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!

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

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.

True or False

For a particular metal, it is possible that green light will eject electrons, but ultraviolet waves will not.

Matter-Wave Duality

(

)

a

a

a

a

†

†

†

=

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.

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

[

]

hf

E

photon

=

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,

t

i

V

m

¶

Y

¶

=

Y

+

Y

Ñ

-

h

h

2

2

2

.

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

L

,

,

,

3

2

1

Y

Y

Y

.

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

2

Y

(much like the energy of an electromagnetic wave is given by the square of its fields).

question

ú

û

ù

ê

ë

é

=

l

mv

h

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

Y

and

2

Y

are eigenfunctions for two possible states, then the system may exist with a wavefunction

Y

that is a combination like

(

)

2

1

2

1

Y

+

Y

and

2

5

4

1

5

3

Y

+

Y

. However, the only values that we can obtain from a single measurement are eigenvalues associated with

1

Y

or

2

Y

, and the probability that the system will be found in condition 1 or 2 depends on the proportion of

1

Y

and

2

Y

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

r

t

Y

+

Y

=

Y

8

.

0

6

.

0

(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,

2

Y

. (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.

ú

û

ù

ê

ë

é

»

D

D

»

D

D

h

E

t

h

p

x

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.

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:

(1)

f

v

l

=

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)

hf

E

=

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)

K

W

hf

+

=

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:

(5)

p

h

=

l

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

m

p

E

2

2

=

and derive the allowed energy levels

E

n

h

mL

n

=

2

2

2

8

.

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 (:

p

=

2

nh

l

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

(

)

r

n

p

=

l

2

.

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,

m

p

K

2

2

=

The total energy E including the potential energy V(x) is then given by K(p) + V(x) = E or

(6)

(

)

E

x

V

m

p

=

+

2

2

ii) Replace the momentum p in Eq. (6) by a differential operator

dx

d

D

º

:

D

i

p

h

-

®

where

h

(called “h-bar”) is the Planck constant divided by 2(,

(7)

p

=

2

h

h

Equation (6) is then converted into the operator form

(8)

(

)

E

x

V

D

m

=

+

-

2

2

2

h

iii) Produce the 1-dimensional, time-independent Schrödinger equation by applying both sides of Eq. (8) to the wavefunction ( The result is

(9)

(

)

Y

=

Y

+

Y

-

E

x

V

D

m

2

2

2

h

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

(

)

kx

A

sin

=

Y

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

(

)

x

k

i

A

exp

=

Y

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,

2

2

2

1

x

m

V

w

=

. Show by substitution that

(

)

2

exp

ax

A

-

=

Y

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

K

,

,

2

1

E

E

and corresponding wavefunctions

EMBED Equation.3

K

,

,

2

1

Y

Y

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

(

)

(

)

kx

A

x

sin

=

Y

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 wavefunction,

2

n

Y

. Consequently, if a particle is confined to a box of length L,

L

x

£

£

0

,

it must have zero probability of being outside and we require that

(10)

(

)

(

)

0

and

0

0

=

Y

=

Y

L

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

(

)

kx

A

sin

=

Y

.

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

n

E

x

2

1

2

1

problem 1.

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

Y

the event of finding it in a condition

a

is characterized by an amplitude written as

Y

a

.

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

(1)

(

)

2

|

Y

a

=

Y

a

P

( discrete

When ( is a continuous (instead of discrete) quantity,

(

)

Y

a

|

P

is a probability density and

(

)

Y

a

|

P

d( is the probability of a measurement yielding a value in a neighborhood d( of (.

(2)

(

)

a

Y

a

=

a

Y

a

d

d

P

2

|

( continuous

For the particle-in-a-box amplitude , the probability of a particle with energy En being found in a neighborhood dx around the point x is ||2dx. Amplitudes of the form that are probability densities for position and are called wave functions. Not all amplitudes are wave functions;

and are counterexamples.

problem 2.

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 3.

Given that the wave function for a bead on a ring of circumference L is

÷

ø

ö

ç

è

æ

p

=

L

inx

L

p

x

2

exp

1

,

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 4.

