Unit
Introduction to Quantum Physics
01
Quantum Basics
02
Amplitudes & Probabilities
1
Vectors & Superposition
2
Operators & Observables
3
Harmonic Oscillator
4
Atoms, Molecules, and Chemistry
03
Higher Dimensions & Angular Momentum
5
The Hydrogen Atom
6
Matrix Mechanics
7
Time Development
8
Summary of Quantum Theory
S
APPENDICES
Plane Waves
A1
Complex Numbers
A2
On Differential Equations
A3
Simplifying Constants
A4
Classical Dynamics
A5
Hydrogen Eigenfunctions
A6
Identical Particles
A7
Delta Function
A8
Perturbation Theory
A9
Scattering
A10
Band Theory of Solids
A11
Fundamental Particles
A12
01 INTRODUCTION TO QUANTUM PHYSICS
The fundamental theories of classical physics were developed before the nineteen hundreds. The
usefulness of these theories has not diminished with age. However, classical theories do not
describe the machinery of the universe at the fundamental levels of atoms, molecules, and
elementary particles. This is the province of modern physics or quantum physics.
Quantum theory is the ultimate physical theory that evolved after centuries of scientific
endeavor. Many physicists believe that an appropriate combination of quantum physics with
relativity will produce a “theory of everything" to describe all phenomena.
Mechanics, electrodynamics, and thermodynamics, are approximations to quantum theorywe
say that they are “included” in quantum theory.
Light is a Wave
We know that light is a wave of electric and magnetic fields. One demonstration that shows light
is a wave is the double slit experiment. Here light comes through two parallel slits. When the
light reaches the screen, it forms a pattern of bars. Waves come from both slits and interfere at
the screen; when two crests meet, they form a higher crest; a bright line. When a crest and
trough meet, they cancel and produce a dark line. The results clearly demonstrate that light is a
wave.
Light is a particle
In order to explain the photoelectric effect, Einstein had to interpret light as little lumps or
particles of energy -- photons.
The photoelectric effect works this way: blue light hits a particular metal and electrons fly out.
Red light hits metal and no electrons are ejected -even when the red light is intensely bright!
Einstein's explanation was that blue light (high frequency) has very
energetic photons that can tear electrons out of the metal. Red light
(low frequency) has less energetic photons that are unable to eject
electrons.
E
photon
 hf

True or False
For a particular metal, it is possible that green light will eject electrons, but ultraviolet waves will
not.
Matter-Wave Duality
De Broglie proposed that light should not be the only thing that exhibits
both wave and particle properties. He said all waves have a particle aspect
and that all particles have a wave aspect. In short, an object is both a
particle and a wave and it reveals one property or the other depending upon
the circumstances.
h 

  mv 


This conjecture was verified by shooting a beam of electrons through a crystal that serves as a
kind of “double slit.” A pattern appeared on a screen showing characteristic wave interference.
True or False
a) Light exhibits both wave and particle behaviors.
b) Protons exhibit both wave and particle behaviors
Bohr Atom
Bohr first calculated the energy levels of a hydrogen atom by imposing a non-classical
requirement on the orbital angular momentum of the electron. The angular momentum (rp or
rmv) was restricted to integer multiples of a basic number called Planck’s constant, . [rmv = n]
The theory predicted that the atom could realize only discrete orbits and energy levels.
The restriction on orbits may be
more readily seen as the result of an
electron-wave fitting in standing
patterns around the nucleus at just the
appropriate distances. Waves may be
imagined to look something like a
string of sausages so that more
sausages (wavelengths) can fit around
larger orbits as seen in the diagram.
.
Schrödinger Equation
The de Broglie equation is correct for free particles, but was found to be inadequate for particles
affected substantially by interactions. Schrödinger constructed an equation to fully describe the
“matter wave” for any system,
.
The Schrödinger Equation is the basic equation of non-relativistic quantum theory. There are
two kinds of information given by the equation, eigenvalues and eigenfunctions:
a)
eigenvalues
When applied to bound atoms and molecules, the Schrödinger equation predicts that the
system has definite, discrete, energies:
El. E2, E:3, ....
No solutions exist for any energies “in-between.” This is in sharp contrast with classical
dynamics where there is a continuum of allowed energy values. The equation also
determines that other quantities such as angular momentum and spin have similar discrete
values. However solutions may also include quantities that are not discrete such as the
momentum of a free particle.
The above quantities that are given by the Schrödinger equation are examples of eigenvalues; the
only values which the system can exhibit. In general, we look to the Schrödinger equation to
provide the appropriate eigenvalues for a specific system.
True or False
(a) Eigenvalues are numbers representing physical properties.
(b) Eigenvalues are always discrete.
(c) Eigenvalues represent all possible characteristics of a system.
(d) Only two energies can be determined from the Schrödinger equation when it is applied to an
ammonia molecule in a magnetic field. This means that these two energies and no others are
possible for this molecule
b)
eigenfunctions
A state is characterized by a set of eigenvalues. For each state of the system, the
Schrödinger equation gives a corresponding eigenfunction
.
Each eigenfunction is associated with specific set of observables like energy, angular
momentum, and spin. The eigenfunctions belong to a more general category of waves
called wave functions that are just combinations of eigenfunctions. We have now to
interpret the wavefunctions.
Born Interpretation
Physicist Max Born was first to correctly interpret the wavefunction. Schrödinger and others
thought  was just a measure of the spread of a wavy particle. This belief had flaws that Born
pointed out. Suppose that a single electron is shot through double slits. If the pattern that results
has bright and dark fringes, then the electron has been divided into pieces.
However, electrons are never seen in parts. It must be that the particle itself is not actually an
extended wavy lump. Rather, it is always a point particle when it is detected and wave patterns
must be the result of a statistical accumulation of many particles. The wave must be interpreted
as the probability that the particle is in a particular position or state. (The larger the wave
function at a point, the more likely is the particle to be at that location.)
True or False
Born interpreted eigenfunctions as extended wavy particles.
More on Wavefunctions
When Schrödinger invented the wavefunction, he expected it to be something physical like the
density of an extended particle. It turned out not to be a material substance. Rather, it is a
ghostly probability wave that dictates the positions and outcomes of physical systems.
 is not the probability itself. The probability of a state is found from the square of the wave
function (much like the energy of an electromagnetic wave is given by the square of its fields).
question
The diagram shows the eigenfunction for the location x of
an electron in a particular molecule.
(a) Which labeled position is the most likely
neighborhood for the electron?
(b) The electron is never at which of the labled points?
Superposition of Waves
An important property of waves is that they “add” (or “subtract”). This is called
superposition of waves. Superposition has profound implications for quantum physics
If 1 and 2 are eigenfunctions for two possible states, then the system may exist with a
wavefunction  that is a combination like 12 1  2  and 53 1  54 2 . However, the only
values that we can obtain from a single measurement are eigenvalues associated with 1 or 2 ,
and the probability that the system will be found in condition 1 or 2 depends on the proportion of
1 and 2 represented in the combination (superposition).
Example (Schrödinger’s Cat):
A cat spends several minutes in a sound proof box with an atomic device that has a 50%
chance to kill the cat. Consequently, the cat is 50% live and 50% dead According to the
conventional quantum interpretation, the cat becomes alive or dead (according to chance)
when the box is opened for observation.
questions
A photon strikes a glass pane. The eigenfunctions for the photon to be transmitted through or
reflected from the pane are t and r, respectively.
The wavefunction of the photon is
  0.6t  0.8r
(a) Is this photon more likely to be transmitted or reflected?
(b) In the event that the photon was transmitted, what has its wavefunction become?
Summary of Quantum Theory
Our exposition may obscure the simple fundamentals of quantum physics, so it is desirable
to summarize them here as a kind of “how to” list.
a)
Given a physical system like an atom or a crystal, write the Schrödinger equation.
This is done with a simple set of rules that transforms an energy relation for the
system into a differential equation.
b)
Solve the Schrödinger equation to find the eigenvalues and eigenfunctions for the
system. The eigenvalues are the only physical quantities that the system can exhibit
and corresponding eigenfunctions determine the likelihood that a specific state (set of
eigenvalues) will be found at a given position.
c)
Two specifics determine most applications of the wavefunctions:
(i) systems may be described by wavefunctions that are superpositions of eigenfunctions and (ii)
probabilities depend on the square of a wavefunction, . (The first of these allows alternative
possibilities to exist and the second allows the alternatives to interfere.)
Some Philosophical Features of Quantum Physics
Quantum mechanics has a profound influence on the philosophy of nature. Indeed, it has altered
our view of objective reality and classical determinism.
In quantum theory, what you know is what you measure (or what some physical system
“records”). The acts of measurement and observation can create the resulting state. A system
does not have a definite value for a quantity until it is observed. Thus an electron is given a
specific spin by an observation; before this, it had only potential spins. A photon in the double
slit interference experiment does not pass through a single slit unless we try to detect that slit
passage and Schrödinger's cat is not dead or alive until an observation makes it so. Quantum
physics does not accept objective reality independent of observation or interaction.
We see that quantum physics differs from classical physics In another basic philosophical sense.
Classical physics is deterministic in that when enough initial information is specified, the
consequences can be predicted with certainty (the clockwork universe). Quantum theory shows
that for a given initial situation, nature can “choose” among alternatives according to predictable
probabilities. The theory asserts that nature is indeterministic.
A related deduction in quantum physics is the uncertainty principle that
says that a particle’s position and velocity can not both be measured with
complete accuracy at the same time. One must sacrifice accuracy in one
to increase accuracy in the other. A similar uncertainty exists between
the energy and lifetime of a particle.
x p  h 
t E  h


