Physics barriers and tunneling

Ain Shams University
Mathematics and Engineering Physics Department

Pre-Junior Communication Systems Engineering Students

Lecture 11
Modern Physics and Quantum Mechanics Course (EPHS 240)
9 December 2009

Dr. Hatem El-Refaei

Contents
Infinite barrier
Finite barrier
Quantum tunnelling

hatem@oilpeakinarabic.org Dr. Hatem El-Refaei 1

Note
All problems today are unbounded problem, i.e. the
particle is not confined in a certain region, so:
We will not be able to do the normalization condition.
Therefore, we will not be able to solve for all unknowns.
Therefore, we will not get a characteristic equation.
Therefore, energy levels are not quantized, and all energies
are possible.
But still there are a lot of important characteristics to
understand and learn today.


Infinite barrier


Potential step of infinite height and infinite width

Energy
∞

∞
E

x
Since the barrier height is infinite, incident particles can’t
penetrate through it, and particles reflect back.
So, there is zero probability of finding the particle inside
the step barrier.
Here, the QM solution leads to the same classical solution.


Potential step of infinite height and infinite
width

Energy, ψ
∞

∞
2mE
k=
h E

x
d ψ I 2m
2 0
2
+ 2 E ψI = 0
dx h
U =∞
ψ ( x ) = Ae jkx
+ Be − jkx

 E 
j  kx − t 
 E 
j  − kx − t  ψ (x ) = 0
Ψ (x, t ) = Ae  h 
+ Be  h 

Forward Backward
propagating wave propagating wave

width

Energy, ψ
∞

∞
2mE
k= E
h
x
d ψ I 2m
2 0
2
+ 2 E ψI = 0
dx h
U =∞
ψ ( x ) = Ae jkx
+ Be − jkx

ψ (x = 0 − ) = ψ (x = 0 + ) ψ (x ) = 0
A+ B = 0


width

Energy, ψ
∞

∞
Sin(x) E

x
d ψ I 2m
2 0
2
+ 2 E ψI = 0
dx h
B = −A U =∞
ψ (x ) = Ae jkx − Ae − jkx ψ (x ) = 0
ψ (x ) = A(e jkx − e − jkx )
ψ ( x ) = 2 jA sin (kx )

width

Energy, ψ
∞

∞
Sin(x) E

x
0
B = −A
2
Re flected Amplitude B∗B
Reflectivity R= = ∗ =1
Incident Amplitude A A

All the incident particle stream is reflected back.

width
Energy, ψ
∞

∞
Sin(x) E

x
0
Since the barrier extends to infinity in the x direction, no particle can penetrate
through the whole barrier. From phenomenological understanding, as x→∞,
ψ→0.


Finite barrier


Potential step of finite height and infinite
width
Energy
E<Uo

∞
E
Uo
x
0

As the step height Uo gets smaller (but still E<Uo),
the penetration of the particles inside the step
increases, but finally no particles will succeed to
travel through the whole step to x→∞.

width
2mE Energy 2m(U o − E )
k= E<Uo α= >0
h h
∞
E
Uo
x
0
d 2ψ I 2m d 2ψ II 2m
2
+ 2 EψI =0 2
+ 2 (E − U o ) ψ II = 0
dx h dx h
ψ I ( x ) = Ae jkx + Be − jkx ψ II ( x ) = C e −α x + Deα x
We have 4 unknowns (A,B,C, and D) and 3 equations:
Finiteness of ψ at x=∞
Continuity of ψ at x=0
Continuity of ∇ψ at x=0

width
k= E<Uo α= >0
h h
∞
E
Uo
x
0
2
+ 2 EψI =0 2
+ 2 (E − U o ) ψ II = 0
dx h dx h

The condition that ψ(x) must be finite as x→∞, leads to D=0


width
k= E<Uo α= >0
h h
∞
E
Uo
x
0
2
+ 2 EψI =0 2
+ 2 (E − U o ) ψ II = 0
dx h dx h
ψ I ( x ) = Ae jkx + Be − jkx ψ II (x ) = C e −α x



width
k= E<Uo α= >0
h h
∞
E
Uo
x
0
2
+ 2 EψI =0 2
+ 2 (E − U o ) ψ II = 0
dx h dx h
We have 3 unknowns (A,B,C) and 2 equations:
Continuity of ψ
Continuity of ∇ψ
Thus, the best we can get is the ratio between parameters.

