An unbound state occurs when the energy is sufficient to take the particle to infinity, E > V().
Free particle in 1D (1)2 2
22
dE
m dx
0V x
cos
sinikx
x kx
x kx
x e
•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
2 2
2
kE
m
2 2
2k
kE
m
, kikx iE tx t e
e+ikx
e-ikx
Re() Im()
Comments:•Note we get two independent solutions, and solutions for every energy•Normalization is a problem
Free particle in 1D (2) ikxx e
2dx
ikx ikxe e dx
1dx
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
, kikx iE tk
k
x t c e , kikx iE tx t c k e dk
Re() Im()
Wave packets are a pain, so we’ll use waves and ignore
normalization
The Step Potential (1)
0
0 if 0
if 0
xV x
V x
2 2
22
dE V x
m dx
•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?
incidenttransmitted
reflected
Solve the equation in each of the regions•Assume E > V0
•Region I•Region II
I II
2 2
22
dE
m dx
ikxI e
2 2
2
kE
m
2 2
022
dE V
m dx
ik x
II e 2 2
0 2
kE V
m
Most general solution is linear combinations of these•It remains to match boundary conditions•And to think about what we are doing!
ikx ikxI
ik x ik xII
x Ae Be
x Ce De
The Step Potential (2)
What do all these terms mean?•Wave A represents the incoming wave from the left•Wave B represents the reflected wave•Wave C represents the transmitted wave•Wave D represents an incoming wave from the right
•We should set D = 0
ikx ikxI
ik x ik xII
x Ae Be
x Ce De
Boundary conditions:•Function should be continuous at the boundary•Derivative should be continuous at the boundary
0 0I II A B C
0 0I II ikA ikB ik C
kA kB k C k A k B k k A k k B
2kC A B A
k k
k kB A
k k
2 2
2
kE
m
2 2
0 2
kE V
m
incidenttransmitted
reflected
I II
2kC A
k k
The Step Potential (3)
80 9V E
•The barrier both reflects and transmits•What is probability R of reflection?
2
2B
A
R
2
2
B
A
2k k
k k
k kB A
k k
0
2
2
mEk
m E Vk
2
0
0
E E VR
E E V
•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:
0
2
0
41
E E VT R
E E V
incidenttransmitted
reflected
I II
The Step Potential (4)•What if V0 > E?•Region I same as before•Region II: we have
ikxI e 2 2
2
kE
m
2 2
022
dV E
m dx
x
II x e 2 2
0 2V E
m
•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
ikx ikxI
xII
Ae Be
Ce
0 0I II
A B C
0 0I II
ikA ikB C
ik A ik B
ikB A
ik
•Calculate the reflection probability2
2
BR
A
2ik
ik
2 2
2 2
k
k
2ik
C Aik
1 R
incident
reflected
I IIevanescent
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!
0V E
R
T
2
00
0
0
if
1 if
E E VV E
R E E V
V E
0 2V E
incident
reflected
I IIevanescent
Sample ProblemElectrons are incident on a step potential V0 = - 12.3 eV. Exactly ¼ of the electrons are reflected. What is the velocity of the electrons?
2
0
0
E E VR
E E V
1
4
0
0
1
2
E E V
E E V
0 0
0 0
2 2 or
2 2
E E V E E V
E E V E E V
0 03 or 3E E V E E V 0 09 or 9E E V E E V
9 10 08 8orE V E V •Must have E > 0
108 1.54 eVE V 21
2E mv
2Ev
m
6 2
2 1.54 eV
0.511 10 eV/c
0.00245c 57.35 10 m/s
0 0
0 otherwise
V x LV x
I II III
ikx ikxI
x xII
ikxIII
x Ae Be
x Ce De
x Fe
The Barrier Potential (1)
2 2
2 20
2
2
E k m
V E m
•Assume V0 > E•Solve in three regions
•Match wave function and derivative at both boundaries
A B C D
ikA ikB C D
•Work, work . . .
2 2
2
2 cosh sinh
ikLke AF
k L i k L
•Let’s find transmission probability
2
2
FT
A
02 2
0 0
4
4 sinh
E V E
E V E V L
L L ikL
L L ikL
Ce De Fe
Ce De ikFe
0 0
0 otherwise
V x LV x
I II III
The Barrier Potential (2)
•Solve for :•Assume a thick barrier: L large
•Exponentials beat everything
02 2
0 0
4
4 sinh
E V ET
E V E V L
2 20 2V E m
1 12 2sinh L L LL e e e
2
0 0
16 1 LE ET e
V V
02m V E
Sample ProblemAn 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?
202 2 .511 MeV/ 20.0 eV 10.0 eVm V E c
6
8 16
2 .511 10 eV 10 eV
3.00 10 m/s 6.582 10 eV s
9 10 110.0 eV 10.0 eV16 1 exp 2 0.20 10 m 1.62 10 m
20.0 eV 20.0 eVT
6.484.00 0.61%e
2
0 0
0
16 1
2
LE ET e
V V
m V E
10 11.62 10 m
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