Transcript Tunneling

Tunneling Phenomena
Potential Barriers
Tunneling

Unlike attractive potentials which traps
particle, barriers repel them.
Tunneling


Unlike attractive potentials which traps
particle, barriers repel them.
Hence we look at determining whether
the incident particle is reflected or
transmitted.
Tunneling



Unlike attractive potentials which traps
particle, barriers repel them.
Hence we look at determining whether
the incident particle is reflected or
transmitted.
Tunneling is a purely QM effect.
Tunneling



Unlike attractive potentials which traps
particle, barriers repel them.
Hence we look at determining whether
the incident particle is reflected or
transmitted.
Tunneling is a purely QM effect. It is
used in field emission, radioactive decay,
the scanning tunneling microscope etc.
Particle Scattering and Barrier
Penetration
Potential Barriers
The Square Barrier

A square barrier is represented by a
potential energy U(x) in the barrier
region (between x=0 and x=L).
U
0
L
The Square Barrier

Using classical physics a particle with
E<U are reflected while those with E>U
are transmitted with same energy.
The Square Barrier


Using classical physics a particle with
E<U are reflected while those with E>U
are transmitted with same energy.
Therefore particles with E<U are
restricted to one of the barrier.
The Square Barrier

However according to QM there are no
forbidden regions for a particle
regardless of energy.
The Square Barrier


However according to QM there are no
forbidden regions for a particle
regardless of energy.
This is because the associated matter
wave is nonzero everywhere.
The Square Barrier



However according to QM there are no
forbidden regions for a particle
regardless of energy.
This is because the associated matter
wave is nonzero everywhere.
A typical waveform is shown:
Potential Barriers
E>U
Potential Barriers (E>U)

Consider the step potential below.
E
U(x)=U0
U=0
x=0
Potential Barriers (E>U)

Consider the step potential below.
E
U(x)=U0
U=0
x=0

Classical mechanics predicts that the
particle is not reflected at x=0.
Potential Barriers (E>U)

Quantum mechanically,

2
2m
d 
2
dx
2
 E
x  0
Potential Barriers (E>U)

Quantum mechanically,

2
2m

2
2m
d 
2
dx
d 
2
dx
2
2
 E
  E  U 0 
x  0
x  0
Potential Barriers (E>U)

Quantum mechanically,

2
2m

2
2m
d 
2
dx
d 
2
dx
2
2
 E
  E  U 0 
x  0
x  0
Potential Barriers (E>U)

Considering the first equation,

2
2m
d 
2
dx
2
 E
x  0
Potential Barriers (E>U)

Considering the first equation,

2
2m

d 
2
dx
2
 E
x  0
The general solution is
 1  x   Ae iK
1x
 Be
 iK 1 x
Potential Barriers (E>U)

Considering the first equation,

2
2m

d 
2
dx
 E
2
The general solution is
 1  x   Ae iK

x  0
where K
1

1x
 Be
2 mE

 iK 1 x
Potential Barriers (E>U)

For the 2nd equation,
 d 
2
2
2 m dx
2
  E  U 
x  0
Potential Barriers (E>U)

For the 2nd equation,
 d 
2
2
2 m dx

2
  E  U 0 
x  0
The general solution is
 2  x   Ce iK
2
x
 De
 iK 2 x
Potential Barriers (E>U)

For the 2nd equation,

2
2m

d 
2
dx
 E
2
The general solution is
 2  x   Ce iK

x  0
where K
2

2
x
 De
 iK 2 x
2 m E  U

0

Potential Barriers (E>U)

However D=0 since there is no reflection
as there is only a transmitted wave for
x>0. We have nothing to cause reflection!
Potential Barriers (E>U)

However D=0 since there is no reflection
as there is only a transmitted wave for
x>0. We have nothing to cause reflection!
iK



x

Ce
2
 Therefore
2
x
Potential Barriers (E>U)

The wave equations represent a free
particle of momentum p1 and p2
respectively.
Potential Barriers (E>U)

The behaviour is shown in the diagram
below.
C
A
U(x)=U0
B
U=0
x=0
Potential Barriers (E>U)

The constants A, B and C must be
chosen to make and continuous at
x=0.
Potential Barriers (E>U)


The constants A, B and C must be
chosen to make and continuous at
x=0.
Satisfying the 1st condition we get that
A B  C
..... 1 
Potential Barriers (E>U)


The constants A, B and C must be
chosen to make and continuous at
x=0.
Satisfying the 1st condition we get that
A B  C

..... 1 
To satisfy the 2nd requirement, we
differentiate.
K 1  A  B   K 2C
.....  2 
Potential Barriers (E>U)

Substituting 1  into 2  we get
K 1  A  B   K 2  A  B  ..... 3 
Potential Barriers (E>U)