Another remarkable feature of quantum physics is that the outcome of a process is influenced by
other possible outcomes that are not themselves realized.
02 QUANTUM BASICS
Before you study quantum mechanics in a more formal and systematic way, it helps to learn a
few basic equations, rules, and examples that are used throughout. Here the topics of the last
unit are repeated, but with associated equations and problems.
Light is a Wave
For the present, the most important expression associated with the wave nature of light is the
relation between wave velocity v, wavelength  and frequency f:
v f
(1)
problem 1.
Find the frequency of light with 4.5 X 10-7 m wavelength. (c = 3.0 X 108 m/s.)
Background Note
Constructive interference of two coherent waves causes the bright fringes in the double slit
experiment. This occurs when the two light rays differ only by an integer number n of whole
wavelengths:
path difference = n  constructive interference
Destructive interference causes the dark fringes in the double slit experiment. In this case the
two rays differ by odd half-integer numbers of wavelengths:
path difference = (n + ½)  destructive interference
Light is a particle
Einstein's interpretation of light as a particle depended in part on Planck’s theory of blackbody
radiation. Planck was able to describe how the amount of radiation from a heated object depends
upon the wavelength and temperaturebut only when he assumed that energy was emitted and
absorbed in “lumps” proportional to the frequency of the radiation,
(2)
E  hf
where the proportionality constant h is a universal constant now called Planck's constant.
(3)
h = 6.626196 X 10-34 J-s
Equation (2) was taken by Einstein to be the expression of photon energy, and this is the
accepted view today.
problem 2.
Calculate the frequency of a photon emitted when an electron falls from a hydrogen
energy level of -3.4 eV to -13.6 eV. (1 eV = 1.60 X 10-19 J)
In the photoelectric effect, the photon energy hf went into the work W needed to rip the electron
from the metal, and into the kinetic energy K of the freed electron,
(4)
hf  W  K
W is called the work function and is different for different metals. K in Eq. (4) represents the
maximum possible kinetic energy of the electron; as it may be slowed by collisions in the metal.
problem 3.
The work function of Aluminum is 4.25 eV. Calculate the maximum kinetic energy of
electrons ejected from Aluminum when irradiated by an ultraviolet beam of wavelength
2.10 X 10-7 m. (1 eV = 1.60 X 10-19 J)
Matter-Wave Duality
De Broglie expressed the wavelength  of the matter-wave in terms of the momentum of the
particle, p = mv:
h
(5)

p
problem 4.
The momentum of a photon is known to be hf/c. Use this fact and the de Broglie relation
to find an expression for c in terms of f and 
The following “particle-in-a-box” is a favorite model problem in quantum mechanics.
problem 5.
A particle of mass m is confined in a box of length L such that its de Broglie wave must
form a standing wave. This can be done only when half-integer wavelengths fit the box.
Use Eq. (5) and the fact that the energy (all kinetic) can be written as
p2
2m
n2h2
and derive the allowed energy levels E n 
.
8mL2
E
Niels Bohr found the correct energy levels of hydrogen in 1913 by proposing that the electron
can only exist in orbits allowed by his “Bohr quantization rule” for angular momentum :
nh

2
This rule is easily derived from the deBroglie relation.
problem 6.
Derive the Bohr quantization rule for angular momentum by requiring that an integer
number n of de Broglie wavelengths fits around the perimeter of a circular orbit
n  2r  .
Schrödinger Equation
The de Broglie relation is accurate for particles that are not subjected to external forces, but it
was found to be a special solution of the more general equation of quantum mechanics, the
Schrödinger equation.
To construct the Schrödinger equation for 1-dimensional systems, follow these rules:
i) Write the energy relation for the system of interest. The kinetic energy K must be written in
terms of the momentum p rather than velocity v,
p2
K
2m
The total energy E including the potential energy V(x) is then given by K(p) + V(x) = E or
p2
(6)
 V x   E
2m
ii) Replace the momentum p in Eq. (6) by a differential operator D 
p   i D
where  (called “h-bar”) is the Planck constant divided by 2,
h
(7)

2
Equation (6) is then converted into the operator form
d
:
dx
(8)

2 2
D  V x   E
2m
iii) Produce the 1-dimensional, time-independent Schrödinger equation by applying both sides of
Eq. (8) to the wavefunction  The result is
(9)

2 2
D   V x   E
2m
Schrödinger Wave Equation
The following three problems are exercises in writing the wave equation. It will be important to
solve these and other differential equations in later chapters. Here you are given trial solutions to
verify.
problem 7.
Follow the rules above to write the Schrödinger equation for a free particle, V = 0. Show
by substitution that   Asin kx is a solution and determine the relation between E and
k. (A and k are constants.)
problem 8.
Follow the rules above to write the Schrödinger equation for a particle in a constant
potential V = V0, where E > V0. Show by substitution that   Aexp i k x  is a solution
and determine the relation between E, V0, and k. (A and k are constants.)
problem 9.
Write the Schrödinger equation for a simple harmonic oscillator, V  12 m 2 x 2 . Show by
substitution that   A exp  ax 2  is a solution and determine the value of E that
accompanies this wavefunction. (A and a are constants.)
Particle in a Box
The particle-in-a-box problem illustrates how the solution of the Schrödinger equation produces
a set of energy eigenvalues E1 , E2 ,and corresponding wavefunctions 1 , 2 , This is
done by writing the particle's wave equation and finding solutions for  that satisfy conditions
imposed by the probability interpretation.
There is no potential energy inside the box, so the wave equation is the same as in problem b-7.
The trail solution x  A sin kx will satisfy the equation, but it must also be made to satisfy
the probability conditions.
The probability that a particle is at position x with energy En is proportional to the square of the
2
wavefunction, n . Consequently, if a particle is confined to a box of length L,
0  x  L,
it must have zero probability of being outside and we require that
0  0 and L  0
(10)
These are the boundary conditions that must be met by the trial solution.
problem 10.
Write the wave equation for a particle-in-a-box and calculate the energy eigenvalues and
wavefunctions. Use a trial solution of the form   Asin kx .
1. AMPLITUDES & PROBABILITIES
The notation and principles of quantum theory are introduced in the next few units. We illustrate
these with two simple examples; a one-dimensional particle-in-a-box and a bead on a circular
ring.
Dirac Notation and Amplitudes
Dirac notation embraces the quantum philosophy that what you know is what you measure.
Consider the one-dimensional particle-in-a-box with allowed energy levels En. The state of this
system is completely characterized by the energy level and the Dirac notation for the nth state is
|En>.
Writing the symbol |En> infers that we have complete knowledge of the particle's nth energy
level, but given this knowledge, what is known about the position x of the particle? The Dirac
symbol for what we want to measure is <x| and the notation that expresses that we want to know
the value of x when we have prior knowledge that the system has energy En is called an
amplitude and is written
x En
“bra”
“ket”
what we want to know
what we know
problem 11.
The Dirac notation above represents the amplitude for a particle with energy En to have
position x. (a) Write the Dirac notation representing the amplitude for the particle with
energy En to have momentum p. (b) A bead on a ring can be completely specified by its
momentum p. Write the Dirac notation representing the amplitude for the bead with
momentum p to have position x. (c) For the bead in part b, write the Dirac notation
representing the amplitude for the bead with momentum p to have energy E.
Summary:
When a system is known to be in a state  the event of finding it in a condition  is
characterized by an amplitude written as   .
Probability Principle
The probability for an event to occur is the absolute square of the amplitude for that
event. When a system is characterized by |>, the probability of it being found in the
condition |> is
(11)
P |     
2
 discrete
When  is a continuous (instead of discrete) quantity, P |   is a probability density
and P |   d is the probability of a measurement yielding a value in a neighborhood
d of .
(12)
P |   d   
2
d
 continuous
For the particle-in-a-box amplitude <x|En>, the probability of a particle with energy En being
found in a neighborhood dx around the point x is |<x|En>|2dx. Amplitudes of the form <x|> that
are probability densities for position and are called wave functions. Not all amplitudes are wave
functions; <p|E> and <E1|E2> are counterexamples.
problem 12.
Use probabilities or probability densities where appropriate and identify any amplitudes
that are wave functions:
(a) Write the Dirac notation representing the probability for a particle with energy En to
have momentum p.
(b) Write the Dirac notation representing the probability for a particle with energy E2 to
be found with different energy E1.
(c) A bead on a ring can be completely specified by its momentum p. Write the Dirac
notation representing the probability for the bead with momentum p to have position x.
The amplitude for position (a wave function) must be a continuous, and single-valued function of
position x.
problem 13.
Given that the wave function for a bead on a ring of circumference L is
1
 2inx 
x p 
exp 
,
L
 L 
find (a) the probability of an n=-3 particle being between x=0 and x=L/4 and
(b) the probability of particle with arbitrary n being in the interval between x=0 and x=L.
problem 14.
Given that the wave functions for a particle in a box of length L are
2
 nx 
x En 
sin 
,
L  L 
find (a) the probability of an n=2 particle being in the “middle” between x=L/4 and
x=3L/4 and (b) the probability of an n=1 particle being in the same interval. Sketch the
probabilities for both cases and comment on the differences. ans:1/2, 1/2 + 1/
2. VECTORS & SUPERPOSITION
Kets describe that which is known about systems and the mathematics obeyed by kets is that of
vectors. This unit describes the vector properties and their physical consequences. For the
present, we continue to treat amplitudes like <x|p> as “given” information instead of deriving
them.
Review of Vectors
A linear vector space is a set of objects called vectors a, b, c,... for which the following
operations are defined:
1. Addition: the sum of two vectors is a vector and addition is commutative and associative
ab ba
a  b   c  a  b  c 
2. Multiplication by a Scalar: multiplication by any complex number is distributive and
associative.
a  b   a  b
   a  a  a
3. A null vector 0 exists in the space such that
a0 a
4. Every vector a has an additive inverse -a such that
a   a  0
problem 15.
Show that the following are vector spaces: (a) displacements in 2-D space and
(b) the set of all continuous functions f(x) defined on the interval (0,L).
Base Vectors
Any two-dimensional vector v can be written as a sum of the form v = vxi + vyj. Generally, when
any vector in a vector space can be expanded similarly with a minimum of N vectors e1, e2, ...,
eN,
(13)
v   ci e i ,
the space is said to be N-dimensional. The c's are called components and the e's are called base
vectors. When Eq.(1) is satisfied for all vectors in the space, we say the base vectors are
complete. A set of base vectors is a basis of the space.
problem 16.
A vector in 2-dimensional Euclidean space has a magnitude of 10 units and is inclined at
60o above the x-axis. (a) Expand this vector in unit vectors i and j. (b) Write this vector
in column matrix form and expand it in column vectors representing i and j.
The Fourier theorem shows that a function f(x) defined on the interval (0,L) can be expanded in
an infinite series of sine and cosine functions. In general,
  nx 
  nx 
f x    a n cos
  bn sin 

