The Wavelike Properties of Particles

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The Wavelike Properties of Particles

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Title: The Wavelike Properties of Particles

1
Chapter 5

The Wavelike Properties of Particles

2
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

3
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.

4
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
? h/p
f E/h
where E is the total energy, p is the momentum,
and
? is called the de Broglie wavelength of the
particle.

5
The de Broglie Hypothesis

For photons these same equations results
directly from Einsteins quantization of
radiation E hf and equation for an energy of a
photon with zero rest energy E pc
E pc hf hc/?
Using relativistic mechanics de Broglie
demonstrated, that this equation can also be
applied to particles with mass and used them to
physical interpretation of Bohrs hydrogen-like
atom.

6
The de Broglie Wavelength

Using de Broglie relation lets find the
wavelength of a 10-6g particle moving with a
speed 10-6m/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.
7
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

Its wavelength is than
8
The de Broglie Wavelength

If we multiply the numerator and denominator by c
we obtain
Where mc20.511MeV for electrons, and K in
electronvolts.
9
The de Broglie Wavelength
We obtained the electron wavelength

Similarly, for proton (mc2 938 MeV for protons)
10
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 Maxwells Law
Than the momentum of the molecule is
Knowing that the mass of He atom, for instance,
is 6.7x10-24g, (kB1.38x10-23J/K) we obtain for
He wavelength
11
The de Broglie Wavelength
Similarly, for the molecule of hydrogen

and for the thermal neutrons
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
12

Show that the wavelength of a nonrelativistic
neutron is
where Kn is the kinetic energy of the neutron in
electron-volts. (b) What is the wavelength of a
1.00-keV neutron?

Show that the wavelength of a nonrelativistic
neutron is
where Kn is the kinetic energy of the neutron in
electron-volts. (b) What is the wavelength of a
1.00-keV neutron?

Kinetic energy, K, in this equation is in Joules
(a)
14

Show that the wavelength of a nonrelativistic
neutron is
where Kn is the kinetic energy of the neutron in
electron-volts. (b) What is the wavelength of a
1.00-keV neutron?

(b)
15

The nucleus of an atom is on the order of 1014
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 1014
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)
17

The nucleus of an atom is on the order of 1014
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)
With its
18
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 Davisson-Germer experiment.
Low energy electrons scattered at angle F 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 F 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 F50º,
as predicted for Bragg scattering of waves having
wavelength ? h/p.

Scattering of electron by crystal. Electron
waves are strongly scattered if the Bragg
condition n? D SinF is met.

Test of the de Broglie formula ? 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 ? h/(2mE)-1/2

Test of the de Broglie formula ? 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 ? h/(2mE)-1/2

A series of a polar graphs of Davisson and
Germers data at electron accelerating potential
from 36 V to 68 V. Note the development of the
peak at F 50º to a maximum when V0 54 V.

Variation of the scattered electron intensity
with wavelength for constant F. 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 n? D Sin F.

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 ) .

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.

In the DavissonGermer experiment, 54.0-eV
electrons were diffracted from a nickel lattice.
If the first maximum in the diffraction pattern
was observed at f 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 DavissonGermer experiment, 54.0-eV
electrons were diffracted from a nickel lattice.
If the first maximum in the diffraction pattern
was observed at f 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 DavissonGermer experiment, 54.0-eV
electrons were diffracted from a nickel lattice.
If the first maximum in the diffraction pattern
was observed at f 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.)

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
For a photon
38

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)
(b)
39

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?

As
(c)
becomes nearly equal to ?.
and
,
(d)
40
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
Harmonic waves of amplitude y0, frequency f and
period T where the angular frequency ? and the
wave number k are defined by and the wave or
phase velocity vp is given by
41

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 dsin? 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
Bohrs 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
for n integer
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 3? 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.
50
(b) The sum of these waves. The spatial extend of
the group ?x is inversely proportional to the
difference in wave numbers ?k, where k is
related to the wavelength by k 2p/?.
51
BEATS

Consider two waves of equal amplitude and nearly
equal frequencies and wavelengths.
The sum of the two waves is (superposition)
F(x) F1(x) F2(x) Fsin(k1x ?1t)sin(k2x
?2t)

52
BEATS

F(x) F1(x) F2(x) Fsin(k1x ?1t)sin(k2x
?2t)
using the trigonometric relation
Sina Sinß 2Cos(a-ß)/2 Sin(aß)/2
with a (k1x ?1t)

ß (k2x ?2t), we get

53
BEATS

with

54
BEATS

This is an equivalent of an harmonic wave
whose amplitude is modulated by
We have formed wave packets of extend ?x and
can imagine each wave packet representing a
particle.