Given that the wave functions for a particle in a box of length L are

÷

ø

ö

ç

è

æ

p

=

L

nx

L

E

x

n

sin

2

,

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 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

(

)

(

)

c

b

a

c

b

a

a

b

b

a

+

+

=

+

+

+

=

+

2. Multiplication by a Scalar: multiplication by any complex number is distributive and associative.

(

)

(

)

a

a

a

b

a

b

a

m

+

l

=

m

+

l

l

+

l

=

+

l

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

(

)

0

a

a

=

-

+

problem 1.

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,

(1)

å

=

i

i

c

e

v

,

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 2.

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,

(2)

(

)

å

å

÷

ø

ö

ç

è

æ

p

+

÷

ø

ö

ç

è

æ

p

=

L

nx

b

L

nx

a

x

f

n

n

sin

cos

This can be viewed as a special case of Eq.(1).

problem 3.

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 and has the following properties:

1. = *

2- = +

3. > 0 and = 0 implies |a>=0

Two vectors whose scalar product is zero are said to be orthogonal. The length of vector |a> is defined as ½

problem 4.

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 5.

Calculate 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)

(

)

L

x

u

/

sin

p

=

and

(

)

L

x

v

/

2

sin

p

=

on the interval (0,L) (e)

(

)

L

ix

u

/

2

exp

p

=

and

(

)

L

ix

v

/

4

exp

p

=

on the interval (0,L)

problem 6.

Calculate the lengths of the u vectors in the previous problem.

problem 7.

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; =0 and =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,

(3)

0

=

m

n

e

e

and normalized to a unit length,

(4)

1

=

n

n

e

e

.

When the base vectors satisfy both Eqs.(3) and (4), we say the basis is orthonormal.

problem 8.

Show that i, j, k are orthonormal.

problem 9.

Show that the set of functions

÷

ø

ö

ç

è

æ

p

=

L

ix

n

L

p

x

2

exp

1

with n=0, 1, 2,... are orthonormal.

problem 10.

Show that the set of functions

÷

ø

ö

ç

è

æ

p

=

L

nx

L

E

x

n

sin

2

with n = 1, 2,... are orthonormal.

Linear Vector Space Principle

Physical states |(> are vectors in a linear vector space and amplitudes

Y

a

are associated scalar products.

This principle has profound consequences. First, we note from the definition of the scalar product that

(5)

= <*,

indicating that forward and reverse events are closely related.

problem 11.

Given that

÷

ø

ö

ç

è

æ

p

=

L

ix

n

L

p

x

n

2

exp

1

for a particle on a ring, write the amplitude for the particle to be found with momentum pn when its position is known to be x.

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 left- and 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

(6)

å

=

Y

n

n

n

u

c

where we will assume the base states are orthonormal.

problem 12.

A special case of Eq.(6) is |(> = 0.8|u1> + 0.6|u2>.Show that the u1 component equals and the u2 component equals . 2-14 Show that Eq.(6) can be written as

(7)

Y

=

Y

=

Y

å

å

n

n

n

n

n

n

u

u

u

u

problem 13.

Write the state |(> of Schrödinger’s cat given that the amplitude for it to be alive is

3

2

. Use the symbols |L> and |D> for the live and dead states.

The most succinct expression of the superposition principle follows from Eq.(7),

(8)

å

=

n

n

n

u

u

1

problem 14.

Use the orthonormal functions

÷

ø

ö

ç

è

æ

p

=

L

ix

n

L

n

x

2

exp

1

and Eq.(8) to expand an arbitrary function f(x)= in terms of these functions and coefficients . The result is a Fourier series.

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:

70

10

100

5

50

4

0

1

=

×

+

×

+

×

=

g

Another way to look at the same average is:

100

10

5

50

10

4

0

10

1

+

+

=

g

.

Notice that the probability of a student getting 0 is 1/10 and the probability of getting 50 is 4/10, etc. We write

100

50

0

100

50

0

×

+

×

+

×

=

p

p

p

g

.

This is a special case of a general way to take an average over g1, g2,( when the corresponding probabilities p1, p2, (are known:

j

j

j

p

g

g

å

=

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 (:

(

)

dx

x

x

dx

x

+

=

and

between

being

system

of

y

probabilit

r

In this case, average g is ,

(

)

(

)

dx

x

x

g

g

b

a

ò

=

r

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:

(

)

dx

x

g

g

b

a

Y

Y

=

ò

*

This form works even when g represents an operator like momentum (

dx

d

i

h

-

).