 L 
 L 
(14)
This can be viewed as a special case of Eq.(1).
problem 17.
Identify the base vectors in Eq.(2). What is the dimension of the function space?
Scalar Products
A scalar product of two vectors |a>, |b> is written <a|b> and has the following properties:
1. <a|b> = <b|a>*
2- <a|b+c> = <a|b>+<a|c>
3. <a|a> > 0 and <a|a> = 0 implies |a>=0
Two vectors whose scalar product is zero are said to be orthogonal. The length of vector |a> is
defined as <a|a>½
problem 18.
Write definitions of the specific scalar products for the following: (a) Euclidean vectors,
(b) column vectors, (c) functions f(x). In each case, demonstrate that the scalar product
properties are satisfied.
problem 19.
Calculate <u|v> for the following: (a) u=5i cos60-5j sin60 and v=10i (b) u= (4, -1, -5)
and v= (2, -3, 2) (c) u= (1+i, 1-2i) and v= (2, -i) (d) u  sin x / L and v  sin 2x / L
on the interval (0,L) (e) u  exp 2ix / L and v  exp 4ix / L on the interval (0,L)
problem 20.
Calculate the lengths of the u vectors in the previous problem.
problem 21.
Show that u=5i cos60-5j sin60 is orthogonal to v=i cos30+j sin30. Which of the vector
pairs in Prob.2 are orthogonal?
Unit Vectors
Ordinary Euclidean vectors are easily treated when they are expanded in terms of unit vectors i,
j, k. Similar useful expansions are also available for more general vectors—including functions.
Notice that i, j, k are orthogonal and have unit lengths. In the scalar product notation; <i|j>=0
and <i|i>=1, etc. In order to have base vectors that are as convenient as these, it is usual to use a
basis that is both orthogonal,
en em  0
(15)
and normalized to a unit length,
en en  1.
(16)
When the base vectors satisfy both Eqs.(3) and (4), we say the basis is orthonormal.
problem 22.
Show that i, j, k are orthonormal.
problem 23.
Show that the set of functions x p 
 2n ix 
exp 
 with n=0, 1, 2,... are
L
 L 
1
orthonormal.
problem 24.
Show that the set of functions x E n 
2
  nx 
sin 
 with n = 1, 2,... are orthonormal.
L  L 
Linear Vector Space Principle
Physical states |> are vectors in a linear vector space and amplitudes   are
associated scalar products.
This principle has profound consequences. First, we note from the definition of the scalar
product that
(17)
<a|b> = <<b|a>*,
indicating that forward and reverse events are closely related.
problem 25.
 2n ix 
exp 
 for a particle on a ring, write the amplitude for the
L
 L 
particle to be found with momentum pn when its position is known to be x.
Given that x p n 
1
A second, more remarkable consequence is the fact that a state |> of a system can be expanded
in terms of any base states |u1>, |u2>,... Thus the state of polarization of a photon is a sum of leftand right-polarized photons and the vitality of Schrödinger’s cat is a sum over the live and dead
cat states. This feature is called the superposition principle. It can be written symbolically as
   cn u n
(18)
n
where we will assume the base states are orthonormal.
problem 26.
A special case of Eq.(6) is |> = 0.8|u1> + 0.6|u2>.Show that the u1 component equals
<u1|> and the u2 component equals <u2|>. 2-14 Show that Eq.(6) can be written as
   un  un
(19)
n
  un
un 
n
problem 27.
Write the state |> of Schrödinger’s cat given that the amplitude for it to be alive is
Use the symbols |L> and |D> for the live and dead states.
2
3
.
The most succinct expression of the superposition principle follows from Eq.(7),
1   un
(20)
un
n
problem 28.
 2n ix 
exp 
 and Eq.(8) to expand an
L
 L 
arbitrary function f(x)=<x|f> in terms of these functions and coefficients <n|f>. The result
is a Fourier series.
Use the orthonormal functions x n 
1
summary:
Quantum states are vectors. As a consequence, an arbitrary state |> can be expanded in
base vectors |un> that represent a complete set of known states. It is then said that |> is
a superposition of |un> states. See Eqs.(6) and (7) for expressions of superposition.
Averages
As an example of averaging with probabilities, consider averaging quiz grades for an imaginary class of
10 students. The possible grades are 0, 50, and 100. One student got 0, four got 50, and five got 100. Here
is the class average:
g 
1  0  4  50  5  100
 70
10
Another way to look at the same average is:
g 
1
4
5
0  50  100 .
10
10
10
Notice that the probability of a student getting 0 is 1/10 and the probability of getting 50 is 4/10, etc. We
write
g  p0  0  p50  50  p100  100 .
This is a special case of a general way to take an average over g1, g2, when the corresponding
probabilities p1, p2, are known:
g  g j pj
j
In quantum mechanics we often have a continuous range of possible outcomes (think 0 to 100 with any
fraction between). Then we modify the average above using a probability density :
 x dx  probabilit y of system being between x and x  dx
In this case, average g is ,
b
g   g x   x  dx
a
We expect  in quantum mechanics to be *. A slight wrinkle occurs in quantum theory where we
make a “sandwich” of the thing being averaged so that it looks like this:
b
g    * g x   dx
a
This form works even when g represents an operator like momentum (  i
d
).
dx
3. OPERATORS & OBSERVABLES
Objects that transform vectors according to some rule are called operators. In quantum physics,
operators represent the instruments or objects that measure or otherwise record physical
quantities. An operator  (measuring device) is applied to a state  (system in a specific
condition) in order to render a value of the observable .
(21)
  
For example, the momentum operator P is applied to the bead-on-a-wire wavefunction in order
to measure the momentum of the corresponding system.
In quantum physics, the acts of measurement and observation create the resulting state. A
system does not have a definite value of a quantity until it is observed. Thus a bead-on-a-wire is
given a specific momentum by an observation; before this, it had only potential momenta.
Schrödinger’s cat is not dead or alive until an observation makes it so.
Linear Operators
A linear operator  transforms a vector |a> into another vector |a> according to a rule that
obeys the following linear relation:
(22)
 (c1|a> + c2|b>) = c1|a> + c2|b>
Operators may also operate “backward” on a bra-vector
(23)
when |a> = c|b> then <a| = c*<b|
problem 29.
A translation operator T is defined by T|x> = |x-b> or <x|T = <x-b|. Using the notation
x   x  , show that x T   x  b.
problem 30.
Given x p  exp ixp /   / L , show that the effect of T can be expressed in momentum
language as
1
pT  
 exp ixp /  x  bdx
L
Eigenvalues and Eigenvectors
In the special case that an operator  acts on a vector  and the result is a constant times the
original vector as in Eq.(21), the vector is called an eigenvector and the constant  is called an
eigenvalue. The eigenvector is said to belong to the eigenvalue.
Most often, the operator is known and the “eigenvalue problem” is to find the possible
eigenvalues and eigenvectors. Note that the eigenvector in Eq.(21) could be multiplied by any
scalar, so eigenvectors are arbitrary by a constant factor (unless they are normalized to unit
length).
problem 31.
Find the eigenvalues and eigenvectors of the matrix operator
0 1


1 0
problem 32.
Find the eigenvalues p and eigenvectors (x) (eigenfunctions) for
d
 i  x   p x 
dx
where we require that 0) =  (L).
problem 33.
Find the eigenvalues E and eigenvectors (x) (eigenfunctions) for
2 d 2

x   E x 
2m dx 2
where we require that (0) =  (L) = 0.
Since the eigenvalues and eigenvectors are linked, it is conventional to label the ket with the
eigenvalue. In this notation, the fundamental eigenvalue problem (21) is written as
(24)
 
Hermitian Operators
A class of linear operators called Hermitian operators has special properties that are important in
quantum mechanics.
An Hermitian operator  is defined by the property
(25)
ab  ba

The right hand side of Eq.(25) is called the Hermitian adjoint and is denoted with a dagger,


ab  ba
The following results obtain for Hermitian operators:
i) The eigenvalues of  are real numbers.
ii) The eigenvectors are orthogonal.
iii) The eigenvectors are a complete basis for the vector space.
problem 34.
Confirm that the matrix of problem 3 is Hermitian. Show directly that the eigenvectors
are orthogonal and are a complete basis of the vector space.
problem 35.
Confirm that the operator of problem 4 is Hermitian. Note that the eigenvectors form the
basis for expanding functions in Fourier series.
problem 36.
Prove that the eigenvalues of an Hermitian operator are real and that the eigenvectors are
orthogonal.
Operator Principle
Hermitian operators represent the act of measuring physical observables and the
corresponding eigenvalues are the only possible results of measuring these observables.
problem 37.
The y-component of spin angular momentum is represented by
 0  i

s y  
2  i 0 
Find all possible values for the y-component of spin.
The operator principle does not say how to construct the operators for physical quantities, and it

is convenient to simply assert that the momentum operator p is given in x-space by
(26)
or, in bra-ket notation,
(26’
)


p  i
x


x p  i
x
x
This enables us to construct many operators by analogy with the classical counterparts. For
example, the energy operator corresponds to the Hamiltonian H of the system,
(27)
H
1 2
p V
2m
where V is an operator corresponding to the system potential energy.
Equations (26) and (27) are restricted to one dimension and to non-relativistic systems. A threedimensional form will be given later. We find it simple and direct to accept Eq.(26) as a
subordinate principle.
problem 38.
 