width
k= E<Uo α= >0
h h
∞
E
Uo
x
ψ I ( x ) = Ae jkx + Be − jkx 0
ψ II (x ) = C e −α x
Continuity of ψ Continuity of dψ/dx
dψ I dψ II
ψ I ( x = 0) = ψ II (x = 0) =
dx dx
jk ( A − B ) = −α C
x =0 x =0
A+ B = C
1 α  1 α 
A= 1 + j  C B = 1 − j  C
2 k  2 k 

width
k= E<Uo α= >0
h h

E

0
1  α  1  α 
A=  1+ j  C B =  1− j  C
2  k  2  k 

 α  α
2 1 −
∗ j  1 + j 
Re flected Amplitude BB  k  k
Reflectivity R= = ∗ = =1
Incident Amplitude AA  α  α
1 + j  1 − j 
 k  k


width
Energy
Uo
∞
C
x
0
1 α  jkx 1  α  − jkx
ψ I (x ) = Ae + Be
jkx − jkx
= 1 + j Ce + 1 − j  Ce
2 k 2 k
 e jkx + e − jkx  α  e jkx − e − jkx 
ψ I (x ) = C 
  + j C
  

 2  k  2 
 α 
C cos(kx ) − C sin (kx ) x≤0 
ψ (x ) =  k 
C e −α x x ≥ 0
 

width
Energy
Uo
∞
C
x
0

Since the barrier height is finite, particles can penetrate partially
in the vicinity of the potential step, and then they reflect back.
So, there is a finite probability of finding the particle in the
classically forbidden position.
Here, the QM solution is different from the classical one.


width
Energy
E<Uo Case 3 2m(U o − E )
α= >0
Case 2 h

Case 2 ψ ( x ) = C e −α x
x
0
Case 3

As the potential barrier height increases (Uo
increases) (from case 2 to case 3), α also increases,
and thus the exponential function dies quicker inside
the barrier. Thus it becomes less probable to find the
particle inside the barrier.

width
Energy
E<Uo 2m(U o − E )
α= >0
h

ψ ( x ) = C e −α x
x
0
If the barrier height (Uo) is kept constant, but the particle
energy increases (provided E<Uo), thus α decreases, and
the particle exponential function dies slower inside the
barrier. Hence, it becomes more probable to find the
particle in the vicinity of the barrier edge.

Tunneling through a potential
barrier


Potential barrier of finite height and finite
width
Energy E<Uo
Uo
E

x

A stream of particles incident on a finite width and height
potential barrier with E<Uo.
Part of the incident stream will succeed to penetrate through the
barrier and appear on the other side, this is the transmitted stream.
The other part will reflect back forming the reflected stream.
Note, a single particle doesn’t split into two.

width
Energy E<Uo
2mE Uo 2m(U o − E )
k= γ= >0
h h

0 a x

d 2ψ I 2m d 2ψ II 2m d 2ψ III 2m
+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0
dx 2 h dx 2 h dx 2 h

ψ I ( x ) = Ae jkx + Be − jkx ψ II (x ) = C eγ x + D e −γ x ψ III (x ) = Ge jkx + Fe − jkx
We have 6 unknowns (A,B,C,D,G,F) and 5 boundary conditions
Continuity of ψ and ∇ψ at x=0.
Continuity of ψ and ∇ψ at x=a.
Only a forward propagating wave on the right hand side of the barrier.

width
Energy E<Uo
Uo

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0


width
Energy E<Uo
Uo

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0

ψ I ( x ) = Ae jkx + Be − jkx ψ II (x ) = C eγ x + D e −γ x ψ III ( x ) = Ge jkx
We have 5 unknowns (A,B,C,D,G) and 4 boundary conditions


width
Energy E<Uo
Uo

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0


We are interested in the transmission probability (T)
2 2
Transmitte d Amplitude G ∗G G
T= = ∗ =
Incident Amplitude A A A

width
Energy E<Uo
Uo

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0


And reflection probability (R)
2 2
Re flected Amplitude B ∗B B
R = = ∗ =


width
First set of boundary conditions at x=0

dψ I dψ II
ψ I (x = 0) = ψ II (x = 0) =
dx x =0 dx x =0

A+ B = C + D jkA − jkB = γC − γD

Second set of boundary conditions at x=a
dψ II dψ III
ψ II (x = a ) = ψ III (x = a ) =
dx x =a dx x =a

C eγa + D e −γa = G eika γ C eγa − γ D e −γa = ik G eika


width
A+ B = C + D (1)

jkA − jkB = γC − γD (2)

γa −γa (3)
C e +De =Ge ika

γ C eγa − γ D e −γa = ik G eika (4)

Eq. (3) × γ + Eq. (4) to eliminate D.
γ + jk ( jk −γ )a
C= e G (5)
2γ
Eq. (3) × (-γ) + Eq. (4) to eliminate C.
γ − jk ( jk +γ )a
D= e G (6)
2γ

width
A+ B = C + D (1)


γa −γa (3)
C e +De =Ge ika

γ C eγa − γ D e −γa = ik G eika (4)

Eq. (1) × jk + Eq. (2) to eliminate B
2 jkA = ( jk + γ )C + ( jk − γ ) D (7)

Substitute from eq. (5) and (6) into (7), we get an equation of A
and G only.