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 (.

(1)

Y

w

=

WY

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:

(2)

( (c1|a> + c2|b>) = c1(|a> + c2(|b>

Operators may also operate “backward” on a bra-vector

(3)

when (|a> = c|b> then

problem 1.

A translation operator T is defined by T|x> = |x-b> or

(

)

x

x

Y

=

Y

, show that

(

)

b

x

T

x

-

Y

=

Y

.

problem 2.

Given

(

)

L

ixp

p

x

/

/

exp

h

=

, show that the effect of T can be expressed in momentum language as

(

)

(

)

dx

b

x

ixp

L

T

p

-

Y

=

Y

ò

h

/

exp

1

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.(1), 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.(1) could be multiplied by any scalar, so eigenvectors are arbitrary by a constant factor (unless they are normalized to unit length).

problem 3.

Find the eigenvalues and eigenvectors of the matrix operator

÷

÷

ø

ö

ç

ç

è

æ

0

1

1

0

problem 4.

Find the eigenvalues p and eigenvectors ((x) (eigenfunctions) for

(

)

(

)

x

p

x

dx

d

i

Y

=

Y

-

h

where we require that (0) = ( (L).

problem 5.

Find the eigenvalues E and eigenvectors ((x) (eigenfunctions) for

(

)

(

)

x

E

x

dx

d

m

Y

=

Y

-

2

2

2

2

h

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 (1) is written as

(4)

w

w

=

w

W

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

(5)

*

W

=

W

a

b

b

a

The right hand side of Eq.(5) is called the Hermitian adjoint and is denoted with a dagger,

*

+

W

=

W

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 6.

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 7.

Confirm that the operator of problem 4 is Hermitian. Note that the eigenvectors form the basis for expanding functions in Fourier series.

problem 8.

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 9.

The y-component of spin angular momentum is represented by

÷

÷

ø

ö

ç

ç

è

æ

-

=

0

0

2

i

i

s

y

h

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

(6)

x

i

p

¶

¶

-

=

h

)

or, in bra-ket notation,

(6’)

x

x

i

p

x

¶

¶

-

=

h

)

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,

(7)

V

p

m

H

+

=

2

2

1

)

where V is an operator corresponding to the system potential energy.

Equations (6) and (7) are restricted to one dimension and to non-relativistic systems. A three-dimensional form will be given later. We find it simple and direct to accept Eq.(6) as a subordinate principle.

problem 10.

Demonstrate that

h

)

)

)

)

i

x

p

p

x

=

-

.

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 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

(8)

w

w

=

w

W

ii) construct the wave mechanical operator

The classical form of the observable ( is usually a function of the position x and momentum p,

(

)

p

x

classical

,

W

=

W

then the appropriate quantum mechanical operator is obtained by replacing p by the operator

D

i

p

h

)

-

=

. (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

)

.)

(9)

(

)

D

i

x

h

-

W

=

W

,

where ( is a differential operator

iii) write the wave mechanical eigenvalue problem

The wave mechanical counterpart of Eq. (8) is a differential equation,

(10)

(

)

Y

=

Y

Y

w

=

Y

W

x

D

i

x

where

,

h

When the operator is the Hamiltonian, ( = H, then the eigenvalue is energy, ( = E, and Eq. (10) is Schrödinger’s equation,

Y

=

Y

E

H

.

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

(

)

(

)

L

x

x

+

Y

=

Y

whereas a particle bound to a potential source must have

(

)

0

=

¥

±

Y

(no probability of escape to infinity). A particle in a box of length L with impenetrable walls satisfies the condition

(

)

(

)

0

0

=

Y

=

Y

L

. Reflection and transmission from a boundary requires a smooth connection at the boundary;

right

left

Y

=

Y

and

right

left

D

D

Y

=

Y

.

Applications: Particle on a Loop and Particle in a Box

problem 11.

Calculate the allowed values of momentum p for a bead constrained to a loop of perimeter L.

problem 12.

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 13.

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

2

1

The relative probability that the particle will be reflected is the ratio R called the reflection coefficient,

R = |B exp(-ikx)|2 / |A exp(ikx)|2

or

(11)

2

2

A

B

R

=

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 (11) 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;

right

left

Y

=

Y

and

right

left

D

D

Y

=

Y

.

problem 14.

