Reflection and Transmission
at a Potential Step
Outline
-  Review: Particle in a 1-D Box
-  Reflection and Transmission - Potential Step
-  Reflection from a Potential Barrier
-  Introduction to Barrier Penetration (Tunneling)
Reading and Applets:
.Text on Quantum Mechanics by French and Taylor
.Tutorial 10 – Quantum Mechanics in 1-D Potentials
.applets at http://phet.colorado.edu/en/get-phet/one-at-a-time
1
Schrodinger: A Wave Equation for Electrons
∂
Eψ = ωψ = −j ψ
∂t
p2
E=
2m
∂
px ψ = kψ = j ψ
∂x
(free-particle)
∂
2 ∂ 2 ψ
−j ψ = −
∂t
2m ∂x2
(free-particle)
..The Free-Particle Schrodinger Wave Equation !
Erwin Schrödinger (1887–1961)
Image in the Public Domain
2
Schrodinger Equation and Energy Conservation
The Schrodinger Wave Equation
2 ∂ 2 ψ(x)
Eψ(x) = −
+ V (x)ψ(x)
2
2m ∂x
The quantity |  |2 dx is interpreted as the probability that the particle can be
found at a particular point x (within interval dx)
|ψ|2
2
P (x) = |ψ| dx
n=3
0
L
x
© Dr. Akira Tonomura, Hitachi, Ltd., Japan. All rights reserved. This content is
excluded from our Creative Commons license. For more information, see
http://ocw.mit.edu/fairuse.
3
Schrodinger Equation and Particle in a Box
2 ∂ 2 ψ(x)
Eψ(x) = −
+ V (x)ψ(x)
2
2m ∂x
ψ(x) =
nπ x 2
2
sin
P (x) = |ψ(x)| dx
L
L
EIGENSTATES for
1-D BOX
EIGENENERGIES for
1-D BOX
PROBABILITY
DENSITIES
ψ3
x
x
x
x
ψ2
ψ1
x
L
0
4
0
x
L
Semiconductor Nanoparticles
(aka: Quantum Dots)
Determining QD energy
using the Schrödinger Equation
Core Shell
Core
Quantum
Dot
2
E1 = n E 1
2 k12
2 π 2
=
E1 =
2m
2mL2
9E1B
Red: bigger dots!
Blue: smaller dots!
9E1R
4E1B
E1B
3KRWRE\-+DOSHUW&RXUWHV\RI0%DZHQGL*URXS&KHPLVWU\0,7
LBLUE QD
5
4E1R
E1R
LRED QD
Solutions to Schrodingers Equation
2 ∂ 2 ψ
−
= (E − V (x)) ψ
2
2m ∂x
The kinetic energy of the electron is related to the
curvature of the wavefunction
Tighter confinement
Higher energy
Even the lowest energy bound state requires some wavefunction
curvature (kinetic energy) to satisfy boundary conditions
Nodes in wavefunction
Higher energy
The n-th wavefunction (eigenstate) has (n-1) zero-crossings
6
The Wavefunction
•  |ψ|
2
dx corresponds to a physically meaningful quantity –
- the probability of finding the particle near x
∗ dψ dx is related to the momentum probability density •  ψ
dx - the probability of finding a particle with a particular momentum
PHYSICALLY MEANINGFUL STATES MUST HAVE THE FOLLOWING PROPERTIES:
 (x) must be single-valued, and finite
(finite to avoid infinite probability density)
 (x) must be continuous, with finite d /dx
(because d /dx is related to the momentum density)
In regions with finite potential, d /dx must be continuous
(with finite d2 /dx2, to avoid infinite energies)
There is usually no significance to the overall sign of  (x)
(it goes away when we take the absolute square)
(In fact,  (x,t) is usually complex !)
7
Solutions to Schrodingers Equation
ψ(x)
In what energy level is the particle? n = …
(a) 7
(b) 8
x
(c) 9
What is the approximate shape of the
potential V(x) in which this particle is
confined?
L
V(x)
V(x)
V(x)
(a)
(b)
(c)
E
E
L
L
8
E
L
WHICH WAVEFUNCTION CORRESPONDS TO WHICH POTENTIAL WELL ?
