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I-1
CHAPTER I. INTRODUCTION TO QUANTUM THEORY
A. Historical perspective
The topics that will be covered during approximately the first 70% of this course focus on
the microscopic properties of matter. The objective is to gain a rudimentary understanding of the
behavior of atoms and molecules, as well as the theoretical framework used in the description of
microscopic particles. The historical development of humankind's understanding of the physical
principles involved is briefly sketched below. (See Table I-1).
The description of macroscopic properties utilizes the tools of classical mechanics and
thermodynamics. Classical mechanics evolved during the 1600's, beginning with the
astronomical observations of Tycho Brahe and their interpretation by Kepler. The additional
information provided by the mechanical experiments of Galileo enabled Newton (in 1687) to
devise an elegant theory of mechanics.
A similar revolution involving the unification of a large variety of experimental facts
concerning electricity and magnetism took place in the nineteenth century. All of these
phenomena, including the wave propagation theory of light, were brought together within one
conceptual framework by the electromagnetic theory of Maxwell in 1864.
The beginning of the twentieth century brought an unprecedented series of experiments
which unveiled new phenomena that defied explanation in terms of the universally accepted laws
of classical physics. Ultimately, these events led to one of the most stimulating periods of
development ever to occur in the physical sciences. As a result of this activity, intuitive ideas
concerning the nature of microscopic systems and space/time relationships underwent drastic
revision. These new concepts are embodied in what is now called "modern physics", which is
comprised of the quantum theory and the theory of relativity.
One of the most important experiments leading to the development of quantum theory
measured the spectral distribution of black body radiation. The theoretical explanation of this
phenomenon introduced the crucial concept of quantization.
I-2
Table I-1. Major events in the evolution of physical concepts
Classical Mechanics: 1600's
1600 - Tycho Brahe's astronomical observations - Interpretation by Kepler - Mechanical experiments of Galileo
1687 - Newton's theory of mechanics
Classical electromagnetism: 1800's
1800 - Experiments on electricity and magnetism1864 - Maxwell's electromagnetic theory
Relativity: 1900's
1905 - Einstein's theory of relativity
Quantum theory: 1900's
1896 - Discovery of radioactivity by Becquerel1897 - Thompson's experiments with electrons1911 - Rutherford model of the atom1913 - Bohr model of the atom1924 - Wave properties of particles (de Broglie/Davisson
& Germer) 1927 - Quantum mechanics (Schrödinger & Heisenberg)
A(x,t) ' A0 sin (kx ! ω t)
I-3
B. Failures of Classical Physics
1. Black Body Radiation
When an object is heated, it emits electromagnetic radiation. A black body has the ability
to absorb and emit all radiation regardless of its frequency. It can be approximated by an oven
with a small hole through which the radiation is sampled. The frequency distribution of black
body radiation is shown in Fig. I-1. Notice that the radiation distribution has a maximum, ν ,max
that increases with temperature (e.g., ν 6 red 6 orange 6 yellow 6 white).max
In the late 1800's, Rayleigh and Jeans examined the problem of black body radiation.
Using classical concepts, they were able to obtain a relatively simple formula for the spectrum of
black body radiation. They began by assuming the electromagnetic radiation was generated by
the vibrations of atomic oscillators in the walls of the oven, much as radio waves are generated
by the oscillation of an electric current in a dipole antenna. Under the steady state conditions of
thermal equilibrium, the radiation inside the oven must exist in the form of standing waves with
nodes at the walls. Before delving into the details of the analysis by Rayleigh and Jeans, a short
digression into the general properties of waves will be beneficial.
When a wave propagates through an elastic medium (e.g., air, water), each particle of the
medium vibrates in simple harmonic motion. Therefore, the displacements of the particles are
given by a sine or cosine function of the space coordinates. The distance between any two
successive particles whose motion is in phase is defined to be the wavelength λ of the wave.
Since the wave has moved a distance of one wavelength during the period T, the wave velocity is
v = λ /T = λ ν , where ν is the frequency. Therefore
.
The general equation for a plane wave propagating to the right along the x-axis is
.
Frequency (1013 s-1)
0 10 20 30 40 50
U(ν
) (1
0-17 J-
s/m
3 )
0
2
4
6
8
2000 K
1750 K
visible light
I-4
Figure I-1. Frequency distribution of black body radiation at two temperatures.
2A0 sinα%
β2
cosα !
β2
A ' 2A0 sin (kx!ω t) cos !dkx%dω t2
I-5
where A is the amplitude, ω = 2π ν (angular frequency) and k = 2π /λ (propagation constant). o
When two waves traverse the same region of space at the same time, the resultant amplitude will
be given by the sum of the two waves. This superposition principle leads to wave interference.
Consider the sum of two waves having the same amplitude but differing by an amount dω in
angular frequency and dk in propagation constant;
A = A sin (kx!ω t) = A sin α1 0 0
A = A sin [(k+dk)x ! (ω +dω )t] = A sin β .2 0 0
Their sum is A = A + A = A sin α + A sin β
1 2 0 0
= (trig. identity)
where α +β = (2k+dk)x!(2ω +dω )t
. 2(kx!ω t)α !β = !dkx+dω t .
Hence
.
