6.007 Lecture 42: Tunneling  MIT OpenCourseWare · Example: Barrier Tunneling . Question: What...
Embed Size (px)
Transcript of 6.007 Lecture 42: Tunneling  MIT OpenCourseWare · Example: Barrier Tunneling . Question: What...

Tunneling
Outline  Review: Barrier Reflection  Barrier Penetration (Tunneling)  Flash Memory
1

A Simple Potential Step
Region 1 Region 2
CASE I : Eo > V
In Region 1: In Region 2:
ψA = Ae−jk1x
ψB = Be−jk1x
ψC = Ce−jk1x
E = Eo
E = 0
x = 0x
Eoψ = − �2
2m
∂2ψ
∂x2
(Eo − V )ψ = − �2
2m
∂2ψ
∂x2
k21 =2mEo�2
k22 =2m (Eo − V )
�2
V
2

A Simple Potential Step
CASE I : Eo > V
is continuous: is continuous:
ψ1 = Ae−jk1x +Bejk1x ψ2 = Ce−jk2x
ψ1(0) = ψ2(0) A+B = C
∂ ∂ψ(0) =
∂x
kψ2(0) A
∂x− 2B =
ψ
∂ψC
k1∂x
Region 1 Region 2
ψA = Ae−jk1x ψC = Ce−jk1x
ψB = Be−jk1x
E = 0
x = 0x
V
3

A Simple Potential Step
CASE I : Eo > V
B
A=
1− k2/k11 + k2/k1
=k1 − k2k1 + k2
C
A=
2
1 + k2/k1
=2k1
k1 + k2
A+B = C
A−B = k2k1
C
Region 1 Region 2
ψA = Ae−jk1x ψC = Ce−jk1x
ψB = Be−jk1x
E = Eo
E = 0
x = 0x
V
4

Example from: http://phet.colorado.edu/en/getphet/oneatatime 5

Quantum Electron Currents
Given an electron of mass that is located in space with charge density and moving with momentum corresponding to
… then the current density for a single electron is given by
m
ρ = q ψ(x)2
< p > < v >= �k/m
J = ρv = q ψ2 (�k/m)
6

A Simple Potential Step
CASE I : Eo > V
JtransmittedTransmission = T =
Jincident=
JCJA
=ψC 2(�k2/m)ψA2(�k1/m) =
∣∣∣∣CA∣∣∣∣2k2
JreflectedReflection = R =
k1
Jincident=
JBJA
=ψB 2(�k1/m)ψA2(�k1/m) =
∣∣∣∣BA∣∣∣∣2
B
A=
1− k2/k11 + k2/k1
C
A=
2
1 + k2/k1
Region 1 Region 2
ψA = Ae−jk1x
ψB = Be−jk1x
ψC = Ce−jk1x
E = Eo
E = 0
x = 0x
V
7

A Simple Potential Step
CASE I : Eo > V
1
1
BReflection = R =
∣∣∣∣2
A
∣∣∣ =∣∣k1 − k2∣ ∣∣
2
k1 + k2
∣∣∣
Transmission = T = 1
∣−R
4k1k2=
T
R k1 + k22T +R = 1
k2k1
=
√1− V
EoEo = V Eo = ∞
Region 1 Region 2
ψA = Ae−jk1x ψC = Ce−jk1x
ψB = Be−jk1x
E = 0
x = 0x
V
8

IBM Almaden STM of Copper
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.
9

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.
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
CASE II : Eo < V
In Region 1: In Region 2:
2�
Eoψ = − ∂2ψ
2m ∂x2
(Eo − V )ψ = − �2
2m
∂2ψ
∂x2
k21 =2mEo�2
κ2 =2m (Eo − V )
ψC = Ce−κx
�2
Region 1 Region 2
ψA = Ae−jk1x
ψB = Be−jk1x
E = EoE = 0
x = 0x
V
12

A Simple
ψ2 = Ce−κx
Potential Step
is continuous: is continuous:
CASE II : Eo < V
ψ1 = Ae−jk1x +Bejk1x
ψ1(0) = ψ2(0) A+B = C
∂ ∂ψ(0) =
∂x
ψ
∂ψψ2(0)
∂x
κA
∂x−B = −j
ψC = Ce−κx
Ck1
Region 1 Region 2
ψA = Ae−jk1x
ψB = Be−jk1x
E = EoV
E = 0x
x = 0
13

A Simple Potential Step
CASE II : Eo < V
B
Total reflection Transmission must be zero
A=
1 + jκ/k1 C
1− jκ/k1 A=
2
1− jκ/k1
R =
∣∣∣∣BA∣∣∣∣2
= 1 T = 0
A+B = C
A−B = −j κ
ψC = Ce−κx
Ck1
Region 1 Region 2
ψA = Ae−jk1x
ψB = Be−jk1x
E = EoE = 0
x = 0x
V
14