(A)
(1)
(B)
(2)
(C)
(3)
NOTICE THAT FOR FINITE POTENTIAL WELLS WAVEFUNCTIONS ARE NOT ZERO AT THE WELL BOUNDARY
9
A Simple
Potential Step
ψA = Ae−jk1 x
ψC = Ce−jk1 x
ψB = Be−jk1 x
E = Eo
V
CASE I : Eo > V
Region 1
E=0
Region 2
x=0
x
In Region 1:
2 ∂ 2 ψ
Eo ψ = −
2m ∂x2
k12
2mEo
=
2
In Region 2:
2 ∂ 2 ψ
(Eo − V ) ψ = −
2m ∂x2
k22
2m (Eo − V )
=
2
10
A Simple
Potential Step
ψA = Ae−jk1 x
ψC = Ce−jk1 x
ψB = Be−jk1 x
E = Eo
V
CASE I : Eo > V
Region 1
E=0
ψ1 = Ae−jk1 x + Bejk1 x
ψ is continuous:
∂ψ is continuous:
∂x
ψ1 (0) = ψ2 (0)
∂
∂
ψ(0) =
ψ2 (0)
∂x
∂x
11
x=0
Region 2
x
ψ2 = Ce−jk2 x
A+B =C
k2
A−B = C
k1
A Simple
Potential Step
ψA = Ae−jk1 x
ψC = Ce−jk1 x
ψB = Be−jk1 x
E = Eo
V
CASE I : Eo > V
Region 1
E=0
x=0
Region 2
x
A+B =C
B
1 − k2 /k1
=
A
1 + k2 /k1
C
2
=
A
1 + k2 /k1
k1 − k 2
=
k1 + k2
2k1
=
k1 + k2
12
k2
A−B = C
k1
Example from: http://phet.colorado.edu/en/get-phet/one-at-a-time
13
Quantum Electron Currents
Given an electron of mass m
that is located in space with charge density
and moving with momentum
ρ = q |ψ(x)|
2
< p > corresponding to < v >= k/m
… then the current density for a single electron is given by
2
J = ρv = q |ψ| (k/m)
14
A Simple
Potential Step
ψA = Ae−jk1 x
ψC = Ce−jk1 x
ψB = Be−jk1 x
E = Eo
V
CASE I : Eo > V
Region 1
E=0
x=0
Region 2
x
2
Jref lected
JB
|ψB | (k1 /m) B = Reflection = R =
=
=
2
JA
|ψA | (k1 /m)
A
Jincident
2
2
Jtransmitted
JC
|ψC | (k2 /m) C k2
= Transmission = T =
=
=
2
Jincident
JA
|ψA | (k1 /m)
A k1
2
C
2
=
1 + k2 /k1
A
B
1 − k2 /k1
=
1 + k2 /k1
A
15
ψA = Ae−jk1 x
A Simple
Potential Step
ψC = Ce−jk1 x
ψB = Be−jk1 x
E = Eo
V
CASE I : Eo > V
Region 1
E=0
x=0
2
2 k1 − k2 B Reflection = R = = A
k1 + k2 1
T
Transmission = T = 1 − R
T +R=1
=
R
1
Eo = V
x
Region 2
Eo = ∞
k2
=
k1
1−
V
Eo
16
4k1 k2
|k1 + k2 |
2
IBM Almaden STM of Copper
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license. For more information, see http://ocw.mit.edu/fairuse.
17
IBM Almaden
Image originally created by the IBM Corporation.
© IBM Corporation. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
18
IBM Almaden
Image originally created by the IBM Corporation.
© IBM Corporation. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
19
A Simple
Potential Step
ψA = Ae−jk1 x
ψC = Ce−κx
ψB = Be−jk1 x
V
CASE II : Eo < V
Region 1
E=0
In Region 1:
2 ∂ 2 ψ
Eo ψ = −
2m ∂x2
In Region 2:
2 ∂ 2 ψ
(Eo − V ) ψ = −
2m ∂x2
20
E = Eo
Region 2
x=0
k12
x
2mEo
=
2
2m (Eo − V )
κ =
2
2
ψA = Ae−jk1 x
A Simple
Potential Step
ψC = Ce−κx
ψB = Be−jk1 x
V
CASE II : Eo < V
Region 1
ψ1 = Ae
ψ is continuous:
∂ψ is continuous:
∂x
−jk1 x
+ Be
E=0
x=0
jk1 x
E = Eo
Region 2
x
ψ2 = Ce−κx
ψ1 (0) = ψ2 (0)
A+B =C
∂
∂
ψ(0) =
ψ2 (0)
∂x
∂x
κ
A − B = −j C
k1
21
A Simple
Potential Step
ψA = Ae−jk1 x
ψC = Ce−κx
ψB = Be−jk1 x
V
CASE II : Eo < V
Region 1
E=0
x=0
E = Eo
Region 2
x
A+B =C
B
1 + jκ/k1
=
1 − jκ/k1
A
2
B R = = 1
A
C
2
=
1 − jκ/k1
A
κ
A − B = −j C
k1
T =0
Total reflection Transmission must be zero
22
Quantum Tunneling Through a Thin Potential Barrier
Total Reflection at Boundary
T =0
R=1
Frustrated Total Reflection (Tunneling)
T = 0
23
KEY TAKEAWAYS
CASE I : Eo > V
A Simple Potential Step
2 B k 1 − k 2 2
Ref lection = R = = A
k1 + k2 T ransmission = T = 1 − R =
4 k1 k2
|k1 + k2 |
Region 1
Region 2
Region 1
Region 2
2