2 jkA =
1
2γ
[ ]
(γ + jk )2 e( jk −γ )a − (γ − jk )2 e( jk +γ )a G (8)


Transmission and Reflection Coefficients
It is an assignment to show that the transmission coefficient is
given by
G ∗G 1
T= ∗ = 2
AA 1 U
1+ o
sinh 2 (γa )
4 E (U o − E )
Also it is an assignment to show that the reflection coefficient
is given by
1 U o2
sinh 2 (γa )
B∗ B 4 E (U o − E )
R= ∗ =
AA 1 U o2
1+ sinh 2 (γa )
4 E (U o − E )

After proving both relations, it will be clear to you that
R +T =1
Which is logical as an incident particle is either reflected or
transmitted.
You may need to use the following relations
e z − e− z
sinh ( z ) =
2
e z + e− z
cosh ( z ) =
2
cosh 2 ( z ) − sinh 2 ( z ) = 1

Transmission coefficient versus barrier width
An electron is tunneling through 1.5 eV barrier

0.35
Transmission Coffecient

0.3
0.25
a=0.5 nm
0.2 a=1 nm
0.15 a=2 nm
a=4 nm
0.1
0.05

0
0 0.25 0.5 0.75 1 1.25 1.5
Energy (e.V)

One notices that shrinking the barrier width by half results in a dramatic
increase in the transmission coefficient. It is not a linear relation.
Also doubling the particle energy results in exponential increase in the
transmission coefficient.


Reflection coefficient versus barrier width

1.2

1
Reflection Coffecient

0.8 a=0.5 nm
a=1 nm
0.6
a=2 nm
0.4 a=4 nm

0.2

0
0 0.25 0.5 0.75 1 1.25 1.5
Energy (e.V)

Notice that R=1-T


Plotting the wave function
Now we have found all
constants in terms of A.
G
We can plot the shape of
the wave function in all
regions.
One would expect that as the barrier gets smaller in height
and/or narrower in width more particles will be able to cross
the barrier to the other side.
This results in a higher transmission coefficient “T”, and also a
larger amplitude for the transmitted wave “G”.
Check: http://phys.educ.ksu.edu/vqm/html/qtunneling.html


Contradiction with Classical Mechanics

G

This results are in contradiction with the classical mechanics
which predicts that is the particle’s energy is lower than the
barrier height, the particle overcome the barrier and thus can
not exist in the right hand side.


Remember classical mechanics
Will never be able to
A man at rest here pass this point
So he can’t
exist here


Ain Shams University
Mathematics and Engineering Physics Department

1st Year Electrical Engineering

Lecture 11
Modern Physics and Quantum Mechanics Course

Dr. Hatem El-Refaei

Contents
Infinite barrier
Finite barrier
Quantum tunnelling

Dr. Hatem El-Refaei 1

Note
All problems today are unbounded problem, i.e. the
particle is not confined in a certain region, so:
We will not be able to do the normalization condition.
Therefore, we will not be able to solve for all unknowns.
Therefore, we will not get a characteristic equation.
Therefore, energy levels are not quantized, and all energies
are possible.
But still there are a lot of important characteristics to
understand and learn today.


Infinite barrier


Potential step of infinite height and infinite width

Energy
∞

∞
E

x
Since the barrier height is infinite, incident particles can’t
penetrate through it, and particles reflect back.
So, there is zero probability of finding the particle inside
the step barrier.
Here, the QM solution leads to the same classical solution.