Demonstrate that x p  p x  i  .
This completes the presentation of the basic quantum principles for several chapters. A principle
of time development will be added later, but we now have the apparatus to analyze many
standard systems and processes.
Wave Mechanics
Wave mechanics is the form of quantum mechanics that is obtained when the amplitudes being
used are wave functions <x|> and the operators  are differential operators. Here we outline a
sequence of conceptual steps in treating systems with wave mechanics:
i) problem statement
The fundamental problem is, for a specified system, to find the allowed values  of an
observable  and the corresponding states |>. The abstract form of the associated eigenvalue
problem is
 
(28)
ii) construct the wave mechanical operator
The classical form of the observable  is usually a function of the position x and momentum p,
 classical  x, p 
then the appropriate quantum mechanical operator is obtained by replacing p by the operator


p  i  D . (It is not necessary to replace x by an operator x in position space because


x x  x x and the value x may be substituted for x .)
(29)
  x,i  D
where  is a differential operator
iii) write the wave mechanical eigenvalue problem
The wave mechanical counterpart of Eq. (28) is a differential equation,
x, iD     
(30)
where   x 
When the operator is the Hamiltonian,  = H, then the eigenvalue is energy,  = E, and Eq. (30)
is Schrödinger’s equation, H  E .
v) invoke boundary conditions and solve
A variety of methods must be used to solve the different differential equations that arise.
Boundary conditions play an important role in these equations. A particle on a loop of length L
must satisfy a single-valued condition x  x  L whereas a particle bound to a potential
source must have    0 (no probability of escape to infinity). A particle in a box of length
L with impenetrable walls satisfies the condition 0  L  0 . Reflection and transmission
from a boundary requires a smooth connection at the boundary; left  right and
D left  D right .
Applications: Particle on a Loop and Particle in a Box
problem 39.
Calculate the allowed values of momentum p for a bead constrained to a loop of
perimeter L.
problem 40.
Calculate the energy eigenvalues and eigenfunctions of a bead of mass m constrained to a
loop of perimeter L. Compare your result with that of the last problem.
problem 41.
Calculate the energy eigenvalues and eigenfunctions for a particle-in-a-box.
Application: Reflection From a Barrier
Quantum mechanics describes how light is reflected or transmitted upon striking a transparent
material. The simplest model of this process has a free particle (V=0) hit a “barrier” material
with a constant potential V=Vo. The particle (wave) can then be reflected back from the barrier
or transmitted through the barrier (when E > Vo). The probability of reflection will depend on
the initial energy E and the barrier “height” Vo
transmitted
incident
A exp(ikx)
reflected
B exp(-ikx)
The relative probability
C exp(iKx)
Vo
that the particle
will be reflected is the ratio R called the reflection coefficient,
R = |B exp(-ikx)|2 / |A exp(ikx)|2
or
2
B
(31)
R 2
A
Since the particle must be either reflected or transmitted, the sum of R and the relative
probability for transmission T, the transmission coefficient, must add to unity:
R+T=1.
Equation (31) requires a relation between A and B in order to evaluate the ratio R. This is
established from the boundary conditions for a smooth connection at the boundary; left  right
and D left  D right .
problem 42.
Calculate reflection and transmission coefficients for a particle incident on a potential
step Vo < E.
4. HARMONIC OSCILLATOR
Here we analyze a system of great importance in physics, the quantum mechanical harmonic
oscillator. The physics and mathematics of the oscillator are relevant to the emission and
absorption of light by matter (blackbody radiation), the analyses of radiation and fields, the
treatment of systems of identical particles, and other basic problems.
The methods used in solving the harmonic oscillator problem are illustrative of the standard
techniques for treating other bound systems. These include i) wave mechanics, where an
eigenvalue problem is represented as a differential equation (usually the Schrödinger equation),
ii) matrix mechanics, where an eigenvalue problem is represented with matrices applied to
column vectors, and iii) abstract operator mechanics, where an eigenvalue problem is solved
using abstract symbols for operators and states. The last approach is the most sophisticated, but
we will emphasize it because, with some modifications, it can be used to solve basic angular
momentum problems and the hydrogen atom relatively easily.
The Classical Oscillator
The kinetic energy of a particle of mass  and momentum p is p2/2 and the potential energy of
an oscillator can be written as 2x2/2, where  is related to the spring constant k by the relation
k
(32)
2 

The classical Hamiltonian is the sum of kinetic and potential energies,
(33)
H
p2 1
  2 x 2
2 2
and conservation of energy E can be expressed as
H E
The quantum analogue of this expression is an eigenvalue problem for energy:
H E E E ,
where H is an operator having the same form as Eq.(2), but with scalars p and x replaced by


operators p and x .
(34)
Wave Mechanics for Oscillator
When the state in Eq.(3) is taken to be in the x-representation, <x|E>, the expression becomes a
differential equation called the (time-independent) Schrödinger equation:
(35)

2 2
1
D x E  2 x 2 x E  E x E
2
2
problem 43.
Construct Eq.(4) from Eqs.(2) using the momentum operator given in Unit 3.
Equation (4) is the time-independent Schrödinger equation for the harmonic oscillator. Its
solution is shown in all quantum mechanics texts. Bound state solutions exist only when
En   n  12 
where n is an integer. Each value of En has an associated wavefunction n = <x|En> (this is
written more simply as <x|n>), and some solutions are graphed below.
(36)
n=1
n=3
n = 10
classical case corresponds to n
Abstract Operator Approach (Dirac)--Eigenvalues
The eigenvalues of the Harmonic oscillator can be found from the following information alone,
without the use of the Schrödinger equation:
(37)
x, p  i
(38)
H n En
where H is given by Eq.(2). The core idea is to construct “raising and lowering” operators a+
and a with the ability to produce one state from another:
a+|n> = constant |n+1>
a|n> = constant |n-1>
These are then applied repeatedly to Eq.(7) with the result that all the possible energy levels are
produced.
In the following problem sequence, let       1 . These quantities can be restored in the
results by dimensional analysis. (Since we are dealing with three fundamental quantities—mass,
time, and distance—three units may be arbitrarily chosen to specify the values of three
quantities.)
problem 44
Define the operator a 
1
2
 x  ip
(39)
and derive the relation
[a, a+] = 1
problem 45
Show that H = = a+a + ½.:
(40)
H = = a+a + ½.
problem 46
Demonstrate that a+a is the “number operator,” that is,
(41)
a+a |n> = n |n>.
problem 47
Use the result of the last problem to show the eigenvalue spectrum of the Harmonic
oscillator is given by Eq.(5). (Use dimensional arguments to restore  and  .)
Eigenfunctions in the Dirac Approach
Eigenfunctions can be generated by first finding the lowest eigenfunction <x|0> and then
“raising” it repeatedly to <x|1>, <x|2>, etc.
problem 48
Use the fact that <<x|0> corresponds to the lowest n value to calculate the ground state
wavefunction.
problem 49
Use the result of the previous problem to generate the wavefunction of the first excited
state.
Solution 4.2
The operators x and p obey the basic commutation relation,
 x, p  i
or in expanded form and setting   1
xp  px  i
.
Substitute operators a 
1
2
 x  ip
(1)
and a † 
1
2
 x  ip into the expression
aa †  a †a
and preserve the order of the x and p factors because they do not commute like ordinary
numbers. Using Eq. (1), we obtain
 a, a   1
†
(2)
Solution 4.3
The purpose of this problem is to write the Hamiltonian in terms of the creation and annihilation
operators a and a † . This will later help us with algebraic manipulations that will give the energy levels.
It is easiest to introduce a 
1
2
 x  ip and a †  12  x  ip into the expression
H  a †a 
1
2
After reducing the result with Eq. (1), we get
H
1
2
 p2  x 2  .
(3)
Solution 4.4
First we note that any operator with the form a†a is Hermitian* so the eigenvalues
real. We write
a †a n = n n
of a†a are
(4)
where n is real, but not yet known to be an integer. Apply the operator a to both sides of Eq. (4). Label
the quantity a n as  , so that
aa † |   n|  
(5)
We would like the operator to be converted to a†a so, with the help of the commutation relation (2),
Eq. (5) becomes
a †a|   (n  1)|  
(6)
Two important observations are made from Eq. (6). Clearly, if n is an eigenvalue of a†a, then so
is n-1. The process can be repeated to produce (n-any integer) and since the oscillator energy cannot
become negative, the process must produce zero for some integer. Consequently, n must be an integer.
A second observation is that
a n  n 1
(7)
The reason we use a proportionality sign is because we cannot guarantee that  is normalized even
when n is. Expression (7) shows why a is called a lowering operator (and often, in the context of
particle physics, an annihilation operator).
Subordinate Problem
Alter this solution to demonstrate the effect of the raising or creation operator a†:
a† n  n  1
*
It is always true that (AB...)† = ...B†A† and that (A† ) † = A. Then
(8)
 a † a †  a † a
Solution 4.5
From Equations (3) and (4) we have
 a † a  21  n   n  21  n
or
H n   n  21  n
(9)
Now we have to restore the constants m, , . The dimensions of these are respectively [M],
[M][L]2[T]-1, and [T]-1. The dimensions of H (energy) are [M] [L]2[T]-2. The latter dimension must be
obtained by choosing x, y, and z so that
 M  x   M  L 2  T  1    T  1     M  L 2  T  2 
y
z
This is satisfied only when x  0, y  1, z  1 so that Eq. (9) becomes
H n   n  21  n
(10)
Solution 4.6
The lowest energy corresponds to n = 0. This state cannot be lowered so
a0 0
(11)
The wave function equivalent of this is
1
2
 ip  x0  0
(12)
where p is the usual differential operator with = 1. The resulting differential equation is
d0
  x0
dx
with the solution
0  const exp  21 x 2 
The exponent must be dimensionless, so the constants m, ,  must be restored.
(13)
(14)
03 ATOMS AND CHEMISTRY
Hydrogen Eigenvalues
The simple Bohr picture of hydrogen gives the right energy levels, but modern quantum mechanics gives
much more information. In fact, all of chemistry is explained in principle by quantum theory. Here we
will see that the Schrödinger solution for hydrogen gives us quantum numbers that reveal the basic plan
behind the periodic table.
When we apply quantum theory to hydrogen, four eigenvalues (or associated quantum numbers) are
needed to fully describe each state. These are
(1) the electron's energy level En characterized by the principal quantum number n, an integer
corresponding to the levels in Bohr theory;
(2) the orbital angular momentum L of the electron characterized by an integer  such that
L      1 .
(3) the azumuthal angular momentum Lz that is the z-component of L. It is characterized by a
positive or negative integer quantum number m, Lz  m .
(4) the intrinsic angular momentum of the electron spinning on its axis called the spin, sz.
The eigenvalues (or the corresponding quantum numbers) are related to each other. For a given energy,
only a limited amount of angular momentum is possible and this restricts the values for  and m. The
rules restricting the quantum numbers are given below:

n principal quantum number is unrestricted
n = 1, 2, 3, ...