Substituting 1  into 2  we get
K 1  A  B   K 2  A  B  ..... 3 

Writing A in terms of B we get after some
algebra that
K  K A
B 
1
K 1
2
 K2
Potential Barriers (E>U)

Substituting 1  into 2  we get
K 1  A  B   K 2  A  B  ..... 3 

Writing B in terms of A we get after some
algebra that
K  K A
B 

1
K 1
2
 K2
Similarly, writing C in terms of A
C 
2K1A
K 1
 K2

Potential Barriers (E>U)

Substituting these expressions into 1 
we have
 1  x   Ae

x  
2
A
iK 1 x
 A
2K1
K1  K 2
K1  K 2
K1  K 2
e
iK
2
x
e
 iK 1 x
Potential Barriers (E>U)

Substituting these expressions into 1 
we have
 1  x   Ae


x  
2
A
iK 1 x
 A
2K1
K1  K 2
K1  K 2
K1  K 2
e
iK
2
e
 iK 1 x
x
As usual we normalize to find A.
Potential Barriers (E>U)

The probability that the particle is
reflected is given by the Reflection
coefficient R.
Potential Barriers (E>U)

The probability that the particle is
reflected is given by the Reflection
coefficient R. The ratio of intensities of
the reflected to incident.
R 
v1 B  B
v1 A  A
Potential Barriers (E>U)

The probability that the particle is
reflected is given by the Reflection
coefficient R. The ratio of intensities of
the reflected to incident.
R 
v1 B  B
v1 A  A
 K1  K 2
 R  
 K  K
1
2





2
Potential Barriers (E>U)

The probability that the particle is
transmitted is given by the Transmission
coefficient T.
Potential Barriers (E>U)

The probability that the particle is
reflected is given by the Reflection
coefficient R. The ratio of intensities of
the transmitted to incident.
T 
v2C  C
v1 A  A
Potential Barriers (E>U)

The probability that the particle is
reflected is given by the Reflection
coefficient R. The ratio of intensities of
the transmitted to incident.
T 
 T 
v2C  C
v1 A  A
4 K 1K 2
K 1
 K2
2
Potential Barriers (E>U)

It is easy to show that R  T  1
Potential Barriers (E>U)

A similar case to the previous example is
given below
A
B
C
U(x)=U0
U=0
x=0
Potential Barriers (E>U)

Applying the same logic as the previous
example we can show that
 d 
2
2
2 m dx

2
2m
2
  E  U 0 
d 
x  0
2
dx
2
 E
x  0
Potential Barriers (E>U)

Applying the same logic as the previous
example we can show that
 d 
2
2
2 m dx

2
2m

2
  E  U 0 
d 
x  0
2
dx
2
 E
x  0
The solution to these equations are
 1  x   Ae iKx  Be
iK
 2  x   Ce
1x
 iKx
Potential Barriers (E>U)

Where
K 

K1 
2 m E  U
0


2 mE

respectively.
Potential Barriers (E>U)

Applying the conditions of continuity we
get:
A B  C
..... 1 
Potential Barriers (E>U)

Applying the conditions of continuity we
get:
A B  C

..... 1 
and
iKA  iKB  iK 1 C
.....  2 
Potential Barriers (E>U)

As an exercise show that the
transmission and reflection coefficients
are the same.
Potential Barriers
E<U
Potential Barriers (E<U)

Consider a stream of particles incident to
barrier below.
U(x)=U0
E
U=0
x=0

Total energy E is constant.
Potential Barriers (E<U)

Classically a particle can not enter the
region x>0.
Potential Barriers (E<U)


Classically a particle can not enter the
region x>0.
Consider what happens quantum
mechanically.
Potential Barriers (E<U)

The time independent Schrödinger's
equation for the regions are:
 d 
2
2
2 m dx
2
  E  U 0 
x  0
Potential Barriers (E<U)

The time independent Schrödinger's
equation for the regions are:
 d 
2
2
2 m dx

2
2m
2
  E  U 0 
d 
x  0
2
dx
2
 E
x  0
Potential Barriers (E<U)

Again in the region x<0, the solution to
iK x
 iK x



x

Ae

Be
SE is 1
1
1
Potential Barriers (E<U)

Again in the region x<0, the solution to
iK x
 iK x



x

Ae

Be
SE is 1
1

where K 1 
2 mE

1
Potential Barriers (E<U)

For the region x>0, the solution to SE is
 2  x   Ce
K2x
 De
K2x
Potential Barriers (E<U)

For the region x>0, the solution to SE is
 2  x   Ce

where K 2
K2x

 De
K2x
2 m E  U


Potential Barriers (E<U)

Choose A, B, C and D so that conditions
are satisfied.
Potential Barriers (E<U)


