Chapter 5
The Wavelike Properties of
Particles
The Wavelike Properties of Particles
•
•
•
•
The de Broglie Hypothesis
Measurements of Particles Wavelengths
Wave Packets
The Probabilistic Interpretation of the Wave
Function
• The Uncertainty Principle
• Some Consequences of Uncertainty Principle
• Wave-Particle Duality
The de Broglie Hypothesis
As it was showed by Thomson the rays of a
cathode tube can be deflected by electric and
magnetic fields and therefore must consist of
electrically charged particles.
Thomson showed that all the particles have
the same charge-to-mass ratio q/m. He also
showed that particles with this charge-to-mass
ratios can be obtained using any materials for
the cathode, which means that these particles,
now called electrons, are a fundamental
constituent of matter.
The de Broglie Hypothesis
Since light seems to have both wave and particle
properties, it is natural to ask whether matter (electrons,
protons) might also have both wave and particle
characteristics.
In 1924, a French physics student, Louis de Broglie,
suggested this idea in his doctoral dissertation.
For the wavelength of electron, de Broglie chose:
λ = h/p
f = E/h
where E is the total energy, p is the momentum, and
λ is called the de Broglie wavelength of the particle.
The de Broglie Hypothesis
For photons these same equations results
directly from Einstein’s quantization of radiation
E = hf and equation for an energy of a photon
with zero rest energy E = pc :
E = pc = hf = hc/λ
Using relativistic mechanics de Broglie
demonstrated, that this equation can also be
applied to particles with mass and used them to
physical interpretation of Bohr’s hydrogen-like
atom.
The de Broglie Wavelength
Using de Broglie relation let’s find the
wavelength of a 10-6g particle moving with a
speed 10-6m/s:
h
h
6.63  1034 J  s
19
 


6
.
63

10
m

9

6
p mv (10 kg)(10 m / s )
Since the wavelength found in this example is so
small, much smaller than any possible apertures,
diffraction or interference of such waves can not
be observed.
The de Broglie Wavelength
The situation are different for low energy electrons
and other microscopic particles.
Consider a particle with kinetic energy K. Its
momentum is found from
p2
K
2m
or
p  2mK
Its wavelength is than
h
 
p
h
2mK
The de Broglie Wavelength
h
 
p
h
2mK
If we multiply the numerator and denominator by c
we obtain:

hc
2mc2 K

1240eV  nm
2(0.511  106 eV ) K

1.226
nm
K
Where mc2=0.511MeV for electrons, and K in
electronvolts.
The de Broglie Wavelength
We obtained the electron wavelength:
 1.226 
nm,
  

K


K in eV
Similarly, for proton (mc2 = 938 MeV for protons)
 0.0286 
nm
 p  

 K 
The de Broglie Wavelength
For the molecules of a stationary gas at the absolute
temperature T, the square average speed of the
molecule v2 is determined by Maxwell’s Law
3k BT
v 
m
Than the momentum of the molecule is:
2
p  3mkBT
Knowing that the mass of He atom, for instance, is
6.7x10-24g, (kB=1.38x10-23J/K) we obtain for He
wavelength:
 1.26 
nm
He  

 T 
The de Broglie Wavelength
Similarly, for the molecule of hydrogen
H 2
 1.78 
nm
 

T


and for the thermal neutrons
 2.52 
nm
n  

 T 
.
This calculations show, that for the accelerated electrons, for
atoms of helium, hydrogen molecules under the room
temperature, for thermal neutrons and other “slow” light
particles de Broglie wavelength is on the same order as for
soft X-rays. So, we can expect, that diffraction can be
observed for this particles
(a) Show that the wavelength of a nonrelativistic
neutron is
11
λ
2.86 10
m
Kn
where Kn is the kinetic energy of the neutron in
electron-volts. (b) What is the wavelength of a
1.00-keV neutron?
(a) Show that the wavelength of a nonrelativistic
neutron is
11
λ
2.86 10
m
Kn
where Kn is the kinetic energy of the neutron in
electron-volts. (b) What is the wavelength of a
1.00-keV neutron?
(a)

h
 
p
h

2m K

h
2m K
Kinetic energy, K, in this
equation is in Joules
6.626  1034 J s


2 1.67  1027 kg 1.60  1019 J eV K n
2.87  1011

m
Kn
(a) Show that the wavelength of a nonrelativistic
neutron is
11
λ
2.86 10
m
Kn
where Kn is the kinetic energy of the neutron in
electron-volts. (b) What is the wavelength of a
1.00-keV neutron?
(b)
K n  1.00 keV  1000 eV
2.87  1011

m  9.07  1013 m  907 fm
1000
The nucleus of an atom is on the order of 10–14 m in
diameter. For an electron to be confined to a nucleus, its
de Broglie wavelength would have to be on this order of
magnitude or smaller. (a) What would be the kinetic energy
of an electron confined to this region? (b) Given that typical
binding energies of electrons in atoms are measured to be
on the order of a few eV, would you expect to find an
electron in a nucleus?
The nucleus of an atom is on the order of 10–14 m in
diameter. For an electron to be confined to a nucleus, its
de Broglie wavelength would have to be on this order of
magnitude or smaller. (a) What would be the kinetic energy
of an electron confined to this region? (b) Given that typical
binding energies of electrons in atoms are measured to be
on the order of a few eV, would you expect to find an
electron in a nucleus?
(a)
6.6  1034 J s
19
p ~

10
kg  m s
14

10 m
h
 ~ 1014 m
2 2
E p c
 m e2c4
~
10   3 10   9 10  3 10 
19 2
8 2
E ~ 1011 J~ 108 eV

K  E  m ec2 ~ 108 eV  0.5  106 eV

~ 108 eV
31 2
8 4
The nucleus of an atom is on the order of 10–14 m in
diameter. For an electron to be confined to a nucleus, its
de Broglie wavelength would have to be on this order of
magnitude or smaller. (a) What would be the kinetic energy
of an electron confined to this region? (b) Given that typical
binding energies of electrons in atoms are measured to be
on the order of a few eV, would you expect to find an
electron in a nucleus?
(b)
keq1q2
Ue
r
With its
9  10

