量子力学学习课件第二章英文版

Chapter

2

Time-independentSchrödinger

EquationStationary

States

24The

infinite

Square

well

30The

Harmonic

Oscillator

40The

Free

Particle

59The

Delta-Function

Potential

68The

Finite

Square

Well

782.1

Stationary

StatesHow

to

get

wave

function?To

solve

the

Schrödinger

Equationwhere

V

is

a

specified

potential

and

independent

of

t.In

this

case

the

Schrödinger

Equation

can

be

solved

bythe

method

of

separation

of

variables,

looking

forthefollowing

solutions:Substituting

above

equations

into

the

Schrödinger

Equation:and,

dividing

through

bySeparation

of

variables

has

turned

a

partial

differentialequation

into

two

ordinary

differential

equations.②①Called

time-independentSchrödinger

Equation.The

rest

of

this

chapter

will

be

devoted

to

solving

the

time-independent

Schrödinger

Equation,

for

a

variety

of

simplepotentials.What

are

the

solutions

and

why

the

solutions

of

time-independent

Schrödinger

Equation

is

so

important.Three

properties

of

the

solutions:①:②:(1)

They

are

stationary

states.solutions:The

probability

density

is

time-independent.Every

expectation

value

is

constant

in

time.(2)

They

are

states

of

definite

total

energyIn

classical

mechanics,

the

total

energy

(kinetic

andpotential

energy)

is

called

the

Hamiltonian:Conclusion:A

separable

solution

has

the

property

that

everymeasurement

of

the

total

energy

is

certain

to

return

thevalue

E.(3)

The

general

solution

is

a

linear

combination

of

separablesolutions.A

more

general

solution

can

be

constructed

by

alinear

combination:Determine

the

constantsc

to

match

the

specifiedinitial

conditions:Example

2.1:2.2

The

Infinite

Square

WellOutside

the

well:Inside

the

well:whereSimple

harmonicoscillator

equationSolutionBoundary

conditions:

Both

and are

continuous.But

if

the

potentialgoes

toinfinity,

only

the

is

continuous.

(2.5)Boundary

points:0and

aAt

point

a:Discrete

energyTo

find

A,

using

the

normalization

condition:Phase

has

no

physical

significancen

=

3n

=

2n

=

10x0xGround

stateExcited

statesThe

functions have

some

interesting

andimportant

properties:They

are

alternately

even

and

odd,

with

respect

to

the

center

ofthe

well:

ψ1

is

even,

ψ2

is

odd,

ψ3

is

even,

and

so

on.Each

successive

state

has

one

more

node:

ψ1

has

0,

ψ2

has

one,

ψ3

has

two,

and

so

on.(3).

They

are

mutually

orthogonal,

in

the

sense

thatKronecker

deltaare

orthonormal.Proof:

see

the

book

as

a

homework.(4).

They

are

complete,

in

the

sense

that

any

other

function

cabe

expressed

as

a

linear

combination

of

them:Fourier

’s

seriesThe

stationary

states

of

the

infinite

square

well

areThe

most

general

states

of

the

infinite

square

well

arewhere

theyield

the

value

Enis

the

probability

that

a

measurement

of

the

energyThereforeThe

expectation

value

of

the

energy

must

beProof:Which

is

independent

of

time,

and

is

manifestation

of

conservation

of

E2.3

The

Harmonic

OscillatorClassical

harmonic

oscillator:mxkoThe

motion

is

governed

by

Hooke’s

Law:The

solution:The

potential

energy:Of

course,

there’s

no

perfect

harmonic

potential,

but

practically

anypotential

is

approximately

parabolic,

that

isWhich

describes

simple

harmonic

oscillation

aboutthe

point

x0,

with

aneffective

spring

constant

k=V”

(x0)

.Therefore

any

oscillatorymotion

is

approximately

simple

harmonic,

aslong

as

the

amplitude

is

small.The

quantum

problem

is

to

solve

the

Schrödinger

Equation

for

thepotentialTwo

approaches:Power

series

methodAlgebraic

technique:

using

ladder

operators2.3.1

Algebraic

MethodRewrite

Equation

:We

define

:Then

:Commutator

of

x

andpIn

general,

the

commutator

of

operators

A

and

B

is

defined

by?That

is:Canonical

commutation

relationSimilarly:Then

:(1)

IfTherefore

the

operators

a-

and

a+

are

called

ladder

operators.Raising

operatorLowering

operatorLowest

rungDetermineground

stateDifferential

equation:Determine

A:

normalizeSoGround

statewhere

An

is

the

normalization

constant.Therefore,

we

haveDetermine

A1

by

normalizingBut

the

method

to

determine

An

by

normalizing

is

so

difficult!To

develop

an

algebraic

method

to

find

AnExample2.4:We

suppose

thatFor

any

functions

f(x)

and

g(x)Proof:In

particular:(1):(2):

