Quantum Mechanics 2023-2024 Chapter Three

Chapter Three
Simplified quantum systems

3.1 Introduction
The present and the following chapters deals with applications of the quantum
mechanics rules mentioned in last chapter for different systems. Indeed this
analysis are set such that, the considered systems are gradually regarded from
simple to complicate. In this chapter a simplified systems that mainly concerns
with a free particle will investigated.

3.2 Free particle

Free particle defined as that particle which moves free from a constraint force,
which means it has not bound by an external force or equivalently moves in a
region where its potential energy is vanished. However, such definition implies
that this particle should have a constant velocity. Therefore, free particle that has
zero potential energy (V (x) = 0) is the simplest system that can consider in
quantum mechanics. Consequently, Schrödinger equation (2-10) for this system
becomes;
  2 d 2 ( x)
 E ( x)
2m dx 2
2mE
Considering that, k 2  , this equation can be written as follows;
2
d 2 ( x)
2
 k 2 ( x)  0 …………..(3-1)
dx
Since equation (3-1) is a seconed order, linear and homogenous differential
equation, its admits a special solution of the forms;

𝜓(𝑥) = 𝑒 𝑖𝑘𝑥 …………..(3-2)

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and;
𝜓(𝑥) = 𝑒 −𝑖𝑘𝑥 …………..(3-3)

Equation (3-2) represent the wave function that describe the free particle that has
2k 2
momentum k and energy ( ) and moves along the positive direction of the
2m
x-axis. While equation (3-3) represent the wave function that describe the free
particle that has a same momentum and energy and moves along the negative
direction of the x-axis. The general solution of equation (3-1) is a linear
combination of the two solutions (3-2) and (3-3). i.e,
 ( x)  A eikx  B e ikx …………..(3-4)
Where, A and B are two arbitrary constants. The probability density for a free
particle is similar at any point along the coordinate-x. i.e, |𝜓|2 = 𝜓 ∗ 𝜓 = 1, see the
figure below. In other word, the wave function 𝜓 = 𝑒 ∓𝑖𝑘𝑥 describe the situation in
which we have a complete uncertinity about position. This is in agreement with the
uncertainty principle because the wave function 𝜓 = 𝑒 ∓𝑖𝑘𝑥 describe a particle
whose momentum is precisely known (𝑝 = ℏ𝑘), hence we exactly know that; Δ𝑝 =
0 which require that Δ𝑥 = ∞.

|𝜓|2

-x x

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Quantum Mechanics 2023-2024 Chapter Three

Example: By using the uncertainty Principle show that the variance in defining the
position of a free particle is infinite.

Solution:
p x  
 p x  k For free particle
 p x  0
 
 x   
p zero
 x  

3-3 Potential step (barrier)

Let’s here try to find the energy and wave function for a particle moves in a
potential varies as follows;
𝑧𝑒𝑟𝑜, 𝑥<0 𝑟𝑒𝑔𝑖𝑜𝑛 𝐼
𝑉(𝑥) = {
𝑉𝑜 𝑥≥0 𝑟𝑒𝑔𝑖𝑜𝑛 𝐼𝐼
Or graphically;

It should be mention that, the step potential argued above is an approximated case
for the dashed curve in the last figure, where free electrons in metals suffer from.
However, the reason behind such an approximation is to get simplifying the

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mathematical procedure. Here we will investigate the cases where the energy of the
incident particle is less and greater than the potential step respectively.
a) E  V

In sense of classical mechanics, this region considered to forbidden for the

particle to be in because it has a kinetic energy less than the potential of the step.
Anyway, the situation is differs in quantum mechanics, as we shall see.
Schrodinger equation (2-10) can be written as follows;
𝑑2𝜓 2𝑚
+ (𝐸 − 𝑉(𝑥̂))𝜓 = 0 …………..(3-5)
𝑑𝑥 2 ℏ2

i)Region-I.
For this region Schrodinger equation (3-5) becomes;

𝑑2𝜓
+ 𝑘2𝜓 = 0 …………..(3-6)
𝑑𝑥 2

2mE
Where k 2  . The general solution of the such equation is given by;
2

 1 ( x)  A eikx  B e ikx …………..(3-7)

Where A e ik x is the wave function for the incident particle and B e ikx is its
counterpart for the reflected one.
ii) Region-II
Schrödinger equation (3-5) for this region converts to;
d 2 2
2
  2 2  0 …………..(3-8)
dx
2m
Where  2  (V  E ) . The general solution for this equation is;
2

