Download lecture 10

Quantum Mechanics
D
(a)
Start here from rest
C
E
A
B
V(x)
(b)
Vo
(x)
E < Vo
A1
A2
(x)
Incident
(x)
Reflected
Transmitted
I
III
II
x=0
x=a
x
(a) The roller coaster released from A can at most make to C, but not to E. Its
PE at A is less than the PE at D. When the car is the bottom its energy is
totally KE. CD is the energy barrier which prevents the car making to E. In
quantum theory, on the other hand, there is a chance that the car could tunnel
(leak) through the potential energy barrier between C and E and emerge on the
other side of the hill at E .
(b) The wavefunction of the electron incident on a potential energy barrier (Vo).
The incident and reflected waves interfere to give yI(x). There is no reflected
wave in region III. In region II the wavefunction decays with x because E < Vo.
Fig 3.1
From Principles of Electronic
Materials and Devices, Third
MOTION OF AN ELECTRON IN ONE DIMENSIONAL
POTENTIAL WELL (PARTICLE IN A BOX)
Consider an electron of mass m, moving along positive x-axis between two
walls of infinite height, one located at x=0 and another at x=a. Let potential
energy of the electron is assumed to be zero in the region in-between the two
walls and infinity in the region beyond the walls.
V  0
V 
for 0  x  a
for x  0 & x  a
V=
V=0
X=0
V=
X=a
X-axis
3
Region beyond the walls:
The Schrodinger’s wave equation representing the motion of the particle in the region beyond the
two walls is given by
d 2 8  2 m

2
dx
h2

E     0
The only possible solution for the above equation is ψ=0.
Since ψ=0 , the probability of finding the particle in the region x<0 and x>a is Zero . i.e.,
particle cannot be found in region beyond the walls.
Region between the two walls:
The Schrodinger’s wave equation representing the motion of the particle in the region
between the two walls is given by
d 2 8  2 m

dx 2
h2
 E  0 
 0
V  0
4
d 2 8  2 mE

  0
2
2
dx
h
d 2
2


 0
2
dx
8  2 mE
2
where  
h2
1
1a
Solution of the equation 1 is of the form
  A sin x  B cos x
2
Where A and B are unknown constants to be determined. Since particle cannot be foun
inside the walls
at x  0 ,   0
and at x  a ,   0
I
II
5
The equations are called boundary conditions. Using the I boundary condition in
equation 2, we get
0  A sin 0  B cos 0
B  0
Therefore equation 2 becomes
  A sin x
3
Using condition II in equation 3 we get
0  A sin a
 A  0 sina  0
 a  n
where n  0,1,2,......
n
 
a
6
Therefore correct solution of the equation 1 can be written as
 n  A sin
n
x
a
4
The above equation represents Eigenfunctions. Where n=1,2,3,..
(n=0 is not acceptable because, for n=o the wavefunction ψ becomes zero for all values
of x. Then particle cannot be found anywhere)
Substituting for  in equation 1a we get
n2h2
En 
8ma 2
5
2
n

8

mE



 
2
a
h


2
Therefore energy Eigenvalues are represented by the equation
n2h2
En 
8ma 2
5
7
Particle in the Infinite Potential Well
2
 x 
n 
sin
n
L
 L 
n defines states,we take n  0
Energy for the n th state is
En 
 2 2
2mL 2
n2
Where n=1,2,3,..
It is clear from the above equation that particle can have only desecrated values of
energies. The lowest energy that particle can have corresponds to n=o , and is called
zero-point energy. It is given by
n 
2
n
sin
x
a
a
Ezero  po int
h2

8ma 2
6
Normalization of wave function:
We know that particle is definitely found somewhere in space
a
  n dx  1
2
0
Therefore Normalised wavefunction is given by
9
PHYSICAL INTERPRETATION OF WAVE FUNCTION
The state of a quantum mechanical system can be completely
understood with the help of the wave function ψ. But wave function ψ can be
real or imaginary. Therefore no meaning can be assigned to wavefunction ψ as
it is.
According to Max Born’s interpretation of the wavefunction, the only quantity
that has some meaning is

2

2
which is called as probability density.

2
10
Thus if ψ is the wavefunction of a particle within a small region of volume dv,
then

2
dV
gives the probability of finding the particle within the region
dv at the given instant of time.
We know that electron is definitely found somewhere in the space. The
wavefunction ψ, which satisfies the above condition, is called normalized
wavefunction.
dv

V
 dv  1
2
11
Similarly for n=2
4h 2
En 
 4K
2
8ma
x
0
a
4
a
2
n 
2
n
sin
x
a
a
0
a
0
2
a
0
2
2
n
sin 2
x
a
a
0
2
a
3a
4
p  n 
2
a
0
2
a
0
12
Wavefunctions, probability density and energies are as shown in the figure.
P2
n=2
2
P1
1
n=1
x=0
x=a/2
x=a
13
Free Particle
Consider a particle of mass m moving along positive x-axis. Particle is said to be
free if it is not under the influence of any field or force. Therefore for a free particle
potential energy can be considered to be constant or zero. The Schrodinger wave
equation for a free particle is given by.
d 2 8  2 m

2
dx
h2

E  0   0
d 2 8  2 mE

  0
2
2
dx
h
d 2
2


 0
2
dx
2
8

mE
where  2 
h2
1
2
14
The solution of the equation 1 is of the form
  A sin x  B cos x
3
Where A and B are unknown constants to be determined. Since there are no
boundary conditions A, B and  can have any values.
Energy of the particle is given by
 2h2
E
8 2 m
4
Since there is no restriction on  there is no restriction on E.
Therefore energy of the free particle is not quantised. i.e., free particle can have
any value of energy.
15
Probability to Find particle in the Right
Half of the Well
1 sinn sin2n 1
Pn  


2
2n
2n
2
Average Momentum of Particle in a
Box (Infinite Potential Well)
• Can be shown that
2

L cos (n ) 
 2n  L
0
p  ( i) 2 


2n
 L  2n

• Reflects the vector nature of the momentum
– a particle have the same probability of moving
to the right or to the left