SimilarlyHenceThusThe

stationary

state

of

theharmonic

oscillator

areorthogonalExample2.5:

Find

the

expectation

value

of

the

potential

energy

in

the

nthstate

of

the

harmonic

oscillator.Solution:Evaluating

the

integral

!2.3.2

Analytic

MethodWe

return

to

the

Schrödinger

Equation

of

the

harmonic

oscillatorSolveit

directly

by

series

method:First,

we

introduce

a

dimensionless

variable:Then

the

equation

reads:where

K

is

the

energy,

in

unit

of:Approximation

solution:

whenAsSoTherefore,

we

suppose

the

form

of

the

solution

iswhere

we

supposeis

a

power

series

inSubstituting

above

solution

into

the

differential

equation,

we

haveSecond,

we

look

forby

series

method,

in

whichis

a

power

series

in,

that

isWe

have

thefollowing

equations:Putting

above

equations

intoWe

findItfollowsthat

the

coefficient

of

each

power

ofmust

be

zero.And

henceThis

is

a

recursion

formulawhich

can

generates

allcoefficients.(1)

Generates

the

even-numbered

coefficient

from

a0Then

we

have:(2)

Generates

the

odd-numbered

coefficient

from

a1TwoarbitraryconstantAnalysis

of

the

solution:Question:

If

the

sum

is

infinite

series?If

j

is

large

enough,

that

isThe

approximate

solution

will

be:Above

asymptotic

solution

is

not

the

physical

one

which

will

be

infinite

asTherefore

j

will

not

be

so

large,

that

is

the

power

series

must

terminateat

certain

value

j=n.Analysis

of

the

solution:Above

conditionrequiresthatThereforeQuantization

of

energy:For

allowed

values

of

K,

the

recursion

formula

readsFor

certain

n,

we

consider

the

solutions:(1)

For

n=0(2)

For

n=1(3)

For

n=2Conclusion

of:Hermite

polynomials

:By

tradition,

an

arbitrary

multiplicative

factor

is

chosen

forthe

coefficient

of

the

highest

power

of

is

2n,

that

isso

thaNormalize

to

determine

CnFigure

of

solutions-3

-2

-1

0

1

2

3E0

E1E2n

=

0n

=

1n

=

22-4

-24|10|2Example

2.1Problem

2.1Problem

2.2Example

2.2normalizing:the

nth

coefficient:wave

function

attimet:conservation

of

energy:Example

2.3Problem

2.3Problem

2.4Problem

2.5(homework)Problem

2.6(homework)Problem

2.7Problem

2.8

?Problem

2.9Example

2.5Problem

2.10Problem

2.11Problem

2.12Problem

2.13Problem

2.14Problem

2.15Problem

2.16Problem

2.172.4

The

Free

ParticlewhereThe

general

solution

of

above

equation

can

be

written

in

exponentialform:everywhereAbove

function

represents

a

combination

of

two

waves:Represents

a

wave

traveling

positive

in

the

x-directionRepresents

a

wave

traveling

in

the

negative

x-directionWe

can

cover

all

the

above

waves

by

letting

k

run

from

negative

to

positive:traveling

to

the

rigwheretraveling

to

the

lefSolution

whenk

is

fixed:The

“stationary

states”

of

the

free

particle

are

propagating

waves.The

wave

length

is:The

speed

of

these

waves

is

:The

classical

speed

of

a

free

particle

with

energyis1.

Properties

of

the

Solution

of

free

particle(1)

Normalization

of

separable

solution

with

certain

k

:This

wave

function

is

not

normalizable.That

means

the

separable

solutions

do

not

represent

physicalrealizable

states.

A

free

particle

cannot

in

a

stationarystate;

or,toput

it

in

another

way,

there

is

no

such

thing

as

a

free

particle

with

adefinite

energy.(2)

General

solution:As

a

free

particle can

carry

any

energy,

that

means

k

can

takecontinues

values.

Therefore

the

sum

of

the

separable

solutionsshould

be

replaced

by

integral

of

k.The

final

general

solution

of

a

free

particle:The

final

solution

of

a

free

particle

is

a

wave

packet

which

is

asuperposition

of

propagating

waves.