 2 ( x)  C e x  D e x …………..(3-9)

The second term in the last equation must be neglected due to tow reasons, the first
one is the finite condition and second one is that there is no reason make the
penetrated particle to reflected back. Hence, equation (3-9) becomes;

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Quantum Mechanics 2023-2024 Chapter Three

 2 ( x)  C e x …………..(3-10)
The fact that  2 not equal to zero means that there is a certain probability for the
particle to be exist in this region, although it being forbidden region in sense of
classical mechanics. Furthermore, this probability get rise decreases as the particle
going deeply in this region. In order to find the constants A, B, C we must start
applying the continuity condition for the wave function and its first derivative at x
= 0 as follows;
1.  1 ( x  0)   2 ( x  0)

d 1 d 2
2. 
dx x 0 dx x 0

Therefore, from the first condition we have;

A BC
However, from the second one get;
ik ( A  B)   C

This leads to;

 2ik   ik   
H.W C   A , B   A
 ik     ik   
So equations (3-7) and (3-10) becomes;

 1 ( x) 
A
ik  

(ik   ) e ik x  (ik   ) e ik x  …………..(3-11)

2ikA x
 2 ( x)  e …………..(3-12)
ik  
2 2
If we consider that A is the intensity of the incident wave function then B is
the intensity of the reflected wave function, which is given as;

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Quantum Mechanics 2023-2024 Chapter Three

2   ik     ik    *
B  B*  B     A  A
  ik     ik   
2   ik      (ik   )  *
B  B*  B        A  A
  ik      (ik   ) 
2 2
B  A

Therefore, the intensity of the incident wave function is equal to the intensity of the
reflected one. The physical meaning of this result is that all of the incident particles
that reach the potential barrier (potential step) when E  V are reflected back
including those that penetrate the region II. Anyway, the using of Euler’s theorem
can leads to rewrite equation (3-11) to be as follows;
2ikA 
 1 ( x)  (cos kx  sin kx ) …………..(3-13)
ik   k
2ikA x
 2 ( x)  e …………..(3-12)
ik  
2ik
By neglect the constant , the last two equations can be drawn as follows.
ik  

Now as the value of potential step V increases the value of  increases too and so
for very high values ofV , the last two equations becomes;
𝜓1 (𝑥) = 2𝑖𝐴 𝑠𝑖𝑛𝑘𝑥 ≈ 𝑠𝑖𝑛𝑘𝑥 …………..(3-14)
𝜓2 (𝑥) = 0 …………..(3-15)
Indeed, this mean that no particle can be penetrate region-II for this case and hence
all of the incident particle being reflects back at x=0.

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Quantum Mechanics 2023-2024 Chapter Three

b) E  V
For such a case the results of classical physics include that all of the incident
particle are going through the region-II. However, result of Q.M. shows different
outcomes as the following section reveals.
i) Region-I
Schrodinger equation in this region takes the following form;
d 2 1 ( x)
2
 k 2  1 ( x)  0 …………..(3-16)
dx
2mE
Where k is expressed as; k 2  , and the solution of equation (3-16) is given
2
by;

 1 ( x)  A eikx  B e ikx …………..(3-17)

The presence of the second term in the last formula refers to there some particles
are reflects back at x=0.
ii) Region-II
Schrödinger equation in this region must take the form;
d 2 2
2
 k 2  2  0 …………..(3-18)
dx
2m
Where k  2  ( E  V ) , and the solution for this equation given by;
2

 2 ( x)  C eikx  D e ikx …………..(3-19)

The second term must be neglected because there are no reason makes the particle
to reflecting back towards the negative part of x-axis. Therefore;
 2 ( x)  C eikx …………..(3-20)
In order to evaluates the constants A, B and C we have to making use continuity

condition for the wave function and its derivatives at x=0 as shown below.

