量子力學(xué)學(xué)習(xí)課件第五章英文版

Chapter

5

Identical

particleTwo-particle

System

201Atoms

210Solids

218Quantum

Statistical

Mechanics

2305.1

Two-particle

SystemsFor

a

single

particle,

is

a

function

of

the

spatialcoordinates

r

and

the

time

t

(here

now

we

ignore

the

spin).The

wave

function

for

a

two-particle

system

is

a

function

ofthe

coordinates

of

particle

one

(r1),

the

coordinates

of

par-ticle

two

(r2),

and

the

time:The

Hamiltonian

for

the

whole

system

is:Its

time

evolution

is

always

determined

by

the

Schr?dingerequation:Normalization:For

time-independent

potentials,

we

obtain

a

complete

set

ofsolutions

by

separation

of

variables:The

time-independent

Schr?dinger

equation:5.1.1

Bosons

and

FermionsQuantum

mechanics

neatly

accommodates

the

E.

Fermiexistence

of

particles

that

are

indistinguishablein

principle:

We

simply

construct

a

wave

functionthat

is

noncommittal

as

to

which

particle

is

inwhich

state.

There

are

actually

two

ways

to

do

it:S.

N.

BoseSuppose

particle

1

is

in

the

(single-particle)

state,

and

particle

2

is

in

the

state .

Inthatcaseis

a

simple

conduct:+:

Bosons;

-:Fermionsones

with

integer

spins

are

bosons

(meson,

photon)ones

with

half-integer

spins

are

fermions

(proton,

electroFurther,

bosons

and

fermions

have

quite

different

statisticalproperties.

The

connection

between

spin

and

“statistics”

canbe

proved

in

relativistic

quantum

mechanics;

in

non-relativisttheory

it

must

be

taken

as

an

axiom.if ,

then:two

identical

fermions

(for

example,

two

electrons)

cannotoccupy

the

same

state,

this

is

the

famous

Pauli

ExclusionPrinciple!solidliquidgasplasmaBose-Einstein

condensationthe

fifth

state

of

matterA

more

general

formulation:define

the

exchange

operator,

P:bosonsfermionsFor

stationary

states:5.1.2

Exchange

Forcestwo

distinguishable

particles

in

one-dimensional

motion:two

identical

bosons:two

identical

fermions:How

about

the

expectation

value

of

the

separation

distancebetween

the

two

particles?Case

1:

distinguishable

particles:Case

2:

Identical

particles:Similarly:whereso:Comparing

the

two

cases,

we

have:This

is

called

exchange

forceP.

A.

M.

Dirac,1928T

h

e

s

y

s

t

e

m

b

e

h

a

v

e

s

a

s

t

h

o

u

g

h

t

h

eo

f

a

t

t

r

a

c

t

i

o

n

”

