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
  r1, r2 , t 
2-particle state :
i
Schrodinger eq.:

H 
t
H 
  r1 , r2 , t 
Normalization :
V  V r1, r2 
2
2
2 m1

2
2m 2
22  V  r1 , r2 , t 
 Probability of finding particle 1 & 2 within
d 3 r1 & d 3 r2 around r1 & r2 , resp.
d 3r1 d 3r2
3
 d r1
 
2
1

d 3r2
  r1 , r2 , t 
2
1
 i

  r1 , r2 , t     r1 , r2  exp   E t 


H  r1, r2   E  r1, r2 
Read Prob 5.1,
Do Prob 5.3.
5.1.1. Bosons & Fermions
 r1, r2    a r1   b r2 
Distinguishable particles 1 & 2 in states a & b, resp. :
1 & 2 indistinguishable (identical) :
bosons
fermions
   r1 , r2   A  a  r1   b  r2    b  r1   a  r2  
Spin statistics theorem :
Integer spin

bosons
Total  :
Symmetric
Half-integer spin

fermions
Anti-symm
   r1 , r2   A  a  r1   a  r2   a  r1   a  r2    0
 No two fermions can occupy the same state.
Pauli exclusion principle
Exchange Operator
Exchange operator :
P f r1, r2   f r2 , r1 
P2 f r1, r2   P f r2 , r1   f r1, r2 
f

Let g be the eigenfunction of P with eigenvalue  :
P2  1
P g  g
P2 g   P g   2 g  g

  1

For 2 identical particles,

P  H  r1 , r2    r1 , r2    H  r2 , r1   r2 , r1   H r1, r2  P  r1, r2 
PH H P
or
 P, H   0

 H & P can, & MUST, share the same eigenstates.
i.e.
 r1, r2     r2 , r1 
for
bosons
fermions
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
2
 n
sin 
a
 a
 n  x 

x

 
2
En  n
  n K
2m  a 
2
2
2
for
Distinguishable particles :
n
1 n2
E.g., Ground state:
1st
excited state :
 x1, x2    n  x1  n  x2 
1
2
En1 n 2   n12  n22  K
 11  x1 , x2  
2
    
sin  x1  sin  x2 
a
a  a 
 12  x1 , x2  
2
    2 
sin  x1  sin 
x2 
a
a
a

 

 21  x1 , x2  
2
 2    
sin 
x1  sin  x2 
a
 a  a 
E11  2 K
E12  5K
Doubly degenerate.
E21  5K
 n  x 
2
 n
sin 
a
 a

x

En  n 2 K
Bosons :
 0  x1 , x2  
Ground state:
2
    
sin  x1  sin  x2 
a
a  a 
E0  2 K
1st excited state :
 1  x1 , x2  
2
a
  
 2 
 2 
  
sin
x
sin
x

sin
x
sin
2
1


 x2  
  a 1
a
a





 a 
 
E1  5K
Nondegenerate
Fermions :
Ground state:
 0  x1 , x2  
2
a
  
 2 
 2 
  
sin
x
sin
x

sin
x
sin
2
1


 x2  
  a 1

 a

 a 
 a 
 
E0  5K
5.1.2. Exchange Forces
Consider 2 particles, one in state a, the other in state b.
Particles distinguishable :
Bosons :
Fermions :
   x1, x2  
   x1, x2  
  x1, x2    a  x1  b  x2 
( p’cle 1 in a, 2 in b. )
1
 a  x1  b  x2   b  x1  a  x2 
2
1
 a  x1  b  x2   b  x1  a  x2 
2
( All states are
normalized )
We’ll calculate for each case the standard deviation of particle separation
 x1  x2 
2
 x12  x22  2 x1 x2
Distinguishable Particles
 x1  x2 
 D  x1, x2    a  x1  b  x2 
x12
2
 x12  x22  2 x1 x2
  d x1  d x2  *  x1 , x2  x12   x1 , x2 
D
2
  d x1  a*  x1  x12  a  x1   d x2  b*  x2  b  x2   x
x22
Similarly,
x1 x2
D
D
 x2
(, a , b normalized )
a
b
  d x1  d x2  *  x1 , x2  x1 x2   x1 , x2 
  d x1  a*  x1  x1  a  x1   d x2  b*  x2  x2  b  x2   x

 x1  x2 
 x2
2
D
a
 x2
b
2 x
a
x
b
a
x
b
Identical Particles
   x1 , x2 
x12

2



x22


2
1 2
x

2
  d x1  d x2    x1 , x2 
1
 x
2
a
 x12  x22  2 x1 x2
2
2
2
2
2
1
 a  x1   b  x2    b  x1   a  x2  

