Transcript Potential Step

Quantum
Physics
2002
Potential Step
Recommended Reading:
Harris Chapter 5, Section 1
Transmission and Reflection of Light
incident wave
Consider a ray of light incident on a
glass slide. Some of the incident light
will be reflected and some
transmitted.
transmitted wave
reflected wave
If I0 is the intensity of the incident light (number of photons pre
unit area per second), IR theintensity of the reflected light and IT
the intensity of the transmitted light then we define.
IR
Re flectionCoefficient, R 
I0
IT
Transmission Coefficient, T 
I0
and from conservation of energy R+T = 1 assuming no
absorption in the glass.
Transmission and Reflection of Matter Waves
According to deBroglie hypothesis matter behaves as a wave. So
does a matter wave divide into a reflected wave and a transmitted
wave when it encounters an abrupt change in potential?
Need to distinguish between particles moving in different directions.
In 1-Dimension we need to be able to distinguish between particles
moving to the left and those moving to the right
For a free particle of mass m, the solution to the Schrodinger equation
are plane waves (see previous lecture)
Ψx, t   Aeikxiωt
particle moving to the right, +x direction
Ψx, t   Aeikxiωt
particle moving to the left, -x direction
the time dependent part ( f(t) = exp(-it) )is the same for both particles
so we can leave it out of the analysis
Potential Step
Two possibilities: (1) total energy E of particle is greater than the
potential energy U0.
I
II
K.E.
U = U0
E = K.E.
E = K.E + U0
U
U=0
x
0
(2) total energy E of particle is less than the potential energy U0.
I
II
U = U0
U
E = K.E.
0
x
U=0
E<U
x > 0 is classically
forbidden region
Experimental Realisation of a Potential Step
-e
If E is > U0 then
electron is transmitted
through the electrodes
If E < U0 then electron
will be reflected
electron with total
energy E
0 Volts
-U0 Volts
I
II
K.E.
U = U0
E = K.E.
U
0
x
U=0
E = K.E + U0
Energy Bands in Solids
In a semiconductor we have allowed energy bands, I.e. Valence band
and Conduction band separated by a Band gap.
Energy
Conduction Band
Band gap
Valence band
Different semiconductors have
different band gaps.
For example:
Si
EG = 1.11 eV
Ge
EG = 0.66 eV
GaAs EG = 1.43 eV
InAs EG = 0.33 eV
If we grow one semiconductor
material on top of another we can
get a discontinuity in the potential.
An electron moving through the
material may be reflected or
transmitted when it encounters
this discontinuity. This will effect
the ‘transport properties’ of the
device.
InAs
GaAs
Potential Step E > U0
E
I
II
K.E
U = U0
E = K.E.
U
0
x
0 x  0
U(x)  
U0 x  0
U=0
Solution in Region I:
2 2ψ

x 
1
 d2

2

 Eψ1x   
 k1  ψ 1 x   0
 dx2

2m x2


2 2mE
k1 
where
2
2
Solution
ψ1x   Aeik1x  Beik1x
Region II:
 2  2 ψ2 x 

 U0 ψ2 x   Eψ2 x 
2
2m x
 d2

2
 
 k2  ψ 2 x   0
 dx2



3
2 2mE  U0 
k2 
2
where

ψ 2 x   Ceik2 x  Deik2x
Solution
I
II
Ceik2 x
Be
U = U0
 ik1x
Aeik1x
0
x
4
No Deik2 x term
because there is
no particle incident
from the right.
Continuity Conditions
Only one boundary, at x = 0.
Match wavefunction and derivative at x = 0.
ψ10   ψ2 0  
Aeik10  Beik10  Ceik2 0
5
AB  C
dψ1
dψ2

 ik1Aeik10  ik1Beik10  ik2Ceik2 0
dx x  0 dx x  0
k1A  B   k2C
6
Solve for B and C in terms of A.
k1  k 2
B
A
k1  k 2
7
C
2k1
A
k1  k 2
8
Reflection and Transmission Coefficients
What happens a particle when it encounters a potential step?
The particle does not split into two fragments, but rather it has a
certain probability of being transmitted and a probability of being
reflected.
A better way to view this problem is to consider a beam of particles
incident on the potential discontinuity. Then a certain fraction of
them will be transmitted and the rest will be reflected.
For example if there are 1000 particles per second in the incident
beam and 400 are reflected per unit time while 600 are transmitted
per unit time then:
400
Re flectionCoefficient, R 
1000
Transmission Coefficient, T 
R  T  0.4 0.6  1
 0.4
600
 0.6
1000
Total number of particles is conserved.
Transmission Probability
To find the reflection and transmission probabilities, we need to find
the ratios of the numbers per unit time in the corresponding beams
The number per unit time follows directly from 2.
Number number dis tance
2

