Transcript An unbound state occurs when the energy is sufficient to take the particle to infinity, E > V(). 2 2  x 

An unbound state occurs when the energy is sufficient to take the particle to infinity,
E > V().
2 2
 x  cos kx
 
 
Free particle in 1D
(1)
  x   sin  kx 
2
2
d
V  x  0
E  
  x   e  ikx
2
2m dx
•In this case, it is easiest to understand results if we use complex waves
•Include the time dependence, and it is clear these are traveling waves
  x, t   eikxiEk t
e+ikx
e-ikx
Re() Im()
k
E 

2m
2
k2
Ek 
2m
Free particle in 1D (2)
  x   eikx
Comments:
•Note we get two independent solutions, and solutions for every energy


2
•Normalization is a problem
eikx e ikx dx 
 dx 




Wave packets solve the problem of normalization
•The general solution of the time-dependant Schrödinger
equation is a linear combination of many waves
•Since k takes on continuous values, the sum becomes an integral
  x, t    ck eikx iEk t
k
Wave packets are a
pain, so we’ll use
waves and ignore
normalization



1dx  
  x, t    c  k  eikx iEk t dk
Re() Im()
I
The Step Potential (1)
 0 if x  0
d
E  
 V  x  V  x   
2
2m dx
V0 if x  0
2
2
II
incident
transmitted
reflected
•We want to send a particle in from the left to scatter off of the barrier
•Does it continue to the right, or does it reflect?
Solve the equation in each of the regions
2
2 2
•Assume E > V0
d 2
k
ikx
E



I  e
E
•Region I
2
2m dx
2m
•Region II
d 2
E  
 V0
2
2m dx
2
 II  e
Most general solution is linear combinations of these
•It remains to match boundary conditions
•And to think about what we are doing!
 ik x
k 2
E  V0 
2m
2
 I  x   Ae  Be
ikx
 ikx
 II  x   Ceik x  De  ik x
I
The Step Potential (2)
 I  x   Ae  Be
ikx
 II  x   Ce
ik x
 ikx
 De
 ik x
k2
E
2m
incident
II
transmitted
reflected
2
2

k
E  V0 
2m
What do all these terms mean?
2
 I  0  II  0
A B  C
•Wave A represents the incoming wave from the left
•Wave B represents the reflected wave
 0  II 0

I
•Wave C represents the transmitted wave
ikA  ikB  ik C
•Wave D represents an incoming wave from the right
•We should set D = 0
kA  kB  k C  k A  k B
Boundary conditions:
k  k A  k  k B
•Function should be continuous at the boundary
•Derivative should be continuous at the boundary
 

k  k
B
A
k  k
2k
C  A B 
A
k  k
 


2k
C
A
k  k

I
The Step Potential (3)
k  k
A
•The barrier both reflects and transmits B 
k  k
II
incident
transmitted
reflected
•What is probability R of reflection?
R
k
B
2
A
2
2

k

k

 2 

A
 k  k 
B
2
V0 
8
9
E
2mE
2m  E  V0 
 E  E  V0
k 
R
 E  E V
0

•The transmission probability
is trickier to calculate because the speed changes
•Can use group velocity, wave packets,
probability current
•Or do it the easy way:




2
T  1 R 

4 E  E  V0 
E  E  V0

2
The Step Potential (4)
•What if V0 > E?
I
•Region I same as before
•Region II: we have
e
d
 V0  E 
2
2m dx
2
2
ikx
2
2
k
E
2m
 II  x   e
 x
 I  0    II  0 
ikA  ikB   C
A B  C
•Calculate the reflection probability
2
2

 k2
ik


R 2 
 2
 1 R
2
 k
A
ik  
B
2
II
incident
evanescent
reflected
•Don’t want it to blow up at infinity, so e-x
•Take linear combination of all of these
•Match waves and their derivative at boundary
 I  0    II  0 
I
V0  E 

2
2
2m
 I  Aeikx  Beikx
 II  Ce x
ik    A  ik    B
ik  
B
A
ik  
2ik
C
A
ik  
The Step Potential (5)
•When V0 > E, wave totally reflects
•But penetrates a little bit!
•Reflection probability is non-zero unless V0 = 0
•Even when V0 < 0!
 E  E  V
0

R   E  E  V0


1





I
incident
reflected
II
evanescent
V0  2E
2
if V0  E
if V0  E
T
R
V0 E
Sample Problem
Electrons are incident on a step potential V0 = - 12.3 eV. Exactly ¼ of
the electrons are reflected. What is the velocity of the electrons?
 E  E  V0
1
 R
 E  E V
4
0

E  E  V0
1

2
E  E  V0




2
•Must have E > 0
E   V0  1.54 eV
1
8
2 E  2 E  V0  E  E  V0 or
2 E  2 E  V0   E  E  V0
E  3 E  V0 or 3 E  E  V0
E  9  E V0  or 9E  E V0
E  98 V0 or E   81 V0
E  12 mv2
2E
2 1.54 eV 
5
v


0.00245c

7.35

10
m/s
6
2
m
0.51110 eV/c
The Barrier Potential (1)
•Assume V0 > E
•Solve in three regions
 I  x   Aeikx  Beikx
 II  x   Ce  De
x
 III  x   Fe
ikx
 x
I
II
E
2
V0  E 
2
III
V0 0  x  L
V  x  
0 otherwise
k 2 2m
 2 2m
•Let’s find transmission
probability
•Match wave function and derivative at both boundaries
A B  C  D
Ce L  De L  FeikL
ikA  ikB   C   D  Ce L   De L  ikFeikL
•Work, work . . .
2 ke  ikL A
F
2 k cosh  L   i  2  k 2  sinh  L 
T
F
A
2
2


4E V0  E 
4E V0  E   V02 sinh 2  L 
The Barrier Potential (2)
V0  E  2 2 2m
I
4E V0  E 
T
4E V0  E   V02 sinh 2  L 
•Solve for :
•Assume a thick barrier: L large
•Exponentials beat everything
II
III
V0 0  x  L
V  x  
0 otherwise
  2m V0  E 
sinh  L  
1
2
L
 L
1 L
e

e


 2e
E  E  2 L
T  16 1   e
V0  V0 
Sample Problem
An electron with 10.0 eV of kinetic energy is trying to
leap across a barrier of V0 = 20.0 eV that is 0.20 nm
wide. What is the barrier penetration probability?
  2m V0  E 

 2 .511 MeV/c
2 .511106 eV  10 eV 
3.00 10
8
m/s  6.582 10
16
2
E
E  2 L
T  16 1   e
V0  V0 
  2m V0  E 
  20.0 eV 10.0 eV
eV  s 
 1.62 1010 m1
 10.0 eV  10.0 eV 
9
10
1


T  16 
1

exp

2
0.20

10
m
1.62

10
m







 20.0 eV  20.0 eV 
 4.00e6.48  0.61%
The Scanning Tunneling Microscope
•Electrons jump a tiny barrier between the tip and the sample
•Barrier penetration is very sensitive to distance
•Distance is adjusted to keep current constant
•Tip is dragged around
•Height of surface is then mapped out