2
1
  a*  x1  b  x1  b*  x2  a  x2    b*  x1  a  x1  a*  x2  b  x2  
2
  d x1  d x2    x1 , x2 
Similarly,
x1 x2
 x1  x2 
1
 a  x1  b  x2   b  x1  a  x2 
2
   x1, x2  
x b x
b
x12 
1 2
x
2
a
 x2 
b
a ,b orthonormal
 x2 
b
a
2
x
x1 x2
1
x a    x
2
ab
x
ba
 x
ba
x
ab
  x
a
ab
  d x  a*  x  x  b  x 
x b x
ab
x
ba
x
 x
x1x2

2
1 
2
2 
 x
a
1
  x2
2
x b x
 x1  x2 

b
2

 x
ab
x
 x2
 x 
2


a
 x1  x2 


 x 
2
 x12  x22  2 x1 x2
ba
a
 x2
 x1  x2 

or
2
2
b
2
2 x
D
2 x
a
2 x
D
x
ab
b
x
2 x
ab
x
ba
ba
2
ab
 Bosons are closer & fermions are further apart than the distinguishable case.

Note :
attractive
repulsive
effective
 x 
2


 x 
2
D
exchange force for
bosons
fermions
if particles are far apart so a ,b don’t overlap.
Simplified Derivation
 x1  x2 
A   A
 D  x1, x2    a  x1  b  x2 
 x1  x2 
   x1, x2  
A
x12



2
D
~
2
 x12  x22  2 x1 x2
D  a b
 ab x12 ab  ab x22 ab  2 ab x1x2 ab  x 2
1
 a  x1  b  x2   b  x1  a  x2 
2
~
 
a
 x2
b
2 x
1
ab  ba

2
a
x
b

1
 ab A ab  ba A ba  ab A ba  ba A ab 
2

1
 ab x12 ab  ba x12 ba  ab x12 ba  ba x12 ab 
2

1 2
x

2
a
 x2
b
 x2
A
ab
ab
b a  x2
 a Ab
ba
1
a b    x2

2
a
 x2 
b
if
a b 0
x12


1 2
x

2
Similarly,
x1 x2

a
x22

 x1  x2 

1
 x
2
2

 x 
or
 x2 
b

1 2
x

2
x b x
a
 x2
2

a
 x2 
b

a
 x2
 x 
x
b
b
a
 x
2 x
2
2 x
D
a
x
ab
x
b
ba
 x
2 x
ab
ba
x
x
ab
ba
  x

a
x b x
 x1  x2 
2
ab
x
2 x
D
ba
ab
x
2
ab
 Bosons are closer & fermions are further apart than the distinguishable case.

Note :
attractive
repulsive
effective
 x 
2


 x 
2
D
exchange force for
bosons
fermions
if particles are far apart so a ,b don’t overlap.
ba
H2
Let the electrons be spinless & in the same state :
If e’s were bosons, form bond.
e’s are fermions, H2 dissociates
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
Z e2  1 Z
e2
H   
j  k
 k 
rj  2 j k r j  rk
j 1 
 2m
Z
 1

k   4 0
 1

Nucleonic DoF dropped.
See footnote, p.211.
SI unit
Gaussian
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.
Non-Interaction e Model
2

Z e2 
2
H   
j  k

2
m
r
j 1 
j 


  h rj 
Z
Z
Drop all e-e terms :
2
where
Thus
Z e2
h r   
 k
2m
r
j 1
2
( Hydrogenic hamiltonian )
h r  nl m r   En  nl m r 
where nlm and En are obtained from the hydrogen case by setting e2  Ze2 .
In particular :
Z2
En   13.6 eV  2
n
and
a
a0
Z
a0 = Bohr radius
Z
(Unsymmetrized) solutions to
H   E
n
with
E   En j
j 1
are
  r1 , , rZ     n l
j 1
j j
mj
r 
j
Z2
En   13.6 eV  2
n
5.2.1. Helium
a
2
2
2e 2
2e 2
e2
2
H 
 k

