Finite potential well and

Potential steps:
Clear deviations from
results of classical
physics
Particle in a Finite Potential
Well
( ) 0 for 0
for x 0 or
o
V x x L
V x L
= < <
= < >
0 V =
o
V V =
x=0
x=L
o
V V =
RI RII
RIII
We take E<V
o
for the bound state
problem.
| |
2
2 2
( ) 2
( ) 0
I
o I
d x m
E V x
dx

+ =
h
| |
2
( ) 2
( ) 0
II
d x m
E x

+ =
RI
RII
| |
2 2
( ) 2
( ) 0
II
II
d x m
E x
dx

+ =
h
| |
2
2 2
( ) 2
( ) 0
III
o III
d x m
E V x
dx

+ =
h
Note the sign of quantity in square bracket
for different regions.
RII
RII
I
General Solutions
( ) ( )
( ) sin cos ; x C kx D kx = +
( )
2
2
2
( ) ;
o x x
I
m V E
x Ae Be

= + =
h
RI
( ) ( )
2
2
( ) sin cos ;
2
II
x C kx D kx
mE
k
= +
=
h
( )
2
2
2
( ) ;
o x x
III
m V E
x Ge Fe

= + =
h
RII
RIII
Since it is not possible to have
large probability density when x
tends to - or +, the
coefficients B and G should coefficients B and G should
definitely vanish.
Well-behaved Wave Functions
0
0
B
G
=
=
( ) ( )
( )
( ) sin cos
( )
x
I
II
x
III
x Ae
x C kx D kx
x Fe

=
= +
=
Boundary Conditions Applied
RI- RII
( ) ( )
0
(0) (0)
sin 0 cos 0
I II
Ae C D
A D
=
= +
=
( ) ( )
0
(0) (0)
cos 0 sin 0
I II
d d
dx dx
A D
A Ck
A e Ck Dk

=
=
=
=
RII- RIII
( ) ( )
sin cos
( ) ( )
( ) ( )
II III
II III
L
L L
d d
L L
C kL D kL Fe

=
=
+ =
( ) ( )
cos si
( ) ( )
n
II III
L
d d
L L
dx
Ck kL Dk kL F e
dx

=
=
( ) ( )
sin cos
L
A D
A Ck
C kL D kL Fe

=
=
+ =
( ) ( )
( ) ( )
sin cos
cos sin
L
C kL D kL Fe
Ck kL Dk kL F e

+ =
=
Solving the Equations
Express all constants in terms of A.
D A

=
=
( ) ( )
( ) ( )
sin cos
cos sin
L
L
C A
k
A kL A kL Fe
k
Ak kL Ak kL F e
k

=
+ =
=
Energy Eigen Values
Divide the last two equations.
( ) ( )
( ) ( )
cos sin
;
sin cos
kL k kL
kL k kL k

=
+
( ) ( )
( )
sin cos
2
2
o
kL k kL k
m V E
k mE

+
=
=
h
h
The above equation governs the
allowed energy levels.
Classically Forbidden Region
There is a non zero probability of
finding the particle in the classically
forbidden region.
This probability decreases This probability decreases
exponentially.
However the uncertainty principle
prohibits one to localize the particle in
the classically forbidden region and
measure its kinetic energy. measure its kinetic energy.
From wave nature we can
understand the finite penetration.
Barrier penetration in
classical optics too.
Frustrated total internal
reflection
Kronig Penney model
A simple application of finite well.
One dimensional periodic potential of a solid
Step potential
E
A particle approaching from left on a
Step Potential (E>V
0
)
x=0
V=0
V=V
o
E
RI RII
( )
1 1
2 2
2
1
2
2
2
2
2
( ) ;
2
( ) ;
ik x ik x
I
ik x ik x o
II
mE
x Ae Be k
m E V
x Ce De k

= + =

= + =
h
General Solution
RI
RII
2
2
( ) ;
II
x Ce De k = + =
h
RII
Physical interpretation of different
components yields
Boundary Conditions yield
0 D =
(nothing to reflect the beam in the right)
Boundary Conditions yield
1 2
( )
A B C
ik A B ik C
+ =
=
Solving the equations
2
1
A B C
k
A B C
k
+ =
=
1
2k C
A k k
=
+ 1
2
1
2
1
1
2
1
2
k
k C
A
k
k C
B
k
| |
= +
|
\
| |
=
|
\
1 2
1 2
1 2
A k k
k k B
A k k
+

=
+
Comments
We have one less number of
equations than in the bound state.
Hence no quantization of energy. All
values of energy are allowed. values of energy are allowed.
Strictly speaking non-normalizable
wave function.
Interesting from the point of view of
scattering of particles.
We can evaluate Transmission and
Reflection coefficients. Reflection coefficients.
What would one expect classically?
Transmission and Reflection
coefficients
2
2
1 2
1 2
2 2
4
k k B
R
A k k
v k k k C C
(

