Chap 5. Identical Particles

1. Two-Particle Systems

2. Atoms

3. Solids

4. Quantum Statistical Mechanics

Two particles having the same physical attributes are equivalent.

They behave the same way if subjected to the same treatment.

CM: Equivalent particles are distinguishable since one can

keep track of each particle all the time.

QM: Equivalent particles are indistinguishable since one cannot keep

track of each particle all the time due to the uncertainty principle.

Indistinguishable particle are called identical.

5.1. Two-Particle Systems

2-particle state :

i H

t

1 2, , tr r

2 2

2 21 2 1 2

1 2

, ,2 2

H V tm m

r r

Schrodinger eq.:

2 3 31 2 1 2, , t d dr r r r Probability of finding particle 1 & 2 within

d 3 r1 & d 3 r2 around r1 & r2 , resp.

Normalization : 23 31 2 1 2, , 1 d d tr r r r

1 2,V V r r

1 2 1 2, , H Er r r r

1 2 1 2, , , exp

it E tr r r r

Read Prob 5.1,Do Prob 5.3.

5.1.1. Bosons & Fermions

1 & 2 indistinguishable (identical) :

1 2 1 2, a br r r rDistinguishable particles 1 & 2 in states a & b, resp. :

1 2 1 2 1 2, a b b aAr r r r r rbosonsfermions

Spin statistics theorem :Integer spin bosons

Half-integer spin fermions

No two fermions can occupy the same state.

1 2 1 2 1 2, 0 a a a aAr r r r r r

Pauli exclusion principle

Total :Symmetric

Anti-symm

Exchange Operator

Exchange operator : 1 2 2 1, ,P f fr r r r

21 2 2 1, ,P f P fr r r r 1 2, f r r f 2 1P

Let g be the eigenfunction of P with eigenvalue : P g g

2 P g P g 2 g g

1

For 2 identical particles, 1 2 1 2 2 1 2 1, , , , P H Hr r r r r r r r 1 2 1 2, ,H Pr r r r

P H H P or , 0P H

H & P can, & MUST, share the same eigenstates.

i.e.

1 2 2 1, , r r r r forbosonsfermions

Symmetrization requirement

Example 5.1. Infinite Square Well

Consider 2 noninteracting particles, both of the same mass m, inside an infinite square well of width a.

1-particles states are 2sin

n

nx x

a afor

222

2

nE n

m a2n K

Distinguishable particles :

1 2 1 21 2 1 2, n n n nx x x x

1 2

2 21 2 n nE n n K

E.g., Ground state: 11 1 2 1 2

2, sin sin

x x x xa a a

11 2E K

1st excited state : 12 1 2 1 2

2 2, sin sin

x x x xa a a

21 1 2 1 2

2 2, sin sin

x x x xa a a

12 5E K

21 5E K

Doubly degenerate.

Bosons :

2sin

n

nx x

a a2nE n K

Ground state: 0 1 2 1 2

2, sin sin

x x x xa a a

0 2E K

1st excited state :

1 1 2 1 2 1 2

2 2 2, sin sin sin sin

x x x x x x

a a a a a1 5E K

Fermions :

Ground state:

0 1 2 1 2 1 2

2 2 2, sin sin sin sin

x x x x x x

a a a a a0 5E K

Nondegenerate

5.1.2. Exchange Forces

Particles distinguishable : 1 2 1 2, a bx x x x

Consider 2 particles, one in state a, the other in state b.

