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