b

e

t

w

e

e

n

i

d

e

n

t

icc

l

o

s

e

r

t

o

g

e

t

h

e

r

,

a

n

d

a

“

f

o

r

c

e

oi

d

e

n

t

i

c

a

l

f

e

r

m

i

o

n

s

,

p

u

s

h

i

n

g

t

ht

h

a

t

w

e

a

r

e

f

o

r

t

h

e

m

o

m

e

n

t

i

g

n

o

r

a

n

e

x

c

h

a

n

g

e

f

o

r

c

e

,

a

l

t

h

o

u

g

h

i

t

’Hydrogen

moleculeCovalent

bonding

requiresthe

singlet

state!5.2

AtomsA

neutral

atom,

of

atomic

number

Z,

its

Hamiltonian

is:kineticpotential

in

nucleus

mutual

repulsionof

electronsSchr?dinger

equation:Solution

is

antisymmetric

with

respect

to

interchange

of

anytwo

electrons,

no

two

electrons

can

occupy

the

same

state:5.2.1

HeliumAfter

hydrogen,

the

simplest

atom

is

helium,

theHamiltonian

is:If

we

simply

ignore

repulsion

of

two

electrons,

the

solutioncan

be

written

as

products

of

hydrogen

wave

functions:The

total

energy

would

be:Then

the

ground

state

energy

is:The

experimentally

determined

energy

is

78.975eVThe

other

excited

states

are

as: ,

there

are

not

doubleexcited

state

for

helium.In

this

case,

we

can

construct

both

symmetric

and

antisymmetriccombinations,

in

the

usual

way:those

go

with

antisymmetric

spin

configuration

are

paraheliumones

go

with

symmetric

spin

configuration

are

orthohelium.The

parahelium

energiesAre

uniformly

higher

thantheir

orthohelium

ones.But

why

there

is

not

1Striplet

state?5.2.2

The

periodic

tableorbital=1

2

3

4

….n-1m=-

-

+1

…..shelln=1

shell

is

called

Kn=2

shell

is

called

Ln=3

shell

is

called

Mn=4

shell

is

called

Nsubshelll=0

is

called

s

(for

“sharp”)l=1

is

called

p

(for

“principle”)l=2

is

called

d

(for

“diffuse”n)ot

listedl=3

is

called

f

(for

“fundamental”)l=4

is

called

g

(out

of

imagination)l=5

is

called

hl=6

is

called

il=7

is

called

kmagneticTotal

angular

momentum

numberTotal

spin

momentum

numberThe

total

orbital

angular

momentum

numberelectronic

configuration:Pauli

exclusion

principle:Hund’s

RulesHund’s

first

rule:

total

spin

SHund’s

second

rule:

total

orbital

angular

momentum

LHund’s

third

rule:

total

momentum

J√ХF.

Hund5.3

SolidsIn

the

solid

state,

outermost

valence

electron

become

detachedand

roam

throughout

the

material.

Two

primitive

models:The

electron

gas

theory:Bloch’s

theory:

Periodic

potential5.3.1

The

Free

Electron

Gas

TheoryS

u

p

p

o

s

e

t

h

e

o

b

j

e

c

ti

n

q

u

e

s

ta

r

e

c

t

a

n

g

u

l

a

r

s

o

l

i

d

,w

i

t

hd

i

m

e

n

s

i

o

n

s

l

x

,

l

y

,

l

z

,

a

n

d

i

mt

h

a

t

a

n

e

l

e

c

t

r

o

n

i

n

s

i

d

ee

x

p

e

r

i

e

n

c

e

s

n

o

f

o

r

c

e

s

a

t

a

le

x

c

e

p

t

a

t

t

h

ei

m

p

e

nA.e

Stomrmearfbelld

e81

nominationsThe

Schrodinger

equation:where

k

is

the

magnitude

of

the

wave

vector,

k=(kx,ky,kz).Boundary

condition:Solution

of

wave

function:The

total

energy:If

you

imagine

a

three-dimensional

space,

with

axes

kx,ky,kz,

aplanes

drawn

in

atspace:(kx,ky,kzo)ne

stationary

statone

state

for

every

blockEach

block

in

this

grid,

and

henceeach

state,

occupies

a

volume

in

k-lxlylzSuppose

our

sample

contains

N

atoms,

and

each

atom

contributes

q

free

electronsAvogadro"s

Constant

=

6.0221415

×

1023

mol-1q

is

small

number:

1

or

2If

electrons

were

bosons

(

or

distinguishable

particles),

they

would

all

settlto

the

ground

state, .

But

electrons

are

in

fact

identical

fermions,

subjethe

Pauli

exclusion

principle,

so

only

two

of

them

can

occupy

any

given

state.They

will

fill

up

one

octant

of

a

sphere

in

k-space,

whose

radius,

kF,

is

determby

the

fact

that

each

pair

of

electrons

requires

a

volume

:Volume

of

k-space:Fermi

surfaceVolume

of

all

electron

require:therefore:where:It

is

free

electron

densityThe

boundary

separating

occupied

and

unoccupied

states,

in

k-space,

is

called

the

Fermi

surface

(hence

the

subscript

F).

Thecorresponding

energy

is

called

the

Fermi

energy,

EF;

for

a

freeelectron

gas:Volume

element

in

k-space:Number

of

states

in

v-element:the

energy

which

each

state

carries

is:the

energy

in

the

volume

element:the

total

energy

is:defining

quantum

pressure

P:Degeneracy

pressurePauli

exclusion

pressureElectron-electron

repulsionThermal

motion5.3.2

Band

StructureThe

periodic

potential

here

determines

thequalitative

behavior

of

solids.

Now

we

developthe

simplest

possible

model:

a

one-dimensionalDirac

comb,

consisting

of

evenly

spaced

deltafunction

spikes.F.

Bloch(1).