 orbital angular momentum quantum number is limited to a value below the value of n,
 = 0, 1, 2, ..,n-1

Lz=m
m the azumuthal quantum number can be negative or positive, but may not
exceed the magnitude of the total orbital angular momentum,
m  0,  1,  2,.., 
The situation can be pictured as in the diagram where the L vector precesses
around the z-axis. The projection of L on the axis is Lz.

sz spin quantum number can only have one of two values,
sz   21
L
1
Use the rules to find how many Hydrogen electron states are possible when
(a) n = 1, and (b) n = 2.
2
Calculate and sketch the possible angles between L and the z-axis for
(a)  = 1, (b)  = 2
Many-Electron Atoms
Pauli's exclusion principle asserts that no two electrons can occupy the same quantum state. For atoms
with many electrons, this implies that only 2 electrons can be in the n=1 ground state. Similarly, there are
8 different states with the next higher energy level (n=2). We say that a maximum of 8 electrons can exist
in this energy level. At the n=3 level, the simple counting rules for hydrogen's electron must be modified
to account for the complicating effects of additional electrons in many-electron atoms. The result for n=3
is that 8 states are allowed.
Quantum mechanics allows 2 electrons in the lowest energy level, n=1.
Quantum mechanics allows 8 electrons in the second energy
level, n=2.
Quantum mechanics allows 8 electrons in the third energy level,
n=3.
We have treated each n level or shell as if all the electrons therein
have exactly the same energy. This is not strictly true for manyelectron atoms. Very small differences in energy occur for the
different  values due to various interactions between electrons.
For example, when these complications are included for n = 2,
the  =0 state is found to be slightly lower in energy than the  = 1 state. Electrons with different values
of  within the same shell are therefore called subshells.
The number of electrons in each subshell is the product of the number of m values (2  +1) and the
number of spin states, 2:
maximum number of electrons in a subshell = 2 (2  +1)
Some historic names for the  values are
s wave
= 0
p wave
= 1
d wave
= 2
f wave
= 3
and the subshells are denoted by the combination “n  ” so that the n=1,  =0 subshell is written 1s and
the n=2,  =1 subshell is written 2p. (This is often called “spectral notation.”)
3
(a) What is the maximum number of electrons in a p wave subshell?
(b) Write the spectral notation for the n =4,  =3 subshell.
Periodic Table
The rules we've established for three energy levels can be used to construct a simplified periodic table.
The elements in each row or period have their outer electrons occupying the same shell or n-quantum
number.
Each column or group has similar chemical properties because the elements in the group have the same
number of outer electrons (called valence electrons). These are the electrons that encounter the world
outside and account for most chemical interactions.
1
2
1
2
2
1
8
3
2
2
8
2
2
4
2
3
8
2
5
2
4
8
2
6
2
5
8
7
2
2
4
The diagram depicts the orbital
electrons of Lithium. Without reference to
the above table, make similar sketches for the
elements just to the right and just below
Lithium on the periodic table.
6
8
8
2
7
8
2
2
2
8
8
2
1
2
Calculations that include the complexities of many-electron atoms give the following sequential order for
the filling of subshells from lowest energy to highest:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d,...
Based on this notation, the orbital structure of Boron (atomic number Z=5) is: 1s 1s 2s 2s 2p or
1s2 2s2 2p.
5
Write the orbital structure of Carbon (Z=6) in spectral notation.
Simplified View of Bonding for Organic Molecules
A full description of a chemical bond is given in terms of electron wavefunctions merging together. We
do not need this kind of detail here. A covalent bond is due to the sharing a pair of electrons (more
accurately, overlap of electron wavefunctions) between elements. The bond is depicted with a simple
stick so that two hydrogen atoms bound together as H2 is depicted as
H—H.
The number of such bonds formed by an element is usually equal to the number of outer electrons it must
share to obtain a closed shell. Here are the most ubiquitous elements depicted with unconnected bonds:
H
O
C
N
Carbon forms the most bonds with its neighbors and it is the most characteristic chemical of life. In fact,
the term organic chemistry is synonymous with carbon chemistry. The sticks representing bonds can be
shared by elements. For example, water can be drawn as
H—O—H and methane (CH4) is
H
H
C
H
H
problem
An amino acid is drawn here without hydrogen atoms. Complete
picture using H’s and sticks.
problem
A sugar is drawn here without hydrogen atoms. The corners or
are assumed to be carbons. Complete the picture using C’s, H’s
sticks.
O
N
C
C
vertices
and
C O
O
O
the
O
O
5. HIGHER DIMENSIONS & ANGULAR MOMENTUM
A few new complexities arise when we consider systems with more than one dimension. In
one dimension, we used a single observable like p or H to describe the system (although it is
possible to use more than one in some cases). In higher dimensions, we must begin with a
sufficient set of such observables  a “complete commuting set of observables.”
The motion of a particle in two or three dimensions is usually much more complex than
motion on a line, but the complexity can be reduced by dealing with conserved quantities like
angular momenta. An analysis of the quantum mechanical properties of angular momentum is
therefore addressed here.
Commuting Observables
It is a central theorem in quantum physics that when operator-observables A and B commute,
[A,B] = AB  BA = 0, then both quantities can be measured simultaneously. For example, it is
easy to see that p and p2/2 commute, so momentum and kinetic energy can be observed at the
same time.
Conversely, when A and B do not commute,
[A,B]  0,
these quantities cannot be measured simultaneously; it is said that the measurement of one
destroys any knowledge of the other. The important commutation relation [x,p]=i shows that x
and p do not commute and indicates that position and momentum cannot have simultaneous
values.
To prove the theorem, consider the simultaneous measurement of A and B. This means that
applying A or B to the same state |> results in an eigenvalue for each:
A|> = a|>
B|> = b|>.
Premultiply the first by B and the second by A. Subtracting gives (AB-BA)|> = 0, and the
result [A,B]=0 follows. If the eigenstates were different at the outset, we could not get this
result.
5-1 Show that if any two observables can be measured simultaneously, their operators must
commute.
5-2 For the following, indicate whether the operators in brackets can or cannot have
simultaneous values:
(a) [p, p2/2m], (b) [x, y]=0, (c) [x, p]=i, (d) [Jx, Jy]= iJz (this
defines angular momentum), (e) [Jz, J2]=0
States and Probabilities in 2-D
A particle's position in two dimensions can be specified by two values, x and y. The position
state can therefore be symbolized by |x, y>. Similarly, momentum has two components in two
dimensions and a momentum state can be indicated as |px ,py>. The wavefunction for the latter
state is written as <x, y|px,py>.
These examples illustrate two general requirements. First, the quantities in a bra or ket must
be able to be specified simultaneously. This means that their operators must commute. The
examples above originate with the relations [x,y]=0 and [px,py]=0. Secondly, the number of
observables specifying a state must be equal to or greater than the number of dimensions. (In
most practical cases these entries are equal in number to the number of dimensions.) In two
dimensions we need two eigenvalue entries like |x,y> or |px,py>.
The requirements can be summarized by saying we must have a complete set of commuting
observables for a given system. This means that for an n-dimensional system we must specify at
least n operators that are linearly independent and mutually commuting. Consider, for example,
a two-dimensional harmonic oscillator with the following commutation relations: [x,H]0,
[px,H]0, [J,H]=0. The first two expressions show that there cannot be states like |x,E> or |px,E>.
The last expression shows that angular momentum and total energy commute so J and H are a
complete commuting set and |j,E> is a legitimate state.
5-3 Consider a free particle in space. Using the information in the following commutation
relations, select a convenient set of complete commuting observables and write a corresponding
ket label:
[x,px]=[y,py]=[z,pz]=i,
[px, py]=[py, pz]=[px, pz]=0.
5-4 A particle is confined to the surface of a sphere. Using the information in the following
commutation relations, select a convenient set of complete commuting observables and write a
corresponding ket label:
[Jx, Jy]=i Jz
[Jy, Jz]=i Jx
[Jz, Jx]=i Jy
[Jx, J2]=[Jy, J2]=[Jz, J2]=0
(Note that in the last problem the linear momenta are related because the particle is constrained
to move on a sphere. This causes the relations given in problem 5-4, [px,py]=[py,pz]=[px,pz]=0,
not to hold in this case.)
5-5 An electron is bound to a stationary proton (Hydrogen atom). Using the information in the
following commutation relations, select a convenient set of complete commuting observables
and write a corresponding ket label:
[Jx, Jy]=i Jz
[Jy, Jz]=i Jx
[Jz, Jx]=i Jy
[Jx, J2]=[Jy, J2]=[Jz, J2]=0
[Jx, H]=[Jy, H]=[Jz, H]=0
[J2, H]=0
Probabilities in Two Dimensions
Calculating probabilities from amplitudes in higher dimensions is practically the same as for
one dimension. The quantity
P(m,En|) = |<<m,En|>|2 is the probability of
measuring the system represented by |> to be in a discrete state with angular momentum m and
energy En.
Probability densities are only slightly more involved. The quantity
P(x,y|) =
|<<x,y|>|2 is the probability of measuring the system represented by |> to be in a
neighborhood dx dy of point (x,y). In three dimensions, the neighborhood is dx dy dz. If two
particles are involved with labels like (x1,y1,z1,x2,y2,z2), then the "volume element" to be
integrated over is dx1dy1dz1dx2dy2dz2. If the bra is labeled by, say, two continuous momenta,
<<px,py|, then the volume element is dpxdpy.
When wavefunctions of multidimensional systems can be "separated," into essentially onedimensional factors, the problem is made tractable. An example of a separable wavefunction is
<<x,y|px,py> = <<x|px> <y|py>. This separation is not always possible. It takes a judicious
choice of representation to accomplish it. As a counterexample, the wavefunction
<<r,<2>q<1>|px,py> cannot be separated.
5-5 A particle is constrained to move on a flat tabletop but is otherwise free. Calculate an
appropriate (unnormalized) wavefunction.
Angular Momentum
Energy and angular momentum are the most familiar conserved quantities in standard
problems. Angular momentum is essentially a two-dimensional quantity because it describes
angular motion for two angles, the polar angle <2>q<1> and the azimuthal angle <2>f<1>.
Earlier in this unit we saw that a 2-dimensional system must be specified by 2 mutually
commuting operators and 2 associated compatible eigenvalues. The two most convenient
commuting observables were seen in problem 5-4 to be the z-component Jz and the total angular
momentum,
(1)
J2 = (Jx)2 + (Jy)2 + (Jz)2 (definition)
Before we evaluate the eigenvalues of angular momentum, we want to review the classical
definition and state the formal quantum definition of angular momentum. In classical
mechanics, angular momentum J is written as
(2)
J = rXp = i(ypz-pyz) +
j(zpx-pzx) +
k(xpy-pxy)
= iJx + jJy + kJz
5-6 Verify that rXp gives the result shown in Eq.(2).
When the operator equivalent of Eq.(2) is used, the following commutation relations can be
proved:
(3)
[Jx,Jy]=iJz
[Jy,Jz]=iJx
[Jz,Jx]=iJy
These commutation relations are taken to be the definition of quantum mechanical angular
momentum.
5-7 Write the defining commutation relations for the components of angular momentum.
5-8 Derive [Jx,Jy]=iJz from the operator equivalent of Eq.(2).
Finally, from Eqs.(1) and (3), we can show that each component of angular momentum
commutes with total angular momentum J2:
(4)
[Jx,J2]=[Jy,J2]=[Jz,J2]=0
5-9 Prove Eq.(4).
Eigenvalues for Jz
We can generate the eigenvalues for Jz using only the defining relations, Eq.(3). Construct
"raising and lowering" operators J+ and J-,
(5)
J+ = Jx + iJy,
J- = Jx - iJy
5-10 Given the raising and lowering operators of Eq.(5), demonstrate the following commutation
relations:
(6)
[Jz,J+] = hJ+, [Jz,J-] = -hJ-
Now we write an eigenvalue equation for Jz in a convenient form, Jz|j,m>=hm|j,m> and apply
the J+ and J- operators to generate a spectrum of eigenvalues (in much the same way that it was
done for the harmonic oscillator).
5-11 Given the relations (6), show that the eigenvalues of Jz are separated by integer multiples of
h. That is, show that if Jz|j,m>=hm|j,m>, then h(m+1) and h(m-1) are also eigenvalues of Jz
(assuming higher and lower values of m exist).
Notation below here must be corrected from old computer notation
In the next section we will establish that the eigenvalues of Jz range from +<3>l<1> to <3>l<1> , where <3>l<1> is a positive integer or half-integer.
Eigenvalues of J2
The following sequence of problems will establish the familiar eigenvalues of total angular
momentum h2<3>l<1> (<3>l<1> +1).
5-12 Given the expressions (5), show
(7a)
J2 = J-J+ + (Jz)2 + hJz (7b)
J2 = J+J- + (Jz)2 - hJz
5-13 Given Eqs.(7), demonstrate that the eigenvalues of J2 are h2<3>l<1> (<3>l<1> +1), where
<3>l<1> is the maximum possible value for m associated with this total angular momentum.
5-14 Given Eqs.(7), demonstrate that the eigenvalues of J2 are h2K(K-1), where K is the
minimum possible value for m associated with this total angular momentum.
5-15 Given the results of problems 5-13 and 5-14, show that K=-<3>l<1> . Use the fact that
values of m differ by integer amounts to show that <3>l<1> must be an integer or half integer.
The results are in; eigenvalues of total angular momentum J2 are (8)
(<3>l<1> + 1), where <3>l<1> is integer or half-integer
J2 = h2<3>l<1>
and eigenvalues of the z-component Jz are
(9)
m = -<3>l<1> , -<3>l<1> +1, ...., <3>l<1> -1, <3>l<1>
5-16 What is the eigenvalue spectrum for angular momentum?
When states |<3>l<1> ,m> are expressed in wavefunction form <<<2>q,f<1>|<3>l<1> ,m>,
they are called <ib>spherical harmonics<ie> and often written in the notation Y<3>l <1>,m.
6. THE HYDROGEN ATOM
The hydrogen atom is the model upon which we base our concepts of atomic physics and the
periodic table. We will find that an isolated hydrogen atom is characterized by discrete energy
levels En and each of these levels has subordinate orbitals having associated angular momenta
 and m.
Hydrogen Eigenvalue Problem
The hydrogen atom problem is most tractable when it is expressed in spherical
coordinates r , ,  . Let pr and J be the operators for radial and angular momenta, respectively.
Then the Hamiltonian is
1  2 J 2  ke2
 pr  2  
H
(42)
2 
r  r
where  ke2 r is the potential associated with the Coulomb force.
It can easily be shown that H commutes with J2 and Jz so these three operators are a
complete commuting set of observables. The hydrogen states can therefore be written as
|E,  ,m> and the energy eigenvalue problem is
(43a)
H E, , m  E E, , m
Schrödinger’s equation for hydrogen results when this is converted to wave-mechanical form
with differential operators and with kets replaced by wave functions. Usually, textbooks find the
allowed eigenvalues and eigenfunctions by solving the Schrödinger differential equation. It is a
tedious process and here we will find solutions by an easier operator approach closely akin to our
harmonic oscillator solution.
Equation (2a) can be simplified by eliminating the J 2 operator. Since
J 2 E, , m  2  1 E, , m , we can write
where
H  E, , m  E E, , m
(2b)
1  2  2   1  ke2
 pr 
 