This is a wave moving to the right which is characterized by beats caused by the second term
(the modulating factor). An illustration of the interference of two waves and the resulting beat
pattern is shown in Fig. I-2. By the appropriate choice of dω and dk, total constructive or total
destructive interference can result.
A special case of interest is a standing wave formed by two waves having the same
amplitudes, ω 's, and k's traveling in opposite directions;
A = A sin (kx!ω t) A = A sin (kx+ω t) .1 0 2 0
By performing the same exercise as above, the sum is found to be
A = 2A sin kx cos ω t .0
This is the equation of a stationary wave, as may be seen by noting that at values of x for which
kx = nπ (where n is an integer), A = 0. Hence the wave has stationary nodes (see Fig. I-3).
Am
plitu
de
-40
-20
0
20
40A
mpl
itude
-40
-20
0
20
X
0.0 0.4 0.8 1.2 1.6 2.0
Am
plitu
de
-80
-40
0
40
I-6
Figure I-2. A wave formed by the superposition of two other waves having slightly different
propagation constants. Note the beat pattern resulting from the cosine term in the
sum equation on p. I-5.
X/λ
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Am
plitu
de
-6
-4
-2
0
2
4
6
I-7
Figure I-3. A standing wave shown for three different values of t. Note that the nodes do not
move and hence the wave is stationary.
2Lλ ' n2x % n2
y % n2z ' n
I-8
Now return to the Rayleigh-Jeans derivation. Consider a cubical oven of dimension L.
Since the waves inside the oven must consist of standing waves having nodes at the walls, a
condition for their existence is that the path length must be an integer number of half-
wavelengths. Therefore standing waves in the x, y, and z directions must obey the following
relations: nλ /2 = nλ /2 = nλ /2 = L where n , n , and n are integers. For a standing wave inx x y y z z x y z
any arbitrary direction having components λ = λ /cos α , λ = λ /cos β, and λ = λ /cos γ , where α ,
β,x y z
and γ are the angles between the propagation direction
and the coordinate axes, it then follows that the allowed
wavelengths must satisfy the relation
where n , n , and n are integers (0,1,2,...).x y z
In order to predict the spectral distribution of black
body radiation, we must calculate the average energy per
unit volume (i.e., the energy density) radiated in the
wavelength interval λ to λ +dλ . It may be expressed as follows:
U(λ )dλ = ε&η (λ )dλwhere ε& = average energy per wave and η (λ )dλ = number of waves per unit volume in the
interval λ to λ +dλ .
The number of waves in the interval λ to λ +dλ can be counted by visualizing a three-
dimensional space containing all combinations of the numbers n , n , and n . Part of the n -nx y z x y
plane of such a space is shown in Fig. I-4. Each lattice point in this space defines a permissible
set of n , n , n values and hence corresponds to an allowed wave. Now let n be a vector from thex y z
origin to a particular set of n , n , n . It follows that the number of waves having wavelengthsx y z
between λ and λ +dλ is equal to the number of points within the spherical shell defined by radius
n and radius n+dn. This number is equal to the volume of the spherical shell. Actually, since n ,x
n , and n must all be positive numbers, the number of points is just equal to the volume of one y z
nx
0 1 2 3 4 5 6 7 8 9 10
n y
0
1
2
3
4
5
6
7
8
9
10
dn
n
I-9
Figure I-4. The n -n plane in n-space. All points lying on an octant of a sphere of radius n arex y
combinations of n , n , n that correspond to a particular value of λ .x y z
I-10
octant of the spherical shell;
N(n)dn = number of points in one octant of a spherical shell of inner radius n and
outer radius n+dn
= (1/8)4π n dn .2
Now transform n to λ :
n = 2L/λ , so
dn = !(2L/λ )dλ2
Also N(λ )dλ = !N(n)dn (Note that the negative sign here is because according to the
first equation, n decreases as λ increases).
Hence N(λ )dλ = (4π /8)(4L /λ )(2L/λ )dλ2 2 2
= (4π L /λ )dλ .3 4
Now η (λ )dλ = (N(λ )/V)dλ = (4π /λ )dλ ,4
but for each λ there are two independent standing waves corresponding to the two
unique directions of polarization. Thereforeη (λ )dλ = (8π /λ )dλ .4
Exercise: Ultimately, we will want the energy density as a function of frequency. Transform the
above expression for η (λ )dλ to get the corresponding expression for η (ν )dν .η (ν )dν = !η (λ )dλ λ = c/ν so dλ = !(c/ν )dν 2η (ν )dν = !8π /(c/ν ) (!c/ν )dν4 2η (ν )dν = (8π ν /c )dν .2 3
The other quantity needed to complete the calculation is ε&, the average energy per wave.
Rayleigh and Jeans assumed that the waves were radiated by atomic oscillators in the walls of the
cavity, which could be treated as classical harmonic oscillators. An example of a classical
harmonic oscillator is a weight attached to a spring. The equilibrium position of the weight
defines the zero value of the displacement coordinate (e.g., x = 0). When the weight is displaced
in the positive direction, the spring is stretched and it exerts a force on the weight given by
Hooke’s law, F = !k x, where k is the force constant. Once released, the weight undergoesf f
F ' ma, so !kfx ' md2x
dt2
d2x
dt2%
kf
mx ' 0
d2x
dt2%
ω 2x ' 0
T '2πω
dV ' !Fdx ' kf xdx
gk ' ½mv2 ' ½m dxdt
2
' ½mω 2A 2sin2ω t ' ½kf A2sin2ω t
I-11
simple harmonic motion and its displacement as a function of time may be obtained by solving
the equation of motion given by Newton’s second law of motion, namely
or
.