Choose A, B, C and D so that conditions
are satisfied.
1st note that C=0 otherwise the TISE
diverges as x  
Potential Barriers (E<U)



Choose A, B, C and D so that conditions
are satisfied.
1st note that C=0 otherwise the TISE
diverges as x  
Considering continuity conditions at x=0
leads to:
A B  D
iK 1 A  iK 1 B   K 2 D
.... 1 
....  2 
Potential Barriers (E<U)

Writing the constants A and B in terms of
D we get:
Potential Barriers (E<U)

Writing the constants A and B in terms of
D we get:
D 
iK 2

A
1

2 
K1
D 
iK 2
1 
B 
2 
K1








.... 3 
....  4 
Potential Barriers (E<U)

Therefore the waveform for each region
can be rewritten in terms of one
constant. And we get,
 1 x  
D
2
1  i K 2
 2  x   De
K2x
K 1 e
iK 1 x

D
2
1  i K 2
K 1 e
 iK 1 x
Potential Barriers (E<U)

Therefore the waveform for each region
can be rewritten in terms of one
constant. And we get,
 1 x  
D
2
1  i K 2
 2  x   De

K 1 e
iK 1 x

D
2
1  i K 2
K 1 e
 iK 1 x
K2x
where we can normalize to find D.
Potential Barriers (E<U)

The ratio of the reflected probability flux
to the incident probability flux gives
Potential Barriers (E<U)

The ratio of the reflected probability flux
to the incident probability flux gives
R 
vB * B
vA * A
Potential Barriers (E<U)

The ratio of the reflected probability flux
to the incident probability flux gives
R 
vB * B
vA * A

1  iK 2
1  iK 2
K 1  1  iK
2
K1 
K 1  1  iK
2
K1
*
*
Potential Barriers (E<U)

The ratio of the reflected probability flux
to the incident probability flux gives
R 
vB * B
vA * A

1  iK 2
1  iK 2

1  iK 2
1  iK 2
K 1  1  iK
2
K1 
K 1  1  iK
2
K1
*
*
K 1 1  iK 2 K 1 
K 1 1  iK 2 K 1 
Potential Barriers (E<U)

The ratio of the reflected probability flux
to the incident probability flux gives
R 
vB * B
vA * A

1  iK 2
1  iK 2

1  iK 2
1  iK 2
K 1  1  iK
2
K1 
K 1  1  iK
2
K1
*
*
K 1 1  iK 2 K 1 
K 1 1  iK 2 K 1 
1
Potential Barriers (E<U)

That is the probability of being reflected
is unity.
Potential Barriers (E<U)


That is the probability of being reflected
is unity.
These results are in agreement with
classical physics.
Potential Barriers (E<U)
Step potential
Potential Barriers (E<U)
A
U(x)=U0
B
C
U=0
x=0
x
Potential Barriers (E<U)

The TISEs are
Potential Barriers (E<U)

The TISEs are

2
d 
2
2m
dx
 d 
2
2
2 m dx
2
2
 E
  E  U 0 
x  0
x  0
Potential Barriers (E<U)

The TISEs are

2
d 
2
2m
dx
 d 
2
2
2 m dx

2
2
 E
x  0
  E  U 0 
x  0
The solution to these equations are
 1  x   Ae

x  
2
Ce
iK 1 x
 Be
K2x
 iK 1 x
 De
K2x
Potential Barriers (E<U)

Choose A, B, C and D so that conditions
are satisfied.
Potential Barriers (E<U)


Choose A, B, C and D so that conditions
are satisfied.
In this case D=0 otherwise the TISE
diverges as x  
Potential Barriers (E<U)



Choose A, B, C and D so that conditions
are satisfied.
In this case D=0 otherwise the TISE
diverges as x  
Considering continuity conditions at x=0
leads to:
AB C
iK 1 A  iK 1 B   K 2 C
.... 1 
....  2 
Potential Barriers (E<U)

Eliminating B we get
2A 
C iK 1  K 2 
iK 1
Potential Barriers (E<U)

Eliminating B we get
2A 
C iK 1  K 2 
iK 1

C
A

2 iK 1
iK 1 
K2
Potential Barriers (E<U)

Eliminating B we get
2A 
C iK 1  K 2 
iK 1

C

A

2 iK 1
iK 1 
K2
Dividing the 1st equation by B we get
C
A
 1
B
A
Potential Barriers (E<U)

Eliminating B we get
2A 
C iK 1  K 2 
iK 1

C

A

2 iK 1
iK 1 
K2
Dividing the 1st equation by B we get
C
 1
A

B
A
B
A

iK 1  K 2
iK 1  K 2
Potential Barriers (E<U)

