Download Tunnelling

Quantum
Physics
2002
Quantum Tunnelling
Recommended Reading:
R.Harris, Chapter 5 Sections 1, 2 and 3
Potential Barrier: E < U0
I
II
III
U = U0
E = K.E.
x =L
x =0
Region I
 d2

2

 k1 φ 1  0
 dx2



Potential
0 x  0

U(x)  U0 0  x  L
0 L  x

E = K.E.
U
x
Region III
Region II
 d2

2

 α φ 2  0
 dx2



 d2

2

 k1 φ 3  0 1
 dx2



where
2
k1

2mE

2
and
α2 
2mU0  E 
2
Wavefunctions
Region I:
φ1 x   Aeik1x  Beik1x
Incident
Region II:
2
Reflected
φ 2 x   Ce
αx
 De
 αx
Must keep both terms.
Do you see why?
3
Region III:
φ3 x   Feik1x  Geik1x
Transmitted
Left
Moving
4
No Ge ik1x
term
because there is no
particle incident from
the right.
Boundary Conditions
Match wavefunction and derivative at x = 0.
φ10   φ 2 0  
Aeik10  Beik10  Ceα0  Deα0
AB  CD
dφ1
dφ 2


dx 0 dx 0
5
ik1Aeik10  ik1Beik10  αCeα0  αDeα0
ik1A  B  αC  D
6
Match wavefunction and derivative at x = L.
φ2 L   φ3 L  
dφ
dφ 2
 3 
dx L dx L
CeαL  DeαL  Feik1L
7
αCeαL  αDeαL  ik1Feik1L
8
Boundary Conditions
We now have 4 equations and 5 unknowns, Can solve for B, C, D and
F in terms of A. This is left as an exercise, A lot of algebra but
nothing complicated!!
Again we define a Reflection and Transmission coefficient:
R
*
B
2
reflcurrent B B


*
inc current A A A 2
2
trans.current k 3 F *F F
T


*
inc.current
k1 A A A 2
Since k1 = k3.
Substituting for B and F in terms of A gives:
sinh2 αL 
Reflection Coefficient R R 

sinh2 αL   4 α 2k12 k12  α 2


 
2
9
R and T Coefficients:
Transmission Coefficient T
T

4 α 2k12


k12
α

sinh αL   4 k12k 22

2

 
2 2
k12
α
 
2 2
Recall that sin(i) =sinh().
We can write k1 and k2 in terms of E and U0.
2
k1

2mE
2
k2
2 
2mE  U0 
2
This then gives
 E 
E 
4
 1 

U0  U0 

T
 E 
E 
2  2mU0  E  
sinh 
L   4
 1 


 U0  U0 


11
10
Dividing across by
 E 
E 
4
 1 

 U0  U0 
α L

4EU0  E  
 U20 sinh2
1
T

gives
1
11a
similarly we can find an expression for the Reflection coefficient
 2mU0  E  
sinh2 
L



R
 E 
E 
2  2mU0  E  
sinh 
L   4
 1 


 U0  U0 


12
or rearranging

4EU0  E  
R  1 

2
2
 U0 sinh αL 
-1
12a
Graph of Transmission Probability
100
100
L = 0.1 nm
U0 = 0.1 eV
U0 = 1.0 eV
10-5
L = 0.5 nm
10-2
T
T
L = 1.0 nm
10-10
U0 = 5.0 eV
10-4
U0 = 10.0 eV
10-15
0
0.2
0.4
0.6
0.8
1.0
E/U0
Transmission curves for a
barrier of constant width 1.0 nm
with different heights U0
10-6
0
0.2
0.4
0.6
0.8
1.0
E/U0
Transmission curves for a
barrier of constant height 1.0
eV for a series of different
widths L.
Wavefunction
φ 2 x   Ceαx  Deαx
φ1 x   Aeik1x  Beik1x
φ3 x   Feik1x
U0
E
Optical Analog
evanescent wave
If reflection angle is greater than the critical angle then the light ray
will be totally internally reflected
If second prism is brought close to
the first there is a small probability
for part of the incident wave to
couple through the air gap and
emerge in the second prism.
Limiting Case
Tunnelling through wide barriers:
Inside the barrier the wavefunction is proportional to exp(-x) or
exp(-x/), where  = 1/ is the penetration depth (see Potential step
lecture). If L   then very little of the wavefunction will survive to
x = L. The condition for a ‘wide barrier’ is thus
2mU0  E 
1
L  αL 
L  1 13
δ

The barrier can be considered to be wide if L is large or if E << U 0.
Making this approximation we see that
e y  e  y y  1 e y
sinhy  
 
2
2
so for a thick barrier
equation 11 reduces to
2y
e
and then sinh y  
4
2
 E 
E   2 αL
T  16
 1 
e
 U0  U0 
14
The probability of tunnelling is then dominated by the exponentially
decreasing term.
Example
An electron (m = 9.11  10-31kg) encounters a potential barrier of
height 0.100eV and width 15nm What is the transmission probability
if its energy is (a) 0.040eV and (b) 0.060 eV?
We first check to see if the barrier is thick (equation 13). for E = 0.04eV




2  9.11 1031kg 0.10  0.40 1.602  1019 J
L

15  109 m
δ
1.055  1034 J.s
= 18.8 >> 1  thick barrier
and for E = 0.060: L/ = 15.5 >> 1  thick barrier
 we can use equation 14 for the transmission probability
 0.04  0.04  218.8
(a) E  0.04eV  T  16
 1.8  1016
 1 
e
0.10 
 0.10 
 0.06  0.06  215.4
13
(b) E  0.06eV  T  16
1

e

1
.
8

10


0
.
10
0
.
10



Very small in both cases!! Can we observe this in a real stuation
Field Emission
+V0
metal
Cathode
electrons bound by
potential step at
surface
Anode
Tunnelling through
potential barrier
Field Emission Displays (FED)
Scanning Tunnelling Microscope (STM)
Scanning Tunnelling Microscope (STM)
Scanning Tunnelling Microscope (STM)
Si (111) Surface
Pt Surface
Pentacene molecules on Silicon
Sample negative
Sample positive
The Tunnel Diode
see http://mxp.physics.umn.edu/s98/projects/menz/poster.htm
Conduction
Band
eV
donors
Conduction
Band
0
EF
EF
EF
Valence
Band
acceptors
Valence
Band
p-type
n-type
p-type
-- ++
-- ++
-- ++
n-type
The Tunnel Diode
reversed biased
forward biased
Conduction
Band
eV0 + Vext
eV0- eVext
EF
EF
Valence
Band
- +
- +
- +
p-type
-
+
Vext
n-type
p-type
--- +++
--- +++
--- +++
-
+
Vext
n-type
The Tunnel Diode
Alpha Decay of Nuclei
Alpha
particle
Uranium
238
Thorium 234
Strong Nuclear
Force
Electrostatic
repulsion
To escape the nucleus the -particle
must tunnel.