding values of

2

2 2





Eb



 _.

Solution3: If we numerically solve the equation

2

cot 2 

  

 (17)

We will find that there are three bound states with

8.33877 and

64146 .

5 , 83595 .

 2



Question4: Show that the function ^ ^{x  A exp}



^{ x}



^;



^{ 0}



satisfies the one- dimensional Schrodinger equation corresponding to^{V x}  ^{  S}  ^x . Find the value of S and the corresponding value of the energy.

Solution4:

 

^{x A exp}



^x



^;

  



^{ 0}



Thus

  ^{for 0}

for 0

x

x

x A e x

A e x





  

 

Thus

 

^{for 0}

for 0

x

x

x A e x

A e x





   

  

The function ^ ^x has a discontinuity of 2A at x = 0. Thus

 ^{x 2 2A}  ^x

      

Question5: Solve the one-dimensional Schrodinger equation for

 

| 2

| 0

| 2

0 | x a x a V x

V











(18) and derive the transcendental equations which would determine the energy

eigenvalues.

(b) Show that if we let a  0 and V0   such that

S V

a ₀  (19)

we would obtain only one bound state with energy as given in the previous problem.

Solution 5: The transcendental equation determining the energy eigenvalues corresponding to symmetric states is given by

2

tan  

2



  

₍₂₀₎

where



₀



^{2 1/2}

2 2 

 

 V E a



  (21)

and

2 2 0 2

2 V a



   (22)

Notice that for bound states E is negative with |E| < V0. When V0   and a  0 such that V0a S we obtain

S a

2 2

2

   (23)

which tends to zero. Thus the root of the equation

2

tan 2 

  

will correspond to a very small value of  so that we may replace tan  by  to obtain

2 2

2  

  

or

2 0

2

4   



or ²



^{1 1 4 2}



2

1 

    

We neglect the minus sign and make a bionomial expansion to obtain

4 2

2  

  

or



₀



^{2 0}₂^{2 2 0}₄^{2 4}

2 2 4

2  

a V a

a V E

V  

   

or

2 2

2

E _S

Question6: Determine the normalized eigenfunctions of the momentum operator

op

p i d

   dx

(24)

and write the orthonormality and completeness conditions.

Solution 6: The eigen value equation for the operator

op

p i d

  dx will be

   

op p p

p u x  p u x

where p (on the RHS) is now a number. Thus

   

p p

i d u x p u x

 dx 

Simple integration will give us

 

^{1 ;}

2

i p x

up x e p

  ^     



where the factor ¹

2 is introduced so that

   

^{1   1  }

2 2

i p p x i p p

*

p p

u x u x dx e dx e ^ d

 

        



  

 



_



^



   p p   

which is the ortho-normaility condition. Similarly,

    ^{1   1  }

2 2

i x x p i x x

*

p p

u x u x dp e dp e ^ d

 

        

  

  



_



^



^



^{x x }



is the completeness condition.

Question7: In continuation of the previous problem show that the eigenfunctions of the operator

2 2 2

2 2 2

pop d

H ^  ^{ }  dx

 (25)

are the same as that of the operator

op

p i d

   dx

Solution7:

2 2 2

2 2 2

p d

H  m   m dx

It is easy to see that

  ¹

2

i p x

up x e

  ^



are eigenfunctions of H. Thus the functions up x are also simultaneous eigenfunctions p , p_{x x}² and H .