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

(1)

m

=

w

k

2

The classical Hamiltonian is the sum of kinetic and potential energies,

(2)

2

2

2

2

1

2

x

p

H

mw

+

m

=

and conservation of energy E can be expressed as

E

H

=

The quantum analogue of this expression is an eigenvalue problem for energy:

(3)

E

E

E

H

=

,

where H is an operator having the same form as Eq.(2), but with scalars p and x replaced by operators

p

)

and

x

)

.

Wave Mechanics for Oscillator

When the state in Eq.(3) is taken to be in the x-representation, , the expression becomes a differential equation called the (time-independent) Schrödinger equation:

(4)

E

x

E

E

x

x

E

x

D

=

mw

+

m

-

2

2

2

2

2

1

2

h

problem 1.

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

(5)

(

)

2

1

+

w

=

n

E

n

h

where n is an integer. Each value of En has an associated wavefunction (n = (this is written more simply as ), and some solutions are graphed below.

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:

(6)

[

]

h

i

p

x

=

,

(7)

n

E

n

H

=

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

=

=

w

=

m

h

. 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 2

Define the operator

(

)

a

x

ip

=

+

1

2

and derive the relation

(8)

[a, a+] = 1

problem 3

Show that H = = a+a + ½.:

(9)

H = = a+a + ½.

problem 4

Demonstrate that a+a is the “number operator,” that is,

(10)

a+a |n> = n |n>.

problem 5

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

w

and

h

.)

Eigenfunctions in the Dirac Approach

Eigenfunctions can be generated by first finding the lowest eigenfunction and then “raising” it repeatedly to , , etc.

problem 6

Use the fact that < corresponds to the lowest n value to calculate the ground state wavefunction.

problem 7

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

,

=

h

or in expanded form and setting

h

=

1

xp

px

i

-

=

(1)

.Substitute operators

(

)

a

x

ip

=

+

1

2

and

(

)

a

x

ip

†

=

-

1

2

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

a

and

†

. This will later help us with algebraic manipulations that will give the energy levels.

It is easiest to introduce

(

)

a

x

ip

=

+

1

2

and

(

)

a

x

ip

†

=

-

1

2

into the expression

H

a

a

=

+

†

1

2

(3)

After reducing the result with Eq. (1), we get

(

)

H

p

x

=

+

1

2

2

2

.

Solution 4.4

First we note that any operator with the form a†a is Hermitian* so the eigenvalues

of a†a are real. We write

a

a

n

n

n

†

=

(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

l

, so that

aa

n

†

|

|

l

l

>=

>

(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

†

|

(

)|

l

l

>=

-

>

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

µ

-

1

(7)

The reason we use a proportionality sign is because we cannot guarantee that

l

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

†

µ

+

1

(8)

Solution 4.5

From Equations (3) and (4) we have

(

)

(

)

a

a

n

n

n

†

+

=

+

1

2

1

2

or

(

)

H

n

n

n

=

+

1

2

(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

M

L

T

T

M

L

T

x

y

z

2

1

1

2

2

-

-

-

=

This is satisfied only when

x

y

z

=

=

=

0

1

1

,

,

so that Eq. (9) becomes

(

)

H

n

n

n

=

+

h

w

1

2

(10)

Solution 4.6

The lowest energy corresponds to n = 0. This state cannot be lowered so

a

0

0

=

(11)

The wave function equivalent of this is

(

)

1

2

0

0

ip

x

+

=

j

(12)

where p is the usual differential operator with = 1. The resulting differential equation is

d

dx

x

j

j

0

0

=

-

(13)

with the solution

(

)

j

0

1

2

2

=

-

const

exp

x

(14)

The exponent must be dimensionless, so the constants m, , ( must be restored.

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

l

such that

(

)

L

=

+

h

l

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,

L

m

z

=

h

.

(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

l

and m. The rules restricting the quantum numbers are given below:

SYMBOL 183 \f "Symbol" \s 10 \hn principal quantum number is unrestricted

n = 1, 2, 3, ...

SYMBOL 183 \f "Symbol" \s 10 \h

l

orbital angular momentum quantum number is limited to a value below the value of n,

L

z

=

h

m

L

l

= 0, 1, 2, ..,n-1

SYMBOL 183 \f "Symbol" \s 10 \hm 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

,

,

,

.

