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

[email protected] Dr. Hatem El-Refaei 1

Contents

� Infinite barrier

� Finite barrier

� Quantum tunnelling


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

� 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.

Energy

∞ ∞

x

E


Potential step of infinite height and infinite

width

( ) jkxjkx BeAex −+=ψ

( )

−−

−

+=ΨtE

kxjtE

kxj

BeAetx hh,

02

22

2

=+ II E

m

dx

dψ

ψh

Backward

propagating wave

Forward

propagating wave

∞=U

( ) 0=xψ

h

mEk

2=

E

Energy, ψ

∞ ∞

x0



width

( ) ( )+− === 00 xx ψψ

0=+ BA

Energy, ψ

∞ ∞

x0

∞=U

( ) 0=xψ

E


02

22

2

=+ II E

m

dx

dψ

ψh

h

mEk

2=



width

AB −=

( ) ( )kxjAx sin2=ψ

Energy, ψ

∞ ∞

x0

∞=U

( ) 0=xψ( ) jkxjkx AeAex −−=ψ

( ) ( )jkxjkx eeAx −−=ψ

Sin(x) E

02

22

2

=+ II E

m

dx

dψ

ψh



width

AB −=

Reflectivity 1Re

2

===∗

∗

AA

BB

AmplitudeIncident

AmplitudeflectedR

All the incident particle stream is reflected back.

Energy, ψ

∞ ∞

x0

Sin(x) E



width

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

Energy, ψ

∞ ∞

x0

Sin(x) E


Finite barrier


Potential step of finite height and infinite

width

� 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→∞.

Energy

∞

x

Uo

0

E<Uo

E



width

0

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx

II DeeCx ααψ += −

Energy

∞

x

Uo

� 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

E<Uo

( )0

2>

−=

h

EUm oαh

mEk

2=

E



width

0

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


Energy

∞

x

Uo

E<Uo

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

( )0

2>

−=

h

EUm oαh

mEk

2=

E



width

0

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) x

II eCx αψ −=

Energy

∞

x

Uo

E<Uo


( )0

2>

−=

h

EUm oαh

mEk

2=

E



width

0

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) x

II eCx αψ −=

Energy

∞

x

Uo

� 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.

E<Uo

( )0

2>

−=

h

EUm oαh

mEk

2=

E



width

( ) ( )00 === xx III ψψ

Continuity of ψ

CBA =+

Continuity of dψ/dx

00 ==

=x

II

x

I

dx

d

dx

d ψψ

( ) CBAjk α−=−

CkjA

+=α

12

1C

kjB

−=α

12

1

∞

x

Uo

EnergyE<Uo

( )0

2>

−=

h

EUm oαh

mEk

2=

0

E

( ) jkxjkx

I BeAex −+=ψ ( ) x

II eCx αψ −=



width

CkjA

+=α

12

1C

kjB

−=α

12

1

Reflectivity 1

11

11Re

2

=

−

+

+

−

=== ∗

∗

kj

kj

kj

kj

AA

BB

AmplitudeIncident

AmplitudeflectedR

αα

αα

EnergyE<Uo

( )0

2>

−=

h

EUm oαh

mEk

2=

0

E



width

jkxjkx CekjCe

kj −

−+

+=αα

12

11

2

1

( )( ) ( )

≥

≤−=

− 0

0sincos

xeC

xkxCk

kxCx

xα

αψ

( ) jkxjkx

I BeAex −+=ψ

∞

x

Uo

0

C

( )

−+

+=

−−

22

jkxjkxjkxjkx

I

eeC

kj

eeCx

αψ

Energy



width

� 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.

∞

x

Uo

0

C

Energy



width

Case 2

Case 2

x0

Energy

Case 3

Case 3E<Uo( )

02

>−

=h

EUm oα

� 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.