~
9
N,  m
2


C 2 1019 C   e
1014 m
~ 105 eV
K  U e  0
the electron w ould im m ediately escape the nucleus
Electron Interference and Diffraction
The electron wave interference was discovered in 1927
by C.J. Davisson and L.H.Germer as they were studying
electron scattering from a nickel target at the Bell
Telephone Laboratories.
After heating the target to remove an oxide coating that
had accumulate after accidental break in the vacuum
system, they found that the scattered electron intensity is a
function of the scattered angle and show maxima and
minima. Their target had crystallized during the heating,
and by accident they had observed electron diffraction.
Then Davisson and Germer prepared a target from a
single crystal of nickel and investigated this phenomenon.
The DavissonGermer experiment.
Low energy electrons
scattered at angle Φ from
a nickel crystal are
detected in an ionization
chamber. The kinetic
energy of electrons could
be varied by changing
the accelerating voltage
on the electron gun.
Scattered intensity vs detector angle for 54-ev
electrons. Polar plot of the data. The intensity at each angle
is indicated by the distance of the point from the origin.
Scattered angle Φ is plotted clockwise started at the vertical
axis.
The same data plotted on a Cartesian graph. The
intensity scale are the same on the both graphs. In each plot
there is maximum intensity at Φ=50º, as predicted for Bragg
scattering of waves having wavelength λ = h/p.
Scattering of electron by crystal. Electron waves are
strongly scattered if the Bragg condition nλ = D SinΦ
is met.
Test of the de Broglie formula λ = h/p. The wavelength is
computed from a plot of the diffraction data plotted against
V0-1/2, where V0 is the accelerating voltage. The straight line
is 1.226V0-1/2 nm as predicted from λ = h/(2mE)-1/2
Test of the de Broglie formula λ = h/p. The wavelength is
computed from a plot of the diffraction data plotted against
V0-1/2, where V0 is the accelerating voltage. The straight line
is 1.226V0-1/2 nm as predicted from λ = h/(2mE)-1/2

h
1
2m
E0

6.6  1034 J  s
1
2  9.22  1031kg  1.6  1019 J / eV
E0
 1.226
1
E0
A series of a polar graphs of Davisson and Germer’s
data at electron accelerating potential from 36 V to 68 V.
Note the development of the peak at Φ = 50º to a
maximum when V0 = 54 V.
Variation of the scattered electron intensity with wavelength
for constant Φ. The incident beam in this case was 10º from
the normal, the resulting diffraction causing the measured
peaks to be slightly shifted from the positions computed from
nλ = D Sin Φ.
Schematic arrangement used for producing a diffraction
pattern from a polycrystalline aluminum target.
Diffraction pattern produced by x-rays of wavelength
0.071 nm and an aluminum foil target.
Diffraction pattern produced by 600-eV electrons and an
aluminum foil target ( de Broigle wavelength of about 0.05 nm: )
Diffraction pattern produced by 600-eV electrons and an
aluminum foil target ( de Broigle wavelength of about 0.05 nm: )
.
h
h
hc
1240eV  nm
e  


 0.05nm
p
2me Ek
2mc2V0
2  0.511 106 eV  600eV

Diffraction pattern produced by 0.0568-eV neutrons (de Broglie
wavelength of 0.120 nm) and a target of polycrystalline copper.
Note the similarity in the pattern produced by x-rays, electrons,
and neutrons.

Diffraction pattern produced by 0.0568-eV neutrons (de Broglie
wavelength of 0.120 nm) and a target of polycrystalline copper.
Note the similarity in the pattern produced by x-rays, electrons,
and neutrons.
n 
h
2mn Ek

1240eV  nm
2  939.57  106 eV  0.0568eV
 0.120nm
In the Davisson–Germer experiment, 54.0-eV
electrons were diffracted from a nickel lattice. If the
first maximum in the diffraction pattern was observed
at φ = 50.0°, what was the lattice spacing a between
the vertical rows of atoms in the figure? (It is not the
same as the spacing between the horizontal rows of
atoms.)
In the Davisson–Germer experiment, 54.0-eV
electrons were diffracted from a nickel lattice. If the
first maximum in the diffraction pattern was observed
at φ = 50.0°, what was the lattice spacing a between
the vertical rows of atoms in the figure? (It is not the
same as the spacing between the horizontal rows of
atoms.)
 
m   2dsin   2dcos 
 2
 
d  asin  
 2
m 1
 
 
  2asin   cos   asin 
 2
 2
In the Davisson–Germer experiment, 54.0-eV
electrons were diffracted from a nickel lattice. If the
first maximum in the diffraction pattern was observed
at φ = 50.0°, what was the lattice spacing a between
the vertical rows of atoms in the figure? (It is not the
same as the spacing between the horizontal rows of
atoms.)
 
 
  2asin   cos   asin 
 
 2
h
h2


p


2m eK
6.626  1034 J s


2 9.11 1031 kg 54.0  1.60  1019 J
 1.67  1010 m
1.67  1010 m
a

 2.18  1010  0.218 nm
sin 
sin 50.0

A photon has an energy equal to the kinetic energy of a
particle moving with a speed of 0.900c. (a) Calculate the
ratio of the wavelength of the photon to the wavelength of
the particle. (b) What would this ratio be for a particle having
a speed of 0.00100c ? (c) What value does the ratio of the
two wavelengths approach at high particle speeds?(d) At low
particle speeds?
A photon has an energy equal to the kinetic energy of a
particle moving with a speed of 0.900c. (a) Calculate the
ratio of the wavelength of the photon to the wavelength of
the particle. (b) What would this ratio be for a particle having
a speed of 0.00100c ? (c) What value does the ratio of the
two wavelengths approach at high particle speeds? (d) At
low particle speeds?
For a particle:
K    1 m c
2
For a photon:
h
h
m  
p  mv
c ch ch
ch
E  K   f  E  K    1 m c2



ch m v
 v


2
m
  1 m c h   1 c
A photon has an energy equal to the kinetic energy of a particle moving with
a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to
the wavelength of the particle. (b) What would this ratio be for a particle
having a speed of 0.00100c ? (c) What value does the ratio of the two
wavelengths approach at high particle speeds?(d) At low particle speeds?

ch m v
 v


2
m
  1 m c h   1 c
(a)
(b)

m
1 0.9


m


1 0.92  1 1 0.92  1



1 0.001

2 
1  0.001  1

1  0.001
2

  2.294
 1.60

 1

 2.00  103
A photon has an energy equal to the kinetic energy of a particle moving with a
speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the
wavelength of the particle. (b) What would this ratio be for a particle having a
speed of 0.00100c ? (c) What value does the ratio of the two wavelengths
approach at high particle speeds?(d) At low particle speeds?
(c) As
and
(d)
v
1
c
 1
v
0
c
 


becomes nearly equal to γ.
 1 1
m

,
2  1 2

v
 1 2 
c 

2
1 v2
 1 v
 1  1    2  1 
 2 c
2 c2

vc
2c
1



2 2
m
v
1 2 v c


What is “waving”? For matter it is the probability of finding the
particle that waves. Classical waves are the solution of the
classical wave equation
d2y
dx
2

