Tunneling
Outline
-  Review: Barrier Reflection
-  Barrier Penetration (Tunneling)
-  Flash Memory
1
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
2
A Simple
Potential Step
ψA = Ae−jk1 x
ψC = Ce−jk1 x
ψB = Be−jk1 x
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
3
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
4
k2
A−B = C
k1
Example from: http://phet.colorado.edu/en/get-phet/one-at-a-time
5
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)
6
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
Jincident
JA
|ψA | (k1 /m)
A
2
2
Jtransmitted
JC
|ψC | (k2 /m) C k2
= Transmission = T =
=
=
2
Jincident
JA
|ψA | (k1 /m)
A k1
2
C
2
=
A
1 + k2 /k1
B
1 − k2 /k1
=
A
1 + k2 /k1
7
ψA = Ae−jk1 x
A Simple
Potential Step
ψC = Ce−jk1 x
ψB = Be−jk1 x
V
CASE I : Eo > V
E=0
Region 1
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
8
4k1 k2
|k1 + k2 |
2
IBM Almaden STM of Copper
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license. For more information, see http://ocw.mit.edu/fairuse.
9
IBM Almaden
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© IBM Corporation. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
10
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.
11
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
12
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
13
A Simple
Potential Step
ψA = Ae−jk1 x
ψC = Ce−κx
ψB = Be−jk1 x
V
CASE II : Eo < V
E=0
Region 1
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
=
A
1 − jκ/k1
κ
A − B = −j C
k1
T =0
Total reflection Transmission must be zero
14
Quantum Tunneling Through a Thin Potential Barrier
Total Reflection at Boundary
T =0
R=1
Frustrated Total Reflection (Tunneling)
0
T =
2a = L
15
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+
16
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)
2a = L
A Rectangular
Potential Step
Real part of Ψ for Eo < V,
shows hyperbolic
(exponential) decay in the
barrier domain and decrease
in amplitude of the
transmitted wave.
x=0
for Eo < V :
2
F T = =
A
1+
2
Transmission Coefficient versus Eo/V
for barrier with 2m(2a)2 V / = 16
x=L
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 )
Tunneling Applet: http://www.colorado.edu/physics/phet/dev/quantum-tunneling/1.07.00/
17
Flash Memory
Stored
Electrons
Programmed
0
Erased
1
Image is in the public domain
Insulating
Dielectric
CONTROL GATE
Tunnel Oxide
Floating
Gate
FLOATING GATE
SOURCE
DRAIN
CHANNEL
Channel
Substrate
Electrons tunnel preferentially when a voltage is applied
18
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((&6
0,7 3KRWRE\/*RPH]
Tunneling causes thin insulating layers
to become leaky !
Image is in the public domain
19
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 ?
20
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?
21
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
22
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
Image originally created
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.
23
Image is in the public domain
Reflection of EM Waves and QM Waves
photons
P = ω ×
s cm2
|E|
P =
η
electrons
J =q×
s cm2
2
Pref lected
R=
Pincident
2
J = ρv = q |ψ| (k/m)
r 2
Eo = i Eo
2
|ψB |
Jref lected
=
R=
2
Jincident
|ψA |
Then for optical material when μ=μ0:
2
2 k1 − k 2 B 2
2 k1 − k 2 B R = = A
k1 + k2 =
A
k1 + k2 2
n1 + n2 = n 1 + n2 R=
= probability of a particular
photon being reflected
= probability of a particular
electron being reflected
24
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6.007 Electromagnetic Energy: From Motors to Lasers
Spring 2011
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