H 
2 
r2
 r
(44)
In the following sections, we solve the eigenvalue problem (2b) by rewriting H  in a
“factored” form
1
2
A A



 b  . The A operator will raise the  eigenvalue and A will lower it.
H in Factored Form
Define an operator A with the form

A  pr  i   1r 1  b

(45)
where
b 
ke2
 2   1
(46)
The problem sequence that follows will determine H  in terms of the A+ and A operators.

6-1 Show that A A  pr2    1 / r  b   i  1 r 1 , pr
2

6-2 Given the required commutation relation between r and its conjugate momentum,
r, pr   i , show that r 1, pr  ir 2 .


6-3 Use the results of problems 6-1 and 6-2 to derive the form
1
A A  b 
H 
2
(47)
.
Raising and Lowering Properties
The next two problems determine the raising and lowering properties of the A operators.


6-4 Show that A , A  2 2   1 / r 2 . (Use the results of problems 6-1 and 6-2.)
6-5 Demonstrate that A E,, m is proportional to E,   1, m ; that is, A is a raising operator
for  . Similarly, show that A E,, m is proportional to E,   1, m such that A is a
lowering operator for  . Note that these operations do not change the value of E.
Eigenvalues and Eigenfunctions of Hydrogen
The energy of the hydrogen atom must dictate upper and lower bounds for angular
momentum, so the A operators cannot raise or lower without limit. Let  * be the maximum
angular momentum at a particular energy E so that E , * , m cannot be “promoted” to higher  :
A E , * , m  0
This condition enables us to determine eigenvalues & eigenfunctions of hydrogen.
6-6 Show that the energy eigenvalues for operator H  in Eq. (6) are given by
(48)
En  
e 4 1
2 2 n 2
(49)
where n = 1, 2, 3,...
Hydrogen eigenfunctions can be written in separated form,
(50)
nm  Rn  r Ym , 
where Rn(r) is the radial wavefunction. It is easily determined when operator pr is expressed in
differential operator form:
(51)
 d 1
pr  i  
 dr r 
(This expression can be verified from the commutation relation r, pr   i .)
6-7 Find the hydrogen radial eigenfunctions (un-normalized) for a maximum allowed value of
angular momentum. (Use Eqs. (7) and (10).)
6-8 Calculate the radial ground state R10 of the hydrogen atom. Demonstrate that the probability
2
density 4 r 2 R10 is a maximum at the Bohr orbit. (Use A0 R  0 )
6-9 Find the normalized radial wavefunction for hydrogen R20.
Selection Rules for Hydrogen
Transitions between Hydrogen energy levels can only occur when the change in total angular momentum changes
by 1:
  1 

As a corollary, this requires m  0,  1 .
Background
Classical electrodynamics says an accelerating charge e radiates power P according to
P
where a is the acceleration and
2 ke2 2
a
3 c3
k  40  .
1
problem
ke2 2
Use dimensional analysis to find P  3 a .
c
Quantum Requirement for Radiation
The acceleration of an oscillating electron is given in three dimensions by a = –2 r , so average power can be
written as
P
k 2
2
er .
3
c
The quantity er is the (average) dipole moment of the atom. Only non-vanishing averages
nm r 2 n' ' m' allow
transitions between the primed and unprimed levels. It can be shown that the integrations over vanish unless
  1 

problem
Show that
100 r 2 201  0 and 100 r 2 210  0 using only integrations over .
7. MATRIX MECHANICS
We saw that matrices are one kind of operator and column vectors are the associated kind of
state. Here we will show that all linear operators have a corresponding matrix and all ket vectors
have a corresponding column vector. As a consequence, all quantum mechanical problems can,
in principle, be cast into matrix and column vector form. This formulation is called matrix
mechanics to distinguish it from wave mechanics where differential operators and wavefunctions
are used. In practice, both forms of quantum mechanics are used side by side along with the
abstract Dirac approach. Usually the form of quantum mechanics chosen is the one that makes
the current problem easiest.
Matrices Column Vectors
A matrix associated with an operator  is an ordered array of numbers i,j given by the
expression
i , j  i  j ,
(52)
where i and j represent sets of quantum numbers specifying the states. Note that the specific form
of the matrix depends on the base states used to describe it. We say the description in a particular
basis is a representation of the operator.
7-1. Spin ½ corresponds to angular momentum (J2=S2, Jz=Sz) with j = ½ and m = –½ ,+½ . (a)
Use the definition (1) to derive a matrix for Sz . (b) Find the eigenvectors and show that
 0 0
0 1

 and S   
S   
1 0
 0 0
are the respective raising and lowering operators and use these to construct Sx and Sy.
7-2. Evaluate the matrix elements x0,0 and x0,1 for the simple harmonic oscillator. Indicate an infinite
matrix for the Hamiltonian of the harmonic oscillator (Hi,j).
Column Vectors
A general vector | has a column vector form given by the array
 1 


