Chapter 5
The Wavelike Properties of
Particles
CHAPTER 5
Wave Properties of Matter and Quantum Mechanics
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De Broglie Waves
Electron Scattering
Wave Motion
Waves or Particles?
Uncertainty Principle
Probability, Wave Functions,
and the Copenhagen
Interpretation
Particle in a Box
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both
matter and radiations, light in particular, it is necessary to introduce the
corpuscle concept and the wave concept at the same time.
- Louis de Broglie, 1929
The Wavelike Properties of Particles
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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
De Broglie Waves
• In his thesis in 1923, Prince
Louis V. de Broglie suggested
that mass particles should
have wave properties similar
to electromagnetic radiation.
• The energy can be written as:
If a light-wave could
also act like a particle,
why shouldn’t matterparticles also act like
waves?
hf = pc = plf
• Thus the wavelength of a
matter wave is called the De
Broglie wavelength:
Louis V. de Broglie
(1892-1987)
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.
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
l
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
l 


6
.
63

10
m
p mv (109 kg)(10 6 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.
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 then
h
l 
p
h
2mK
The de Broglie Wavelength
h
l 
p
h
2mK
If we multiply the numerator and denominator by c
we obtain:
l
hc
2mc2 K

1240eV  nm
2(0.511  106 eV ) K

1.226
nm
K
Where mc2=0.511MeV for electrons, and K in
electron-volts.
The de Broglie Wavelength
We obtained the electron wavelength:
 1.226 
nm,
l  

K


K in eV
Similarly, for proton (mc2 = 938 MeV for protons)
 0.0286 
 nm
l 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
Then 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:
lHe
 1.26 
nm
 

T


The de Broglie Wavelength
Similarly, for the molecule of hydrogen
lH 2
 1.78 
nm
 

 T 
and for the thermal neutrons
 2.52 
 nm
ln  

 T 
.
This calculations shows, 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.
(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)
l
h
h
l 
p
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
l
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
l
10 m
h
l ~ 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
The 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
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.
In 1925, Davisson and Germer
experimentally
observed
that
electrons were diffracted (much like
x-rays) in nickel crystals.
George P. Thomson (1892–1975),
son of J. J. Thomson, reported
seeing
electron
diffraction
in
transmission
experiments
on
celluloid, gold, aluminum, and
platinum.
A randomly oriented polycrystalline
sample of SnO2 produces rings.
Test of the de Broglie formula λ = h/p. The wavelength is
computed from a plot of the diffraction data for electrons
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
l
h
2m
1

E0
6.6 10 34 J  s
2  9.11 10 31 kg 1.6 10 19 J / eV
1
1
 1.226
E0
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
le  


 0.05nm
p
2me Ek
2mc2V0
2  0.511 106 eV  600eV
l
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.
l
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.
ln 
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 l  2dsin  2dcos 
 2
 
d  asin  
 2
m 1
 
 
l  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.)
 
 
l  2asin   cos   asin 
 2
 2
h
h
l
l
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
l
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
lm  
p  mv
c ch ch
ch
E  K l  f  E  K    1 m c2
l
ch m v
 v


2
lm
  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?
l
ch m v
 v


2
lm
  1 m c h   1 c
(a)
(b)
l
lm
1 0.9

l
lm

 1.60

1 0.92  1 1 0.92  1



1 0.001

2 
1  0.001  1

1  0.001
2
  2.294


 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 speed? (d) At low particle speed?
(c) As
and
(d)
v
1
c
 1
v
0
c
 
l

becomes nearly equal to γ.
 1 1
lm

,
2  1 2

v
 1 2 
c 

2
1 v2
 1 v
 1  1    2  1 
 2 c
2 c2
l
vc
2c
1



2 2
lm
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:
x t 
y  y0 cos(kx  t )  y0 cos 2   
l T 
where the angular frequency ω and the wave number k are
defined by
2
  2f 
T
and
k
2
l
and the wave or phase velocity vp is given by
v p  fl
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
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 

 nl  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)  2F cos
x
t   sin(k x  t )
2 
  2

BEATS
  k

 
F ( x)  2F cos
x
t   sin(k x  t )
2 
  2

This is an equivalent of an harmonic wave
F  y0 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.
l  2 x
Now
and