If

we

choose

proper

Φ(k)

,

thewave

function

can

be

normalized!Question:Plancherel’s

theorem:Inverse

Fouriertransformation

of

F(k)Fourier

transformation

of

f(x)Generic

quantum

problem:ThereforeExample

2.6:Solution:(1)

Normalization

to

determine

A:(2)

Calculate:(3)

Calculate:The

integral

can

not

be

solved

in

terms

of

elementary

functions,

thoughit

can

be

evaluated

numerically.(3)

DiscussionWhen

a<<1:When

a>>1:2.

Paradox:

speed

(phase

velocity

and

group

velocity)A

realphysicalstate

is

a

wave

packet

which

is

a

superposition

ofOn

sinusoidal

state

,

we

haveHow

to

determine

the

group

velocity

of

any

wave

packet

?as

a

function

of

,Generally,

dispersion

relation

is

a

formula

forthat

is

.For

example,

for a

wave

packet

of

free

particle:where

is

a

dispersion

relation.For

anywave

packet:(1)

Let

us

assume

thatvalue

.is

a

narrowly

peaked

about

some

particularor(2)

Changing

variables

fromto,

we

haveAt

t=0At

tThe

total

wave

packet

evidently

moves

along

x

at

a

speedthat

is

actuallythe

group

velocity

of

whole

wave

packet:,In

our

case

offree

particle:the

group

velocity:the

phase

velocity:Example

2.6Phase

velocity

and

group

velocityProblem

2.18Problem

2.19Problem

2.20(homework)Problem

2.21Problem

2.22(homework)2.5

The

Delta-Function

Potential2.5.1

Bound

states

and

scattering

statesDifferent

potentials:Infinite

square

wellHarmonic

oscillatorsolutionsFreeparticleNormalizableDiscrete

nNon-normalizableContinuous

kTwo

different

solutionsSumIntegralRead

book

!Properties

of

Delta-function:2.5.1

The

Delta-Function

Well1.

Delta-Function:2.

A

potential

of

Delta-Function:A

well

potential

of

Delta-Function:(1)

Bound

statesIn

the

region

x<0,

V(x)=0,

soThe

general

solution

to

above

equation

isAsThereforeIn

the

regionx>0,

V(x)=0,

so

Stitch

these

two

functions

togetherby

using

theboundary

conditions

as

folderivativeNow

we

use

it:The

first

condition:

the

function

is

continuous

at

any

pointThen,

we

haveThe

second

condition:

the

derivative

of

function

is

continuous

exceptis

not

continuous

at

point

x=0Generally,

integrate

the

Schrödinger

EquationIf

we

setIfWe

haveBy

using

it

on

delta-potentialThereforeThe

allowed

energy

is:NormalizationAtlast,

the

bound-state

solution

isone

bound-state

solution(2)

Scattering

statesIn

the

region

x<0,

V(x)=0,

soThe

general

solution

to

above

equation

isSimilarly,

in

the

region

x>0,

V(x)=0,

soStitch

these

two

functions

together

at

x=0

by

using

the

boundary

conditiThe

continuity

of

at

x=0

requires

thatThe

second

condition:

the

derivative

of

function

is

continuous

exceIntegrate

the

Schrödinger

EquationUnknown

variable:Two

equations,

and

the

scattering

states

can

not

be

normalized.How

to

determine

the

5

unknown

variablesIn

a

realizable

scattering

experiment,particles

are

fired

in

from

onedirection.

Here

we

suppose

a

leftcoming.Therefore:is

the

amplitude

of

the

incident

wave.is

the

amplitude

of

the

reflected

wave.is

the

amplitude

of

the

transmitted

wave.By

usingWe

findReflection

rate

and

transmission

rate:

R

and

TThe

probability

that

an

incident

particlewillbereflected

back

is

defined

byReflectioncoefficientMeanwhile,

the

probability

that

an

incidentparticle

will

be

transmitted

is

defined

byTransmissioncoefficientA

barrier

potential

of

Delta-Function:Read

the

book:2.6

The

Finite

Square

WellFinite

square

well:Bound-state

Solutions:

E<0In

the

region

of

:Or

:Where(1)(2)

In

the

region

of:Or

:WhereAsThe

general

solution

is(2)(3)

In

the

region

of:Or

:Where(3)Symmetric

problem:The

advantage

of

the

symmetry

is

that

we

need

only

imposethe

boundary

conditions

on

one

side

(say,

at

+a);

the

otherside(say,

at

–a)

is

then

automatically

satisfied.Even

solutions(Or

odd

solutions)Therefore

the

following

form

ofsolutionis

looking

forEven

solutionsIf

E

can

be

determined,

then

the

k

and

l

will

be

obtained,

and

the

wavefunction

will

be

plotted.EQuantum

m