2  k
k
2m
r1
2m
r2
r1  r2
2
1
Non-interacting e model :

a0
Z
H    h  r1   h  r2     E 
 r1, r2    nl m r1   nlm r2 
E  En  E n 
Ground state :
 0 r1, r2   100 r1  100 r2  
 2

8
exp

r

r


 a 1 2 
 a03
 0

E0   13.6 eV   22 1  1   13.6 eV   8  109 eV
R10  2a 3/2 e  r / a
Y00 
1
4
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
Singlet
Long-living excited states :
 r1, r2    nl m r1  100 r2 
( both e in excited states quickly
turns into an ion + free e. )
Spatial part of parahelium is symmetric
 higher e-e interaction
 higher energy than orthohelium counterpart
Do Prob 5.10
Triplet
5.2.2. The Periodic Table
l – degeneracy lifted by e-e interaction (screening).
Filling order of the periodic table.
n\ l
0
s
1
p
2
d
3
f
4
g
1
2
3
4
5
6
7
2(2l+1)
2
6
10
14
18
Ground State Electron
Configuration
2 S 1
Spectral Term :
LJ
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
sz
 S = 1, L = 1, J = 0
0
1


3
P0
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
V r   


Set
2
2m
2  E 
  r    X i  xi 
i
for
Let
0  xi  li

1 d 2 Xi

E

2
2m i X i d xi

E   Ei
i
d 2 Xi
 ki2 X i
2
d xi
ki 
2mEi

X i  Ai sin ki xi  Bi cos ki xi

i =1, 2, 3
unless stated otherwise
2
2
Set
d 2Xi

 Ei X i
2m d xi2
0  xi  li
otherwise
E
i
2 2
i
k
2m
Boundary Conditions
X i  Ai sin ki xi  Bi cos ki xi
X i  0  0
X  li   0

Bi  0
X i  Ai sin ki xi

ki li  ni 
ni  1, 2, 3,
li
Normalization :
 d xi X i  Ai
2
2
0

Ai 

li
2
d
x
sin
ki xi  Ai
i

2
0
2
li
Xi 
2
n
sin i xi
li
li
 n n n r   
1 2 3
i
En1 n2 n3  
i
2 2
i
2
k

2m
2m
2mEi
ki 
li
2
X i  Ai sin
ni
xi
li
2
d
x
sin
x

x sin 2 x

2
4
2
n
sin i xi
li
li
n 
 n 
n 
8
sin  1 x1  sin  2 x2  sin  3 x3 
l1 l2 l3
 l1 
 l2

 l3 
 ni 
i  l 
 i 
2

2
2m
k2
k   k1, k2 , k3 
k-Space Density
ki 

ni 
li
 ki 

ni  1, 2, 3,
 ni 
li
 Volume occupied by one allowed (stationary)
state in the 1st quadrant of the k-space is
 k   ki  
i
i

li

3
V
with ni  1.
V   li = Volume of solid
i


k
1 d 3k
V
3
 

d
k
3 
8  k
 2 
V
 2 
3
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
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.
Since
En 
2
2m
these states occupy a sphere centered at the
origin of the k-space.
k2
Let the radius of this sphere be kF .


2
V
 2 
3
4 3
k F  qN
3
1/3
qN 

k F   3 2

V 

EF 
2
2m
k
2
F

( Factor 2 comes from spin degeneracy. )
  3 2  
1/3
2
 3  
2m
2

qN
V
2/3
Fermi energy
= free electron density
Total Energy
Total energy of the ground state :
Etot 

smallest n
V
k
En  2
d
k
3
2m
 2  k kF
k F   3  
2
2 2
1/3

3
V 2
Etot 
10  2m
d Etot  
2
kF
V
3
k
4  k d k
2m
0
5/3
2
 2 
 2 qN 
 3

V 

2 2
2

10  2m
V 2

k F5
2
10  m
 3 2qN  V 2/3
5/3
2 Etot
dV
3 V
 Compressing the solid increases its engergy.
Work need be done to compress it.
Electrons exerts outward pressure on the solid boundary.
P
d Etot
2 Etot

dV
3 V
2
2
5


k
F
2
3 10  m
 3 2 
2/3
5m
2
 5/3
degeneracy pressure
( Caused by Pauli exclusion )
Miscellaneous
V 2
Etot 
k F5
2
10  m
EF 
V k F3 2k F2