 2 
 
 


 n 


(53)
Again, note that the column vector depends on a specific representation.
7-3. Calculate normalized column eigenvectors for spin Sz directly from the eigenvalue
equations. Show that the matrix for Sz gives the appropriate eigenvalues when it is applied
to these vectors.
7-4. Given a state of Schrödinger’s cat as  
eigenvector in the “live” and “dead “basis.”
2
3
L 
1
3
D , calculate the corresponding
8. TIME DEVELOPMENT
Most often, it is convenient and desirable to use eigenstates | that correspond to conserved
quantities of operator . Since conserved quantities do not change with time, we have not yet
had occasion to see any explicit time development. Here we give the time development principle.
Time Development Principle

The Hamiltonian operator H determines the time development of a quantum mechanical system:


(54)
H  i
t
or


(1’)
H   i
 .
t
8-5. Show that a formal solution to Eq.(1’) is
|(t) = |(0) exp(-itH/)
8-6. Show that when observable  commutes with H, [,H] = 0, then  is conserved.
8-7. Write the time-dependent Schrödinger equation for a simple harmonic oscillator.
APPENDICES
A6 HYDROGEN EIGENFUNCTIONS
The following is a list of some normalized wavefunctions for hydrogen  n  m for various
values of n, , and m. a0 denotes the Bohr radius.
4 0 2
 0.529  1010 m .
2
e
The separated form for the wavefunctions is
a0 
 nm r , ,   Rn r  m  m 
where
(55)
 m   e im
 m   sin
m
polynomial in cos 
Rn r   e constantr / n r  polynomial in r 
(56)
A7 Identical Particles
The Eigenvalues of Exchange ..........................................................................................1
Fermi and Bose Particles .................................................................................................2
Fermi Particles .....................................................................................................2
Bose Particles .......................................................................................................2
Quantum Statistics ...........................................................................................................3
Applications .....................................................................................................................4
Free Electron Model for Metals ...........................................................................4
Density of States for Electrons ................................................................5
Black-Body Radiation ..........................................................................................6
However similar they are, the members of a set of twins or a pair of billiard balls are
clearly different objects. Fundamental particles, however, are identical in all respects so that
when two electrons in the same state are interchanged, there is no possible way to distinguish the
exchanged system from the original.
The simple fact that the interchange of identical particles cannot change a system has
profound consequences. We will find that only two kinds of particles can exist. For one kind,
bosons, an unlimited number can coexist in the same state (including the same place). The other
kind, fermions, can have only one particle per state.
The Eigenvalues of Exchange
Consider a system with two identical particles at positions x1 and x2 . The system is
governed by a wavefunction x1 , x2  . We can exchange the particles so that the positions are
reversed producing a wavefunction x2 , x1  . This is expressed mathematically by having an
exchange operator  applied to the original wavefunction,
x1 , x2   x2 , x1  .
(57)
Apply  to both sides of Eq. (57) to produce
 2 x1 , x2   x1 , x2  .
(58)
If there is a physical consequence of indistinguishability upon exchange, there must be an
eigenvalue problem of the form  =  where  is some observable. Clearly, Eq. (58) must
correspond to 2  = 2  and we find that
  1
(59)
Particles that exhibit the + sign upon exchange so that  =  are called bosons where those
that exhibit the - sign with  =  are called fermions.
problem 1.
Derive   1 starting from Eq. (57).
Fermi and Bose Particles
When two particles are not interacting, their composite wavefunction  is a product of
the independent wavefunctions u a x1  and ub x2  , where a and b refer to different quantum
numbers. To see this, write
Setting
H1  H 2   E .
(60)
  u a x1 ub x2  ,
(61)
we can separate Eq. (60) into two parts,
H1ua  Ea ua
H2 ub  Eb ub
where
(62)
E  Ea  Eb
problem 2.
Derive Eqs. (62) from Eq. (60).
Fermi Particles
Notice that the product function   u a x1 ub x2  does not obey    for either
eigenvalue   1. However, it is easy to check that the exchanged product   u a x2 ub x1  is
also a solution to Eq. (60). The difference of both products is a solution that satisfies the Fermi
condition,
  
(63)
with
  12 u a x1 ub x2   u a x2 ub x1 
(64)
problem 3.
Show that Eq. (63) is satisfied by expression (64).
Now note that when a=b. That is, there is no chance that two identical fermions can
coexist in the same quantum state! This stuff is familiar matter. It includes electrons, protons,
and neutrons—in fact, all particles with half-integer spins, 21 , 23 , 25 ,...,are Fermi particles.
Bose Particles
The eigenstate with   1,
   ,
applies to bosons and is easily constructed from the sum of the two products:
  12 u a x1 ub x2   ua x2 ub x1  .
(65)
(66)
problem 4.
Verify that expression. (66) satisfies    .
You can see immediately that  does not vanish when a=b. An unlimited number of
these particles can occupy the same state. These particles are usually associated with fields and
the most familiar examples are photons and mesons. Bosons all have integer spins, 0, 1, 2,..
problem 5.
Identify the following particles as fermions or bosons:
 , n, e - ,  0 , p, quark , gluon
Quantum Statistics
The indistinguishability of particles has consequences for the macroscopic behavior of
particle aggregates. For instance, we can describe conductivity with a free electron model and
blackbody radiation with a photon gas model. The passage from quantum mechanical
information to macroscopic physics is effected by the methods of statistical mechanics. Here we
introduce some results of statistical mechanics without proof:
Individual particles with energy j have an average occupation number nj given by the
following expressions:
j 

n j  exp 
Maxwell - Boltzmann
kT


1
nj 
Fermi - Dirac
j 
  1
exp 
kT


1
nj 
Bose - Einstein
j 
  1
exp 
 kT 
(The Maxwell-Boltzmann distribution applies to distinguishable particles)
Applications
There are many applications of the distributions above. Here we focus on the free
electron model and the Planck formula for blackbody radiation.
Free Electron Model for Metals
(67)
When we plot the Fermi-Dirac
distribution for very low temperatures, we
get the curves shown in the diagram. The
curve drops to half the maximum value at
  0 where 0 is the value of  at T=0.
This energy is termed the Fermi energy and
is roughly the typical energy per particle
near zero temperature. In metals, room
temperature is a good approximation to .T=0.
In order to calculate 0 for the freeelectron model, we need to convert to a
continuous spectrum for . That is,
N   n j   nd
j
(68)
where d is the number of states in energy interval d.  is called the density of states
factor. To proceed with the calculation of N, we need to determine .
Density of States for Electrons
The eigenvalues of momentum for a particle in a cube of length L are
nh
(69)
px  x
2L
with similar expressions for the y and z dimensions. The n’s can be approximated as
continuous variables in a macroscopic volume so an increment of the total number of
states is
dnx dny dnz
We find
8 L3
dnx dn y dnz  3 dpx dp y dpz .
h
The product dpx dpy dpz is a volume element in momentum space in direct analogy with
the standard spatial volume. This can be converted to spherical coordinates using
dpx dpy dpz = 4 p 2 dp where the total momentum magnitude p is the analog of radius r.
Finally, we must (i) divide by 8 because only positive values of n’s (and p’s) are used and
this applies to only 18 of a sphere of radius p; (ii) multiply by 2 to account for the two
possible spin states of an electron. The result is
d (number of states ) 
8V 2
p dp
h3
(70)
problem 6.
Derive Eq. (70).
We want to identify the left hand side of (70) with d so we need to convert
from momentum to energy. Since
p2

,
2m
Eq. (70) becomes
1
8Vm2m  2 12
(71)
 d 
 d
3
h
Now we can complete the calculation of Eq. (68). Note from the diagram that at T=0,
n=1 below   0 and n=0 above   0 . The integral can be solved for :
0 
h2
2m
 2 N
3 V 


2
3
(72)
problem 7.
Complete the integral in Eq. (68).
A numerical estimate for a typical metallic lattice spacing of 3X 10-10 m gives
N
 3. 7  1022 conduction electrons per cm 3 .
V
so that
0  4.1 eV .
This Fermi energy is huge compared with typical thermal energy kT  0. 025 eV. That is, room
temperature is a lot like absolute zero to conduction electrons.
We see that the cathode of a vacuum tube must be heated to increase electron emission.
Likewise, a photomultiplier must be cooled to decrease spontaneous emission of electrons.
Calculations similar to those done here can establish the heat capacity of the metal and detail the
features of thermionic emission.
Black-Body Radiation
Light and radiation in equilibrium with heated material can be treated as a photon gas.
The model material is termed a “black body” because it absorbs all the radiation incident upon it.
The energy of radiation in volume V is given by the expression

E    j n j   n d
j
(73)
0
where n is the Bose-Einstein distribution (with =0 because photons have no ‘rest’ energy) and 
is the energy density of states for photons.
Fortunately, we have done most of the work already. Photons have two polarization
states (corresponding to two spin states for electrons) and Eq. (70) applies to the number of
photon states. Photon momentum and energy are given in terms of frequency  ,
h
(74)
p
and   h
c
so that Eq. (70) becomes
8V
(75)
 d  3  2 d
c
The integral of Eq. (73) can now be assembled:

8hV
3
(76)
 3
d .
h / kT
c e
1
0
The integrand of this expression enables us to write the energy density per unit frequency
interval (),
  
8h
3
c 3 e h / kT  1
(77)
This is the famous Planck formula.
problem 8.
Derive the Planck formula from Eq. (73).
Integrating Eq. (76) is not trivial. It can be accomplished using
x3
4
dx

 e x  1 15
giving the detailed T4 law,
 8 5 k 4  4
VT
E  
3 3 
 15c h 
(78)
A8 Dirac Delta Function
Delta Function Concept ...................................................................................................1
Kronecker Delta ...................................................................................................1
Dirac Delta ...........................................................................................................1
Properties and Representations ........................................................................................2
Properties .............................................................................................................2
Representations of the Delta Function .................................................................3
Fourier Integral Theorem .........................................................................3
Other Representations ..............................................................................3
Applications .....................................................................................................................4
Boundary Conditions ...........................................................................................4
Bound State ..........................................................................................................5
Reflection and Transmission................................................................................5
The Kronecker delta  ij is familiar from relations like i j  ij . In this case the states
are discrete like the eigenfunctions of a one-dimensional well or harmonic oscillator. However,
quantum states may also be continuous like the eigenstate of free-particle momentum, k . The
Dirac delta function x  is the analog of  ij for inner products of continuous states,
k  k  k   k  .
Delta Function Concept
We introduce x  by first writing properties of  ij for discrete sums. We then insist that
the change  ij   x  preserves these properties when we convert the sums to continuous
integrals,
 
.
Kronecker Delta
The Kronecker Delta is defined by
1
 ij  
0
and two central properties are
i j
i j
(79)
 ij  1
j
 f j ij  f i
(80)
j
problem 9.
4
Evaluate the sum
 f j  j3 by expanding it and using the definition (79).
j 1
Dirac Delta
Changing from sums in Eqs. (80) to integrals and changing the notation by the
replacement  ij   x  , we write
 x dx  1
 f x x dx  f 0
(81)
where the integration intervals include the neighborhood
around x=0. These properties can be realized by a curve
like the one shown in the diagram. It is zero everywhere
except for a narrow spike of unit area around x=0.
Imagine now that the rectangle is made narrower
and taller so that the area remains the same but the width
approaches zero and the height approaches infinity. The
resulting Dirac delta function satisfies the desired
conditions and is defined by the relations (81) and

x   
0
x0
x0
problem 10.