( p’cle 1 in a, 2 in b. )

Bosons : 1 2 1 2 1 2

1,

2 a b b ax x x x x x

Fermions : 1 2 1 2 1 2

1,

2 a b b ax x x x x x

We’ll calculate for each case the standard deviation of particle separation

2 2 21 2 1 2 1 22 x x x x x x

( All states are normalized )

Distinguishable Particles

2 * 21 1 2 1 2 1 1 2, , D

x d x d x x x x x x

1 2 1 2, D a bx x x x 2 2 21 2 1 2 1 22 x x x x x x

2a

x * 2 *1 1 1 1 2 2 2 a a b bd x x x x d x x x

(, a , b normalized )

Similarly,2 22

D bx x

*1 2 1 2 1 2 1 2 1 2, , D

x x d x d x x x x x x x

* *1 1 1 1 2 2 2 2 a a b bd x x x x d x x x x

a bx x

2 2 21 2 2

a ba bDx x x x x x

Identical Particles

1 2 1 2 1 2

1,

2 a b b ax x x x x x 2 2 2

1 2 1 2 1 22 x x x x x x

2 2 2 2 2

1 2 1 2 1 2

* * * *1 1 2 2 1 1 2 2

1,

21

2

a b b a

a b b a b a a b

x x x x x x

x x x x x x x x

22 21 1 2 1 2 1,

x d x d x x x x 2 21

2 a b

x x a ,b orthonormal

Similarly,2 2 22

1

2 b a

x x x

2

1 2 1 2 1 2 1 2, x x d x d x x x x x

1 1

2 2 a b b a ab ba ba ab

x x x x x x x x a b ab ba

x x x x

* a babx d x x x x

2 2 21 2 1 2 1 22 x x x x x x2 2 2 2

1 2

1

2 b a

x x x x

1 2

a b ab bax x x x x x

2 2 21 2 2 2

a b ab baa bx x x x x x x x

2

1 2 2 ab baD

x x x x

or 22 22

abDx x x

Bosons are closer & fermions are further apart than the distinguishable case.

2 2

Dx xNote : if particles are far apart so a ,b don’t overlap.

effectiveattractiverepulsive exchange force for

bosonsfermions

Simplified Derivation

A A

2 2 2a ba b

x x x x

2 2 21 2 1 2 1 22 x x x x x x

D a b 1 2 1 2, D a bx x x x ~

2 2 21 2 1 2 1 22

Dx x ab x ab ab x ab ab x x ab

1 2 1 2 1 2

1,

2a b b ax x x x x x 1

2a b b a ~

1

2A ab A ab ba A ba ab A ba ba A ab

2 2 2 2 21 1 1 1 1

1

2x ab x ab ba x ba ab x ba ba x ab

2 2 2 21

2 a b ab bax x x b a x a b

2 21

2 a bx x if 0a b

abA a A b

Similarly,

2 2 22

1

2 a bx x x

1 2

1

2 a b b a ab ba ba abx x x x x x x x x x

a b ab ba

x x x x

2 2 21 2 2 2

a b ab baa bx x x x x x x x

2

1 2 2ab baD

x x x x

2 2 21

1

2 a bx x x

or 22 22

abDx x x

Bosons are closer & fermions are further apart than the distinguishable case.

2 2

Dx xNote : if particles are far apart so a ,b don’t overlap.

effectiveattractiverepulsive exchange force for

bosonsfermions

H2

If e’s were bosons, form bond. e’s are fermions, H2 dissociates

Let the electrons be spinless & in the same state :

In actual ground state of H2 :Spins of the e’s are anti-parallel so the spatial part is symmetric.

Do Prob 5.7

5.2. Atoms

Atom with atomic number Z ( Z protons & Z electrons ) :

2 2 22

1

1

2 2

Z Z

jj j kj j k

Z e eH k k

m r r r

1-e plan of attack :

1.Replace e-e interaction term with single particle potential.

2.Solve the 1-e eigen-problem.

3.Contruct totally anti-symmetric Z-e wave function, including spins.

4.Total energy is just the sum of the 1-e energies.

Nucleonic DoF dropped.See footnote, p.211.