PARTIAL REFLECTION
CASE II : Eo < V
TOTAL REFLECTION
24
A Rectangular
Potential Step
ψA = Ae−jk1 x
ψC = Ce−κx
ψB = Bejk1 x
ψD = Deκx
V
CASE II : Eo < V
E=0
In Region 2:
for Eo < V :
2 ∂ 2 ψ
Eo ψ = −
2m ∂x2
2 ∂ 2 ψ
(Eo − V )ψ = −
2m ∂x2
2
F T = =
A
1+
25
E = Eo
−a 0 a
Region 1
In Regions 1 and 3:
ψF = F e−jk1 x
Region 3
Region 2
k12
2mEo
=
2
2m(V − Eo )
κ =
2
2
1
1
V2
4 Eo (V −Eo )
sinh2 (2κa)
A Rectangular
Potential Step
E
U
for Eo < V :
2
F T = =
A
1+
2
1
1
V2
4 Eo (V −Eo )
2κa
sinh2 (2κa)
−2κa 2
sinh (2κa) = e
−e
≈ e−4κa
2
F 1
−4κa
T = ≈
e
2
A
1 + 14 Eo (VV −Eo )
26
Flash Memory
Stored
Electrons
Programmed
0
Erased
1
Image is in the public domain
CONTROL GATE
Tunnel Oxide
Insulating
Dielectric
Floating
Gate
FLOATING GATE
SOURCE
CHANNEL
DRAIN
Channel
x
V (x)
Substrate
Electrons tunnel preferentially when a voltage is applied
27
MOSFET: Transistor in a Nutshell
Conduction electron flow
Control Gate
Conducting Channel
Semiconductor
,PDJHFRXUWHV\RI-+R\W*URXS((&60,7
3KRWRE\/*RPH]
,PDJHFRXUWHV\RI-+R\W*URXS((&60,7
3KRWRE\/*RPH]
Tunneling causes thin insulating layers
to become leaky !
Image is in the public domain
28
Reading Flash Memory
UNPROGRAMMED
PROGRAMMED
CONTROL GATE
CONTROL GATE
FLOATING GATE
FLOATING GATE
SILICON
To obtain the same channel charge, the programmed gate needs a
higher control-gate voltage than the unprogrammed gate
How do we WRITE Flash Memory ?
29
Example: Barrier Tunneling
•  Lets consider a tunneling problem:
An electron with a total energy of Eo= 6 eV
approaches a potential barrier with a height of
V0 = 12 eV. If the width of the barrier is
L = 0.18 nm, what is the probability that the
electron will tunnel through the barrier?
V0
Eo
κ=
2me
(V − Eo ) = 2π
2
T = 4e
2me
(V − Eo ) = 2π
2
h
−2(12.6 nm−1 )(0.18 nm)
metal
metal
2
F 16Eo (V − Eo ) −2κL
T = ≈
e
2
A
V
0 L air
gap
x
6eV
−1
≈
12.6
nm
1.505eV-nm2
= 4(0.011) = 4.4%
Question: What will T be if we double the width of the gap?
30
L = 2a
Multiple Choice Questions
Consider a particle tunneling through a barrier:
1. Which of the following will increase the
likelihood of tunneling?
a. decrease the height of the barrier
b. decrease the width of the barrier
c. decrease the mass of the particle
V
Eo
0 L
2. What is the energy of the particles that have successfully escaped?
a. < initial energy
b. = initial energy
c. > initial energy
Although the amplitude of the wave is smaller after the barrier, no
energy is lost in the tunneling process
31
x
Application of Tunneling:
Scanning Tunneling Microscopy (STM)
Due to the quantum effect of barrier penetration, the
electron density of a material extends beyond its surface:
material
One can exploit this
to measure the
electron density on a
materials surface:
Sodium atoms
on metal:
STM tip
~ 1 nm
material
E0
STM tip
V
Single walled
carbon nanotube:
STM images
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by IBM Corporation
© IBM Corporation. All rights reserved. This content is excluded from our Creative
Commons license. For more information, see http://ocw.mit.edu/fairuse.
Image is in the public domain
32
MIT OpenCourseWare
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6.007 Electromagnetic Energy: From Motors to Lasers
Spring 2011
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