2a = L
T = 0
Quantum Tunneling Through a Thin Potential Barrier
Total Reflection at Boundary
Frustrated Total Reflection (Tunneling)
R = 1 T = 0
�
15

CASE II : Eo < V
Region 1 Region 2 Region 3
In Regions 1 and 3: In Region 2:
A Rectangular Potential Step
for Eo < V :
κxψA = Ae
−jk1x ψC = Ce−
ψB = Bejk1x
ψF = Fe−jk1x
ψD = Deκx
VE = Eo
E = 0 −a a0
Eoψ = − �2
2m
∂2ψ
∂x2
(Eo − V )ψ = − �2
2m
∂2ψ
∂x2
k21 =2mEo�2
κ2 =2m(V − Eo)
�2
T =
∣∣∣∣FA∣∣∣∣2
=1
1 + 14V 2
Eo(V−Eo) sinh2(2κa)
16

A Rectangular Potential Step
x=0 x=L
2a = L
Tunneling Applet: http://www.colorado.edu/physics/phet/dev/quantumtunneling/1.07.00/
Real part of Ψ for Eo < V, shows hyperbolic
(exponential) decay in the barrier domain and decrease
in amplitude of the transmitted wave.
for Eo < V :
Transmission Coefficient versus Eo/V for barrier with
T =
∣∣∣∣FA∣∣∣∣2
=1
1 + 14V 2
Eo(V−Eo) sinh2(2κa)
T =
∣∣∣∣FA∣∣∣∣2
≈ 11 + 14
V 2
Eo(V−Eo)e−4κa
sinh2(2κa) =[e2κa − e−2κa]2 ≈ e−4κa
2m(2a)2V/� = 16
17

Flash Memory
Electrons tunnel preferentially when a voltage is applied
Erased 1
Stored Electrons
Programmed 0
Image is in the public domain
Insulating Dielectric
CONTROL GATETunnel Oxide
FLOATING GATE
SOURCE CHANNEL
Substrate
Channel
Floating Gate
DRAIN
18

MOSFET: Transistor in a Nutshell
Conducting Channel
Tunneling causes thin insulating layers to become leaky !
Image is in the public domain
Conduction electron flow
Control Gate
Semiconductor
19

Reading Flash Memory
UNPROGRAMMED PROGRAMMED
To obtain the same channel charge, the programmed gate needs a higher controlgate voltage than the unprogrammed gate
How do we WRITE Flash Memory ?
CONTROL GATE
FLOATING GATE
SILICON
CONTROL GATE
FLOATING GATE
20

0 L
V0
x
Eo metal metal
air gap
Example: Barrier Tunneling
Question: What will T be if we double the width of the gap?
• Let s 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?
L = 2a
T =
∣∣∣∣FA∣∣∣∣2
≈ 16Eo(V − Eo)V 2
e−2κL
κ =
√2me�2
(V − Eo) = 2π√
2meh2
(V − Eo) = 2π√
6eV
1.505eVnm2≈ 12.6 nm−1
T = 4e−2(12.6 nm−1)(0.18 nm) = 4(0.011) = 4.4%
21

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
2. What is the energy of the particles that have successfully escaped ?
a. < initial energy b. = initial energy c. > initial energy
0 L
V
x
Eo
Although the amplitude of the wave is smaller after the barrier, no energy is lost in the tunneling process
22

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 STM tip One can exploit this to measure the ~ 1 nm electron density on a material s surface:
Sodium atoms on metal:
STM images
Single walled carbon nanotube:
V E0
STM tip material
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 Image is in the public domain .
23

Reflection of EM Waves and QM Waves
Then for optical material when μ=μ0:
= probability of a particular = probability of a particular photon being reflected electron being reflected
photonsP = �ω ×
s cm2
P =E2η
PreflectedR =
E=
Pincident
∣r∣o∣∣
2
Eio
∣∣∣∣
R =
∣∣B∣∣A∣∣∣∣2
=
∣∣∣∣k1 − k2k1 + k2∣∣∣∣2
=
∣∣∣∣n1 + n2
electronsJ = q ×
2
n1 + n2
∣∣∣∣
s cm2
J = ρv = q ψ2 (�k/m)
JreflectedR =
2ψ
= B 
Jincident ψA2
R =
∣∣B∣∣2
A
∣∣∣ =∣∣k1 − k2∣ ∣∣
2
k1 + k2
∣∣∣∣
24

MIT OpenCourseWarehttp://ocw.mit.edu
6.007 Electromagnetic Energy: From Motors to LasersSpring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
http://ocw.mit.edu/termshttp://ocw.mit.edu