0

1SI unit

4

1 Gaussian

k

Non-Interaction e Model

2 22

1 2

Z

jj j

Z eH k

m rDrop all e-e terms :

1

Z

jj

h r

where 2 2

2

2

Z eh k

m rr ( Hydrogenic hamiltonian )

where nlm and En are obtained from the hydrogen case by setting e2 Ze2 .

nl m n nl mh Er r rThus

In particular : 2

213.6 n

ZE eV

n0

aa

Zand

(Unsymmetrized) solutions to H E are 11

, ,

j j j

Z

Z n l m jj

r r r

with1

j

n

nj

E E

a0 = Bohr radius

5.2.1. Helium

2 2 2 2 22 21 2

1 2 1 2

2 2

2 2

e e e

H k k km r m r r r

1 2 H h h Er rNon-interacting e model :

1 2 1 2, nl m n l mr r r r n nE E E

2

213.6 n

ZE eV

n

Ground state :

0 1 2 100 1 100 2, r r r r 1 230 0

8 2exp

r ra a

3/2 /10 2 r aR a e

00

1

4Y

0a

aZ

20 13.6 2 1 1 E eV 13.6 8 eV 109 eV

Anti-symmetrized total wave function :0 gives only symmetric spatial part spin part must be antisymm (singlet).

Experiment : Total spin is a singlet. E0 79 eV.

Excited States

Long-living excited states :

1 2 1 100 2, nl mr r r r

( both e in excited states quickly turns into an ion + free e. )

Singlet Triplet

Spatial part of parahelium is symmetric

higher e-e interaction

higher energy than orthohelium counterpart

Do Prob 5.10

5.2.2. The Periodic Table

n \ l 0 s 1 p 2 d 3 f 4 g

1

2

3

4

5

6

72(2l+1) 2 6 10 14 18

Filling order of the periodic table.

l – degeneracy lifted by e-e interaction (screening).

Ground State Electron Configuration

Spectral Term :2 1S

JL

Ground state spectral terms are determined by

Hund’s rules (see Prob 5.13 ) :

1.Highest S.

2.Highest L.

3.No more than half filled : J = | L S |.

More than half-filled : J = L+S .

See R.Eisberg,R.Resnick, “Quantum Physics”, 2nd ed., §10-

3.E.g. , C = (2p)2 .

m 1 0 1

sz

30P S = 1, L = 1, J = 0

Do Prob 5.14

5.3. Solids

1. The Free Electron Gas

2. Band Structure

5.3.1. The Free Electron Gas

Solid modelled as a rectangular infinite well with dimensions { li , i = 1, 2, 3 }.

0 0

i ix lV

otherwiser

22

2

E

m

2 2

22

ii i

i

d XE X

m d x

i ii

X xr

ii

E E

Let2

ii

mEk

22

2i

i ii

d Xk X

d x

0 i ix lfor

sin cos i i i i i i iX A k x B k x

2 2

2

1

2 i

i i i

d XE

m X d x

Set

Set

i =1, 2, 3 unless stated otherwise

2 2

2 i

i

kE

m

Boundary Conditions2

ii

mEk

0 0iX 0iB

0iX l

sin cos i i i i i i iX A k x B k x

i i ik l n 1, 2, 3, in

sini i i iX A k x

22 2

0 0

sin i il l

i i i i i id x X A d x k xNormalization :2

2 i

i

lA 2 sin 2

sin2 4

x x

d x x

2i

i

Al

2sin

i

i ii i

nX x

l l

1 2 3

2sin

in n n i

i i i

nx

l lr 1 2 3

1 2 31 2 3 1 2 3

8sin sin sin

n n nx x x

l l l l l l

sin

ii i i

i

nX A x

l

1 2 3

2 2

2 i

n n ni

kE

m

22

2

i

i i

n

m l

22

2

m

k 1 2 3, , k k kk

k-Space Density

1, 2, 3, in

i

ii

nk

l

i i

i

k nl

Volume occupied by one allowed (stationary) state in the 1st quadrant of the k-space is

k ii

k

i il

3

V

= Volume of solid ii

V l

31

8

k

d k

k

33

2

Vd k

k is over allowed k in 1st quadrant of k-space. d 3 k is over all k-space.By symmetry, all 8 quadrants are equivalent.