Now
The particle is in the region ?x, the momentum
of the particle in the range ?k
p ?k ? ?p ??k
?x ?p h - Uncertainty Principle
In order to localize the particle within a
region ?x, we need to relax the precision on the
value of the momentum, ?p .

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 sxsk ½.

57
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.
58

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 ?px psin?
hsin?/ ?. 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 ?x
?/sin?. The product of the uncertainties ?px ?x
is therefore of the order of Plancks constant h.

62
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 ? that obeys a wave equation called the
Schrödinger equation.

63
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
Ehf, 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.

64
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 ??
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.

65
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 ?2(x)dx. If we call this probability P(x)dx,
where P(x) is the probability density, we have
P(x) ?2(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.

66
The Interpretation of the Wave Function

Then, the sum of the probabilities over all the
possible values of x must equal 1. That is,
This equation is called the normalization
condition. If ? is to satisfy the normalization
condition, it must approach zero as x is approach
infinity.

67
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 x2cm?
(c) What is the probability of finding the
particle between x3.0 cm and x3.4 cm?

68
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 ?,
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 ? 0 for x0 and for xL.

69
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( ?n/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/?n,

70
A Particle in a Box

Then the standing wave condition ?n 2L/n gives
the allowed energies
where

71
A Particle in a Box

The equation
gives the allowed energies for a particle in the
box.
This is the ground state energy for a particle
in the box, which is the energy of the lowest
state.

72
A Particle in a Box

The condition that we used for the wave function
in the box
? 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.

73
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
? c/f hc/(Ei Ef)

74
Standing Wave Function

The amplitude of a vibrating string fixed at x0
and xL is given as
where An is a constant and is the wave number.
The wave function for a particle in a box are
the same
Using , we have

75
Standing Wave Function

The wave function can thus be written

The constant An is determined by normalization
condition The result of evaluating the
integral and solving for An is
independent from n. The normalized wave function
for a particle in a box are thus
76
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 particles total
energy En is E1(n1), 4E1(n2), 9E1(n3) ..
77
Wave functions ?n(x) and probability densities
Pn(x) ?n2(x) for n1, 2, and 3 for the infinity
square well potential.
78
Probability distribution for n10 for the
infinity square well potential. The dashed line
is the classical probability density P1/L, which
is equal to the quantum mechanical distribution
averaged over a region ?x containing several
oscillations. A physical measurement with
resolution ?x will yield the classical result if
n is so large that ?2(x) has many oscillations in
?x.
79
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 n3 to a
lower-energy state.

80
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81
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 ?x 0.01L, centered at x ½L (b) in
the region 0ltxlt(1/4)L.

82
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 ?(x).

83
Expectation Values

Since ?2(x)dx is the probability of finding a
particle in the region dx, the expectation value
of x is
The expectation value of any function f(x) is
given by

84
Calculating expectation values

Find (a) x and (b) x2 for a particle in
its ground state in a box of length L.

85
Complex Numbers

A complex number has the form aib, with i2-1 or
iv-1 imaginary unit.
a - real part b imaginary part i imaginary
unit
(a ib) (c id) (ac) i(bd)
m(a ib) ma imb
(a ib) (c id) (ac - bd) i(ad bc)
The absolute value of a ib is denoted by aib
and is given by aib v a2 b2

86
Complex Numbers

The complex conjugate of aib is denoted by
(aib) and is given
(aib) (a-ib)
Then
(aib) (aib) (a-ib) (aib)a2 b2

87
Polar Form of Complex Numbers

p v a2 b2 a ib
We can represent the number (a ib) in the
complex xy plane.
Then the polar coordinates
a ib p(cosf isinf)
Remembering the Euler formula
eif (cosfisinf)
a ib p eif