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1.  1 ( x  0)   2 ( x  0) , this leads to; A  B  C

d 1 d 2
2.  , this leads to; k ( A  B)  k C
dx x 0 dx x 0

Consequently, from the two conditions we get;

 2k   k  k 
H.W: C    A, B   A
 k  k   k  k 

Example: Determine the reflection and transmission coefficients of the potential

step for the case E  V .
Solution:
𝑝
Let us define the following quantities; 𝑣 = = ℏ𝑘/𝑚 is the particle velocity in
𝑚

the region-I. |𝐴|2 is the intensity of the incident beam of particle (or intensity of the
incident particles, i.e. the number or particles per unit volume). 𝑣|𝐴|2 is the flux of
the incident beam (the number of particles that passes from unit area per unit time).
𝑝́
𝑣́ = = ℏ𝑘́/𝑚 is the velocity of the particles in region-II, |𝐵|2 is the intensity of
𝑚

the reflected beam of particle, 𝑣́ |𝐵|2 is the flux of the reflected beam and 𝑣́ |𝐶|2 is
the flux of the transmitted beam.
Therefore, the reflection and transmission coefficients can be set up such that;
2
𝑣|𝐵|2 𝑘−𝑘́
𝑅= =( ) …………..(3-21)
𝑣|𝐴|2 𝑘+𝑘́

𝑣́ |𝐶|2 4𝑘𝑘́
𝑇= = 2 …………..(3-22)
𝑣|𝐴|2 (𝑘+𝑘́ )

H.W: Show that the law of conservation of mass is satisfied for the above
example. i.e. R+T=1.

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Quantum Mechanics 2023-2024 Chapter Three

3-4 Particle in a box

Let’s now try to investigate a free particle moves in a potential distribution as
shown below;

0 0x a
V (x )  
 0x a

It is seen that for such a system the particle is restricted to move in the confined
region 0 ≤ 𝑥 ≤ 𝑎 due to the infinite potential value that prevents the particle to
penetrate the barrier at x = 0 and x = a. This system looks like a molecule in a gas
cylinder that cannot escape out due to the solid walls of the cylinder. In addition,
this system is similar to a free electron in metals.

H.W: Prove mathematically that 𝜓=0 in the regions 0 ≥ 𝑥 ≥ 𝑎 and state the
physical meaning for that.

Anyway Schrodinger equation in the region under consideration is given by;

d 2 ( x)
2
 k 2 ( x)  0 …………..(3-23)
dx
2mE
Keeping in mind that k 2  , the solution of this equation is;
2
 ( x)  A eikx  B e ikx …………..(3-24)
The constants A and B can found as before with aid of the boundary condition as in
below. Indeed, at x=0 the wave function in (3-24) vanishes, i.e.  ( x  0)  0 . So,
A  B  0 and hence A  B , therefore;
 ( x)  A ( eikx  e ikx)

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Using Euler’s method e  i  cos  i sin  last equation becomes;

 ( x)  A (cos kx  i sin kx  cos kx  i sin kx )
 ( x)  2iA sin kx
 ( x)  C sin kx
Where C  2iA , and similarly at x=a the wave function is also vanishes, i.e.
 ( x  a)  0 , therefore;
 ( x  a)  C sin ka  0
This requires that; sin ka  0 , which leads to ka  n and thus;
n
k
a
Consequently, the energy eigen values and eigen wave functions of a particle
moving in a potential box becomes respectively as follows;
 2 2 n 2
En  and  n ( x)  C sin(n x / a)
2m a 2
It is obvious that the energy of this system is quantize because the particle is
confine in a limited region by means of the potential. Accordingly, we may draw
the following;

n4 E 4  16E1

n3 E 3  9E1

n2 E2  4E1

 2 2
n 1 E1 
2ma 2

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 2 2
Indeed, the energy of ground state, E1  call Zero Point Energy. Concerning
2m a 2
with the wave function ( n ( x)  C sin(n x / a) ) the normalization constant C can
found as follows;

 n ( x) n ( x) dx 1
*

C sin(n x / a) C sin(n x / a) dx  1
*

 sin (n x / a) dx  1
2 2
C
0

Using the identity; sin 2   2 (1  cos 2 ) ,we get;

1

1 2 
a a
C  dx   cos(2n x / a) dx  1
2 0 0 

1 2 a a
a


C x 0 ( ) sin(2n x / a)   1
2

 2n 0
C 2 {( a  0)  (a / 2n )( 0  0)}  1
1
2

1
2
C 2a  1 C 2
a

  n ( x)  2
a
sin( n x / a )

The figure below shows the first three eigen wave functions ψ1, ψ2, ψ3.