1 d2y
f 2 dt 2
Harmonic waves of amplitude y0, frequency f and period T:
2
x t 
y  y0 cos( kx  t )  y0 cos 2     y0 cos
( x  t )

 T 
where the angular frequency ω and the wave number k are
defined by
2
  2f 
T
and
k
2

and the wave or phase velocity vp is given by
v p  f
If the film were to be observed at various stages, such
as after being struck by 28 electrons the pattern of
individually exposed grains will be similar to shown here.
After exposure by about 1000 electrons the pattern
will be similar to this.
And again for exposure of about 10,000 electrons
we will obtained a pattern like this.
Two source interference pattern. If the sources are
coherent and in phase, the waves from the sources
interfere constructively at points for which the path
difference dsinθ is an integer number of wavelength.
Grows of two-slits interference pattern. The photo is
an actual two-slit electron interference pattern in which
the film was exposed to millions of electrons. The pattern
is identical to that usually obtained with photons.
Using relativistic mechanics, de Broglie was able
to derive the physical interpretation of Bohr’s
quantization of the angular momentum of electron.
He demonstrate that quantization of angular
momentum of the electron in hydrogenlike atoms is
equivalent to a standing wave condition:
nh
mvr  n 
2
for n = integer
nh nh
2r 

 n  circumference of
mv
p
orbit
The idea of explaining discrete energy states in
matter by standing waves thus seems quite promising.
Standing waves around the circumference of a
circle. In this case the circle is 3λ in circumference. For
example, if a steel ring had been suitable tapped with a
hammer, the shape of the ring would oscillate between
the extreme positions represented by the solid and
broken lines.
Wave pulse moving along a string. A pulse have a
beginning and an end; i.e. it is localized, unlike a pure
harmonic wave, which goes on forever in space and time.
Two waves of slightly different wavelength and frequency
produced beats.
(a) Shows y(x) at given instant for each of the two waves.
The waves are in phase at the origin but because of the
difference in wavelength, they become out of phase and
then in phase again.
(b)
The sum of these waves. The spatial extend of the group
Δx is inversely proportional to the difference in wave numbers
Δk, where k is related to the wavelength by k = 2π/λ.
BEATS
Consider two waves of equal amplitude and nearly equal
frequencies and wavelengths.
F1 ( x )  F sin( k1 x   1 t )
F2 ( x )  F sin( k2 x   2 t )
The sum of the two waves is (superposition):
F(x) = F1(x) + F2(x) = F[sin(k1x – ω1t)+sin(k2x – ω2t)]
BEATS
F(x) = F1(x) + F2(x) = F[sin(k1x – ω1t)+sin(k2x – ω2t)]
using the trigonometric relation
Sinα + Sinβ = 2Cos[(α-β)/2] Sin[(α+β)/2]
with α =(k1x – ω1t)
β = (k2x – ω2t), we get:
 



 

 1  2 
1  2 
 k1  k2
  k1  k2
F ( x)  2 F cos
x
t   sin 
x
t 
2
2
2
 2


 

 


BEATS
 



 

 1  2 
1  2 
 k1  k2
  k1  k2
F ( x)  2 F cos
x
t   sin 
x
t 
2
2
2
 2


 

 


with
k  k1  k2
k1  k2
k
2
   1  2
 
 1 2
2
  k

 
F ( x)  2 F cos
x
t   sin( k x  t )
2 
  2

BEATS
  k

 
F ( x)  2 F cos
x
t   sin( k x  t )
2 
  2

This is an equivalent of an harmonic wave
F sin( k x  t )
whose amplitude is modulated by
 
 k
2 cos
x
t
2 
 2
We have formed wave packets of extend Δx and can
imagine each wave packet representing a particle.
  2x
Now
and

2
2 x 
k
2

xk  2
2

k
2
The particle is in the region Δx, the momentum of the
particle in the range Δk:
p = ћk → Δp = ћΔk
Δx Δp ≈ h - Uncertainty Principle
In order to localize the particle within a region Δx, we
need to relax the precision on the value of the
momentum, Δp .
Gaussian-shaped wave packets y(x) and the
corresponding Gaussian distributions of wave numbers A(k).
(a) A narrow packet. (b) A wide packet. The standard
deviations in each case are related by σxσk = ½.
Wave packet for which
the group velocity is half
of phase velocity. Water
waves whose
wavelengths are a few
centimeters, but much
less than the water
depth, have this
property. The arrow
travels at the phase
velocity, following a point
of constant phase for the
dominant wavelength.
The cross at the center
of the group travels at
the group velocity.
A three-dimensional wave packet representing a particle
moving along the x-axis. The dots indicate the position of
classical particle. Note that the particle spreads out in the x
and y directions. This spreading is due to dispersion ,
resulting from the fact that the phase velocity of the individual
wave making up the packet depends on the wavelength of
the waves.
“Seeing an electron” with a gamma-ray microscope.
Because of the size of the lens, the momentum of the
scattered photons is uncertain by Δpx ≈psinθ = hsinθ/ λ.
Thus the recoil momentum of the electron is also uncertain
by at least this amount.
The position of the electron can not be resolved better than
the width of the central maximum of the diffraction pattern
Δx ≈ λ/sinθ. The product of the uncertainties Δpx Δx is
therefore of the order of Planck’s constant h.
The Interpretation of the Wave Function
Given that electrons have wave-like properties,
it should be possible to produce standing electron
waves. The energy is associated with the frequency
of the standing wave, as E = hf, so standing waves
imply quantized energies.
The idea that discrete energy states in atom
can be explained by standing waves led to the
development by Erwin Schrödinger in 1926
mathematical theory known as quantum theory,
quantum mechanics, or wave mechanics.
In this theory electron is described by a wave
function Ψ that obeys a wave equation called the
Schrödinger equation.
The Interpretation of the Wave Function
The form of the Schrödinger equation of a
particular system depends on the forces acting
on the particle, which are described by the
potential energy functions associated with this
forces.
Schrödinger solved the standing wave
problem for hydrogen atom, the simple harmonic
oscillator, and other system of interest. He found
that the allowed frequencies, combined with
E=hf, resulted in the set of energy levels, found
experimentally for the hydrogen atom.
Quantum theory is the basis for our
understanding of the modern world, from the
inner working of the atomic nucleus to the
radiation spectra of distant galaxies.
The Interpretation of the Wave Function
The wave function for waves in a string is the
string displacement y. The wave function for
sound waves can be either the displacement of
the air molecules, or the pressure P. The wave
function of the electromagnetic waves is the
electric field E and the magnetic field B.
What is the wave function for the electron Ψ?
The Schrödinger equation describes a single
particle. The square of the wave function for a
particle describes the probability density,
which is the probability per unit volume, of
finding the particle at a location.
The Interpretation of the Wave Function
The probability of finding the particle in some volume
element must also be proportional to the size of volume
element dV.
Thus, in one dimension, the probability of finding a
particle in a region dx at the position x is Ψ2(x)dx. If we
call this probability P(x)dx, where P(x) is the probability
density, we have
P(x) = Ψ2(x)
The probability of finding the particle in dx at point x1
or point x2 is the sum of separate probabilities
P(x1)dx + P(x2)dx.
If we have a particle at all the probability of finding a
particle somewhere must be 1.
The Interpretation of the Wave Function
Then, the sum of the probabilities over all the
possible values of x must equal 1. That is,