5  2 2m
2 2
F
k
2m
 

1/3
N 

k F   3 2 e 
V 

Ideal gas :
e gas :
PV 
PV 
2
Etot  N kB T
3
Etot 3
 EF
Ne 5

2
2
Etot  N e EF
5
3
Energy per e.
Te 
2 EF
5 kB
k B  8.617  105 eV / K
 6.582  1016 eV  s
Do Prob 5.16
and for Al
me  0.511 MeV / c 2
5.3.2. Band Structure
1-D crystal with periodic potential
V  x  a  V  x
a = lattice constant
d 2
H  
V  x  E 
2m d x 2
2
For a finite solid with N atoms, strict mathematical periodicity
can be achieved by imposing the periodic boundary condition V  x  Na   V  x 
Simplest model :
1-D Dirac comb
Bloch’s Theorem
Bloch’s theorem :
Proof :
V  x  a  V  x

  x  a   ei K a   x 
D f  x  f  x  a
Let D be the displacement operator :
d  x d  x  a
d  x  a  d

 D  x  
D


d
x
dx
d  x  a
dx
d 2
d2
D 2  2  D 
dx
dx

D V  x   x    V  x  a    x  a   V  x  D  x 

D  H   H  D 
D , H   0
i.e.
DH  HD
or
D , H   0
 D & H can share the same eigenstates.
Let D   x      x 

Since   0, setting   ei K a
K = const
  x  a     x
completes the proof.
Periodic Boundary Condition
V  x  a  V  x

  x  a   ei K a   x 
For a 1-D solid with N atoms, imposing the periodic BC   x  Na     x 
gives

ei K N a  1
K
K is real
2 n
Na
n  0,  1,  2,

  x  a
2
   x
2
As expected
Dirac Comb
1-D Dirac comb :
V  x  
N 1
   x  ja 
j 0
d 2

 E
2m d x 2
2
For x  ja :


d 2
 k 2 
2
dx
  x   Asin k x  B cos k x
k
2mE
Periodic BC
  x   Asin k x  B cos k x
Impose periodic BC :
Bloch’s theorem 
Let

 x  ja
  x  Na     x 
  x  a   ei K a   x 
  x   Asin k x  B cos k x
K
2 n
Na
n  0,  1,  2,
 0<x<a
  x   ei K a   x  a   e  i K a  A sin k  x  a   B cos k  x  a  
 continuous at x = 0
 d   d 
2m


  0
 dx   dx 
2

 
 
 a < x < 0

B  ei K a  Asin ka  B cos ka 

A k  e  i K a  k A cos ka  k B sin ka  
2m
2
B
B  ei K a  Asin ka  B cos ka 

e
i K a
A k  e  i K a  k A cos ka  k B sin ka  
sin ka  A   e  i K a cos ka  1 B  0
2m 

k 1  e  i K a cos ka  A   ke i K a sin ka  2  B  0


Condition for consistency is
e  i K a sin ka
k 1  e  i K a cos ka 
e  i K a cos ka  1
2m  0
i K a
ke
sin ka  2
2m  i K a
 i K a
ke
sin
ka

e
sin ka  k 1  e i K a cos ka  e i K a cos ka  1  0

2 


ke 2i K a sin 2 ka 

2m
2
2m
2
e  i K a sin ka  k 1  2e  i K a cos ka  e 2i K a cos2 ka   0
e  i K a sin ka  k 1  2e  i K a cos ka  e 2i K a   0
2m
2
B

2m
2

e  i K a sin ka  k 1  2e  i K a cos ka  e 2i K a   0
2m
2

sin ka  k  ei K a  2 cos ka  e  i K a   0
2m
2
sin ka  2k  cos Ka  cos ka   0
cos Ka  cos ka 
Let

Since
z  ka

cos Ka  cos z  
cos Ka  1
m
sin ka
2
k
m a
2
sin z
 f  z
z
there’s no solution for | f (z) | > 1 ( band gaps ).
  10
2 2
2 2
k
z
E