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3.5 Particle in potential box in three dimensions

The same procedure used in last section repeats here but in three dimensions x,y
and z instead of one. So, the particle now moves freely in a box of dimensions a,b
and c as shown in figure below.

The Schrödinger equation for this system must written as follows;

 2 ( x, y, z )  2 ( x, y, z )  2 ( x, y, z )
  k 2  ( x, y, z )  0 …………..(3-25)
x 2
y 2
z 2

2mE
Keeping in mind that; k 2  . However, to solve this equation the method of
2
separating of variables must be use. Accordingly, we may write;
 ( x, y, z )   ( x)  ( y)  ( z ) …………..(3-26)
The substitution of equation (3-26) into equation (2-25) and dividing the resultant
expression on equation (3-26) leads to;

1 d 2 ( x) 1 d 2 ( y ) 1 d 2 ( z )
   k2  0 …………..(3-27a)
 ( x) dx 2
 ( y ) dy 2
 ( z ) dz 2

Or;
1 𝑑 2 𝜓(𝑥) 1 𝑑 2 𝜓(𝑦) 1 𝑑 2 𝜓(𝑧)
+ 𝑘𝑥2 + + 𝑘𝑦2 + + 𝑘𝑧2 = 0 ……..(3-27b)
𝜓(𝑥) 𝑑𝑥 2 𝜓(𝑦) 𝑑𝑦 2 𝜓(𝑧) 𝑑𝑧 2

Which is equivalent to;

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Quantum Mechanics 2023-2024 Chapter Three

𝑑 2 𝜓(𝑥)
+ 𝑘𝑥2 𝜓(𝑥) = 0
𝑑𝑥 2
2
𝑑 𝜓(𝑦)
+ 𝑘𝑦2 𝜓(𝑦) = 0 …………..(3-27c)
𝑑𝑦 2
2
𝑑 𝜓(𝑧)
+ 𝑘𝑧2 𝜓(𝑧) = 0
𝑑𝑧 2

The solution of each of thses three equations is as shown below respectively;

𝜓(𝑥) = 𝐴𝑥 𝑒 𝑖𝑘𝑥 + 𝐵𝑥 𝑒 −𝑖𝑘𝑥

𝜓(𝑦) = 𝐴𝑦 𝑒 𝑖𝑘𝑦 + 𝐵𝑦 𝑒 −𝑖𝑘𝑦 …………..(3-28)
𝜓(𝑥) = 𝐴𝑧 𝑒 𝑖𝑘𝑥 + 𝐵𝑧 𝑒 −𝑖𝑘𝑧
By using the boundary conditions [ψ(x=0, x=a) = 0, ψ (y=0, y=b) = 0, ψ(z=0, z=c)
= 0], we can get;
𝜓(𝑥) = 𝐶𝑥 𝑠𝑖𝑛𝑘𝑥 𝑥
𝜓(𝑦) = 𝐶𝑦 𝑠𝑖𝑛𝑘𝑦 𝑦 …………..(3-29)
𝜓(𝑧) = 𝐶𝑧 𝑠𝑖𝑛𝑘𝑧 𝑧
Therefore the wave function of this system is as follows;
𝜓(𝑥, 𝑦, 𝑧) = 𝐶𝑥 𝑠𝑖𝑛(𝑘𝑥 𝑥). 𝐶𝑦 𝑠𝑖𝑛(𝑘𝑦 𝑦). 𝐶𝑧 sin(𝑘𝑧 𝑧) ………..(3-30)
Remember that; 𝑘𝑥 = 𝑛𝑥 𝜋/𝑎, 𝑘𝑦 = 𝑛𝑦 𝜋/𝑏 and𝑘𝑧 = 𝑛𝑧 𝜋/𝑐 one get;
𝑛𝑥 𝜋 𝑛𝑦 𝜋 𝑛𝑧 𝜋
𝜓(𝑥, 𝑦, 𝑧) = 𝐶 sin( 𝑥). sin( 𝑦). sin( 𝑧) …..……..(3-31)
𝑎 𝑏 𝑐