If we let ω = k /m, the above equation becomes2f
,
which has the solution
x = A cos ω t,
where A is the amplitude (i.e., the displacement at t = 0). The period, T is the time it takes for
the weight to travel through a complete cycle, and so ω T = 2π , or
,
Therefore, the oscillation frequency, which is defined as 1/T, is given by
.
The potential energy of an harmonic oscillator is easily found to be
,
.
The kinetic energy is
,
and hence the total energy of an harmonic oscillator is
g ' gk % V ' ½kf A2sin2ω t % ½kf A
2cos2ω t ' ½kf A2
g '
4
0
P(g)gdg
4
0
P(g)dg
'
B
4
0
ge!g/kTdg
B
4
0
e!g/kTdg
4
0
ge!g/kTdg '1
1/(kT)2' (kT)2
4
0
e!g/kTdg ' !kTe!g/kT|40 ' kT
g ' kT
I-12
.
The above result shows that the total energy of a classical harmonic oscillator is proportional to
the square of its amplitude.
Since there is no limitation on the amplitude of a classical harmonic oscillator, its energy
can vary continuously from zero to infinity. To complete the Rayleigh-Jeans derivation, we must
find the average energy of a classical harmonic oscillator. As we shall prove when we cover the
chapter on statistical thermodynamics, the probability that an atom at thermal equilibrium has a
particular energy g is given by the Boltzmann probability distribution function
P(g) = Be where B is a constant and β = (kT) .!
βg !1
(Here, k is the Boltzmann constant). Therefore, the average energy may be calculated as follows:
and
.
Hence .
We now know that the classical prediction for the average oscillator energy is ε& = kT,
and hence the result for the energy radiated per unit volume over the frequency interval ν to ν +dνis obtained by multiplying the expression for η (ν )dν times kT, which gives
U(ν )dν '8π ν 2kT
c3dν
¯ε '
j4
0ε nP(ε n)
j4
0P(ε n)
'
j4
0nhν e!
βnhν
j4
0e!
βnhν
hν j nx n
j x n, where x' e!
βhν
j x n ' 1 % x % x 2 %..... ' 11!x
I-13
Rayleigh-Jeans formula.
This result, which proved to be disastrous for classical physics, was referred to as the ultraviolet
catastrophe because it predicts that U(ν ) increases quadratically with ν and approaches infinity in
the ultraviolet region of the spectrum. The measured spectrum of black body radiation at T =
2000 K is compared with the predictions of the Rayleigh-Jeans formula in Fig. I-5.
We now address the question of what was wrong with the Rayleigh-Jeans derivation. The
answer lies in the assumption that the energy of an atomic oscillator is a continuous function of
the amplitude. Planck found that by assuming the oscillator energies were quantized in
integer units of the quantity hν , he could obtain the correct result. The derivation proceeds
as follows.
According to the Planck quantum postulate, each oscillator may have an energy given byε = nhν , where n = 0,1,2,..., and h is a constant now referred to as Planck’s constant. Now,
when we calculate the average energy, we must use summations instead of integrals to account
for the discrete nature of the oscillator energies. Proceeding as before, the average oscillator
energy is given by
The expression on the right can be rewritten in the form
.
Now note that
(with x < 1), and
Frequency (1013 s-1)
0 10 20 30 40 50
U(ν
) (1
0-17 J-
s/m
3 )
0
2
4
6
8
T = 2000 KRayleigh-Jeans formula
Experiment
I-14
Figure I-5. Comparison of the frequency distribution of black body radiation predicted by the
Rayleigh-Jeans formula with the experimental distribution.
U(ν )dν '8π hν 3
c3
dν(ehν /kT!1)
U(ν )dν '8π ν 2kT
c3dν .
1xj
nx n ' 1 % 2x % 3x 2 %..... ' ddx
(1 % x % x 2 % x 3 %....) ' ddx
11!x
'1
(1!x)2
¯ε '
(xhν )1x j nx n
j x n'
xhν (1!x)
(1!x)2,
¯ε 'xhν1!x
'hν
eβ
hν !1
U(ν )dν ' ¯ε n(ν ) dν 'hν
eβ
hν !1
8π ν 2
c3dν
I-15
Therefore
or
.
Now we can use this result to complete the calculation ;
or
Planck Radiation formula.
Exercise: Show that the Planck formula reduces to the Rayleigh-Jeans formula when hν << kT.
(Hint: use the series expansion of e ).hν /kT
e . 1 + hν /kT + .........hν /kT
hence e ! 1 . hν /kT. Substituting this into the Planck equation giveshν /kT
2. The Photoelectric Effect
Several experimental phenomena involving the interaction of light with matter defied
explanation in terms of a classical wave description. One of these was the photoelectric effect in
which light incident upon a metal plate causes electrons to be ejected from the surface. Classical
I-16
electromagnetic theory predicts the energy deposited by the incident light should be proportional
to the light intensity (wave amplitude squared). However, this prediction was not compatible
with the following characteristics of the process:
a. The number of electrons ejected per second is proportional to the light intensity.
b. The kinetic energy of the ejected electrons is directly proportional to the light
frequency.
c. There is a threshold frequency below which no electrons are ejected.