2

 dx  1

This equation is called the normalization
condition. If Ψ is to satisfy the normalization
condition, it must approach zero as x is approach
infinity.
Probability Calculation for a Classical Particle
It is known that a classical point particle moves back and forth
with constant speed between two walls at x = 0 and x = 8cm.
No additional information about of location of the particle is
known.
(a) What is the probability density P(x)?
(b) What is the probability of finding the particle at x=2cm?
(c) What is the probability of finding the particle between x=3.0
cm and x=3.4 cm?
A Particle in a Box
We can illustrate many of important features of quantum
physics by considering of simple problem of particle of mass
m confined to a one-dimensional box of length L.
This can be considered as a crude description of an
electron confined within an atom, or a proton confined within
a nucleus.
According to the quantum theory, the particle is
described by the wave function Ψ, whose square describes
the probability of finding the particle in some region. Since
we are assuming that the particle is indeed inside the box,
the wave function must be zero everywhere outside the box:
Ψ =0 for x≤0 and for x≥L.
A Particle in a Box
The allowed wavelength for a particle in the box are
those where the length L equals an integral number of
half wavelengths.
L = n( λn/2)
n = 1,2,3,…….
This is a standing wave condition for a particle in the
box of length L.
The total energy of the particle is its kinetic energy
E = (1/2)mv2 = p2/2m
Substituting the de Broglie relation pn = h/λn,
2
 h 


pn2  n 
h2
En 


2
2m
2m
2mn
A Particle in a Box
2
 h 



2
pn  n 
h2
En 


2m
2m
2m2n
Then the standing wave condition λn= 2L/n gives the
allowed energies:
2 

h
  n 2 E1
En  n 2 
 8mL2 


where
E1 
h2
8mL2
A Particle in a Box
The equation
2 

h
  n 2 E1
En  n 2 
 8mL2 


gives the allowed energies for a particle in the
box.
E1 
h2
8mL2
This is the ground state energy for a particle in
the box, which is the energy of the lowest state.
A Particle in a Box
The condition that we used for the wave function in the box
Ψ = 0 at x = 0 and x = L
is called the boundary condition.
The boundary conditions in quantum theory lead to energy
quantization.
Note, that the lowest energy for a particle in the box is not zero.
The result is a general feature of quantum theory.
If a particle is confined to some region of space, the
particle has a minimum kinetic energy, which is called
zero-point energy. The smaller the region of space the
particle is confined to, the greater its zero-point energy.
A Particle in a Box
If an electron is confined (i.e., bond to an atom)
in some energy state Ei, the electron can make
a transition to another energy state Ef with the
emission of photon. The frequency of the
emitted photon is found from the conservation of
the energy
hf = Ei – Ef
The wavelength of the photon is then
λ = c/f = hc/(Ei – Ef)
Standing Wave Function
The amplitude of a vibrating string fixed at x=0 and x=L
is given as
yn  An sin kn x
2
where An is a constant and k n 
is the wave
n
number.
The wave function for a particle in a box are the
same:
yn  An sin kn x
Using
2 L , we have
n 
n
2
2
n
kn 


n 2 L
L
n
Standing Wave Function
The wave function can thus be written
 nx 
n ( x)  An sin 

 L 
The constant An is determined by normalization condition

 n dx 
2

2
2  nx 
An sin 
dx
 L 
1
An An 
The result of evaluating the integral and solving for
is independent from n.
The normalized wave function for a particle in a box are thus
n ( x) 
2
 nx 
sin 

L
 L 
2
L
Graph of energy vs. x for a particle in the box, that we also call
an infinitely deep well. The set of allowed values for the
particle’s total energy En is E1(n=1), 4E1(n=2), 9E1(n=3) …..
Wave functions Ψn(x) and probability densities Pn(x)= Ψn2(x)
for n=1, 2, and 3 for the infinity square well potential.
Probability distribution for n=10 for the infinity square well
potential. The dashed line is the classical probability density
P=1/L, which is equal to the quantum mechanical distribution
averaged over a region Δx containing several oscillations. A
physical measurement with resolution Δx will yield the classical
result if n is so large that Ψ2(x) has many oscillations in Δx.
Photon Emission by Particle in a Box
An electron is in one dimensional box of length 0.1nm.
(a) Find the ground state energy. (b) Find the energy
in electron-volts of the five lowest states, and then
sketch an energy level diagram. (c) Find the
wavelength of the photon emitted for each transition
from the state n=3 to a lower-energy state.
The probability of a particle being found in a
specified region of a box.
The particle in one-dimensional box of length L is in
the ground state. Find the probability of finding the
particle (a) anywhere in a region of length Δx = 0.01L,
centered at x = ½L; (b) in the region 0<x<(1/4)L.
Expectation Values
The most that we can know about the position
of the particle is the probability of measuring a
certain value of this position x. If we measure the
position for a large number of identical systems, we
get a range of values corresponding to the
probability distribution.
The average value of x obtained from such
measurements is called the expectation value and
written ‹x›. The expectation value of x is the same
as the average value of x that we would expect to
obtain from a measurement of the position of a
large number of particles with the same wave
function Ψ(x).
Expectation Values
Since Ψ2(x)dx is the probability of finding a
particle in the region dx, the expectation value of x
is:
x   x ( x)dx
2
The expectation value of any function f(x) is
given by:
f ( x)   f ( x) ( x)dx
2
Calculating expectation values
Find (a) ‹ x › and (b) ‹ x2› for a particle in
its ground state in a box of length L.
Complex Numbers
A complex number has the form a+ib, with
i2=-1 or i=√-1 – imaginary unit.
a - real part; b – imaginary part; i – imaginary
unit
(a +ib) + (c +id) = (a+c) + i(b+d)
m(a +ib) = ma + imb
(a +ib) (c +id) = (ac - bd) + i(ad + bc)
The absolute value of a + ib is denoted by
│a+ib│ and is given by │a+ib│= √ a2 + b2
Complex Numbers
The complex conjugate of a+ib is denoted by
(a+ib)* and is given
(a+ib)* = (a-ib)
Then
(a+ib)*∙ (a+ib) = (a-ib) (a+ib)=a2 + b2
Polar Form of Complex Numbers
p
b
φ
a
Real axis
Euler Identities:
eiφ = cosφ + isinφ
e-iφ = cosφ - isinφ
where i = √-1
p = √ a2 + b2 = │a + ib│
We can represent the
number (a + ib) in the
complex xy plane.
Then the polar coordinates
a + ib ≡ p(cosφ +isinφ)
Remembering the Euler
formula:
eiφ = (cosφ+isinφ)
a + ib = p eiφ
Fourier Transform
In quantum mechanics, our basic function is the pure
sinusoidal plane wave describing a free particle, given in
equation:
i ( kx  wt )
 ( x, t )  Ae
We are not interest here in how things behave in time, so
we chose a convenient time of zero. Thus, our “building
block” is
ikx
e
Now we claim that any general, nonperiodic wave function
ψ(x) can be expressed as a sum/integral of this building
blocks over the continuum of wave numbers:

 ( x) 

A(k )eikxdk

Fourier Transform

 ( x) 

A(k )eikxdk

The amplitude A(k) of the plane wave is naturally a function
of k, it tell us how much of each different wave number goes
into the sum. Although we can’t pull it out of the integral, the
equation can be solved for A(k). The result is:
1
A( k ) 
2


 ( x)e ikxdx

The proper name of for A(k) is the Fourier transform of the
function ψ(x).
ψ(x) =
A(k ) 
 ( x) 
 ( x) 
│x│< a
│x│> a
1
0
1
2

 ( x)eikxdx
a
a
1
2

1  eikxdk 
a
1  eikx 


2  ik   a
(eika  e ika )
ik
2
1
_eika = cosφ+isinφ
e-ika = cosφ-isinφ
eika- e-ika = 2isinka
And we can overwrite the equation for A(k):
Let use Euler identities:
A(k ) 
2
sin ka

2 k
2
1
sin ka
2 k
General Wave Packets
Any point in space can be described as a linear
combination of unit vectors. The three unit vectors î, ĵ,
and k̂ constitute a base that can generate any points in
space.
In similar way: given a periodic function, any value
that the function can take, can be produced by the linear
combination of a set of basic functions. The basic
functions are the harmonic functions (sin or cos). The set
of basic function is actually infinite.
The General Wave Packet
A periodic function f(x) can be represented by the
sum of harmonic waves:
y(x,t) = Σ [Aicos(kix – ωit) + Bisin(kix – ωit)]
Ai and Bi ≡ amplitudes of the waves with wave
number ki and angular frequency ωi.
For a function that is not periodic there is an
equivalent approach called Fourier Transformation.
Fourier Transformation
A function F(x) that is not periodic can be represented by
a sum (integral) of functions of the type
e±ika = Cosφ±iSinφ
In math terms it called Fourier Transformation. Given a
function F(x)
1
ikx
F ( x) 
f
(
k
)
e
dk

2
where
1
f (k ) 
2
ikx
F
(
x
)
e
dx

f(kj) represents the amplitude of base function e-ikx used to
represent F(x).
The Schrödinger Equation
The wave equation governing the
motion of electron and other particles with
mass m, which is analogous to the classical
wave equation
 y 1  y
 2 2
2
x
v t
2
2
was found by Schrödinger in 1925 and is now
known as the Schrödinger equation.
The Schrödinger Equation
Like the classical wave equation, the
Schrödinger equation is a partial differential
equation in space and time.
Like Newton’s laws of motion, the
Schrödinger equation cannot be derived.
It’s validity, like that of Newton’s laws, lies in
its agreement with experiment.
We will start from classical description of the
total energy of a particle:
p2
Etot  KE  U 
 U ( x)
2m
Schrödinger converted this equation into a wave
equation by defining a wavefunction, Ψ. He
multiplied each factor in energy equation with
that wave function:
2
p
E 
  U ( x)
2m
To incorporate the de Broglie wavelength of the particle he
 2 2
introduced the operator,
 2 ,which provides the square of
x
the momentum when applied to a plane wave:
i ( kx t )
e
If we apply the operator to that wavefunction:
  ( x)
2 2
2

 k  p 
2
dx
2
2
where k is the wavenumber, which equals 2π/λ.
We now simple replace the p2 in equation for energy:
 d  ( x)


U

(
x
)

E

(
x
)
2
2m dx
2
2
Time Independent Schrödinger Equation
This equation is called time-independent Schrödinger
equation.
 2 d 2  ( x)

 U ( x) ( x)  E ( x)
2
2m dx
E is the total energy of the particle.
The normalization condition now becomes
∫ Ψ*(x)Ψ(x)dx = 1
A Solution to the Srödinger Equation
Show that for a free particle of mass m moving in
one dimension the function
 ( x)  A sin kx  B cos kx
is a solution of the time independent Srödinger
Equation for any values of the constants A and B.
Energy Quantization in Different Systems
The quantized energies of a
system are generally
determined by solving the
Schrödinger equation for that
system. The form of the
Schrödinger equation
depends on the potential
energy of the particle.
The potential energy for a one-dimensional box from x = 0 to
x = L is shown in Figure. This potential energy function is
called an infinity square-well potential, and is described by:
U(x) = 0,
0<x<L
U(x) = ∞, x<0 or x>L
A Particle in Infinity Square Well Potential
Inside the box U(x) = 0, so the Schrödinger equation
is written:
 2 d 2  ( x)

 E( x)
2
2m dx
where E = ħω is the energy of the particle, or
d 2  ( x)
2

k
 ( x)  0
2
dx
where k2 = 2mE/ħ2
The general solution of this equation can be written
as
ψ(x) = A sin kx + B cos kx
where A and B are constants. At x=0, we have
ψ(0) = A sin (k0) + B cos (0x) = 0 + B
A Particle in Infinity Square Well Potential
The boundary condition ψ(x)=0 at x=0 thus gives B=0
and equation becomes
ψ(x) = A sin kx
We received a sin wave with the wavelength λ related to
wave number k in a usual way, λ = 2π/k. The boundary
condition ψ(x) =0 at x=L gives
ψ(L) = A sin kL = 0
This condition is satisfied if kL is any integer times π, or
kn = nπ / L
If we will write the wave number k in terms of wavelength
λ = 2π/k, we will receive the standing wave condition for
particle in the box:
nλ / 2 = L
n = 1,2,3,……
A Particle in Infinity Square Well Potential
Solving k2 = 2mE/ħ2 for E and using the standing wave
condition k = nπ / L gives us the allowed energy values:
2
 k
  n 
h
2
2
E n


n

n
E1


2
2m 2m  L 
8mL
2
where
2
n
2
2
2
h
E1 
2
8mL
For each value n, there is a wave function ψn(x) given by
 n x 

n ( x)  An sin 
 L 
A Particle in Infinity Square Well Potential
Compare with the equation we received for particle
in the box, using the standing wave fitting with the
constant An = √2/L determined by normalization:
n ( x) 
 n x 
2

sin 
L  L 
Although this problem seems artificial, actually it is
useful for some physical problems, such as a
neutron inside the nucleus.
A Particle in a Finite Square Well
This potential energy function
is described mathematically
by:
U(x)=V0, x<0
U(x)=0, 0<x<L
U(x)=V0, x>L
Here we assume that 0 ≤E≤V0.
Inside the well, U(x)=0, and
the time independent
Schrödinger equation is the
same as for the infinite well
 d  ( x)