Theorem

of

periodic

potential:

Bloch

theoremA

periodic

potential

is

one

that

repeats

itself

aftersome

fixed

distance

a:Let

D

be

the

“displacement”

operator:then

we

are

free

to

choose

eigenfunctions

of

H

that

aresimultaneously

eigenfunctions

of

D:(2).

Band

structureno

real

solid

satisfies

the

condition

of

periodic

potential12N-1However,

for

large

number

ofatoms

containing

in

the

solid,we

can

wrap

the

x-axis

arounda

circle

to

meet

the

periodiccondition

and,

finally,

weimpose

the

following

boundarycondition:12Nthen:Now,

suppose

the

potential

consists

of

a

long

string

ofdelta-function

spikes:In

region

0<x<a

the

potential

is

zero,

sowhere:the

generation

solution:According

to

Bloch’s

theorem,

the

wave

function

in

thecell

immediately

to

the

left

of

the

origin

is:at

x=0,

there

are

two

boundary

condition,

one

is

fromEquation

2.125,

the

other

one

is

continuity:let:The

constant

β

is

a

dimensionless

measure

of

the

“strength”of

the

delta

function.all

values

of

Ka

are

very

closGraph

of

f(z):Graph

of

f(z)

for

β=10,

showing

allowed

bands

(shaded)

separateby

forbidden

gaps

(where

|f(z)|>1).(1)

Electrons

occupation

of

the

band:In

practice

there

will

be

Nq

of

them,

where

q

is

again

the

number

of

“free”

electrons

per

atom.

Because

of

Pauli

exclusion

principle,

only

two

electron

can

occupy

a

given

spatial

state,

so

if

q=1,

they

will

half

fill

the

first

ban

q=2

they

will

completely

fill

the

first

band,

if

q=3

they

half

fill

the

secon

band,

and

so

on——in

the

ground

state.(2)

Conductors,

insulators

and

semiconductorsBand

structure

is

the

signature

of

periodic

potential.If

a

band

is

entirely

filled,

it

takes

a

relatively

large

energy

to

excite

an

electron,

since

it

has

to

jump

across

the

forbidden

zone.

Such

materials

wil

be

electrical

insulators.

On

the

other

hand,

if

a

band

is

only

partly

filled,

takes

very

little

energy

to

excite

an

electron,

and

such

materials

are

typic

conductors.

If

you

dope

an

insulator

with

a

few

atoms

of

larger

or

smaller

q,

this

puts

some

“extra”

electrons

into

the

next

higher

band,

or

creates

som

holes

in

the

previously

filled

one,

allowing

in

either

case

for

weak

electri

currents

to

flow;

such

materials

are

called

semiconductors.5.4

Quantum

Statistical

Mechanics(1).The

fundamental

assumption

of

statistical

mechanicsis

that

in

thermal

equilibrium

every

distinct

state

with

thesame

total

energy,

E,

is

equally

probable!!(2).The

temperature,

T,

is

a

measure

of

the

total

energy

of

asystem

in

thermal

equilibrium

in

classical

mechanics.

Whatis

the

new

in

quantum

mechanics?

How

to

count

the

distinctstates!Problem:

If

we

have

a

large

number

of

N

particles,

inthermal

equilibrium

at

temperature

T

(total

energy

is

E),what

is

the

probability

that

a

particle

would

be

found

tohave

the

specific

energy,

Ej?5.4.1

An

exampleSuppose

we

have

just

have

three

noninteracting

particles,

A,B,

and

C,

(all

of

mass

m)

in

the

one-dimensional

infinitesquare

well.

The

total

energy

is:where

nA,

nB,

and

nC

are

positive

integers.

Now

suppose,

forthe

sake

of

argument,

that

total

energy

is

,which

is

to

say:Then,

there

are

13

combinations

of

three

positive

integers.Thus

(nA,nB,nC

)

can

be

one

of

the

following:configurationnumbercombinations1st1(11,11,11)2nd3(13,13,5),(13,5,13),(5,13,13)3rd3(1,1,19),(19,1,1),(1,19,1)4th6(5,7,17),(5,17,7),(7,5,17)(7,17,5),(17,5,7),(17,7,5)For

example,

(nA,nB,nC

)=(5,7,17)

means

nA=5,

nB=7,

nC=17,If

the

particles

are

distinguishable,

the

three-particle

stateThe

most

important

quantity

is

the

number

of

particles

ineach

state---the

occupation

number,

so

there

are

4configurations:(11,11,11)(13,13,5)(13,5,13)(5,13,13)(1,1,19)(19,1,1)(1,19,1)(5,7,17)(5,17,7)(7,5,17)(7,17,5)(17,5,7)(17,7,5)(1)(3)(3)(6)(1).