A graphical representation of characteristics b and c is shown below. Evidently the energy of
the ejected electrons is given by the formula
This formula can be understood as a simple expression of the law of conservation of energy if
one views the action of the light to be equivalent to that of a massless particle of energy
.
Then the incident photon (particle of light) energy is distributed between the ejected electron
I-17
kinetic energy E and the energy required to remove the electron from the metal surface B (i.e.,e
its binding energy).
Exercise: The 60 keV γ -ray from the radioactive decay of Am is useful for ejecting K-shell241
electrons from intermediate Z atoms via the photoelectric effect. Calculate the
wavelength of the γ -ray and the kinetic energy of a K-electron ejected from a silver
atom if the K-electron binding energy is 25.5 keV.λ = hc/E = 12398 eV-D/60x10 eV = 0.21 D.3
E = (60 ! 25.5) keV = 34.5 keV.e
3. Compton Scattering
Compton scattering is another mechanism by which electromagnetic radiation interacts
with matter and it clearly demonstrates that the scattering of light by an electron is directly
analogous to the scattering of two particles. According to classical electromagnetic theory, the
electron should be accelerated by the field of the light wave and the scattered wave should
contain a distribution of wavelengths. However, the experiment showed that the scattered wave
had a well-defined wavelength that depended on the energy of the incident wave and the
scattering angle. By again treating the light waves as massless particles and applying the laws of
conservation of energy and momentum, the relationship between the change in wavelength and
the scattering angle may be derived, as outlined below.
λ ) ! λ 'h
m0c(1! cos
θ)
I-18
Conservation of energy: hc/λ + m c = hc/λ ' + [(m c ) +(p c) ]o o e2 2 2 2 1/2
Conservation of momentum:
x-direction h/λ = h/λ ' cos θ
+ p cos φe
y-direction 0 = h/λ ' sin θ
! p sin φe
By solving the above simultaneous equations, the final result is obtained:
.
4. Wave Properties of Particles
In 1924, Louis de Broglie suggested that the dual (wave/particle) nature of light should
also apply to the behavior of particles. Under the right circumstances, he reasoned, it should be
possible to see, for example, diffraction phenomena occurring with particles. Since photons of
electromagnetic radiation have momentum
p = E/c
and energy E = hν = hc/λ ,
their wavelengths and momenta are related by the equation
.
de Broglie proposed that this equation also applies to particles and can be used to predict the
wavelength that characterizes a particle's wavelike properties. A year later, Davisson and Germer
showed electrons do indeed undergo diffraction from crystal lattices (see Fig. I-6).
Exercise: Calculate the wavelength of an electron with a kinetic energy of 13.6 eV.
E = ½mv = p /(2m)k2 2
p = [2mE]1/2λ = h/p = h/[2mE] = (hc)/[2mc E] 1/2 2 1/2
= (12398 eV-D))/[(2)(511x10 eV)(13.6 eV)]3 1/2λ = 3.32 D (This is the average circumference of the H-atom)
I-19
Exercise: Calculate the wavelength of a baseball weighing 200 g and traveling at
90 mi/hr.
p = mv = (200 g)(1 kg/1000 g)(90 mi/1 h)(1 h/3600 s)(5280 ft/1 mi)
x(12 in/1 ft)(2.54 cm/1 in)(1 m/100 cm) = 8.05 kg-m/s
= 8.05 J-s/mλ = h/p = (6.62x10 J-s)/(8.05 J-s/m) = 8.23x10 m = 8.23x10D.-34 -35 -25
Note that the reason one does not expect to observe wavelike phenomena with baseballs
is because they have such small wavelengths (i.e., physical size is much larger than λ ).
Figure I-6. Diffraction pattern obtained by passing a beam of electrons through a gold foil.
mv 'Ze2m
r
1/2
r 'L
Ze2mr
1/2'
L 2
Ze2m
E '!Z 2e4m
2L 2
h2π
I-20
5. The Bohr Model
Bohr used the idea of quantization to develop a simple model of the hydrogen atom. This
model allowed Bohr to propose that atomic line spectra originate from electronic transitions
between quantum states.
We shall begin the derivation with a statement of the requirement for a stable circular
orbit of an electron of charge !e about a nucleus of charge Ze;
F = ma = mv /r (cent. force) = Ze /r (Coul. force)2 2 2
so mv = Ze /r . (a)2 2
Now the total energy is
E = E + V = ½mv ! Ze /r = !½(Ze /r) . (b)k2 2 2
The angular momentum for a circular orbit is given by
L = mvr,
so the radius of the orbit may be expressed as
r = L/(mv),
but from Eq. (a), ,
and hence
.
Inserting this expression for r into Eq. (b) gives
.
At this point, Bohr departed from the laws of classical physics by introducing
quantization into the problem. He did so by postulating the angular momentum could only take
on integer multiples of the quantity (usually written as £ , which is referred to as h-bar).
E ' !2π 2e4mZ2
n2h2and r ' n2h2
4π 2e2mZ
I-21
Thus
, where n = integer The Bohr quantum rule.