.

,

l

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.

SYMBOL 183 \f "Symbol" \s 10 \hsz spin quantum number can only have one of two values,

s

z

=

±

1

2

1Use the rules to find how many Hydrogen electron states are possible when

(a) n = 1, and (b) n = 2.

2Calculate and sketch the possible angles between L and the z-axis for

(a)

l

= 1, (b)

l

= 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 many-electron atoms. Very small differences in energy occur for the different

l

values due to various interactions between electrons. For example, when these complications are included for n = 2, the

l

=0 state is found to be slightly lower in energy than the

l

= 1 state. Electrons with different values of

l

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

l

+1) and the number of spin states, 2:

maximum number of electrons in a subshell = 2 (2

l

+1)

Some historic names for the

l

values are

l

= 0

s wave

l

= 1

p wave

l

= 2

d wave

l

= 3

f wave

and the subshells are denoted by the combination “n

l

” so that the n=1,

l

=0 subshell is written 1s and the n=2,

l

=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,

l

=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

1

2

3

4

5

6

7

8

2

1

2

3

4

5

6

7

8

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

8

8

8

8

8

8

8

8

4The 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.

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.

5Write 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:

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

problem

An amino acid is drawn here without hydrogen atoms. Complete the picture using H’s and sticks.

problem

A sugar is drawn here without hydrogen atoms. The corners or vertices are assumed to be carbons. Complete the picture using C’s, H’s and sticks.

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 .

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|() = |<|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|() = |<|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, <

When wavefunctions of multidimensional systems can be "separated," into essentially one-dimensional factors, the problem is made tractable. An example of a separable wavefunction is < = <. This separation is not always possible. It takes a judicious choice of representation to accomplish it. As a counterexample, the wavefunction <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) J2 = h2<3>l<1> (<3>l<1> + 1), where <3>l<1> is integer or half-integer

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 spherical harmonics 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

l

and m.

Hydrogen Eigenvalue Problem

The hydrogen atom problem is most tractable when it is expressed in spherical coordinates

f

q

,

,

r

. Let pr and J be the operators for radial and angular momenta, respectively. Then the Hamiltonian is

r

ke

r

J

p

H

r

2

2

2

2

2

1

-

÷

÷

ø

ö

ç

ç

è

æ

+

m

=

(1)

where

r

ke

2

-

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,

l

,m> and the energy eigenvalue problem is

m

E

E

m

E

H

,

,

,

,

l

l

=

(2a)

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

2

J

operator. Since

(

)

m

E

m

E

J

,

,

1

,

,

2

2

l

l

l

h

l

+

=

, we can write

m

E

E

m

E

H

,

,

,

,

l

l

l

=

(2b)

where

(

)

r

ke

r

p

H

r

2

2

2

2

1

2

1

-

÷

÷

ø

ö

ç

ç

è

æ

+

+

m

=

l

l

h

l

(3)

In the following sections, we solve the eigenvalue problem (2b) by rewriting

l

H

in a “factored” form

(

)

l

l

l

b

A

A

-

+

m

2

1

. The

+

l

A

operator will raise the

l

eigenvalue and

l

A

will lower it.

H in Factored Form

Define an operator

+

l

A

with the form

(

)

[

]

l

l

l

h

b

r

i

p

A

r

-

+

+

=

-

+

1

1

(4)

where

(

)

1

2

2

+

m

=

l

h

l

ke

b

(5)

The problem sequence that follows will determine

l

H

in terms of the A+ and A operators.

6-1 Show that

(

)

[

]

(

)

[

]

r

r

p

r

i

b

r

p

A

A

,

1

/

1

1

2

2

-

+

+

-

-

+

+

=

l

h

l

h

l

l

l

6-2 Given the required commutation relation between r and its conjugate momentum,

[

]

h

i

p

r

r

=

,

, show that

[

]

2

1

,

-

-

-

=

r

i

p

r

r

h

.

6-3 Use the results of problems 6-1 and 6-2 to derive the form

(

)

l

l

l

l

b

A

A

H

-

m

=

+

2

1

(6)

.

Raising and Lowering Properties

The next two problems determine the raising and lowering properties of the A operators.