Distinguishable

particleIf

the

particles

are

distinguishable,

each

of

these

(nA,nB,nC

)represents

a

distinct

quantum

state,

and

the

fundamentalassumption

of

statistical

mechanics

says

that

in

thermalequilibrium

they

are

all

equally

likely.

Then

what

is

theprobability

(Pn)of

getting

a

specific

(allowed)

energy

En

？(2).

Identical

fermions:For

fermions,

no

two

particles

are

in

thesame

state.

This

antisymmetrizationrequirement

exclude

the

configurationswhere

two

particles

are

in

the

same

state.Only

the

fourth

configuration

isavailable

now!(3).

Identical

bosons:For

bosons,

each

configuration

enables

one

state,

soDiscussion:This

example

shows

that

the

nature

of

the

particlesdetermines

the

counting

properties,

or

the

statisticalproperties!

The

number

of

internal

distinct

states

is

differeand

the

probability

of

getting

specific

energy

is

different

tThis

example

gives

a

system

of

three

particles.

If

thenumber

of

particles

in

huge,

we

can

conclude:

Thedistribution

of

individual

particle

energies,

at

equilibriumsimply

their

distribution

in

the

most

probable

configuration5.4.2

The

General

CaseNow

consider

an

arbitrary

potential,

for

which

one

particleenergies

are:degeneracies

is:Suppose

we

put

N

particles

(all

with

the

same

mass)

intothis

potential;

we

are

interested

in

the

configurationfor

which

there

are

N1

particles

with

energy

E1,

N2

particleswith

energy

E2,

and

so

on.Now

we

consider

general

question:

how

many

distinctstates

correspond

to

this

particular

configuration?The

answer:

The

number

of

the

distinct

states

Q(N1,N2,N3,…)depends

on

whether

the

particles

are

distinguishable,identical

fermions,

or

identical

bosons.(1).

Distinguishable

particles:Choose

N1

from

N

for

energy

bin:

the

binomial

coefficient,

and

then

Arrangement

of

the

N1

particles

within

the

bin

on

the

degenerate

d1

states.

Thus

the

number

of

ways

to

putN1

particles,

selected

from

a

total

population

of

N,

into

a

bin

containing

d1

distinct

option,

is:The

same

goes

for

energy

bin

E2,

of

course,

except

thatthere

are

now

only

N-N1

particles

left

to

work

with:and

so

on,

it

follows:(2).

Identical

fermions:The

particles

are

identical.

The

antisymmetrization

requiresthat

only

one

particle

can

occupy

any

given

state,

so:(3).

Identical

bosons:The

particles

are

identical.

Although

the

wave

function

of

theN-particle

state

is

symmetry,

more

than

one

particles

canoccupy

the

draws

in

certain

bin.Example:

dn=5,

Nn=7Why?5.4.3

The

Most

Probable

ConfigurationIn

thermal

equilibrium,

every

state

with

a

given

totalenergy

E

and

a

given

particle

number

N

is

equally

likely.So

the

most

probable

configuration

(N1,

N2,

N3,

……)

isthe

one

that

particular

configuration

for

whichQ(N1,N2,N3,……..)

is

a

maximum,

subject

to

theconstraints:The

problem

of

maximizing

a

function

F(x1,

x2,

x3,……)

ofseveral

variables,

subject

to

the

constraints

f(x1,

x2,x3,……)=0,

f(x1,

x2,

x3,……)=0,

etc.,

is

most

convenientlyhandled

by

the

method

of

Lagrange

multipliers.

We

introduce

the

new

function:and

set

all

its

derivatives

equal

to

zero:In

our

case

it’s

a

little

easier

to

work

with

the

logarithm

oQ,

instead

of

Q

itself.

So

we

letwhere

α

and

β

are

the

Lagrange

multipliers.

Then:(1).

Distinguishable

particles:Assuming

the

relevant

occupation

numbers

(Nn)

are

large,we

can

invoke

stirling’s

approximation:then:Z=1015.1113.04Z=100364.0360.8Z=100059165912Z=30002103721033....It

follows

that:The

most

probable

occupation

numbers,

for

distinguishableparticles,

are:(