8
H.W: Show that; C 
abc
𝑝2 ℏ2 𝑘 2
Concerning with energy eigen values we have that; 𝐸 = = , so;
2𝑚 2𝑚

ℏ2
𝐸= (𝑘𝑥2 + 𝑘𝑦2 + 𝑘𝑧2 ).
2𝑚

The substituation of thevalues of 𝑘𝑥 = 𝑛𝑥 𝜋/𝑎, 𝑘𝑦 = 𝑛𝑦 𝜋/𝑏 and 𝑘𝑧 = 𝑛𝑧 𝜋/𝑐 leads

to;
2
ℏ2 𝜋2 𝑛𝑥2 𝑛𝑦 𝑛𝑧2
𝐸= ( + + ) …..……..(3-32)
2𝑚 𝑎2 𝑏 2 𝑐2

For the case when a=b=c last formula becomes;

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Quantum Mechanics 2023-2024 Chapter Three

 2
E 2
(n x2  n 2y  n z2 ) …..……..(3-33)
2ma
Or equvilently;
𝐸 = 𝐸1 (𝑛𝑥2 + 𝑛𝑦2 + 𝑛𝑧2 ) = 𝐸1 𝑛2 …..……..(3-34)
Equation (3-34) represent the the allowable energies eigen values, however
it indicates that all of the states characterized by (nx,ny,nz) that gives the same
value of N have the same value of energy. Indeed this is directly shows to
the degeneracy where there are more than one wave function describe the
same state, see the table below.

Degeneracy Combination of (nx,ny,nz) Energy

1 (1,1,1) 3E1
3 (1,1,2) (1,2,1) (2,1,1) 6E1
3 (1,2,2) (2,1,2) (2,2,1) 9E1
3 (3,1,1) (1,3,1) (1,1,3) 11E1
1 (2,2,2) 12E1
6 (1,2,3) (3,2,1) (2,3,1) (1,3,2) (2,1,3) (3,1,2) 14E1

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3-6 Density of states

Concerning with equation (3-34) it is obvious that the energy interval between
any two-adjecint states depend on the widgth of the potential cube. Consequently,
these states becomes very much closes to each other and hence indistinguishable as
the box widgth increases. Strictly speaking, this means that as the size of of the box
leaves the microscopic scale their dynamical variables becomes of a continuum
values.

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Anyway, the problem now is to find the number of energy levels for a range dE
of the energy when the potential box being very large. Actually, the number of
states (energy levels) that have energy vales between 0 to E (0 to k) can be deduces
1 4
with aid of the figure below to be;
8
(3 𝜋𝑛3 ), therefore;

1
8𝜋𝑉
𝑁(𝐸) = (2𝑚3 )2 𝐸 3⁄2 …..……..(3-35)
3ℎ3

Where V = a3 is the potential box volume. Consequently, the number of states

(energy levels) with energy between E and E+dE is;
1
4𝜋𝑉(2𝑚3 )2
𝑑𝑁(𝐸) = 𝐸 1⁄2 𝑑𝐸 …..……..(3-36)
ℎ3
1
4𝜋𝑉(2𝑚3 )2
The quantity 𝐸 1⁄2 called density of states and defined as the number of
3ℎ3

states (energy levels) with energy between E and E+dE and mathematically written
𝑑𝑁(𝐸)
as; 𝑔(𝐸) = . However, the integration of the density states curve, see figure
𝑑𝐸

below, from E to E+ε gives the total number of states in that energy interval.

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Quantum Mechanics 2023-2024 Chapter Three

H.W: show that;

𝑑𝑁(𝑃) 𝑑𝐸 4𝜋𝑉
i- = 𝑔(𝑃) = 𝑔(𝐸) = 𝑃2
𝑑𝑃 𝑑𝑃 ℎ3
𝑑𝑁(𝛾) 𝑑𝑃 4𝜋𝑉
ii- = 𝑔(𝛾) = 𝑔(𝑃) = 𝛾2
𝑑𝑃 𝑑𝑣 𝑐3

3.7 Potential Barrier Penetration

The fact that the wave function can be continue extends beyond the classically
forbidden region give rise to expect that could be penetrate a potential barrier. So
let us investigate a particle moves in the following potential distribution.
0 𝑥<0 𝑟𝑒𝑔𝑖𝑜𝑛 − 𝐼
𝑉(𝑥) = 𝑉𝑜 0≤𝑥≤𝑎 𝑟𝑒𝑔𝑖𝑜𝑛 − 𝐼𝐼
0 𝑥>𝑎 𝑟𝑒𝑔𝑖𝑜𝑛 − 𝐼𝐼𝐼
Which graphically equivalent to;