Imaginary axis
p
b
f
Real axis
a
Euler Identities eif cosf isinf e-if
cosf - isinf where i v-1
88
Fourier Transform

In quantum mechanics, our basic function is the
pure sinusoidal plane wave describing a free
particle, given in equation
We are not interest here in how things behave
in time, so we chose a convenient time of zero.
Thus, our building block is
Now we claim that any general, nonperiodic wave
function ?(x) can be expressed as a sum/integral
of this building blocks over the continuum of
wave numbers

89
Fourier Transform
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 cant pull it out of the integral, the
equation can be solved for A(k). The result is
The proper name of for A(k) is the Fourier
transform of the function ?(x).
90

1 xlt a
0 xgt a
Let use Euler identities _eika cosfisinf
e-ika cosf-isinf
eika- e-ika 2isinka
And we can overwrite the equation for A(k)

?(x)
91
General Wave Packets

Any point in space can be described as a linear
combination of unit vectors. The three unit
vectors î, j, and 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.

92
The General Wave Packet

A periodic function f(x) can be represented by
the sum of harmonic waves
y(x,t) S Aicos(kix ?it) Bisin(kix ?it)
Ai and Bi amplitudes of the waves with wave
number ki and angular frequency ?i.
For a function that is not periodic there is an
equivalent approach called Fourier
Transformation.

93
Fourier Transformation

A function F(x) that is not periodic can be
represented by
a sum (integral) of functions of the type
eika CosfiSinf
In math terms it called Fourier Transformation.
Given a function F(x)
where
f(kj) represents the amplitude of base function
e-ikx used to represent F(x).

94
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
was found by Schrödinger in 1925 and is now known
as the Schrödinger equation.

95
The Schrödinger Equation

Like the classical wave equation, the
Schrödinger equation is a partial differential
equation in space and time.
Like Newtons laws of motion, the Schrödinger
equation cannot be derived. Its validity, like
that of Newtons laws, lies in its agreement with
experiment.

We will start from classical description of the
total energy of a particle
Schrödinger converted this equation into a wave
equation by defining a wavefunction, ?. He
multiplied each factor in energy equation with
that wave function

To incorporate the de Broglie wavelength of the
particle he introduced the operator,
,which provides the square of the momentum when
applied to a plane wave

If we apply the operator to that wavefunction
where k is the wavenumber, which equals 2p/?. We
now simple replace the p2 in equation for energy
98
Time Independent Schrödinger Equation

This equation is called time-independent
Schrödinger equation.
E is the total energy of the particle.
The normalization condition now becomes
? ?(x)?(x)dx 1

99
A Solution to the Srödinger Equation

Show that for a free particle of mass m moving
in one dimension the function
is a solution of the time independent Srödinger
Equation for any values of the constants A and B.

100
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, 0ltxltL U(x) 8, xlt0 or xgtL
101

A Particle in Infinity Square Well Potential
Inside the box U(x) 0, so the Schrödinger
equation is written
where E h? is the energy of the particle, or
where k2 2mE/h2
The general solution of this equation can be
written as
?(x) A sin kx B cos kx
where A and B are constants. At x0, we have
?(0) A sin (k0) B cos (0x) 0 B

102

A Particle in Infinity Square Well Potential
The boundary condition ?(x)0 at x0 thus gives
B0 and equation becomes
?(x) A sin kx
We received a sin wave with the wavelength ?
related to wave number k in a usual way, ?
2p/k. The boundary condition ?(x) 0 at xL gives
?(L) A sin kL 0
This condition is satisfied if kL is any integer
times p, or
kn np / L
If we will write the wave number k in terms of
wavelength ? 2p/k, we will receive
the standing wave condition for particle in the
box
n? / 2 L n 1,2,3,

103

A Particle in Infinity Square Well Potential
Solving k2 2mE/h2 for E and using the
standing wave condition k np / L gives us the
allowed energy values
where
For each value n, there is a wave function ?n(x)
given by

104

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 v2/L determined by
normalization

Although this problem seems artificial, actually
it is useful for some physical problems, such as
a neutron inside the nucleus.
105
A Particle in a Finite Square Well
This potential energy function is described
mathematically by U(x)V0, xlt0 U(x)0,
0ltxltL U(x)V0, xgtL Here we assume that 0 EV0.
Inside the well, U(x)0, and the time independent
Schrödinger equation is the same as for the
infinite well

106
A Particle in a Finite Square Well

or
where k2 2mE/h2. The general solution is
?(x) A sin kx B cos kx
but in this case, ?(x) is not required to be
zero at x0, so B is not zero.