= =
(
+

( )
2 2
2 2 1 2
2
1 1
1 2
4 v k k k C C
T
A v A k
k k
= = =
+
We can easily verify that
1 R T + =
v
1
and v
2
stand for the speeds in the first and second regions
Analogy with classical wave
optics
k
k
v
v
(n) index refractive the
language, optics In
2
1
2
1
= =
case. etic electomagn the to identical exactly
,
) 1 (
1) - (n
R and
) 1 (
4
k v
2
2
2
2 2
+
=
+
=
n n
n
T
Step Potential (Case 2)
A particle approaching from left on a
Step Potential, with an energy E less
than the step height V
o
.
x=0
V=0
V=V
o
E
RI RII
( )
2
2
2
2
2
( ) ;
2
( ) ;
ikx ikx
I
o x x
II
mE
x Ae Be k
m V E
x Ce De

= + =

= + =
h
h
General Solution
RI
RII
2
( ) ;
II
x Ce De = + =
h
RII
Finiteness of wave function
0 C =
(as E< V
0
, it is unresonable to
have large probability density
in the right side)
( )
A B D
ik A B D
D

+ =
=
| |
Boundary Conditions and solution
yield
1
2
1
2
D
A i
k
D
B i
k

| |
= +
|
\
| |
=
|
\
Reflection Coefficient
B k i
A k i

=
+
2
1
B k i k i
R
A k i k i


+
= =
+
=
Comments
Reflection Coefficient is one. So all
particles eventually get reflected.
There is still a finite probability for
finding the particle in classically finding the particle in classically
forbidden region.
E
Another example: free state vs. bound state
E>V
0
0 V =
o
V V =
x=0
x=L
RI
RII
V =
( )
2
1 1 1
2
2
( ) sin cos ;
I
mE
x A k x B k x k = + =
h
RI
( )
2 2
2
2
2
2
( ) ;
ik x ik x o
II
m E V
x Ce De k


= + =
h
RII
2 2
1
(0) 0 0
( ) ( ) sin
I
I II
ik L ik L
B
L L A k L
Ce De

= =
=
= +
( )
2 2
2 2
1 1
2
( ) ( ) cos
ik L ik L
I II
ik L ik L
Ce De
L L Ak k L
ik Ce De

= +

=
=
Reflection coefficient
2
C
R
D
=
Bound State
E<V
0
0 V =
o
V V =
x=0
x=L
RI
RII
V =
2
2
( ) sin cos ;
2
( ) ;
I
x x
x A kx B kx
mE
k
x Ce De

= +
=
= +
h
( )
2
2
( ) ;
2
0
0
x x
II
o
x Ce De
m V E
B
C

= +

=
=
=
h
( ) ( ) sin
( ) ( ) cos
L
I II
L
I II
L L A kL De
L L Ak kL De

= =

= =
This gives
cot kL
k

=
( )
2
0
o
m V E

= =
h
Special Case: E=V
o
2
2
o
mV
mE
k = =
h
h h
The boundary conditions now becomes.
sin
cos 0
A kL D
Ak kL
=
=
cos 0 or (2 1)
2
2
(2 1)
o
kL kL n
mV
L n

= = +
= +
Since A can not be zero, hence this
implies
2
2
2
2
(2 1)
2
(2 1)
2
2
o
o
mV
L n
n
V
mL

= +
(
+
(

=
h
h
Bound States
There will be no bound state if
2 2
2
8
o
V
mL

<
h
2
8mL
There will be only one bound state if
2 2 2 2
2 2
9
8 8
o
V
mL mL

< <
h h
Quantum mechanical Tunneling
(E<V
0
)
( ) for 0
0 for x 0 or
o
V x V x L
x L
= < <
= < >
0 V =
0 V =
x=0
x=L
o
V V =
RII
RIII
RI
E
The General Wave Functions
( )
2
2
2
2
( ) ;
2
( ) ;
ikx ikx
I
o x x
mE
x Ae Be k
m V E
x Ce De

= + =

= + =
h
RI
RII
( )
2
2
2
2
( ) ;
2
( ) ;
o x x
II
ikx ikx
III
x Ce De
mE
x Fe Ge k

= + =
= + =
h
h
RII
RIII
Is any coefficient zero?
Boundary Conditions
0
( ) ( )
G
A B C D
ik A B C D
=
+ = +
=
2 2
2
( )
1 1
1 sinh
4
i L i L ikL
i L i L ikL
Ce De Fe
Ce De Fike
A k
L
T F k

+ =
=
| |
= = + +
|
\
i.e., T is non zero in the forbidden region
Experimental evidences for tunneling
1. Alpha decay
2. Ammonia inversion
3. Tunnel diode
Ordinary pn diodes depend on diffusion
current and hence the switching speed is very
low.
A heavily doped pn junction has a very thin A heavily doped pn junction has a very thin
depletion layer and hence has tunneling
current.
The tunneling current can be controlled by
changing the potential barrier (the step) by
applied voltage.
Very high switching frequencies (~ GHz).
Scanning tunneling microscope
Atomic resolution of surfaces
Nobel Prize in Physics (1986)
Summary
Comparison between classical optics and quantum mechanics
Free vs. bound states in wells and steps in quantum
mechanical treatment. mechanical treatment.
Quantum mechanical tunneling and its importance