= k-space density = density of allowed states 32V

with ni 1.

Fermi Energy

Consider a solid of N atoms, each contributing q free electrons.In the ground state, these qN electrons will occupy the lowest qN states.

Since2

2

2

E

mn kthese states occupy a sphere centered at the origin of the k-space.

Let the radius of this sphere be kF .

3

3

42

32

F

Vk qN ( Factor 2 comes from spin degeneracy. )

1/323

F

qNk

V 1/323

qN

V= free electron density

22

2

F FE k

m

22/323

2

m

Fermi energy

Total Energy

Total energy of the ground state :

totsmallest

E Enn

2 23

3222

Fk k

V kd k

m

2 22

30

2 422

Fk

V kk d k

m

25

210

F

Vk

m

1/323 Fk 5/32

22

310

tot

V qNE

m V

25/32 2/3

23

10

qN V

m

Compressing the solid increases its engergy.

Work need be done to compress it.

Electrons exerts outward pressure on the solid boundary.

totd EP

dV

2

3 tot

tot

Ed E dV

V

2

3 totE

V

25

2

2

3 10

Fk

m

2/32 2

5/33

5

mdegeneracy pressure

( Caused by Pauli exclusion )

Miscellaneous

25

210tot F

VE k

m

3 2 2

25 2F FV k k

m

1/323 e

F

Nk

V

2 2

2F

F

kE

m

3

5tot

Fe

EE

N Energy per e.

2

3 totPV EIdeal gas :

BN k T

2

3 totPV E2

5 e FN E

2

5F

eB

ET

k

5

16

2

8.617 10 /

6.582 10

0.511 /

B

e

k eV K

eV s

m MeV c

e gas :

Do Prob 5.16and for Al

5.3.2. Band Structure

1-D crystal with periodic potential V x a V x a = lattice constant

2 2

22

d

H V x Em d x

Simplest model : 1-D Dirac comb

For a finite solid with N atoms, strict mathematical periodicity can be achieved by imposing the periodic boundary condition V x Na V x

Bloch’s Theorem

V x a V xBloch’s theorem : i K ax a e x K = const

Proof : Let D be the displacement operator : D f x f x a

d x d x a

Dd x d x a

d x a

d x

dD x

d x

D V x x V x a x a V x D x

2 2

2 2

d d

D Dd x d x

D H H D i.e.

DH HD , 0D Hor

, 0D H D & H can share the same eigenstates.

Let D x x x a x

Since 0, setting i K ae completes the proof.

Periodic Boundary Condition

V x a V x i K ax a e x

For a 1-D solid with N atoms, imposing the periodic BC x Na x

gives 1i K N ae

2

n

KN a

0, 1, 2, n

K is real 2 2 x a x As expected

Dirac Comb

1-D Dirac comb :