Although this postulate seems quite arbitrary, it is likely that Bohr drew his insight from the
knowledge that frequencies of the lines appearing in the spectrum of hydrogen are accurately
predicted by the empirical equationν = 3.29x10 (n ! n ),15 !2 !21 2
where n and n are integers, as will become apparent below.1 2
The final equations for the total energy of a one-electron atom and the radius of the
(Bohr) orbit are obtained upon substitution of the Bohr quantum rule for L;
.
When all the constants are evaluated, these formulas reduce to
Therefore, according to the Bohr model, one-electron atoms can exist only with quantized
amounts of energy corresponding to the integer n (called a quantum number). These quantized
energies, which are referred to as energy levels, are shown below for hydrogen.
�E ' hν ' Eu ! El ' !E0 Z 2 1
n2u
!1
n2l
I-22
Although the simple formulas given in the above box are strictly applicable to one-electron
atoms and ions, they are very useful for order-of-magnitude estimates of the energies and radii of
many electron atoms.
Bohr suggested that light emission from an atom is the result of an electronic transition
from an upper energy level to a lower energy level. The energy change in such a transition is
therefore given by
Exercise:
Calculate the wavelength of the Lyman alpha transition (n = 2 ro n = 1; see Fig.I-7) in Ne.9+�E = !(13.60 eV)(10) [(1/2) ! (1/1) ] = 1020 eV = hν = hc/λ2 2 2λ = hc/
�E = (12398 eV-D)/1020 eV) = 12.2 D.
The spectrum of hydrogen consists of several series of lines each of which contains all
transitions that terminate at the same energy level. A diagram showing the most prominent
transition series is displayed in Fig. 7.
As a final point of interest, lets consider the electron to be represented by a standing wave
and deduce the condition that will assure this standing wave will exactly fit the circumference of
the electron orbit. Clearly this condition is that an integer number of wavelengths must equal the
orbit circumference;
nλ = 2π r .
According to the de Broglie relationship, λ = h/p and so the above condition becomes
nh/p = 2π r
or L = pr = (nh)/(2π ) , which is the Bohr quantum rule.
Unfortunately, the Bohr model proved to be incapable of describing multielectron atoms
and, as we shall see, it violates the uncertainty principle. Nevertheless, the Bohr radius and
energy equations, which are correct for one-electron atoms, are very useful for estimating atomic
radii and binding energies.
Pfund series
Brackett seriesPaschen series
Balmer series
Lyman seriesn = 1
n = 2
n = 3n = 4n = 5n = 8
- 13.6 eV
- 3.39 eV
- 1.51 eV- 0.85 eV- 0.54 eV
I-23
Figure I-7. Spectral series of the hydrogen atom.
C. The Development of Quantum Theory
1. The Uncertainty Principle
The concept advanced by de Broglie of waves accompanying the motion of a particle
leads inevitably to certain fundamental limitations concerning several combinations of variables,
such as position and momentum. Consider an idealized picture of a "de Broglie wave";
�E
�t $ £/2
I-24
The principal difference between a de Broglie wave and a conventional wave is the former must
be localized around the particle since the particle occupies a particular position in space.
Nevertheless, insofar as the wave-like properties of the particle are concerned, the position of the
particle must be uncertain by an amount equal to the extent of the wave �
x.
Mathematically, a localized wave packet can be constructed by combining harmonic
waves having different wavelengths (recall wave interference). However, the more localized the
wave packet, the larger must be the range of wavelengths used in its construction. Since the
wavelength is inversely proportional to the momentum, the range of wavelengths needed to
localize a wave packet to a length �
q is associated with a range of momenta �
p (or momentum
uncertainty). A detailed mathematical analysis shows the formal relationship between �
p and �
q
is
Heisenberg Uncertainty
Principle.
This relationship imposes a fundamental limitation on the certainty with which both the position
and momentum of a particle can be known simultaneously.
Another form of the uncertainty principle is
where E is energy and t is time.
Exercise: Suppose we wish to define the trajectory of an electron in the 1st Bohr orbit of a
hydrogen atom(r = 0.529D). We must establish both its position and velocity at a
particular instant in time. Assume we make a measurement to establish the location
of the electron with an uncertainty of ±0.05 D (i.e., ±10%), what is the minimum
simultaneous uncertainty in the velocity of the electron?�p = £/(2
�x) = (6.62x10 J-s)/[(4)(3.14)(0.10x10 m)] = 5.0x10 kg-m/s-34 -10 -24�
v = �
p/m = (5.0x10 kg-m/s)/(9x10 kg) = 5.6x10 m/s.-24 -31 6
Since the actual average velocity of the electron in the lowest energy state of the hydrogen
ψ (Pr)
I-25
atom is 2.19x10 m/s, we must conclude that the uncertainty principle rules out the6
possibility of describing atoms in terms of well defined electron trajectories or orbits.
2. The Schrödinger Equation
Once we accept the idea that the behavior of a submicroscopic particle is dictated by the
propagation properties of its associated wave, we are forced to seek a new way of describing
nature which is compatible with the de Broglie relation and the Heisenberg principle. It is not
physically meaningful to continue to describe such particles in terms of precisely defined
trajectories. Instead, an approach that uses the concept of probability and which combines the
particle and wave aspects of matter into one unified description, must be sought.