107
A Particle in a Finite Square Well

Outside the well, the time independent
Schrödinger equation is
or
where

108
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
where ?0 vk/m2pf is the angular frequency of
the oscillator. Classically, the object
oscillates between x A and x-A. Its total
energy is
which can have any nonnegative value, including
zero.

109

Potential energy function for a simple harmonic
oscillator. Classically, the particle with energy
E is confined between the turning points A and
A.

110
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
The classical probability is thus

111
The Harmonic Oscillator

The classical probability is

Any values of the energy E is possible. The
lowest energy is E0, in which case the particle
is in the rest at the origin. The Shrödinger
equation for this problem is
112
The Harmonic Oscillator

In quantum theory, the particle is represented
by the wave function ?(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 ?n(x) occur only for discrete values of
the energy En given by
where f0?0/2p is the classical frequency of the
oscillator.

113
The Harmonic Oscillator

where f0?0/2p is the classical frequency of the
oscillator. Thus, the ground-state energy is ½h?
and the exited energy levels are equally spaced
by h?.

114
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115

Energy levels in the simple harmonic oscillator
potential. Transitions obeying the selection rule
?n1 are indicated by the arrows. Since the
levels have equal spacing, the same energy h? 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.

116
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
The frequency f of the emitted photon is
therefore equal to the classical frequency f0 of
the oscillator.

117
Wave function for the ground state and the first
two excited states of the simple harmonic
oscillator potential, the states with n0, 1, and
2.
118
Probability density for the simple harmonic
oscillator plotted against the dimensionless
value , for n0, 1, 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
119

Molecules vibrate as harmonic oscillators.
Measuring vibration frequencies enables
determination of force constants, bond strengths,
and properties of solids.

120

Verify that , where a
is a positive constant, is a solution of the
Schrödinger equation for the harmonic oscillator

121
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
In general, the expectation value of any
function f(x) is given by

122
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 ?(x), the
distribution function can be found by Fourier
analysis. It can be shown that p can be found
from

123
Operators

Similarly, p2 can be found from
Notice that in computing the expectation value
the operator representing the physical quantity
operates on ?(x), not on ?(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.

124
Expectation Values for p and p2

Find p and p2 for the ground state wave
function of the infinity square well.

125

In classical mechanics, the total energy written
in terms of position and momentum variables is
called the Hamiltonian function

If we replace the momentum by the momentum
operator pop and note that U U(x), we obtain
the Hamiltonian operator Hop
The time-independent Schrödinger equation can
then be written
126
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.
127
Symbol Physical quantity Operator
f(x) Any function of x (the position, x the potential energy U(x), etc. f(x)
px x component of momentum
py y component of momentum
pz z component of momentum
E Hamiltonian (time-independent)
E Hamiltonian (time-dependent)
Ek Kinetic energy
Lz Z component of angular momentum
128
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.
Lets consider a one-dimensional box of length
L. If we know that the particle is in the box, ?x
is not larger than L. This implies that ?p is at
least h/L. Let us take the standard deviation as
a measure of ?p

129
Minimum Energy of a Particle in a Box

If the box is symmetric, p will be zero since
the particle moves to the left as often as to the
right. Then
and the average kinetic energy is

130
Minimum Energy of a Particle in a Box

The average kinetic energy of a particle in a
box is
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.

131
The Hydrogen Atom

The energy of an electron of momentum p a
distance r from a proton is
If we take for the order of magnitude of the
position uncertainty ?x r, we have
(?p2) p2 ?2/r2
The energy is then

132
The Hydrogen Atom

There is a radius rm at which E is minimum.
Setting dE/dr 0 yields rm and Em

rm came out to be exactly the radius of the first
Bohr orbit
The ground state energy
133
The Hydrogen Atom

The potential energy of the electron-proton
system varies inversely with separation distance
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.