Therefore we calculated the reflection
coefficient.
K1  K 2
2
R 
2
K1  K 2
2
2
1
Potential Barriers (E<U)

Therefore we calculated the reflection
coefficient.
K1  K 2
2
R 

2
K1  K 2
2
2
1
Therefore an infinitely wide barrier
reflects all incident particles!
Square Barrier
General Case
Potential Barriers (E<U)

Recall:
The Square Barrier

A square barrier is represented by a
potential energy U(x) in the barrier
region (between x=0 and x=L).
U
0
L
Potential Barriers (E<U)

The infinitely wide step potential is a
special case of the square barrier where
the width L is infinite.
Potential Barriers (E<U)


The infinitely wide step potential is a
special case of the square barrier where
the width L is infinite.
To complete our look at various potential
barriers we summarize a few ideas using
the square barrier.
The Square Barrier

For the general case the stationary wave
may be decomposed into incident,
reflected and transmitted waves.
Ae
ikx
Fe
Be
 ikx
0
L
ikx
Potential Barriers (E<U)

On the LHS of the barrier, the particle is
free. The waveform for free particles is
 1  x   Ae


iK 1 x
 Be
 iK 1 x
The part of the wavefunction A is
interpreted as a wave incident on the
barrier. B is the wave reflected.
The squared-amplitude of intensity of the
reflected wave relative to the incident is
R 
B
A
2
2
Potential Barriers (E<U)

In wave terminology, R is the wave
intensity. For particles, R is the
probability that that the particle incident
on the barrier is reflected.
Potential Barriers (E<U)

On the RHS of the barrier, the particle is
free. The waveform for the free particle



x  
3
Fe
iK 1 x
 Ge
 iK 1 x
The part of the wavefunction F is
interpreted as wave traveling to the right.
G has no interpretation.
The squared-amplitude of intensity of the
transmitted relative to the incident is
T 
F
A
2
2
Potential Barriers (E<U)

The transmission coefficient measures
the likelihood that a particle incident on
the barrier penetrates and emerges on
the other side.
Potential Barriers (E<U)

The sum of the two probabilities (the
probability of transmission and the
probability of reflection) is one.
 T  R 1
Potential Barriers (E<U)

The region of the barrier (0<x<L) where
U=U0 the TISE can be written in the
form,
d 
2
dx

2
 2m

  2 U 0  E  


With E<U, the expression in the braces
is a positive constant. The solution to
which are in the real exp form.
 
1



2 m U  E 
Potential Barriers (E<U)


The parameter  is the penetration
depth.
The distance in the barrier where the
waveform and hence the probability of
finding the particle remains appreciable.
Potential Barriers (E<U)

Therefore the waveform in the barrier is

x  
2
Ce
x
 De
x
Potential Barriers (E<U)

The arbitrary constants are determined
by applying the continuity conditions at
the boundaries x=0 and x=L.
Potential Barriers (E<U)

Applying these conditions gives the
following equations,
AB C D
iK 1 A  iK 1 B    C   D
Ce
L
  Ce

 De
L
L
 Fe
  De
L
iK 1 L
 iK 1 Fe
iK 1 L
Where terms must be written as ratios.
eg B A , C A
Wide Barriers

The solution to the previous set of
equations are tedious and generally too
complicated to be useful.
Wide Barriers



The solution to the previous set of
equations are tedious and generally too
complicated to be useful.
A simplification is possible where L  
In these cases the wide barrier
approximates to the infinitely wide
barrier. This leads to the approximation
2
that
 4 k   2 L 
2 mE  k    E
 e
T  E   
k 
,
2 
U E



1

k



2
2
2
Wide Barriers

The transmission curve is shown on the
handout.
Barrier Penetration (Applications)

We consider a few practical applications
of potential barriers.
Barrier Penetration (Applications)


a.
b.
c.
d.
e.
We consider a few practical applications
of potential barriers.
Some interesting cases are:
In field emissions
Alpha decay
Josephson Junction
Ammonia inversion
Decay of Black Holes
Barrier Penetration (Applications)

We briefly at one case…
Barrier Penetration (Applications)

Ammonia inversion
Barrier Penetration (Applications)



The inversion of ammonia is an example
of tunneling.
There are two configurations with the
same energy.
As shown, the two configurations are on
either side of the hydrogen plane.
Barrier Penetration (Applications)

Ammonia inversion
Barrier Penetration (Applications)




The inversion of ammonia is an example
of tunneling.
There are two configurations with the
same energy.
As shown, the two configurations are on
either side of the hydrogen plane.
The repulsive Coulomb force creates a
barrier.
Barrier Penetration (Applications)

The nitrogen must overcome this force to
get from one configuration to the other.
Barrier Penetration (Applications)

Ammonia inversion