A description which satisfies the above criteria was developed independently by
Schrödinger and Heisenberg in 1926. The methods both involve finding "acceptable" solutions
to an equation of motion deduced by analogy from the mechanics of waves. The formalism
developed by Schrödinger is mathematically simpler to understand.
We shall begin by accepting the Schrödinger equation as the first postulate of quantum
mechanics. (This equation is to quantum mechanics what F=ma is to classical mechanics).
Since we will deal exclusively with time independent forces in this course, it is sufficient to state
the time independent Schrödinger equation:
where L = the Laplacian operator = M /Mx + M /My + M /Mz (in Cartesian coordinates) 2 2 2 2 2 2 2
V(r) = the time independent potential energy
E = the total energy
= the wavefunction.
By solving the Schrödinger equation, one seeks to determine both the wavefunctions and the total
energies of the system.
3. Interpretation of the Wavefunction
The more general form of the Schrödinger equation for a time dependent force contains
&1
I-26
Postulate #2 (the Born postulate): If at instant t a measurement is madeto locate a particle whose wavefunction is ψ (x,y,z), the probabilityP(x,y,z)dτ that the particle will be found within volume dτ = dxdydz isP(x,y,z)dτ = ψ (x,y,z)ψ (x,y,z)dτ .*
Working with complex functions: Consider the function f(x) = e . ix
Its complex conjugate is f (x) = e , and the square of the function is * !ix
f (x) = f (x)f(x) = e e =1.2 * !ix ix
If f(x) = ig(x), then f (x) = !ig(x) and f (x) = f (x)f(x) = !i g (x) =* 2 * 2 2
g (x). Hence, the square of a complex quantity is a positive, real2
quantity.
the term i£ [MΨ
(r ,t)/Mt], where i is the imaginary number . Consequently, its solutionsΨ(r ,t) are complex functions and therefore cannot represent real, physically significant quantities.
By analogy with the mechanics of waves, in which the square of the wave amplitude is
proportional to the intensity, Born proposed that the square of the quantum mechanical
wavefunction is proportional to the probability density.
.
In other words, ψ ψ represents the location probability density.*
In order to properly represent the location probability density, the wavefunction must
satisfy the condition
,
which is simply a statement that the total probability of finding the particle somewhere is unity.
Proper normalization of the wavefunction can be achieved through multiplication by a
normalization constant. Suppose u(r ) is a wavefunction that is not normalized. Then the
normalized wavefunction may be written asψ (r ) = Nu(r ), where N = normalization constant.
all space
ψ ((r)ψ (r)dτ' N 2
all space
u((r)u(r)dτ' 1
φ θr
z
y
xrsin
θ
N '1
all space
u((r)u(r)dτ 1/2
a
0
ψ ((x)ψ (x)dx '
a
0
ψ 2(x)dx ' A 2
a
0
sin2 nπ xa
dx
A 2 x2
!a
4nπ sin 2nπ xa
a
0
' A 2 a2
' 1; so A '2a
1/2ψ (r)' N e&r/a0
I-27
The normalization constant may be determined as follows:
Thus,
.
Exercise: As we shall soon find out, the wavefunctions for a particle constrained to a 1-D box
of length a are ψ (x) = A sin (nπ x/a) where n is an integer.
Determine the normalization constant A.
= 1
.
Exercise: One of the wavefunctions for the hydrogen atom is
, where a = Bohr radius (0.529 D).0
Determine the normalization constant for this wavefunction.
The hydrogen atom is a spherically
symmetric system and so the natural
coordinate system for this problem is a
spherical polar coordinate system. First
we will review the characteristics of this
coordinate system, as indicated below.
x = r sin θ
cos φy = r sin
θ sin φ
z = r cos θ
z
y
x
d θd φdr
N 2
4
0
r 2e&2r/a0dr
π0
sinθ
dθ 2π
0
dφ ' 1
2π0
dφ ' 2ππ0
sinθ
dθ
' [!cosθ
]π0 ' ![!1!1] ' 2
4π N 2
4
0
r 2 e&2r/a0 dr ' 1
I-28
dτ = r sin θ dr dθ dφ2
We must now determine the volume
element dτ in spherical polar coordinates.
Consider the volume between r and r+dr,θ and
θ+d
θ, φ +dφ ; dτ = dR x dw x dh
where dR = rdθ
dw = r sin θ
dφ dh = dr.
Hence
Now we can continue with the normalization problem. Inserting the wavefunction and
the expression for the volume element into the normalization integral gives the triple integral
.
The last two integrals are common to many problems and so it is worthwhile to
remember the following result:
,
so
.
N '1π a3
0
1/2
.
4
0
x n e&ax dx 'n!
an%1for n and a > 0
4π N 2 2
2a0
3' π a3
0 N 2 ' 1
P '1π a3
0
a0
0
r 2 e!2r/a0 dr
π0
sinθ
dθ 2π
0
dφ ,
x ne!axdx ' !e!ax
an%1[(ax n) % n(ax)n!1 % n(n!1)(ax)n!2 % .... n!
P '4
a30
!a0r
2
2%
a20 r
2%
a30
4e!2r/a0
a0
0
P ' !4
a30
5a30
4e!2 !
a30
4' 1 ! 5 e!2 ' 0.323
I-29
(Note: The above integral may be solved by twice applying the method of integration by
parts). The definite integral has the general solution
.