134
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 8 the energy approaches 0.
135
Step Potential

Consider a particle of energy E moving in region
in which the potential energy is the step
function
U(x) 0, xlt0
U(x) V0, xgt0
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 v2E/m

136
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 v2(E U0)/m

137
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 mghV0.
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.

138

The quantum mechanical result is similar when E
is less than V0. If EltV0 the wave function does
not go to zero at x0 but rather decays
exponentially. The wave penetrates slightly into
the classically forbidden region xgt0, but it is
eventually completely reflected.

139
Step Potential

This problem is somewhat similar to that of
total internal reflection in optics.
For EgtV0, the quantum mechanical result differs
from the classical result. At x0, the wavelength
changes from
?1h/p1 h/v2mE
to
?2h/p2 h/v2m(E-V0).
When the wavelength changes suddenly, part of
the wave is reflected and part of the wave is
transmitted.

140
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.

141
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
where k1 is the wave number for the incident
wave and k2 is the wave number for the
transmitted wave.

142
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.

143

Time development of a one dimensional wave
packet representing a particle incident on a step
potential for EgtV0. The position of a classical
particle is indicated by the dot. Note that part
of the packet is transmitted and part is
reflected.

144

Reflection coefficient R and transmission
coefficient T for a potential step V0 high versus
energy E (in units V0).

145

A particle of energy E0 traveling in a region in
which the potential energy is zero is incident on
a potential barrier of height V00.2E0. Find the
probability that the particle will be reflected.

146
Lets consider a rectangular potential barrier of
height V0 and with a given byU(x) 0, xlt0U(x)
V0, 0ltxltaU(x) 0, xgta
147
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 0ltxlta. Upon reaching
the far wall of the barrier (xa), the wave
function must join smoothly to a sinusoidal wave
function to the right of barrier.
148
The potentials and the Schrödinger equations
for the three regions are as follows Region I
(xlt0) V 0, Region II (0ltxlta) V
V0, Region III (xgta) V 0,
149
Barrier Potential

If we have a beam of particle incident from
left, all with the same energy EltV0, the general
solution of the wave equation are, following the
example for a potential step,
where k1 v2mE/h and a v2m(V0-E)/h
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.

150

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 G0, and the only term in region
III is Feik1x. We summarize these wave functions

151
Barrier Potential

For the case in which the quantity
aa v2ma2(V0 E)/h2
is much greater than 1, the transmission
coefficient is proportional to e-2aa, with
a v2m(V0 E)/h2
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.

152

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.

153

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?

154

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.

155
a- Decay
The theory of barrier penetration was used by
George Gamov in 1928 to explain the enormous
variation of the half-lives for a decay of
radioactive nuclei.
Potential well shown on the diagram for an a
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 Coulombs law,
U(r) k(2e)(Ze)/r, where Ze is the nuclear
charge and 2e is the charge of a particle.
156
a- Decay
An a-particle inside the nucleus oscillates back
and forth, being reflected at the barrier at R.
Because of its wave properties, when the
a-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.
157

The probability that an a-particle will tunnel
through the barrier is given by
which is a very small number, i.e., the a
particle is usually reflected. The number of
times per second N that the a particle approaches
the barrier is given by

where v equals the particles speed inside the
nucleus.
The decay rate, or the probability per second
that the nucleus will emit an a particle, which
is also the reciprocal of the mean life time
, is given by
158

The decay rate for emission of a particles from
radioactive nuclei of Po212. The solid curve is
the prediction of equation
The points are the experimental results.

159
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

160
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

161
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

162
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

163
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 samples
surface atoms.

164
SUMMARY

1. Time-independent Schrödinger equation
2.In the simple harmonic oscillator
the ground wave function is given
where A0 is the normalization constant and
am?0/2h.
3. In a finite square well of height V0, there
are only a finite number of allowed energies.

165
SUMMARY

4.Reflection and barrier penetration
When the potentials changes abruptly over a
small distance, a particle may be reflected even
though EgtU(x). A particle may penetrate a region
in which EltU(x). Reflection and penetration of
electron waves are similar for those for other
kinds of waves.

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