6-4 Show that

[

]

(

)

2

2

/

1

2

,

r

A

A

+

-

=

+

l

h

l

l

. (Use the results of problems 6-1 and 6-2.)

6-5 Demonstrate that

m

E

A

,

,

l

l

+

is proportional to

m

E

,

1

,

+

l

; that is,

+

l

A

is a raising operator for

l

. Similarly, show that

m

E

A

,

,

l

l

is proportional to

m

E

,

1

,

-

l

such that

l

A

is a lowering operator for

l

. 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

*

l

be the maximum angular momentum at a particular energy E so that

m

E

,

,

*

l

cannot be “promoted” to higher

l

:

0

,

,

*

=

+

m

E

A

l

l

(7)

This condition enables us to determine eigenvalues & eigenfunctions of hydrogen.

6-6 Show that the energy eigenvalues for operator

l

H

in Eq. (6) are given by

2

2

4

1

2

n

e

E

n

h

m

-

=

(8)

where n = 1, 2, 3,...

Hydrogen eigenfunctions can be written in separated form,

(

)

(

)

f

q

=

Y

,

m

n

m

n

Y

r

R

l

l

l

(9)

where Rn(r) is the radial wavefunction. It is easily determined when operator pr is expressed in differential operator form:

÷

ø

ö

ç

è

æ

+

-

=

r

dr

d

i

p

r

1

h

(10)

(This expression can be verified from the commutation relation

[

]

h

i

p

r

r

=

,

.)

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 density

2

10

2

4

R

r

p

is a maximum at the Bohr orbit. (Use

0

0

=

+

R

A

)

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

±

=

D

l

As a corollary, this requires

1

,

0

±

=

D

m

.

Background

Classical electrodynamics says an accelerating charge e radiates power P according to

2

3

2

3

2

a

c

ke

P

=

where a is the acceleration and

(

)

1

0

4

-

pe

=

k

.

problem

Use dimensional analysis to find

2

3

2

a

c

ke

P

µ

.

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

2

3

2

r

e

c

k

P

w

µ

.

The quantity er is the (average) dipole moment of the atom. Only non-vanishing averages

'

'

'

2

m

n

r

m

n

l

l

allow transitions between the primed and unprimed levels. It can be shown that the integrations over vanish unless

1

±

=

D

l

problem

Show that

0

201

100

2

=

r

and

0

210

100

2

¹

r

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

j

i

j

i

W

=

W

,

,

(1)

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

÷

÷

ø

ö

ç

ç

è

æ

=

+

0

0

1

0

S

and

÷

÷

ø

ö

ç

ç

è

æ

=

-

0

1

0

0

S

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

÷

÷

÷

÷

÷

ø

ö

ç

ç

ç

ç

ç

è

æ

Y

Y

Y

=

Y

n

M

2

1

(2)

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

D

L

3

1

3

2

+

=

Y

, calculate the corresponding eigenvector in the “live” and “dead “basis.”

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:

t

i

H

¶

¶

=

h

)

(1)

or

Y

¶

¶

=

Y

t

i

H

h

)

.

(1’)

8-1. Show that a formal solution to Eq.(1’) is

|(t) = |(0) exp(-itH/()

8-2. Show that when observable commutes with H, [,H] = 0, then is conserved.

8-3. 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

y

n

m

l

for various values of n, SYMBOL 108 \f "MT Extra", and mSYMBOL 108 \f "MT Extra". a0 denotes the Bohr radius.

a

e

0

0

2

2

10

4

0

529

10

=

=

´

-

pe

m

h

.

m

.

The separated form for the wavefunctions is

(

)

(

)

(

)

(

)

j

F

q

Q

=

j

q

y

l

l

l

l

l

l

m

m

n

m

n

r

R

r

,

,

(1)

where

(

)

(

)

(

)

(

)

(

)

(

)

r

r

e

r

R

e

n

r

n

m

m

im

m

in

polynomial

cos

in

polynomial

sin

/

constant

l

l

l

l

l

l

l

-

j

=

q

q

=

q

Q

=

j

F

(2)

A7 Identical Particles

The Eigenvalues of Exchange1

Fermi and Bose Particles2

Fermi Particles2

Bose Particles2

Quantum Statistics3

Applications4

Free Electron Model for Metals4

Density of States for Electrons5

Black-Body Radiation6

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