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Quantum Mechanics 2023-2024 Chapter Three

a) E  V
Results of classical mechanics predicts that a particle of energy E less than the
height of the barrier Vo definitely reflect back at x=0. While the correspondence
one of quantum mechanics show different behavior as shown below;

i) Region-I
Schrodinger equation in this region takes the form;
d 2 1 2mE
 2  1  0 , or
dx 2 
d 2 1 ( x)
2
 k 2  1 ( x)  0
dx
2mE
Where k 2  . The solution of this equation is;
2

 1 ( x)  A eikx  B e ikx ……………(3-37)

The inclusions obtained before are applicable here, where the first term in last
equation refers to the wave function of the incident particle while the second one
represent the wave function of the particles that are reflected back at x=0.

ii) Region-II
Schrodinger equation for this case has the form;
d 2 2 2m
 2 (V  E )  2  0 , or
dx 2 
d 2 2
2
  2 2  0
dx
2m
Where 2  (V  E ) . The solution of this equation is;
2

 2 ( x)  C e x  D e x …….……… (3-38)

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Quantum Mechanics 2023-2024 Chapter Three

The second term cannot be ignored as before due the end edge of the potential at
x=a.

H.W: Verify whether the wave function in equation (3-38) satisfies the boundary
condition or not.

iii) Region-III
Schrodinger equation in this region takes the form;
d 2 3 2mE
 2  3  0 , or
dx 2 
d 2 3 ( x)
2
 k 2 3 ( x)  0
dx
2mE
Where k 2  . The solution of this equation is;
2

 3 ( x)  Aeikx  Be ikx

The second term of the last equation must be ignore because there is no reasons
makes the wave function (particle)) reflected back in this region. Thus;

 3 ( x)  Aeik x ……………. (3-37)

The last equation shows that it is possible for the incident particle to go through the
potential barrier even if its kinetic energy is less than the height of the barrier.
Furthermore, the transmitted particle has energy equal to its incident one but has a
less probability to be in region-III. i.e 𝐴 > 𝐴́, see the figure below.

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Quantum Mechanics 2023-2024 Chapter Three

b) E > Vo
According to results of classical mechanics the incident particles penetrates the
potential barrier and transmitted to the third region for such a case. While results of
quantum mechanics shows a such possibility of some of the particles to reflects
back at the two points x = a, x= 0, as will be shown below.
i) Region-I
Schrodinger equation for this case being as follows;
d 2 1 ( x)
2
 k 2 1 ( x)  0
dx
2mE
Where k 2  2 . Solution of this equation given by;

 1 ( x)  A eikx  B e ikx ……………. (3-38)
ii) Region-II
Schrodinger equation for this case becomes;
d 2 2
2
 k  2 2  0
dx
2m
Where k  2  2 ( E  V ) . Solution of this equation is;

 2 ( x)  C eikx  D e ikx ……………. (3-39)
iii) Region-III
Schrodinger equation for this case being as follows;
d 2 3 2mE
 2 3  0
dx 2 
2mE
Where k 2  2 . Solution of this equation given by;

 3 ( x)  Aeikx  Be ikx
The second term in the last equation must be neglects for the same reasons
mentioned previously. Thus;

 3 ( x)  Aeik x ……………. (3-40)

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Quantum Mechanics 2023-2024 Chapter Three

Solved Problem

1) A particle describe by the following wave function;  ( x)  2

a
sin( nx / a ) . Find
the following; i)  p x  , ii)  p x2  , iii) x , iv)  x 2  andv) p x .
Solution:

i)  p x    n ( x) pˆ x  n ( x) dx



a
 2
sin(nx / a) (i ) 2
sin(nx / a) dx
0
a x a

 2i n
a

a 0  sin(nx / a) 
a
 cos(nx / a) dx

 2i n
a

a2  sin(nx / a) cos(nx / a) dx
0

a
 i
 sin 2 (nx / a)
a 0

0
The physical meaning of this result is that the mumentum of the particles that
moves in –x-axis is exactle similar to that of the particles which moves in the +x-
axis.
ii)
2
a
 p x2   2
sin(nx / a) ( 2
) 2
sin(nx / a) dx
0
a  x2 a

 2 2 2
a

a 0
 sin(nx / a) sin(nx / a) dx
 x2
 2 2  n
a

a 0
 sin( nx / a ) cos(nx / a) dx
x a
 2n   2 
a

a 2  sin(nx / a)
 x
cos(nx / a) dx
0

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Quantum Mechanics 2023-2024 Chapter Three