φ v
time
distance time
where v is the velocity of the particle
Number per unit time
v
p
h k


k
m m m
number
2
φ k
time
The Transmission probability T, is then defined as
2
number transmitted time φ trans k 2 C * C k 2
T


2
number transmitted time φ
A * A k1
inc k1
Reflection Probability
Similarly the Reflection probability R, is defined as
2
φ refl k1 B * B
number reflected time
R


2
number transmitted time φ
A*A
inc k1
Using these definitions and the values of B and C in terms of A we find
R
B
2
A
2

k1  k 2 

k1  k2 2

1 k2

1 k2
Recall
2
4k1k2
k2 C
T

2
k1 A
k1  k2 2
4 k 2 k1

1 k2 k12
2
k1 2
k1 2
2 2mE
k1 
2


1
R
1
2 2mE  U0 
and k2 
2
so

1 U0 E 
2
9
2
1  U0 E 
T
4
1
1 U0 E 
1 U0 E 
2
10
Reflection and Transmission Coefficients
Things to note about equations 9 and 10:
1. R + T = 1
2. R is non-zero, so some particles are reflected even
though they have E > U0  non-classical behaviour,
wavelike property.
3. When E = U0 , T = 0 and R = 1.
k2
 1 U0 E  1  R = 0 and T = 1,
4. When E >> U0 ,
k1
so at very high energy all of the particles are
transmitted
Reflection and Transmission Coefficients
1
R+T=1
0.8
T
R, T
0.6
0.4
R
0.2
0
0
1
2
3
E/U0
4
5
6
Potential Step E < U0
I
II
U = U0
0 x  0
U(x, y)  
U0 x  0
E = K.E.
0
2


Region I:
2m
where
Solution
 2 ψ 1 x 
2
k1

2
x
2mE
x
U=0
 Eψ1x 
 d2

2
 
 k1  ψ 1 x   0
 dx2



11
2
ψ1x   Aeik1x  Beik1x
same as previous case
12
Region II:
 2  2 ψ2 x 

 U0 ψ2 x   Eψ2 x 
2
2m x
 d2

2
 
 α  ψ 2 x   0
 dx2



2
α 
where
2mU0  E 
2
note that  is always
positive
ψ 2 x   Ce αx  Deαx
Solution
I
II
U = U0
Beik1x
Aeik1x
0
Ce αx
x
13
14
No Deαx term
because
ψ2 x   0
as
x
Continuity Conditions
Only one boundary, at x = 0.
Match wavefunction and derivative at x = 0.
ψ10   ψ2 0  
Aeik10  Beik10  Ce-α0
15
AB  C
dψ1
dψ2


dx x  0 dx x  0
ik1Aeik10  ik1Beik10  αCe α0
ik1A  B   -αC
16
Solve for B and C in terms of A.
k1  iα
B
A
k1  iα
17
C
2k1
A
k1  iα
18
Reflection Coefficient
Reflection Coefficient
R
B
2
A
2

k1  α 2

1
2
k1  α 
19
Transmitted current density?
T  1- R
 T0
For E < U0 we get total reflection and no transmitted
‘wave’, this is the same as the classical case BUT there is
a difference between this and the classical case!
7. What happens in the region x > 0
•Quantum mechanically there is a finite probability that the
wavefunction penetrates the classically forbidden region, x > 0
2
2
Px   ψ2 x   C e  2αx
A
2
2
4k1
2
k1
 α2
e  2 αx
20
 finite probability of finding particle in region II (x > 0).
This probability is a maximum at x = 0 and falls off exponentially to
small values as x becomes large.
The extent of the penetration into the classically forbidden region is
governed by the constant . We define the penetration depth as
1

δ 
α
2mU0  E
21
The closer E is to U0 the greater is 
and the slower is the decay of the
wavefunction.
Question: What happens if we have a potential barrier instead
of a potential step?
I
II
III
U = U0
E
x
0
I
III
U = U0
II
E
0
Again 2 cases to consider
1) E > U0. Allowed both in
both classically and
quantum mechanics
x
2) E < U0. Classically
forbidden but allowed in
quantum systems.
QUANTUM TUNNELING.