width

Energy, ψ
∞

∞
2mE
k=
h E

x
d ψ I 2m
2 0
2
+ 2 E ψI = 0
dx h
U =∞
ψ ( x ) = Ae jkx
+ Be − jkx

 E 
j  kx − t 
 E 
j  − kx − t  ψ (x ) = 0
Ψ (x, t ) = Ae  h 
+ Be  h 

Forward Backward
propagating wave propagating wave

width

Energy, ψ
∞

∞
2mE
k= E
h
x
d ψ I 2m
2 0
2
+ 2 E ψI = 0
dx h
U =∞
ψ ( x ) = Ae jkx
+ Be − jkx

ψ (x = 0 − ) = ψ (x = 0 + ) ψ (x ) = 0
A+ B = 0


width

Energy, ψ
∞

∞
Sin(x) E

x
d ψ I 2m
2 0
2
+ 2 E ψI = 0
dx h
B = −A U =∞
ψ (x ) = Ae jkx − Ae − jkx ψ (x ) = 0
ψ (x ) = A(e jkx − e − jkx )
ψ ( x ) = 2 jA sin (kx )

width

Energy, ψ
∞

∞
Sin(x) E

x
0
B = −A
2
Re flected Amplitude B∗B
Reflectivity R= = ∗ =1
Incident Amplitude A A

All the incident particle stream is reflected back.

width
Energy, ψ
∞

∞
Sin(x) E

x
0
Since the barrier extends to infinity in the x direction, no particle can penetrate
through the whole barrier. From phenomenological understanding, as x→∞,
ψ→0.


Finite barrier


width
Energy
E<Uo

∞
E
Uo
x
0

As the step height Uo gets smaller (but still E<Uo),
the penetration of the particles inside the step
increases, but finally no particles will succeed to
travel through the whole step to x→∞.

width
k= E<Uo α= >0
h h
∞
E
Uo
x
0
2
+ 2 EψI =0 2
+ 2 (E − U o ) ψ II = 0
dx h dx h
We have 4 unknowns (A,B,C, and D) and 3 equations:
Finiteness of ψ at x=∞
Continuity of ψ at x=0
Continuity of ∇ψ at x=0

width
k= E<Uo α= >0
h h
∞
E
Uo
x
0
2
+ 2 EψI =0 2
+ 2 (E − U o ) ψ II = 0
dx h dx h



width
k= E<Uo α= >0
h h
∞
E
Uo
x
0
2
+ 2 EψI =0 2
+ 2 (E − U o ) ψ II = 0
dx h dx h
We have 3 unknowns (A,B,C) and 2 equations:
Continuity of ψ
Continuity of ∇ψ
Thus, the best we can get is the ratio between parameters.

width
k= E<Uo α= >0
h h
∞
E
Uo
x
ψ I ( x ) = Ae jkx + Be − jkx 0
ψ II (x ) = C e −α x
Continuity of ψ Continuity of dψ/dx
dψ I dψ II
ψ I ( x = 0) = ψ II (x = 0) =
dx dx
jk ( A − B ) = −α C
x =0 x =0
A+ B = C
1 α  1 α 
A= 1 + j  C B = 1 − j  C
2 k  2 k 

width
k= E<Uo α= >0
h h

E

0
1  α  1  α 
A=  1+ j  C B =  1− j  C
2  k  2  k 

 α  α
2 1 −
∗ j  1 + j 
Re flected Amplitude BB  k  k
Reflectivity R= = ∗ = =1
Incident Amplitude AA  α  α
1 + j  1 − j 
 k  k


width
Energy
Uo
∞
C
x
0
1 α  jkx 1  α  − jkx
ψ I (x ) = Ae + Be
jkx − jkx
= 1 + j Ce + 1 − j  Ce
2 k 2 k
 e jkx + e − jkx  α  e jkx − e − jkx 
ψ I (x ) = C 
  + j C
  

 2  k  2 
 α 
C cos(kx ) − C sin (kx ) x≤0 
ψ (x ) =  k 
C e −α x x ≥ 0
 

width
Energy
Uo
∞
C
x
0

Since the barrier height is finite, particles can penetrate partially
in the vicinity of the potential step, and then they reflect back.
So, there is a finite probability of finding the particle in the
classically forbidden position.
Here, the QM solution is different from the classical one.


width
Energy
E<Uo Case 3 2m(U o − E )
α= >0
Case 2 h

Case 2 ψ ( x ) = C e −α x
x
0
Case 3

As the potential barrier height increases (Uo
increases) (from case 2 to case 3), α also increases,
and thus the exponential function dies quicker inside
the barrier. Thus it becomes less probable to find the
particle inside the barrier.

width
Energy
E<Uo 2m(U o − E )
α= >0
h

ψ ( x ) = C e −α x
x
0
If the barrier height (Uo) is kept constant, but the particle
energy increases (provided E<Uo), thus α decreases, and
the particle exponential function dies slower inside the
barrier. Hence, it becomes more probable to find the
particle in the vicinity of the barrier edge.