2n 2 2  2
a

a3  sin(nx / a) sin(nx / a) dx
0

n 2 2  2
p   2
a2
n 2 2  2
(p)   p    p 
2 2
-0 2
a2
n 2 2  2

a2
n 
 (p) 
a
iii)
a
 x   n ( x) x n ( x) dx
0

a
 2
sin(nx / a) x 2
sin(nx / a) dx
a a
0

a
  x sin 2 (nx / a ) dx
0

This integration is very simple and valuable

a a2
 x    x  2
2 4
It indicates the presence of the particle in half of the left or right box with the same
probability
iv)  x 2 
a
 x    n ( x) x 2  n ( x) dx
2

a
 2
sin(nx / a) x 2 2
sin(nx / a) dx
a a
0

a
2
  x 2 sin 2 (nx / a) dx
a0

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Quantum Mechanics 2023-2024 Chapter Three

This integration is very simple and valuable

a2 a2
x  
2

3 2n 2 2
 (x) 2   x 2    x 2

a2 a2 a2
  
3 2n 2 2 4
1 1 
 a2   2 2 
12 2n  
1 1 
 x  a   2 2 
12 2n  
n 1 1 
p x  a 12  2n 2 2 
a

 n 2 2 1 
p x     
 12 2
The lowest value of the multiplication factor belongs to the ground state n  1

p x  0.567 
This result is consistent with p x  

2) Find the energy of a particle describe by the wave function;

 ( x)  2
a
sin( nx / a ) and restricted to move along the interval 0  x  a .

Solution:
Aˆ  n  an n
Hˆ  n  En n
 2 2
  2 sin(n x / a)
2m x 2 a

 2  n
 { 2  cos(n x / a)}
2m x a a
n 2 2  2 2
  sin(n x / a)
2ma 2 a

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Quantum Mechanics 2023-2024 Chapter Three

n 2 2  2
 n
2ma 2
n 2 2  2
 En 
2ma 2

3) Find the momentum of a particle describe by the wave function;

 ( x)  2
a
sin( nx / a ) and restricted to move along the interval 0  x  a .

Solution:
Aˆ  n  an  n
pˆ 2 n  pn2  n
2 2
  2
sin(nx / a)
x 2 a
 n
  2 { 2 cos(nx / a )}
x a a
n 2 2  2 2
 sin(nx / a)
a2 a

n 2 2  2
 n
a2
n 2 2  2
 pn 2
a2
4) Prove that, wave functions that describe a particle in a potential box are
orthogonal.

Solution:

 n  m dx  0
*

a a

  m dx   sin(n x / a)  sin(m x / a) dx
* 2 2
n a a
0 0

a
  sin(n x / a)  sin(m x / a) dx
2
a
0

sin  x sin  x  2 {cos(   ) x  cos(   ) x}

1

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Quantum Mechanics 2023-2024 Chapter Three

{cos(n  m)x / a  cos(n  m) x / a}dx

1
a
0

a
1 a a
{ sin( n  m)  x / a  sin( n  m)  x / a}
a (n  m) (n  m) 0

1 a a
{ sin(n  m)   sin(n  m)  }
a (n  m) (n  m)
 (n  m) and (n  m) are integer
a
  n m dx  0
*

5) Prove that, wave functions that describe a particle in a potential box are
normalized.

Solution:
a

 n  n dx  1
*

0
a a

  n dx   sin(n x / a)  sin(n x / a) dx
* 2 2
n a a
0 0
a
2
  sin 2 (n x / a) dx
a0

sin 2   2 (1  cos 2 )
1
By using

a
2 1 1
   {  cos(2n x / a)} dx
a0 2 2
a a
1 1
  dx   cos(2n x / a) dx
a0 a0

2n x
a
1 a 1 a
( x)  sin( )
a 0 a 2n a 0

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Quantum Mechanics 2023-2024 Chapter Three

1 1
( a  0)  sin( 2n )
a 2 n
 n is integer  sin 2n  0
a
  n n dx  1
*

H.W: Deduced the wave functions and their correspondence energies for a particle
moves in the following potential distribution.

0 𝑥<0
𝑉(𝑥) = −𝑉𝑜 0≤𝑥≤𝑎
0 𝑥>𝑎

81