( ) xeCx αψ −=



width

� 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.

x0

EnergyE<Uo

( )0

2>

−=

h

EUm oα

( ) xeCx αψ −=


Tunneling through a potential

barrier


Potential barrier of finite height and finite

width

� 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.

x

Uo

E<Uo

E

Energy



width

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx

II eDeCx γγψ −+= ( ) jkxjkx

III FeGex −+=ψ

02

22

2

=+ IIIIII E

m

dx

dψ

ψh

Energy

x

Uo

0 a

E<Uo

( )0

2>

−=

h

EUm oγh

mEk

2=

� 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

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


III FeGex −+=ψ

02

22

2

=+ IIIIII E

m

dx

dψ

ψh

Energy

x

Uo

0 a

E<Uo




� Only a forward propagating wave on the right hand side of the barrier.



width

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx

II eDeCx γγψ −+= ( ) jkx

III Gex =ψ

02

22

2

=+ IIIIII E

m

dx

dψ

ψh

Energy

x

Uo

0 a

E<Uo

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





width

We are interested in the transmission probability (T)

22

A

G

AA

GG

AmplitudeIncident

AmplitudedTransmitteT === ∗

∗

Energy

x

Uo

0 a

E<Uo

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


III Gex =ψ

02

22

2

=+ IIIIII E

m

dx

dψ

ψh



width

And reflection probability (R)

22

Re

A

B

AA

BB

AmplitudeIncident

AmplitudeflectedR === ∗

∗

Energy

x

Uo

0 a

E<Uo

( ) jkxjkx

I BeAex −+=ψ

02

22

2

=+ II E

m

dx

dψ

ψh

( ) 02

22

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


III Gex =ψ

02

22

2

=+ IIIIII E

m

dx

dψ

ψh



width

� First set of boundary conditions at x=0

( ) ( )00 === xx III ψψ

DCBA +=+

00 ==

=x

II

x

I

dx

d

dx

d ψψ

DCjkBjkA γγ −=−

� Second set of boundary conditions at x=a

( ) ( )axax IIIII === ψψax

III

ax

II

dx

d

dx

d

==

=ψψ

ikaaa eGeDeC =+ −γγ ikaaa eGikeDeC =− −γγ γγ



width

DCBA +=+ (1)

DCjkBjkA γγ −=− (2)

ikaaa eGeDeC =+ −γγ (3)

ikaaa eGikeDeC =− −γγ γγ (4)

� Eq. (3) × γ + Eq. (4) to eliminate D.

� Eq. (3) × (-γ) + Eq. (4) to eliminate C.

( ) Gejk

C ajk γ

γγ −+

=2

(5)

( ) Gejk

D ajk γ

γγ +−

=2

(6)



width

� Eq. (1) × jk + Eq. (2) to eliminate B

( ) ( )DjkCjkjkA γγ −++=2 (7)

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

and G only.

( ) ( ) ( ) ( )[ ]GejkejkjkA ajkajk γγ γγγ

+− −−+= 22

2

12 (8)

DCBA +=+ (1)





Transmission and Reflection Coefficients

� It is an assignment to show that the transmission coefficient is

given by

( )( )a

EUE

UAA

GGT

o

o γ22

sinh4

11

1

−+

== ∗

∗

� Also it is an assignment to show that the reflection coefficient

is given by

( )( )

( )( )a

EUE

U

aEUE

U

AA

BBR

o

o

o

o

γ

γ

22

22

sinh4

11

sinh4

1

−+

−== ∗

∗



� After proving both relations, it will be clear to you that

1=+TR

� Which is logical as an incident particle is either reflected or

transmitted.

� You may need to use the following relations

( )2

sinhzz ee

z−−

=

( )2

coshzz ee

z−+

=

( ) ( ) 1sinhcosh 22 =− zz


Transmission coefficient versus barrier width

� One notices that shrinking the barrier width by half results in a dramaticincrease in the transmission coefficient. It is not a linear relation.

� Also doubling the particle energy results in exponential increase in thetransmission coefficient.

An electron is tunneling through 1.5 eV barrier

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.25 0.5 0.75 1 1.25 1.5

Energy (e.V)

Transmission Coffecient

a=0.5 nm

a=1 nm

a=2 nm

a=4 nm


Reflection coefficient versus barrier width


0

0.2

0.4

0.6

0.8

1

1.2

0 0.25 0.5 0.75 1 1.25 1.5

Energy (e.V)

Reflection Coffecient

a=0.5 nm

a=1 nm

a=2 nm

a=4 nm

� Notice that R=1-T


Plotting the wave function

� Now we have found all

constants in terms of A.

� 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

G


Contradiction with Classical Mechanics

� 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.