Hence
and
Exercise: Calculate the probability of finding the electron in the region of space between r = 0
and r = a using the above wavefunction.0
4. Mathematical behavior of the wavefunction
If the square of the wavefunction is to be interpreted as probability density, the
wavefunction must satisfy the following conditions:
I-30
Conditions for an acceptable wavefunction
The wavefunction ψ (q) and its first derivative dψ /dq mustbe finite, continuous and single valued.
Examples: ψ (x) = Ax
This function is not acceptable because ψ 6 4
as x 6 4 . ψ (x) = Ae!x
This function is not acceptable because ψ 6 4 as x 6 !4
What about the function ψ (r) =Ae ?!r
What about the function shown to the right?
5. Operators and observables
An operator performs a mathematical operation on a function. An operator you are
already familiar with is the differential operator, d/dx. An example of an operator equation is the
following:
Pp ' mPv ! i £ PL
!£ 2
2mL2
I-31
Eigenvalue Equation
F f = α fop
where f is the eigenfunction and α is the eigenvalue.
Postulate #3: a). Every observable (i.e., measurable physical quantity such as position, energy,
momentum, etc.)may be represented by an operator.b). If an observable has a precisely defined value, its operator satisfies the eigenvalue
equation F ψ = a ψ ,op
where ψ is the wavefunction and a is the value of the observable.
Let F = d/dx and f(x) = Ax . Then op2
F f(x) = d/dx (Ax ) = 2Ax.op2
An eigenvalue equation is an operator equation having special significance in quantum theory.
Exercise: Determine whether the function e is an eigenfunction of the operator d/dx.ax
Several operators commonly encountered in quantum theory are listed below.
Classical observable Quantum mechanical operator
position x x (multiply by the coordinate)
momentum p = m (dx/dt) !i£ (d/dx)x
kinetic energy E = ½mvK2
The kinetic energy operator given above is easily obtained by expressing the classical kinetic
energy in terms of momentum
E = ½mv = p /(2m) ,K2 2
and then replacing the classical momentum with the corresponding quantum mechanical
operator;
(EK)op '(! i £ PL)2
2m' !
£ 2
2mL2
Hop ' (EK)op % Vop ' !£ 2
2mL2 % V
!£ 2
2mL2 % V ψ
' E ψx '
4
!4
x P(x) dx
<x> '
4
!4
x ψ (ψ dx
I-32
.
In classical mechanics, the total energy expression E = E + V is called the Hamiltonian . LetsK
write the corresponding quantum mechanical Hamiltonian (total energy) operator;
.
(Note that since the potential energy only depends on coordinates, V is the same as V). Usingop
the Hamiltonian operator, we see that the Schrödinger equation is an eigenvalue equation
and the total energy is the eigenvalue;
,
or
.
Next we consider the problem of determining the average value of an observable, such as
position. The probability of finding a particle between x and x+dx in any single measurement is
P(x) dx = ψ ψ dx.*
Suppose we measure the position of the particle a large number of times. The average value of
the position may be calculated in the usual way;
,
but according to the Born postulate, P(x) = ψ ψ , and so the average value of the position, which*
in quantum theory is termed the expectation value and is denoted by brackets, is
.
<�
> 'all space
U ((Pr)�
op U(Pr) dτall space
U ((Pr) U(Pr) dτψ (r) '1π a3
0
e!r/a0 and dτ ' r 2 sin
θdr d
θdφ
so <r> '1π a3
0
4
0
r 3 e!2r/a0 dr
π0
sinθ
dθ 2π
0
dφ<r> '
4
a30
4
0
r 3 e!2r/a0 dr
I-33
Postulate #4: The expectation value of any observable is given by
,
where � is the quantum mechanical operator representing the observable.op
Notice that the operator must operate on the wavefunction and hence it is placed between ψ and*ψ in the above equation. You should also note that if the wavefunction is not normalized, the
above integral must be divided by the normalization integral;
.
Exercise: Calculate the expectation value of the radius of an electron in the ground state of the
hydrogen atom.
<r> = (3/2) a = 0.794 D .0
6. The commutator
In the multiplication of numbers, the order does not matter. For example, 5 x 10 = 50 =
I-34
10 x 5. With operators, however, this is not necessarily true. An example of this is provided by
the operators A = x and B = d/dx. Does A B f(x) = B A f(x)? We proceed as follows:op op op op op op
A B f(x) = A [B f(x)] = x[(d/dx)f(x)] = x(df/dx)op op op op
B A f(x) = B [A f(x)] = (d/dx)xf(x) = f(x) + x(df/dx).op op op op
Hence A B f(x) … B A f(x)op op op op
or
[A B ! B A ]f(x) = !f(x) … 0.op op op op
We may consider [A B ! B A ] to be a new operator, called the commutator. In the presentop op op op
example,
[A B ! B A ] = !1.op op op op
In quantum mechanics, a special case of interest is when f(x) is an eigenfunction of both
operators. Then
A f(x) = α f(x) and B f(x) = βf(x),op op
so
[A B ! B A ]f(x) = α βf(x) !
β α f(x) = 0op op op op
or
[A B ! B A ] = 0.op op op op
When the above condition is true, the two operators are said to commute. Recall that
observables whose quantum mechanical operators and wavefunctions satisfy eigenvalue
equations are exactly determined. Therefore, if we wish to determine the values of two
observables simultaneously and exactly, their operators must commute.