2m 2ma 2
Do Prob 5.19
with   20
5.4. Quantum Statistical Mechanics
Fundamental assumption of statistical mechanics :
In thermal equilibrium, each distinct system configuration
of the same E is equally likely to occur.
1. An Example
2. The General Case
3. The Most Probable Configuration
4. Physical Significance of  and 
5. The Blackbody Spectrum
( 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.
  2
2
2
(
n

n

n
E  E A  EB  EC 
B
C)
  A
2m  a 
2
Let
2
n A2  nB2  nC2  363
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 = 1
N13 = 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 = 2
N19 = 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 = 1
N7 = 1
N17 = 1
Pn : Particles Distinguishable
Particles are distinguishable  each configurations is distinct.
 Each is equally likely when system is in equilibrium.
(nA , nB , nC )
Pn = Probability of finding
a particle with energy En .
(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
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
Pn : Fermions
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 = 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
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,... )
Total = 1 distinct configurations
N5 = 1, N7 = 1, N17 = 1
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
Pn : Bosons
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 = 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
Allowed distinct configurations are :
( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... )
( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... )
( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... )
( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... )
Total = 4 distinct configurations
N11 = 3
N5 = 1, N13 = 2
N1 = 2, N19 = 1
N5 = 1, N7 = 1, N17 = 1
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
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.:
N
1. There’re CN1 
N!
ways, without regard of picking order, to choose
N1 !  N  N1 ! N1 particles from the whole N particles.
2. When putting a particle into the 1st bin, there’re d1 choices of compartments.

N1
N
!
d
1
the number of ways to put N1 particles into the 1st bin is
N1 !  N  N1  !
 N  N1 ! d 2N
N 2 !  N  N1  N 2  !
2
3. For the 2nd bin, one starts with N  N1 particles so one gets

N  N1 ! d 2N2
N  N1  N 2  ! d 3N3


N ! d1N1
Q  N1 , N 2 , N 3 ,  
N1 !  N  N1 ! N 2 !  N  N1  N 2 ! N 3 !  N  N1  N 2  N 3 !
N ! d1N1 d 2N2 d 3N3

N1 ! N 2 ! N 3 !

d iNi
 N!
i 1 N i !
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
di !
Ni !  di  Ni !

Q  N1 , N 2 , N 3 ,


i 1
if
di  Ni
di !
Ni !  di  Ni !
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
 Ni  di  1!
However, the dots & partitions are indistinguishable among themselves. Hence
Q  N1 , N 2 , N 3 ,


i 1
 N i  d i  1!
 d i  1! N i !
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.,

N
i 1
i
Ei  E

N
i 1
i
N
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 constraints
N
i 1
i
Ei  E

and
Using Lagrange’s multiplier method, we maximize, without constraint,






G  ln Q    N   N i     E   N i Ei 
i 1
i 1




i.e., we set
Note :
G
0
 Ni
G
0

G
0

are just the original constraints.
N
i 1
i
N
Distinguishable Particles

d iNi
Q  N!
i 1 N i !


ln Q  ln N !   N i ln d i  ln N i !
i 1
ln z !  z ln z  z
Stirling’s approximation :







G  ln Q    N   N i     E   N i Ei 
i 1
i 1




for z >>1

ln Q  N ln N  N    N i ln d i  N i ln N i  N i 
i 1






G  N ln N  N    N i ln d i  N i ln N i  N i     N   N i     E   N i Ei 
i 1
i 1
i 1





G
 ln d i  ln N i     Ei  0
 Ni
Ni  di exp     Ei 
Fermions

di !
Ni !  di  Ni !
Q
i 1






G  ln Q    N   N i     E   N i Ei 
i 1
i 1





ln Q   ln d i !  ln N i !  ln  d i  N i ! 

i 1

di , Ni >>1 :
ln Q   d i ln d i  N i ln N i   d i  N i  ln  d i  N i  
i 1






G    d i ln d i  N i ln N i   d i  N i  ln  d i  N i      N   N i     E   N i Ei 
i 1
i 1
i 1