 E ( x)
2
2m dx
2
2
A Particle in a Finite Square Well
 d  ( x)

 E ( x)
2
2m dx
2
or
2
d  ( x)
2
 k  ( x)  0
2
dx
2
where k2 = 2mE/ħ2. The general solution is
ψ(x) = A sin kx + B cos kx
but in this case, ψ(x) is not required to be zero at x=0,
so B is not zero.
A Particle in a Finite Square Well
Outside the well, the time independent Schrödinger equation is
 d  ( x)

 U 0  ( x)  E ( x)
2
2m dx
2
2
or
d  ( x)
2


 ( x)  0
2
dx
2
where
2m
  2 (U 0  E )  0

2
The Harmonic Oscillator
More realistic than a particle in a box is the harmonic
oscillator, which applies to an object of mass m on a spring of
force constant k or to any systems undergoing small
oscillations about a stable equilibrium. The potential energy
function for a such oscillator is:
U ( x)  kx  m x
2
1
2
2
0
1
2
2
where ω0 = √k/m=2πf is the angular frequency of the
oscillator. Classically, the object oscillates between x = +A and
x=-A. Its total energy is
E  mv  m A
1
2
2
1
2
2
0
2
which can have any nonnegative value, including zero.
Potential energy function for a simple harmonic
oscillator. Classically, the particle with energy E is
confined between the “turning points” –A and +A.
The Harmonic Oscillator
Classically, the probability of finding the particle in dx
is proportional to the time spent in dx, which is dx/v.
The speed of the particle can be obtained from the
conservation of energy:
E  mv  m x
1
2
2
1
2
2
0
2
The classical probability is thus
dx
PC ( x)dx 

v
dx
2
1
2 2
 E  m x 
m
2

The Harmonic Oscillator
E  12 mv2  12 m02 A2
The classical probability is
dx
PC ( x)dx 

v
dx
2
1
2 2
 E  m x 
m
2

Any values of the energy E is possible. The lowest energy is
E=0, in which case the particle is in the rest at the origin. The
Shrödinger equation for this problem is
 d  ( x) 1
2 2

 m x  ( x)  E ( x)
2
2m dx
2
2
2
The Harmonic Oscillator
In quantum theory, the particle is represented by the
wave function ψ(x), which is determined by solving the
Schrödinger equation for this potential.
Only certain values of E will lead to solution that are
well behaved, i.e., which approach zero as x approach
infinity. Normalizeable wave function ψn(x) occur only for
discrete values of the energy En given by
1
1


En   n  hf 0   n  
2
2


n  0,1,2,3......
where f0=ω0/2π is the classical frequency of the oscillator.
The Harmonic Oscillator
1
1


En   n  hf 0   n  
2
2


n  0,1,2,3......
where f0=ω0/2π is the classical frequency of the
oscillator. Thus, the ground-state energy is ½ħω
and the exited energy levels are equally spaced
by ħω.
Energy levels in the simple harmonic oscillator potential.
Transitions obeying the selection rule Δn=±1 are indicated by
the arrows. Since the levels have equal spacing, the same
energy ħω is emitted or absorbed in all allowed transitions. For
this special potential, the frequency of emitted or absorbed
photon equals the frequency of oscillation, as predicted by
classical theory.
The Harmonic Oscillator
Compare this with uneven spacing of the energy
levels for the particle in a box. If a harmonic
oscillator makes a transition from energy level n to
the next lowest energy level (n-1), the frequency f
of the photon emitted is given by hf = Ef – Ei.
Applying this equation gives:
1
1


hf  En  En 1   n  hf 0   (n  1)  hf 0  hf 0
2
2


The frequency f of the emitted photon is therefore
equal to the classical frequency f0 of the oscillator.
Wave function for the ground state and the first two excited
states of the simple harmonic oscillator potential, the states
with n=0, 1, and 2.
Probability density
for the simple  2
n
harmonic oscillator
plotted against the
dimensionless value
m
u
x , for n=0, 1,
2
n2
and 2. The blue
curves are the
classical probability
densities for the
same energy, and
the vertical lines
indicate the
classical turning
points x = ±A
Molecules vibrate as harmonic
oscillators. Measuring vibration
frequencies enables determination
of force constants, bond strengths,
and properties of solids.
x 2
Verify that 0 ( x)  A0 e
, where α is a positive
constant, is a solution of the Schrödinger equation
for the harmonic oscillator
 d  ( x) 1
2 2

m

x  ( x)  E ( x)
2
2m dx
2
2
2
Operators
As we have seen, for a particle in a state of definite energy
the probability distribution is independent of time. The
expectation value of x is then given by

x 

* ( x) x ( x) dx

In general, the expectation value of any function f(x) is given
by

f ( x)    * ( x) f ( x) ( x)dx

Operators
If we know the momentum p of the particle as function of
x, we can calculate the expectation value ‹p›. However, it is
impossible in principle to find p as function of x since,
according to uncertainty principle, both p and x can not be
determined at the same time.
To find ‹p› we need to know the distribution function for
momentum. If we know Ψ(x), the distribution function can be
found by Fourier analysis. It can be shown that ‹p› can be
found from

p 


 
 
 ( x ) dx
 i x 
 
Operators
Similarly, ‹p2› can be found from

p
2



    
 