2
2 x 
k
2

xk  2
2
l
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 .
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.
Uncertainty Principle
Consider a wave packet Ψ(x,t) representing an
electron. The most probable position of the electron is the
value of x for which IΨ(x,t)I2 is a maximum.
Since IΨ(x,t)I2 is proportional to the probability that the
electron is at x, and IΨ(x,t)I2 is nonzero for a range of
value x, there is an uncertainty in the value of position of
the electron.
This means that if we make a number of position
measurements on identical electrons – electrons with same
wave function – we shall not always obtain the same result.
In fact, the distribution function for the results of such
measurements will be given by IΨ(x,t)I2.
Uncertainty Principle
If the wave packet is very narrow, the uncertainty in the
position will be small.
However, a narrow
wave packet must
contain a wide range
2
2

k


l
2x x
of wave number k:
2
2

k


l
2x x
The momentum is
related to the wave
number by
p  k
So, a wide range of k values means a wide range of
momentum values.
Uncertainty Principle
We can see that for all wave packets the ranges Δx and
Δk are related by
l 2 1
xk 

2 l
2
Similarly, a packet that is localized in time Δt must contain
a range of frequencies Δω, where the ranges are related
by
1
  t 
2
If we multiply these equations by ћ and use p = ћk and
E = ћω we obtain

xp 
2
and

Et 
2
If an excited state of an atom is known to have a lifetime
of 6∙10-7 s, what is the uncertainty in the energy of
photons emitted by such atoms in the spontaneous
decay to the ground state?
An electron (me = 9.11 × 10–31 kg) and a bullet (m = 0.0200 kg)
each have a velocity of magnitude of 500 m/s, accurate to
within 0.0100%. Within what limits could we determine the
position of the objects along the direction of the velocity?
..
An electron (me = 9.11 × 10–31 kg) and a bullet (m = 0.0200 kg)
each have a velocity of magnitude of 500 m/s, accurate to
within 0.0100%. Within what limits could we determine the
position of the objects along the direction of the velocity?
For the electron,




p  m ev  9.11 1031 kg  500 m s 1.00  104  4.56  1032 kg  m s
h
6.626  1034 J s
x 

 1.16 m m

32
4 p 4 4.56  10
kg  m s


For the bullet,


p  m v   0.020 0 kg 500 m s 1.00  104  1.00  103 kg  m s
h
x 
 5.28  1032 m
4 p
Show that the kinetic energy of a nonrelativistic particle can
be written in terms of its momentum as K = p2/2m. (b) Use
the results of (a) to find the minimum kinetic energy of a
proton confined within a nucleus having a diameter of 1.00 ×
10–15 m.
Show that the kinetic energy of a nonrelativistic particle can
be written in terms of its momentum as K = p2/2m. (b) Use
the results of (a) to find the minimum kinetic energy of a
proton confined within a nucleus having a diameter of 1.00 ×
10–15 m.
(a)
p2
1 2  m v
K  mv 

2
2m
2m
2
(b) To find the minimum kinetic energy, think of the minimum momentum
uncertainty, and maximum position uncertainty of 1015 m  x
. We
model the proton as moving along a straight line with

p x 
2

p 
2 x
The average momentum is zero. The average squared momentum is equal to the
squared uncertainty:
p 2 (p) 2
2
h2
(6.631034 J  s) 2
K




2
2
2
2m
2m
4(x) 2m 32 (x) m 32 2 (1015 m) 2 (1.671027 kg)
 8.331013 J  5.21MeV
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 a single 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
approaching 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  ln 
h2
En 


2m
2m
2ml2n
A Particle in a Box
2
 h 



2
pn  ln 
h2
En 


2m
2m
2ml2n
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 
h
2
2
8mL
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
k

where An is a constant and n
ln
number.
is the wave
The wave function for a particle in a box are the
same:
yn  An sin kn x
Using
2 L , we have
ln 
n
2
2
n
kn 


ln 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
  dx  
2
n
2
2  nx 
An sin 
dx  1
 L 
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 plane
wave describing a free particle, given in equation:
( x, t )  Aei (kx wt )
We are not interested here in how things behave in time,
so we chose a convenient time of zero. Thus, our “building
block” is
eikx
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:

ikx
 ( x) 

A(k )e dk

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
ikx
dx

The proper name of for A(k) is the Fourier transform of the
function ψ(x).
A(k ) 
 ( x) 
 ( x) 
│x│< a
│x│> a
1
0
ψ(x) =

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,
ˆconstitute a base that can generate any points in
and k
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 functions 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:
E to t
p2
 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
 2 2
he introduced the operator,  2
,which provides the
x
square of 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
2 m 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 coskx
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
8m L
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:
2  n x 

n ( x) 
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 2  ( x)
2


 ( x)  0
2
dx
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 2
0
1
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 2
0
The classical probability is thus
dx
PC ( x)dx 

v
dx
2
1
2 2
 E  m x 
m
2

The Harmonic Oscillator
E  mv  m A
1
2
2
1
2
2
0
2
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  hf0   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  hf0   (n  1)  hf0  hf0
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)  A0e
, 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