1

0

N

j

V x x ja

For x ja :2 2

22

d

Em d x

22

2

d

kd x

2

mE

k

sin cos x A k x B k x

Periodic BC

sin cos x A k x B k x

Impose periodic BC : x Na x

Bloch’s theorem i K ax a e x2

n

KN a

0, 1, 2, n

Let

i K ax e x a

sin cos x A k x B k x

x ja

continuous at x = 0 sin cos i K aB e A ka B ka

0 < x < a

a < x < 0 sin cos i K ae A k x a B k x a

2

20

d d m

d x d x 2

2cos sin

i K a mA k e k A ka k B ka B

sin cos i K aB e A ka B ka 2

2cos sin

i K a mA k e k A ka k B ka B

sin cos 1 0 i K a i K ae ka A e ka B

2

21 cos sin 0

i K a i K a mk e ka A ke ka B

Condition for consistency is

2

sin cos 102

1 cos sin

i K a i K a

i K a i K a

e ka e ka

mk e ka ke ka

2

2sin sin 1 cos cos 1 0

i K a i K a i K a i K amke ka e ka k e ka e ka

2 2 2 22

2sin sin 1 2 cos cos 0

i K a i K a i K a i K amke ka e ka k e ka e ka

22

2sin 1 2 cos 0

i K a i K a i K ame ka k e ka e

22

2sin 1 2 cos 0

i K a i K a i K ame ka k e ka e

2

2sin 2cos 0

i K a i K amka k e ka e

2

2sin 2 cos cos 0

m

ka k Ka ka

2cos cos sin

m

Ka ka kak

Let z ka 2

m a

sincos cos

zKa z f z

z

Since cos 1Ka there’s no solution for | f (z) | > 1 ( band gaps ).

2 2 2 2

22 2

k zE

m ma

10

Do Prob 5.19with 20

5.4. Quantum Statistical Mechanics

1. An Example

2. The General Case

3. The Most Probable Configuration

4. Physical Significance of and

5. The Blackbody Spectrum

Fundamental assumption of statistical mechanics :In thermal equilibrium, each distinct system configuration of the same E is equally likely to occur.

( Can also be stated in terms of ensemble. )

5.4.1. An Example

Consider 3 non-interacting particles ( all of mass m ) in an 1-D infinite square well.

A B CE E E E22

2 2 2( )2

A B Cn n n

m a

Let 2 2 2 363 A B Cn n n

There’re 13 combinations of (nA , nB , nC ) that can satisfy the condition :

(11, 11, 11),

(13, 13, 5), (13, 5, 13), (5, 13, 13),

(1, 1, 19), (1, 19, 1), (19, 1, 1),

(5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5).

Occupation Number

Specification of system configuration:

(nA , nB , nC ) occupation number

(11, 11, 11) ( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... ) N11 = 3

(13, 13, 5), (13, 5, 13), (5, 13, 13) ( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... ) N5 = 1N13 = 2

(1, 1, 19), (1, 19, 1), (19, 1, 1), ( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... ) N1 = 2N19 = 1

(5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5).

( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1N7 = 1N17 = 1

Pn : Particles Distinguishable

Pn = Probability of finding a particle with energy En .

P1 (3/13) (2/3) = 2/13

P5 (9/13) (1/3) = 3/13

P7 (6/13) (1/3) = 2/13

P11 (1/13) (3/3) = 1/13

P13 (3/13) (2/3) = 2/13

P17 (6/13) (1/3) = 2/13

P19 (3/13) (1/3) = 1/13

Sum = 1

Particles are distinguishable each configurations is distinct.

Each is equally likely when system is in equilibrium.

(nA , nB , nC )

(11, 11, 11)

(13, 13, 5), (13, 5, 13), (5, 13, 13)

(1, 1, 19), (1, 19, 1), (19, 1, 1),

(5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5).

Total = 13 distinct configurations

Pn : Fermions

Pn = Probability of finding a particle with energy En .

P1 0

P5 (1/1) (1/3) = 1/3

P7 (1/1) (1/3) = 1/3

P11 0

P13 0

P17 (1/1) (1/3) = 1/3

P19 0

Sum = 1

No state can be doubly occupied.Allowed distinct configurations are :

( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1, N7 = 1, N17 = 1

occupation number

( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... ) N11 = 3

( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... ) N5 = 1N13 = 2

( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... ) N1 = 2N19 = 1

( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1N7 = 1N17 = 1

Total = 1 distinct configurations

Pn : Bosons

Pn = Probability of finding a particle with energy En .

P1 (1/4) (2/3) = 1/6

P5 (1/4) (1/3) + (1/4) (1/3) = 1/6

P7 (1/4) (1/3) = 1/12

P11 (1/4) (3/3) = 1/4

P13 (1/4) (2/3) = 1/6

P17 (1/4) (1/3) = 1/12

P19 (1/4) (1/3) = 1/12

Sum = 1

Allowed distinct configurations are :

( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... ) N11 = 3( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... ) N5 = 1, N13 = 2( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... ) N1 = 2, N19 = 1( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1, N7 = 1, N17 = 1

occupation number

( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... ) N11 = 3

( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... ) N5 = 1N13 = 2

( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... ) N1 = 2N19 = 1

( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1N7 = 1N17 = 1

Total = 4 distinct configurations

5.4.2. The General Case

Consider a system whose 1-particle energies are Ei with degeneracy di , i = 1,2,3,...