Evaluate (a)
 cosd ,

2
(b)
 cosd .

(82)
Properties and Representations
Properties
Several delta function properties are collected here: Among the most important is
 f  y  y  a dy  f a 
(83)
problem 11.
Prove Eq. (83). Hint: change the variable from y to x = y-a then use one of the basic
relations (81)
problem 12.
2
Evaluate
 cos  d
0
problem 13.
Prove the following properties.
 x    x 
(84)
1
a x    x 
a
Representations of the Delta Function
The delta function is written in many ways. The most important of these is based on the
Fourier integral theorem.
Fourier Integral Theorem
f x  

 g k e
ikx
dk

g k  
(85)

1
f  x e ikx dx

2  
A way to look at this theorem is to suppose you are given a function f(x) that you
want to expand in the base states exp(ikx). The second expression shows how to find the
coefficients of the expansion and the first is the expansion. You can use this view to
prove the most important delta function representation:
 x  
problem 14.
Prove Eq. (86).
Other Representations

1
e ikx dk

2  
(86)
Applications
Delta functions are easy to manipulate and are used here to analyze simplified versions of
standard problems. We treat a bound state of a delta function well and reflection-transmission
from a delta function barrier.
Boundary Conditions
The boundary conditions imposed on the usual square well or square barrier are that at
the interface of regions I and II obey
 I   II and I  II
with similar conditions at the interface of regions II and III.
When the barrier or well is infinitesimally narrow, the wavefunction must still match on
both sides,      , but the slope changes abruptly. This can be seen by writing the
Schrödinger equation with potential Vx and integrating over an infinitesimal interval   ,  
surrounding x=0. Taking the limit   0 gives the result
2mV
(87)
  0    0  2 0

problem 15.
Show that in the neighborhood x=0 of a potential Vx , the boundary condition for
2mV
slopes is given by   0    0  2 0 .

Bound State
The bound state energy of delta function well V x   S x  is calculated as follows:
(a) Find solutions to the Schrödinger equation for regions outside of x=0 that assure
  0 as x   . The form for the wavefunction to the left and right of the potential are then
   ex and    ex where  is a positive real constant, (b) Apply the boundary conditions at
x=0:
  0    0
  0    0 
2mV
0
2
This imposes a condition on the constant  that determines the allowed energy.
problem 16.
Find the bound state energy and wavefunction for V x   S x . Sketch the

mS 2 
wavefunction and well. ans E   2 
2 

Reflection and Transmission
Reflection and transmission from a delta function barrier V x  S x is calculated as
follows: (a) Find a solution to the Schrödinger equation for the region to the left of x=0 that
describes waves traveling both left and right (reflected and incident waves). The form for the
wavefunction here is   x   Aeikx  Be ikx where k is a positive real constant. (b) For the region
to the right of the delta function, find a solution to the Schrödinger equation that represents a
transmitted wave moving to the right,   x   Ceikx . (c) Apply the boundary conditions at x=0:
  0    0
  0    0 
2mV
0
2
These give relations between the coefficients. (d) Determine the reflection and transmission
2
2
B
C
coefficients using any two of R 
,T
, and R + T=1.
A
A
problem 17.
Find reflection and transmission coefficients for the potential energy V x  S x . At
what incident energy are reflection and transmission equally likely?
A9 Perturbation Theory
Basic Problem ..................................................................................................................1
First Order Results ...............................................................................................1
Derivation ............................................................................................................2
Applications .....................................................................................................................4
Alkali Atoms ........................................................................................................4
Normal Zeeman Effect .........................................................................................5
We can find exact solutions to only a few basic systems like the simple harmonic
oscillator and the Hydrogen atom. More complex systems must be treated with approximation
techniques. Foremost among these is the perturbation approach where we neglect relatively
small terms to arrive at a simple system (however artificial) that can be solved. Then we add
back corrections (perturbations) due to the formerly neglected quantities.
As an example, consider finding the ground state energy of He. If we neglect the
interactions between the electrons and consider only the forces between the nucleus and each
electron, the wavefunction is made of Hydrogen wavefunctions. The electron—electron
interaction is then treated as a perturbation and its approximate contribution to energy is added to
the energy of the simplified system.
Basic Problem
The basic time-independent perturbation problem is as follows: Consider a Hamiltonian
H,
H  H0  V ,
where H0 is the unperturbed Hamiltonian for a fully solved system and V is a relatively small
perturbation term. Find corrections to the unperturbed eigenvalues and eigenfunctions to various
degrees of accuracy.
First Order Results
At present, we limit ourselves to the simplest case of first order corrections. Denote the
energy and eigenstate of the unperturbed system as E0 and 0 (here the zero does not necessarily
refer to the ground state). Then we have
H0 0  E 0 0
(88)
and the correction to the energy (to first order) is
E  E0  0 V 0 .
(89)
The first order eigenstate is given by the expression
nV 0
n
n  0 E0  En
0  0  
(90)
where the sum is over the possibly infinite number of unperturbed eigenstates. In practice, only
a few states near the original 0 are included.
problem 18.
Find first order corrections to both energies and wavefunctions for a particle in a rigid
one-dimensional box of length L containing potential V x  x, 0  x  L where is a
small parameter.
Derivation
A clever but simple derivation produces Eqs. (58) and (90). Introduce an artificial
parameter  as a coefficient for the presumed small interaction V so that
H  H0   V .
Now we can think of the solutions as power series in and write
(91)
E  E0   E 1  2 E 2   ...
,
  0  
1
 
2
2 
(92)
 ...
where the superscripts indicate the order of the correction. Substitute these into the exact
expression,
H E
(93)
and since Eq. (62) must hold for an arbitrary value of , the coefficients of each power ofmust
be independent and equal on each side of (62). We obtain
0 : H 0 0  E0 0
1 : H 0 
1
2 : H 0 
2 
 V 0  E0 
V 
1
1
 E 1 0
 E0 
2 
 E 1 
(94)
1
 E 2  0
problem 19.
Derive the intermediate results (63).
The first order approximations can be obtained from the second of Eqs. (63), but first the
1
unknown kets  must be re expressed in terms of known quantities; in particular, they are
expanded in terms of the complete base states of the unperturbed problem n :

1
  an n .
(95)
n0
Substitute and apply the bra 0 to both sides of the expression so that the normalization
condition 0 0  1 gives Eq. (58). The coefficients an are similarly determined by applying the
bra n and using the orthonormality condition n k   nk . Equation (64) then gives the first
order state of Eq. (90).
problem 20.
Derive Eqs. (58) and (90) from Eqs.(63).
Applications
Perturbation theory is applied so widely that almost every branch of modern physics will
provide examples. Here we give two relatively simple physical applications of first order
perturbation theory.
Alkali Atoms
The alkali elements have one outer electron orbiting an inner core of electrons. On
average, this inner core can be treated as spherically symmetric. When the outer electron is far
from the core, it sees only a single charge and has a Hydrogenic energy level. When it penetrates
the inner core, it sees a larger positive charge and is perturbed from the Hydrogen level.
We can consider the outer electron
energy to be a perturbed Hydrogen level.
The situation is shown in the diagram where
the total interaction experienced by the
valence electron is the sum of a single
proton and a combined point charge +(Z-1)e
surrounded by a uniform cloud of negative
charge -(Z-1)e
problem 21.
Show that the potential corresponding to the above diagram is given by
2
 Z  1) 2 e 2

Z  1)  e 2 r 2
 k
k
rR
V 
r
R3

rR
0
The degree to which the outer electron
penetrates the inner core depends on its angular
momentum . This agrees with classical ideas where
the lower the angular momentum, the closer the orbit
2
approaches the nucleus. It follows that  is larger
for smaller .
Note that the perturbation V becomes increasingly
negative for decreasing r. It follows that the
perturbation in energy,
E 1   V  dr
2
(96)
is more negative for lower angular momenta. This explains why the nS level is lower than the
nP level in Alkali metals (n=2 for Lithium and n=3 for Sodium, etc.)..
problem 22.
Briefly explain why the outer electron S wave is lower than the outer electron P wave in
Alkali atoms.
Normal Zeeman Effect
When a hydrogen atom is placed in a strong uniform magnetic field B, certain energy
levels “split” into multiple levels. The interaction of the magnetic field with the orbiting electron
causes perturbations of the original energy levels that depend on the angular momentum. We
note that this “normal” Zeeman effect neglects the contribution of electron spin.
You will recall that a current I experiences a force when it is placed in a magnetic field.
The magnetic moment  for a current loop of area A is defined by   IA . The classical value
for the electron current is
e
(97)

L
2m
where L is the angular momentum and the energy of the interaction with the field is given by
V     B . If the direction of B is chosen to be the z-direction, then
e
(98)
V
BL z
2m
problem 23.
m eB 

Calculate the energy perturbations for the normal Zeeman effect. ans : 
2m 

Notice that each  level is now split into 2  +1 levels by the magnetic field. This is a
special case of a general phenomenon. Levels that are degenerate in energy are often perturbed
differently and they acquire different energies. We say that the degenerate level is split by the
interaction.