G
  ln N i  ln  d i  N i      Ei  0
 Ni

di
 1  exp    Ei 
Ni
Ni 
di
exp    Ei   1






G  ln Q    N   N i     E   N i Ei 
i 1
i 1




Bosons
 N i  d i  1!
Q
i 1  d i  1 ! N i !



ln Q   ln  N i  d i  1 !  ln  d i  1 !  ln N i ! 
i 1

di , Ni >>1 :
ln Q    N i  d i  1 ln  N i  d i  1   d i  1 ln  d i  1  N i ln N i 
i 1






G    N i  d i  1 ln  N i  d i  1   d i  1 ln  d i  1  N i ln N i     N   N i     E   N i Ei 
i 1
i 1
i 1





G
 ln  N i  d i  1  ln N i     Ei
 Ni

1
0
di  1
 exp    Ei 
Ni
Ni 
di
exp    Ei   1
di 1  di
Do Prob 5.26, 5.27
5.4.4. Physical Significance of  and 
Ideal gas in 3-D infinite square well.
 n x  n y  nz 
k
,
,
 l
l
lz
y
 x
2 2
k
Ek 
2m


k
V
 2 
3



3
d
 k
Taking the “bin” as the spherical shell between k and k+dk, the degeneracy dk is
just the number of states in the shell :
dk 
V
 2 
2
4

k
dk 
3
V 2
k dk
2
2
1
   
2
Distinguishable Particles
Ni  di exp     Ei 

N
i 1
i
N


N E
i 1
i
i
E

 dx x e
n
 x 2
0
1  n 1 1
 

2  2   n 1


  2 2
V
V 
2
2
N  2  dk k exp     Ek   2 e  dk k exp  
k 
2 0
2
 2m 
0
V
  2m 
 2 e 


2
4   2


V 2
dk 
k dk
2 2
e 
N  2 
 
V 
m
2



3/2

m
 V e  
 2 
cf. kinetic theory result :
E
3
N k BT
2
3/2
3/2

  2 2
V  2
4
E 2e
dk k exp  
k 
2
2m 0
2
m


3 V  
m
E
e 
2
 2

2 


2 

3/2



2
V

e
4 2m
3   2m 


4   2
3N
2

1
k BT
universal
5/2
Distributions
D
F
B

E 
N i  d i exp    i 
k BT 

di
Ni 

E 
exp    i   1
 k BT 
Ni 
di

E
exp    i
 k BT

 1

Set
 

k BT
N
ni  i
di

  
n     exp  

 k BT 
1
n   
 
exp 
 1
k
T
 B 
n   
1
  
exp 
 1
k
T
 B 
  Chemical potential
n()  Most probable number of particles in a state with energy .
 occupation number
Maxwell –
Boltzmann
Fermi –
Dirac
Bose Einstein
Fermi-Dirac as T  0
n   
T0:

 0
 
exp 
  
k
T
 B 

n  
1
 
0
    0
    0
    0
    0
 (0)  EF
T0
T>0
1
 
exp 
 1
k
T
 B 
Ideal Gas
E
Distinguishable particles
e

N  2 
 
V 
m
2



3/2

3
N k BT
2
 N  2 2 3/2 
 N 3  2 2
  kBT ln  
   k BT ln V  2 ln  m k T
 V  mk BT  
B





Indistinguishable particles :
N
V

V
 2  dk k 2 n  k 
2 0
d k n  

 2 
3
k
3

V
 2  dk
2 0
k2
 1  2k 2

exp 


 1



 k BT  2 m
fixes 
E
V
d k n   

 2 

3
3
Read Prob. 5.29
k
k
2
V
 2
dk
2 2m 0
Do Prob 5.28
k4
 1  2k 2

exp 
 2m      1
k
T

 B 
F
B
gives C 
E
T
F
B
5.4.5. The Blackbody Spectrum
Photon
E  h  
 quantum of EM field
 spin 1, massless boson with v  c.
k
Only m  1 occurs.
2


Number not conserved    0.
N 
d
 
exp 
 k BT
d  d k

 1

Energy density in range d 
   
k   /c
 2
V
2 c
2 3
 2d 
N 
    d 
V
1
3
    2c3
exp 
 1
k
T
 B 
Black body spectrum

c
Blackbody Spectrum
Visible region
6000 K
4000 K
2000 K
Read Prob 5.30
Do Prob 5.31