G


Remember classical mechanics

A man at rest hereWill never be able to

pass this point

So he can’t exist here

Ain Shams University

Mathematics and Engineering Physics Department

1stYear Electrical Engineering

Lecture 11

Modern Physics and Quantum Mechanics Course

Dr. Hatem El-Refaei

Dr. Hatem El-Refaei 1

Contents

� Infinite barrier

� Finite barrier

� Quantum tunnelling


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

� 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.

Energy

∞ ∞

x

E



width


( )

−−

−

+=ΨtE

kxjtE

kxj

BeAetx hh,

0222

2

=+ II E

m

dx

dψ

ψh

Backward

propagating wave

Forward

propagating wave

∞=U

( ) 0=xψ

h

mEk

2=

E

Energy, ψ

∞ ∞

x0



width

( ) ( )+− === 00 xx ψψ

0=+ BA

Energy, ψ

∞ ∞

x0

∞=U

( ) 0=xψ

E


0222

2

=+ II E

m

dx

dψ

ψh

h

mEk

2=



width

AB −=

( ) ( )kxjAx sin2=ψ

Energy, ψ

∞ ∞

x0

∞=U

( ) 0=xψ( ) jkxjkx AeAex −−=ψ

( ) ( )jkxjkx eeAx −−=ψ

Sin(x) E

0222

2

=+ II E

m

dx

dψ

ψh



width

AB −=

Reflectivity 1Re

2

===∗

∗

AA

BB

AmplitudeIncident

AmplitudeflectedR

All the incident particle stream is reflected back.

Energy, ψ

∞ ∞

x0

Sin(x) E



width

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

Energy, ψ

∞ ∞

x0

Sin(x) E


Finite barrier



width

� 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→∞.

Energy

∞

x

Uo

0

E<Uo

E



width

0

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


Energy

∞

x

Uo

� 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

E<Uo( )

02

>−

=h

EUm oαh

mEk

2=

E



width

0

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


Energy

∞

x

Uo

E<Uo


( )0

2>

−=

h

EUm oαh

mEk

2=

E



width

0

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) x

II eCx αψ −=

Energy

∞

x

Uo

E<Uo


( )0

2>

−=

h

EUm oαh

mEk

2=

E



width

0

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) x

II eCx αψ −=

Energy

∞

x

Uo

� 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.

E<Uo( )

02

>−

=h

EUm oαh

mEk

2=

E



width

( ) ( )00 === xx III ψψ

Continuity of ψ

CBA =+

Continuity of dψ/dx

00 ==

=x

II

x

I

dx

d

dx

d ψψ

( ) CBAjk α−=−

CkjA

+=α

12

1C

kjB

−=α

12

1

∞

x

Uo

EnergyE<Uo

( )0

2>

−=

h

EUm oαh

mEk

2=

0

E

( ) jkxjkx

I BeAex −+=ψ ( ) x

II eCx αψ −=



width

CkjA

+=α

12

1C

kjB

−=α

12

1

Reflectivity 1

11

11Re

2

=

−

+

+

−

=== ∗

∗

kj

kj

kj

kj

AA

BB

AmplitudeIncident

AmplitudeflectedR

αα

αα

EnergyE<Uo

( )0

2>

−=

h

EUm oαh

mEk

2=

0

E



width

jkxjkx CekjCe

kj −

−+

+=αα

12

11

2

1

( )( ) ( )

≥

≤−=

− 0

0sincos

xeC

xkxCk

kxCx

xα

αψ

( ) jkxjkx

I BeAex −+=ψ

∞

x

Uo

0

C

( )

−+

+=

−−

22

jkxjkxjkxjkx

I

eeC

kj

eeCx

αψ

Energy



width

� 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.

∞

x

Uo

0

C

Energy



width

Case 2

Case 2

x0

Energy

Case 3

Case 3E<Uo( )

02

>−

=h

EUm oα

� As the potential barrier height increases (Uoincreases) (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.