 ( x)dx
 i x  i x 
 
Notice that in computing the expectation value the
operator representing the physical quantity operates on Ψ(x),
not on Ψ*(x). This is not important to the outcome when the
operator is simply some function of x, but it is critical when the
operator includes a differentiation, as in the case of
momentum operator.
Expectation Values for p and p2
Find ‹p› and ‹p2› for the ground state wave
function of the infinity square well.
In classical mechanics, the total energy written in terms of
position and momentum variables is called the Hamiltonian
function
2
p
H
U
2m
If we replace the momentum by the momentum operator pop
and note that U = U(x), we obtain the Hamiltonian operator
Hop:
H op 
2
pop
2m
 U ( x)
The time-independent Schrödinger equation can then be
written:
H op   E
H op   E
The advantage of writing the Schrödinger equation in this
formal way is that it allows for easy generalization to more
complicated problems such as those with several particles
moving in three dimensions.
We simply write the total energy of the system in terms of
position and momentum and replace the momentum
variables by the appropriate operators to obtain the
Hamiltonian operator for the system.
Symbol
Physical quantity
Any function of x (the position, x; the
f(x)
potential energy U(x), etc.
px
x component of momentum
py
y component of momentum
pz
z component of momentum
E
Hamiltonian (time-independent)
Operator
f(x)
 
i x
 
i y
 
i z
2
pop
2m
E
Hamiltonian (time-dependent)
Ek
Kinetic energy
Lz
Z component of angular
momentum
 U ( x)

i
t
2 2

2m x 2

 i

Minimum Energy of a Particle in a Box
An important consequence of the uncertainty principle
is that a particle confined to a finite space can not have
zero kinetic energy.
Let’s consider a one-dimensional box of length L. If
we know that the particle is in the box, Δx is not larger
than L. This implies that Δp is at least ħ/L. Let us take the
standard deviation as a measure of Δp:
2
(p)  ( p  p)  ( p  2 p p  p ) av  p  p
2
2
av
2
2
2
Minimum Energy of a Particle in a Box
2
(p)  ( p  p)  ( p  2 p p  p ) av  p  p
2
2
av
2
2
If the box is symmetric, p will be zero since the particle
moves to the left as often as to the right. Then

(p )  p   
L
2
2
2
and the average kinetic energy is:
p2
2
E

2
2m 2mL
2
Minimum Energy of a Particle in a Box
The average kinetic energy of a particle in a box
is:
p2
2
E

2m 2mL2
Thus, we see that the uncertainty principle
indicate that the minimum energy of a particle in
a box cannot be zero. This minimum energy is
called zero-point energy.
The Hydrogen Atom
The energy of an electron of momentum p a distance r
from a proton is
p 2 ke2
E

2m r
If we take for the order of magnitude of the position
uncertainty Δx = r, we have:
(Δp2) = p2 ≥ ћ2/r2
The energy is then
2
2

ke
E

2
2mr
r
The Hydrogen Atom
2
2

ke
E

2
2mr
r
There is a radius rm at which E is minimum.
Setting dE/dr = 0 yields rm and Em:
2

rm  2  a0  0.0529nm
ke m
2
2
2 4
rm came out to
be exactly the
radius of the first
Bohr orbit

ke
k em
Em 


 13.6eV
2
2
2mr
r
2
The ground
state
energy
The Hydrogen Atom
The potential energy of the electron-proton
system varies inversely with separation distance
U  qeV  k
qe q p
r
As in the case of gravitational potential energy,
the potential energy of the electron-proton system
is chosen to be zero if the electron is an infinity
distance from the proton. Then for all finite
distances, the potential energy is negative.
The Hydrogen Atom
Like the energies of
a particle in a box
and of a harmonic
oscillator, the
energy levels in the
hydrogen atom are
described by a
quantum number n.
The allowed
energies of the
hydrogen atom are
given by
En = -13.6 eV/n2,
n = 1,2,3,……
Energy-level diagram for the hydrogen
atom. The energy of the ground state is
-13.6 eV. As n approaches ∞ the energy
approaches 0.
Step Potential
Consider a particle of energy E moving in region in which the
potential energy is the step function
U(x) = 0,
x<0
U(x) = V0,
x>0
What happened when
a particle moving from
left to right encounters
the step?
The classical answer is
simple: to the left of the
step, the particle moves
with a speed v = √2E/m
Step Potential
At x =0, an impulsive force
act on the particle. If the
initial energy E is less than
V0, the particle will be
turned around and will then
move to the left at its
original speed; that is, the
particle will be reflected by
the step. If E is greater than
V0, the particle will continue
to move to the right but with
reduced speed given by
v = √2(E – U0)/m
Step Potential
We can picture this classical problem as a ball
rolling along a level surface and coming to a steep
hill of height h given by mgh=V0.
If the initial kinetic energy of the ball is less than
mgh, the ball will roll part way up the hill and then
back down and to the left along the lower surface at
it original speed. If E is greater than mgh, the ball
will roll up the hill and proceed to the right at a
lesser speed.
The quantum mechanical result is similar when E is
less than V0. If E<V0 the wave function does not go to
zero at x=0 but rather decays exponentially. The wave
penetrates slightly into the classically forbidden region
x>0, but it is eventually completely reflected.
Step Potential
This problem is somewhat similar to that of total
internal reflection in optics.
For E>V0, the quantum mechanical result differs
from the classical result. At x=0, the wavelength
changes from
λ1=h/p1 = h/√2mE
to
λ2=h/p2 = h/√2m(E-V0).
When the wavelength changes suddenly, part of
the wave is reflected and part of the wave is
transmitted.
Reflection Coefficient
Since a motion of an electron (or other
particle) is governed by a wave equation, the
electron sometimes will be transmitted and
sometimes will be reflected.
The probabilities of reflection and
transmission can be calculated by solving the
Schrödinger equation in each region of space
and comparing the amplitudes of transmitted
waves and reflected waves with that of the
incident wave.
Reflection Coefficient
This calculation and its result are similar to finding the
fraction of light reflected from the air-glass interface. If R
is the probability of reflection, called the reflection
coefficient, this calculation gives:
(k1  k 2 ) 2
R
2
(k1  k 2 )
where k1 is the wave number for the incident wave and k2
is the wave number for the transmitted wave.
Transmission Coefficient
The result is the same as the result in optics for the
reflection of light at normal incidence from the boundary
between two media having different indexes of
refraction n.
The probability of transmission T, called the
transmission coefficient, can be calculated from the
reflection coefficient, since the probability of
transmission plus the probability of reflection must
equal 1:
T+R=1
In the quantum mechanics, a localized particle is
represented by the wave packet, which has a maximum
at the most probable position of the particle.
Time development of a one dimensional wave packet
representing a particle incident on a step potential for E>V0.
The position of a classical particle is indicated by the dot. Note
that part of the packet is transmitted and part is reflected.
Reflection coefficient R and transmission coefficient T for a
potential step V0 high versus energy E (in units V0).
A particle of energy E0 traveling in a region in
which the potential energy is zero is incident on
a potential barrier of height V0=0.2E0. Find the
probability that the particle will be reflected.
Lets consider a rectangular potential barrier of height V0 and
with a given by:
U(x) = 0,
x<0
U(x) = V0,
0<x<a
U(x) = 0,
x>a
Barrier Potential
We consider a particle of
energy E , which is slightly
less than V0, that is incident
on the barrier from the left.
Classically, the particle
would always be reflected.
However, a wave incident
from the left does not
decrease immediately to
zero at the barrier, but it
will instead decay exponentially in the classically forbidden
region 0<x<a. Upon reaching the far wall of the barrier
(x=a), the wave function must join smoothly to a sinusoidal
wave function to the right of barrier.
The potentials and the Schrödinger equations for the three
regions are as follows:
 2I
Region I (x<0)
Region II (0<x<a)
Region III (x>a)
V = 0,
V = V0,
V = 0,
x
2
 2II
x
2