Now, N particles are put into the system such that Ni particles have energy Ei .

Question :

For a given configuration (N1 , N2 , N3 ,... ), what is the number, Q(N1 , N2 , N3 ,... ),

of distinct states allowed?

Distinguishable Particles

Each energy level Ei corresponds to a bin with di compartments.

The problem is equivalent to finding the number of distinct ways to put N particles,

with Ni of them going into the di compartments of the ith bin.

Ans.: 1. There’re 1

1 1

!

! !N

N

NC

N N N

ways, without regard of picking order, to choose N1 particles from the whole N particles.

2. When putting a particle into the 1st bin, there’re d1 choices of compartments.

the number of ways to put N1 particles into the 1st bin is 1

1

1 1

!

! !

NN d

N N N

321

1 2 1 2 311 2 3

1 1 2 1 2 3 1 2 3

! !!, , ,

! ! ! ! ! !

NNN N N d N N N dN dQ N N N

N N N N N N N N N N N N

3. For the 2nd bin, one starts with N N1 particles so one gets

2

1 2

2 1 2

!

! !

NN N d

N N N N

31 21 2 3

1 2 3

!

! ! !

NN NN d d d

N N N 1

!!

iN

i

i i

dN

N

Fermions

1. Fermions are indistinguishable, so it doesn’t matter which ones are going to which

bin.

2. Each bin compartment can accept atmost one fermion.

So the number of way to place Ni fermions into the ith bin is

!

! !i

i i i

d

N d Nif i id N

1 2 31

!, , ,

! !

i

i i i i

dQ N N N

N d N

Bosons

1. Bosons are indistinguishable, so it doesn’t matter which ones are going to which bin.

2. Each bin compartment can accept any number of bosons.

Consider placing Ni bosons into the di compartments of the ith bin.

Let the bosons be represented by Ni dots on a line.

By inserting di 1 partitions, the dots are “placed” into di compartments.

If the dots & partitions are all distinct, the number of all possible arrangement is 1 ! i iN d

However, the dots & partitions are indistinguishable among themselves. Hence

1 2 3

1

1 !, , ,

1 ! !

i i

i i i

N dQ N N N

d N

5.4.3. The Most Probable Configuration

An isolated system in thermal equilibrium has a fixed total energy E,

and fixed number of particle N , i.e.,

1

ii

N N1

i ii

N E E

Since each possible way to share E among N particles has the same probability to exist,

the most probable configuration (N1 , N2 , N3 ,... )

is the one with the maximum Q(N1 , N2 , N3 ,... ).

Since Q(N1 , N2 , N3 ,... ) involves a lot of factorials, it’s easier to work with lnQ.

Lagrange Multipliers

Problem is to maximize ln Q, subject to constraints1

i ii

N E E1

ii

N Nand

Using Lagrange’s multiplier method, we maximize, without constraint,

1 1

ln

i i ii i

G Q N N E N E

i.e., we set 0

i

G

N

Note : 0

G 0

Gare just the original constraints.

Distinguishable Particles

1

!!

iN

i

i i

dQ N

N

1

ln ln ! ln ln !

i i ii

Q N N d N

Stirling’s approximation : ln ! ln z z z z for z >>1

1

ln ln ln ln

i i i i ii

Q N N N N d N N N

1 1

ln

i i ii i

G Q N N E N E

1 1 1

ln ln ln

i i i i i i i ii i i

G N N N N d N N N N N E N E

ln ln

i i ii

Gd N E

N0

exp i i iN d E

Fermions 1 1

ln

i i ii i

G Q N N E N E

1

!