( ) xeCx αψ −=



width

� If the barrier height (Uo) is kept constant, but the particle

energy increases (provided E<Uo), thus α decreases, andthe particle exponential function dies slower inside the

barrier. Hence, it becomes more probable to find the

particle in the vicinity of the barrier edge.

x0

EnergyE<Uo

( )0

2>

−=

h

EUm oα

( ) xeCx αψ −=


Tunneling through a potential

barrier



width

� 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.

x

Uo

E<Uo

E

Energy



width

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


III FeGex −+=ψ

0222

2

=+ IIIIII E

m

dx

dψ

ψh

Energy

x

Uo

0 a

E<Uo

( )0

2>

−=

h

EUm oγh

mEk

2=


� 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

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


III FeGex −+=ψ

0222

2

=+ IIIIII E

m

dx

dψ

ψh

Energy

x

Uo

0 a

E<Uo


� 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

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


III Gex =ψ

0222

2

=+ IIIIII E

m

dx

dψ

ψh

Energy

x

Uo

0 a

E<Uo

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





width

We are interested in the transmission probability (T)

22

A

G

AA

GG

AmplitudeIncident

AmplitudedTransmitteT === ∗

∗

Energy

x

Uo

0 a

E<Uo

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


III Gex =ψ

0222

2

=+ IIIIII E

m

dx

dψ

ψh



width

And reflection probability (R)

22

Re

A

B

AA

BB

AmplitudeIncident

AmplitudeflectedR === ∗

∗

Energy

x

Uo

0 a

E<Uo

( ) jkxjkx

I BeAex −+=ψ

0222

2

=+ II E

m

dx

dψ

ψh

( ) 0222

2

=−+ IIoII UE

m

dx

dψ

ψh

( ) xx


III Gex =ψ

0222

2

=+ IIIIII E

m

dx

dψ

ψh



width

� First set of boundary conditions at x=0

( ) ( )00 === xx III ψψ

DCBA +=+

00 ==

=x

II

x

I

dx

d

dx

d ψψ

DCjkBjkA γγ −=−

� Second set of boundary conditions at x=a

( ) ( )axax IIIII === ψψax

III

ax

II

dx

d

dx

d

==

=ψψ

ikaaa eGeDeC =+ −γγ ikaaa eGikeDeC =− −γγ γγ



width

DCBA +=+ (1)




� Eq. (3) × γ + Eq. (4) to eliminate D.

� Eq. (3) × (-γ) + Eq. (4) to eliminate C.

( ) Gejk

C ajk γ

γγ −+

=2

(5)

( ) Gejk

D ajk γ

γγ +−

=2

(6)



width

� Eq. (1) × jk + Eq. (2) to eliminate B

( ) ( )DjkCjkjkA γγ −++=2 (7)

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

and G only.

( ) ( ) ( ) ( )[ ]GejkejkjkA ajkajk γγ γγγ

+− −−+= 22

2

12 (8)

DCBA +=+ (1)






� It is an assignment to show that the transmission coefficient is

given by

( )( )a

EUE

UAA

GGT

o

o γ22

sinh4

11

1

−+

== ∗

∗

� Also it is an assignment to show that the reflection coefficient

is given by

( )( )

( )( )a

EUE

U

aEUE

U

AA

BBR

o

o

o

o

γ

γ

22

22

sinh4

11

sinh4

1

−+

−== ∗

∗



� After proving both relations, it will be clear to you that

1=+TR

� Which is logical as an incident particle is either reflected or

transmitted.

� You may need to use the following relations

( )2

sinhzz ee

z−−

=

( )2

coshzz ee

z−+

=

( ) ( ) 1sinhcosh 22 =− zz


Transmission coefficient versus barrier width

� One notices that shrinking the barrier width by half results in a dramaticincrease in the transmission coefficient. It is not a linear relation.

� Also doubling the particle energy results in exponential increase in thetransmission coefficient.


0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.25 0.5 0.75 1 1.25 1.5

Energy (e.V)

Transmission Coffecient

a=0.5 nm

a=1 nm

a=2 nm

a=4 nm


Reflection coefficient versus barrier width


0

0.2

0.4

0.6

0.8

1

1.2

0 0.25 0.5 0.75 1 1.25 1.5

Energy (e.V)

Reflection Coffecient

a=0.5 nm

a=1 nm

a=2 nm

a=4 nm

� Notice that R=1-T


Plotting the wave function

� Now we have found all

constants in terms of A.

� 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

G


Contradiction with Classical Mechanics

� 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.

G


Remember classical mechanics

A man at rest hereWill never be able to

pass this point

So he can’t exist here

Date post:	08-Feb-2017
Category:	Education
Upload:	mohamed-anwar
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Physics barriers and tunneling

Education