2m

2m
 2III
x
2

2


EI  0
2
( E  V0 )II  0
2m

2
EIII  0
Barrier Potential
If we have a beam of particle incident from left, all with the
same energy E<V0, the general solution of the wave equation
are, following the example for a potential step,
1 ( x)  Ae ik1 x  Be ik1 x
x0
2 ( x)  Ce x  Dex
0 xa
3 ( x)  Fe ik1 x  Geik1 x
xa
where k1 =√2mE/ħ and α = √2m(V0-E)/ħ
This implies that there is some probability of the particle (which
is represented by the wave function) being found on the far side
of the barrier even though, classically, it should never pass
through the barrier.
We assume that we have incident particles coming
from the left moving along the +x direction. In this case the
term Aeik1x in region I represents the incident particles.
The term Be-ik1x represents the reflected particles moving
in the –x direction. In region III there are no particles
initially moving along the -x direction. Thus G=0, and the
only term in region III is Feik1x. We summarize these wave
functions:
I (incident )  Ae
ik1 x
I ( reflected )  Be
 ik1 x
III (transmitte d )  Fe
ik1 x
Barrier Potential
For the case in which the quantity
αa = √2ma2(V0 – E)/ħ2
is much greater than 1, the transmission coefficient
is proportional to e-2αa, with
α = √2m(V0 – E)/ħ2
The probability of penetration of the barrier thus
decreases exponentially with the barrier thickness
a and with the square root of the relative barrier
height (V0-E). This phenomenon is called barrier
penetration or tunneling. The relative probability of
its occurrence in any given situation is given by the
transmission coefficient.
A wave packet representing a particle incident on two
barriers of height just slightly greater than the energy of the
particle. At each encounter, part of the packet is transmitted
and part reflected, resulting in part of the packet being
trapped between the barriers from same time.
A 30-eV electron is incident on a square barrier of
height 40 eV. What is the probability that the electron will
tunnel through the barrier if its width is (a) 1.0 nm?
(b) 0.1nm?
The penetration of the barrier is not unique to quantum
mechanics. When light is totally reflected from the glass-air
interface, the light wave can penetrate the air barrier if a
second peace of glass is brought within a few wavelengths of
the first, even when the angle of incidence in the first prism is
greater than the critical angle. This effect can be demonstrated
with a laser beam and two 45° prisms.
α- Decay
The theory of barrier
penetration was used by
George Gamov in 1928 to
explain the enormous
variation of the half-lives for
α decay of radioactive
nuclei.
Potential well shown on the diagram for an α particle in a
radioactive nucleus approximately describes a strong attractive
force when r is less than the nuclear radius R. Outside the
nucleus the strong nuclear force is negligible, and the potential
is given by the Coulomb’s law, U(r) = +k(2e)(Ze)/r, where Ze is
the nuclear charge and 2e is the charge of α particle.
α- Decay
An α-particle inside the nucleus oscillates back and forth, being
reflected at the barrier at R. Because of its wave properties,
when the α-particle hits the barrier there is a small chance that it
will penetrate and appear outside the well at r = r0. The wave
function is similar to that for a square barrier potential.
The probability that an α-particle will tunnel through the
barrier is given by
T e

2 2 m (V0  E )a

which is a very small number, i.e., the α particle is usually
reflected. The number of times per second N that the α
particle approaches the barrier is given by
v
N
2R
where v equals the particle’s speed inside the nucleus.
The decay rate, or the probability per second that the nucleus
will emit an α particle, which is also the reciprocal of the mean
life time
, is given by

decay
1
v
rate  
e
 2R

2 2 m (V0  E )a

The decay rate for emission of α particles from radioactive
nuclei of Po212. The solid curve is the prediction of equation
decay
1
v
rate  
e
 2R

2 2 m (V0  E )a
The points are the experimental results.

Applications of Tunneling
• Nanotechnology refers to the design and application of
devices having dimensions ranging from 1 to 100 nm
• Nanotechnology uses the idea of trapping particles in
potential wells
• One area of nanotechnology of interest to researchers is the
quantum dot
– A quantum dot is a small region that is grown in a silicon
crystal that acts as a potential well
• Nuclear fusion
– Protons can tunnel through the barrier caused by their
mutual electrostatic repulsion
Resonant Tunneling Device
• Electrons travel in the gallium arsenide
semiconductor
• They strike the barrier of the quantum dot from the
left
• The electrons can tunnel through the barrier and
produce a current in the device
Scanning Tunneling Microscope
• An electrically conducting
probe with a very sharp
edge is brought near the
surface to be studied
• The empty space
between the tip and the
surface represents the
“barrier”
• The tip and the surface
are two walls of the
“potential well”
Scanning Tunneling Microscope
• The STM allows
highly detailed
images of surfaces
with resolutions
comparable to the
size of a single atom
• At right is the surface
of graphite “viewed”
with the STM
Scanning Tunneling Microscope
• The STM is very sensitive to the distance from
the tip to the surface
– This is the thickness of the barrier
• STM has one very serious limitation
– Its operation is dependent on the electrical
conductivity of the sample and the tip
– Most materials are not electrically conductive at their
surfaces
– The atomic force microscope (AFM) overcomes this
limitation by tracking the sample surface maintaining
a constant interatomic force between the atoms on
the scanner tip and the sample’s surface atoms.
SUMMARY
1. Time-independent Schrödinger equation:
 2 d 2  ( x)

 U ( x) ( x)  E ( x)
2
2m dx
2.In the simple harmonic oscillator:
1

E n   n    0
2

the ground wave function is given:
0 ( x)  Ae
 ax2
where A0 is the normalization constant and a=mω0/2ħ.
3. In a finite square well of height V0, there are only a finite
number of allowed energies.
SUMMARY
4.Reflection and barrier penetration:
When the potentials changes abruptly over a
small distance, a particle may be reflected even
though E>U(x). A particle may penetrate a
region in which E<U(x). Reflection and
penetration of electron waves are similar for
those for other kinds of waves.