Exercise: Determine if the following (one-dimensional) observables may be determined
simultaneously and exactly: (a) momentum and kinetic energy; (b) position and
momentum.
!i£ddx
!£ 2
2md2
dx 2f(x) ! !
£ 2
2md2
dx 2!i£
ddx
f(x)
'i£ 3
2mddx
d2
dx 2!
d2
dx 2
ddx
f(x) ' 0
!i£ ddx
(x)f(x) ! (x) !i£ ddx
f(x) ' !i£ f(x) % x df(x)dx
! x df(x)dx
' !i£f(x)
I-35
a).
Yes.
b).
No.
I-36
Review Questions
1. Sketch the shape of the frequency distribution for black body radiation at two different
temperatures.
2. What condition must be met by a standing wave in a cubical box of length L?
3. What was the ultraviolet catastrophe and what was its impact on the history of science?
4. What was wrong with the Rayleigh-Jeans derivation of the formula for the energy density of
black body radiation and what postulate did Planck make that led to the correct answer?
5. Transform the Planck radiation formula for energy density as a function of frequency, U(ν )dν= (8π hν /c )(e -1) dν , to energy density as a function of wavelength.3 3 hν /kT -1
6. Draw a sketch illustrating the photoelectric effect and write the equation for the kinetic
energy of a photoelectron.
7. Calculate the binding energy of an electron ejected from an atom with a kinetic energy of 523
eV by a photon having a wavelength of 8.343 D.
8. Draw a sketch illustrating the Compton effect and briefly describe how to go about deriving a
formula for the wavelength of the scattered photon.
9. How much energy does a 4.00 keV x-ray lose when it scatters from an electron at an angle
of 135 ?o
10. Calculate the wavelength of a bullet weighing 5 g and traveling at a speed of 1000 ft/s.
11. Derive an equation for the angular momentum of an electron in a hydrogen atom by
requiring that its orbit be the proper size to accommodate a standing wave having a
wavelength given by the de Broglie formula.
12. (a). Calculate the binding energy of an electron in the n = 3 level of an Ar ion.17+
(b). Calculate the Bohr radius of the electron in part (a).
(c). Calculate the wavelength of a photon emitted in a transition between the n = 3 and n =
1 levels.
13. State the Heisenberg uncertainty principle and explain its meaning.
14. Why is it not meaningful to think of electrons as residing in well defined orbits around
atoms?
I-37
15. Calculate the kinetic energy (in MeV) of an electron having a wavelength equal to the
diameter of an average nucleus; 10 cm. -12
16. Write the general form (i.e. using the Laplacian operator) of the time independent
Schrödinger equation.
17. What is the Born postulate with regard to the wave function of a particle?
18. Determine the normalization constant for a hydrogen atom 2s wave function;
ψ = N (2 - r/a )exp(-r/2a ).2s 2 o o
19. What are the requirements for an "acceptable" wave function?
20. Which of the following could be acceptable wave functions over the range
-4 < x < 4: (a) A exp(-x ), (b) iB/x , (c) -Cx , (d) A cos(x) exp(-x ).2 2 2 2
21. What is an eigenvalue?
22. Calculate the expectation value of r for a hydrogen atom 2s electron (see prob. 19 for ψ ) and
compare it to the value predicted by the Bohr formula.
23. Can the momentum and the total energy of a particle be determined exactly and
simultaneously?
Answers
2. nλ /2 = L
5. U(λ )dλ = (8π hc/λ )(e - 1) dλ 5 hc/(λ kT) -1
7. 963 eV
9. 53 eV
10. 4.34x10 m-34
11. L = n£
12. 489.6 eV, 0.265 D, 3.165 D
15. 1.504x10 MeV4
18. N = (32π a )2 o3 -1/2
20. (a). yes, (b). no, (c). no, (d). yes
22. <r> = 6a , 4ao o
λ)!
λ'
hm0c
(1!cosθ )
!£ 2
2mL2 Ψ % V Ψ ' EΨ
P(Pr) ' Ψ ((Pr) Ψ (Pr)
all space
Ψ ( Ψ dτ' 1
I-38
FORMULAS YOU SHOULD KNOW Chapter 1
Waves:λ = v/ν ; for electromagnetic radiation
λ = c/νω = 2π ν (angular frequency)
k = 2π /λ (propagation constant)
A = A sin (kx ! ω t)o
Photoelectric effect: E = hνE = hν ! Be
Compton scattering:
de Broglie wavelength:λ = h/p
Bohr model:V = !Ze /r (electron - nucleus potential energy)2
r = a n /Z02
E = -E (Z/n)02
Uncertainty principle: � p � q $ £/2� E � t $ £/2
Schrödinger equation:
Probability density:
Normalization:, where dτ = dxdydz (cartesian)
dτ = r sin θ dr dθ dφ (polar)2
N '
all space
U (Udτ !1/2
Hop ' !£ 2
2mL2 % V
<�
> '
all space
Ψ (�
opΨ dτ
I-39
Operators:coordinate; x, y, z, r, etc
momentum; p = !i£ M/Mx, etc.x
Hamiltonian;
Eigenvalue equation: F Ψ = a Ψop
H Ψ = E Ψop
Expectation value:
Commutator: [A B ! B A ]op op op op