Tunneling through a potential
barrier


width
Energy E<Uo
Uo
E

x

A stream of particles incident on a finite width and height
potential barrier with E<Uo.
Part of the incident stream will succeed to penetrate through the
barrier and appear on the other side, this is the transmitted stream.
The other part will reflect back forming the reflected stream.
Note, a single particle doesn’t split into two.

width
Energy E<Uo
2mE Uo 2m(U o − E )
k= γ= >0
h h

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0


width
Energy E<Uo
Uo

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0


width
Energy E<Uo
Uo

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0

We have 5 unknowns (A,B,C,D,G) and 4 boundary conditions


width
Energy E<Uo
Uo

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0


We are interested in the transmission probability (T)
2 2
Transmitte d Amplitude G ∗G G
T= = ∗ =


width
Energy E<Uo
Uo

0 a x

+ 2 EψI = 0 + 2 (E − U o ) ψ II = 0 + 2 E ψ III = 0


And reflection probability (R)
2 2
Re flected Amplitude B ∗B B
R = = ∗ =


width
First set of boundary conditions at x=0

dψ I dψ II
ψ I (x = 0) = ψ II (x = 0) =
dx x =0 dx x =0

A+ B = C + D jkA − jkB = γC − γD

Second set of boundary conditions at x=a
dψ II dψ III
ψ II (x = a ) = ψ III (x = a ) =
dx x =a dx x =a

C eγa + D e −γa = G eika γ C eγa − γ D e −γa = ik G eika


width
A+ B = C + D (1)


γa −γa (3)
C e +De =Ge ika

γa − γa
γ C e − γ D e = ik G e ika (4)

Eq. (3) × γ + Eq. (4) to eliminate D.
γ + jk ( jk −γ )a
C= e G (5)
2γ
Eq. (3) × (-γ) + Eq. (4) to eliminate C.
γ − jk ( jk +γ )a
D= e G (6)
2γ

width
A+ B = C + D (1)


γa −γa (3)
C e +De =Ge ika

γa − γa
γ C e − γ D e = ik G e ika (4)

Eq. (1) × jk + Eq. (2) to eliminate B
2 jkA = ( jk + γ )C + ( jk − γ ) D (7)

Substitute from eq. (5) and (6) into (7), we get an equation of A
and G only.

2 jkA =
1
2γ
[ ]
(γ + jk )2 e( jk −γ )a − (γ − jk )2 e( jk +γ )a G (8)


It is an assignment to show that the transmission coefficient is
given by
G ∗G 1
T= ∗ = 2
AA 1 U
1+ o
sinh 2 (γa )
4 E (U o − E )
Also it is an assignment to show that the reflection coefficient
is given by
1 U o2
sinh 2 (γa )
B∗ B 4 E (U o − E )
R= ∗ =
AA 1 U o2
1+ sinh 2 (γa )
4 E (U o − E )

After proving both relations, it will be clear to you that
R +T =1
Which is logical as an incident particle is either reflected or
transmitted.
You may need to use the following relations
e z − e− z
sinh ( z ) =
2
e z + e− z
cosh ( z ) =
2
cosh 2 ( z ) − sinh 2 ( z ) = 1

Transmission coefficient versus barrier width

0.35
Transmission Coffecient

0.3
0.25
a=0.5 nm
0.2 a=1 nm
0.15 a=2 nm
a=4 nm
0.1
0.05

0
0 0.25 0.5 0.75 1 1.25 1.5
Energy (e.V)

One notices that shrinking the barrier width by half results in a dramatic
increase in the transmission coefficient. It is not a linear relation.
Also doubling the particle energy results in exponential increase in the
transmission coefficient.


Reflection coefficient versus barrier width

1.2

1
Reflection Coffecient

0.8 a=0.5 nm
a=1 nm
0.6
a=2 nm
0.4 a=4 nm

0.2

0
0 0.25 0.5 0.75 1 1.25 1.5
Energy (e.V)

Notice that R=1-T


Plotting the wave function
Now we have found all
constants in terms of A.
G
We can plot the shape of
the wave function in all
regions.
One would expect that as the barrier gets smaller in height
and/or narrower in width more particles will be able to cross
the barrier to the other side.
This results in a higher transmission coefficient “T”, and also a
larger amplitude for the transmitted wave “G”.
Check: http://phys.educ.ksu.edu/vqm/html/qtunneling.html


Contradiction with Classical Mechanics

G

This results are in contradiction with the classical mechanics
which predicts that is the particle’s energy is lower than the
barrier height, the particle overcome the barrier and thus can
not exist in the right hand side.


Remember classical mechanics
Will never be able to
A man at rest here pass this point
So he can’t
exist here


Physics barriers and tunneling

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (12)

Recently uploaded

Recently uploaded (20)

Physics barriers and tunneling