! !

i

i i i i

dQ

N d N 1

ln ln ! ln ! ln !

i i i ii

Q d N d N

di , Ni >>1 : 1

ln ln ln ln

i i i i i i i ii

Q d d N N d N d N

1 1 1

ln ln ln

i i i i i i i i i i ii i i

G d d N N d N d N N N E N E

ln ln

i i i ii

GN d N E

N0

1 exp ii

i

dE

N

exp 1

i

ii

dN

E

Bosons 1 1

ln

i i ii i

G Q N N E N E

1

ln ln 1 ! ln 1 ! ln !

i i i ii

Q N d d N

di , Ni >>1 :

1 1 1

1 ln 1 1 ln 1 ln

i i i i i i i i i i ii i i

G N d N d d d N N N N E N E

ln 1 ln

i i i ii

GN d N E

N0

11 exp

ii

i

dE

N

exp 1

i

ii

dN

E

1

1 !

1 ! !

i i

i i i

N dQ

d N

1

ln 1 ln 1 1 ln 1 ln

i i i i i i i ii

Q N d N d d d N N

di 1 di

Do Prob 5.26, 5.27

5.4.4. Physical Significance of and

Ideal gas in 3-D infinite square well.

2 2

2

k

kE

m, ,

yx z

x y z

nn n

l l lk

3

32

V

d kk

Taking the “bin” as the spherical shell between k and k+dk, the degeneracy dk isjust the number of states in the shell :

2

3 42

k

Vd k d k 2

22

Vk d k

Distinguishable Particles

exp i i iN d E

222

k

Vd k dk

1

ii

N N 22

0

exp2

k

VN dk k E

22 2

20

exp2 2

V

e dk k km

3/2

22

mV e

2

10

1 1 1

2 2

n x

n

ndx x e

1

2

3/2

2 2

2

2 4

V me

3/222

Ne

V m

1

i ii

N E E2 2

4 22

0

exp2 2 2

V

E e dk k km m

5/22

2 2

3 2

4 4

V me

m3/2

2

3

2 2

V mE e

3

2

N

cf. kinetic theory result : 3

2 BE N k T

1 Bk T universal

Distributions

Setexp

exp 1

exp 1

ii i

B

ii

i

B

ii

i

B

EN d

k T

dN

Ek T

dN

Ek T

Bk T

D

F

B

exp

1

exp 1

1

exp 1

B

B

B

nk T

n

k T

n

k T

ii

i

Nn

d

Maxwell – Boltzmann

Fermi – Dirac

Bose - Einstein

Chemical potential

n() Most probable number of particles in a state with energy . occupation number

Fermi-Dirac as T 0 1

exp 1

B

n

k TT 0 :

0 0exp

0

Bk T

1 0

0 0

n (0) EF

T 0

T > 0

Ideal Gas

Distinguishable particles 3

2 BE N k T

3/222

Ne

V m

3/222ln

B

B

Nk T

V mk T

23 2ln ln

2

B

B

Nk T

V mk T

Indistinguishable particles :

3

32

k

VN d k n

2

2 2 202 1

exp 12

B

V kdk

kk T m

22

02

k

Vdk k n F

B

3

32

k k

VE d k n

2 4

2 2 202 2 1

exp 12

B

V kdk

m kk T m

FB

fixes

gives

EC

TRead Prob. 5.29 Do Prob 5.28

5.4.5. The Blackbody Spectrum

Photon quantum of EM field

spin 1, massless boson with v c.

Only m 1 occurs.

Number not conserved 0.

E h

2

kc

exp 1

B

dN

k T/

k k cd d 2

2 32

2

Vd

c

N

dV

Energy density in range d

3

2 3

1

exp 1

B

c

k T

Black body spectrum

Blackbody Spectrum

6000 K

4000 K

2000 K

Visible region

Do Prob 5.31Read Prob 5.30