Quantum Statistics
• Determine probability for object/particle in a group
of similar particles to have a given energy
• derive via:
a. look at all possible states
b. assign each allowed state equal probability
c. conserve energy
d. particles indistinguishable (use classical distinguishable - if wavefunctions do not overlap)
e. Pauli exclusion for 1/2 integer spin Fermions
• classical can distinguish and so these different:
E1  1 E1  2
E2  2 E2  3 Etotal  6
E3  3 E3  1
•
but if wavefunctions overlap, can’t tell “1” from
“2” from “3” and so the same state
P461 - Intro. Quan. Stats.
1
Simple Example
• Assume 5 particles with 7 energy states
(0,1,2,3,4,5,6) and total energy = 6
• Find probability to be in each energy state for:
a. Classical (where can tell each particle from each
other and there is no Pauli exclusion)
b. Fermion (Pauli exclusion and indistinguishable)
c. Boson (no Pauli exclusion and indistinguishable)
 different ways to fill up energy levels (called
microstates)
1 E=6 plus 4 E=0 1 E=5 + 1 E=1 + 3 E=0
1 E=4 + 1 E=2 + 3E=0
* 1 E=4 + 2E=1 + 2E=0
*1E=3 + 1E=2 + 1E=1 + 2E=0
2E=3 + 3E=0
1E=3 + 3E=1 + 1E=0
3E=2 + 2E=0
*2E=2 + 2E=1 + 1E=0
1E=2 + 4E=1
• can eliminate some for (b. Fermion) as do not obey
Paili exclusion. Assume s=1/2 and so two particles
are allowed to share an energy state. Only those *
are allowed in that case
P461 - Intro. Quan. Stats.
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Simple Example
• If Boson or classical particle then can have more
then 1 particle in same state….all allowed
• But classical can tell 1 particle from another
State 1 E=6 + 4 E=0
 1 state for Bosons
 5 states for Classical (distinguishable)
assume particles a,b,c,d,e then have each of
them in E=6 energy level
• for Classical, each “energy level” combination is
weighted by a combinatorial factor giving the
different ways it can be formed
N!
W
N1! N 2 ! N 3! N 4 ! N 5!
5!

5
1  4!1  1  1
P461 - Intro. Quan. Stats.
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Simple Example
• Then sum over all the microstates to get the number
of times a particle has a given energy
• for Classical, include the combinatoric weight
(W=1 for Boson or Fermion)
N EW

Pr ob( E ) 
N W
Boson : 10 states, N  5, W  10  50
Fermion : 3 states, N  5, W  3  15
Classical : 10 states, N  5, W  210  1050
Energy Probability: Classical Boson
0
.4
.42
1
.27
.26
2
.17
.16
3
.095
.08
4
.048
.04
5
.019
.02
6
.005
.02
P461 - Intro. Quan. Stats.
Fermion
.33
.33
.20
.07
.07
0
0
4
Distribution Functions
• Extrapolate this to large N and continuous E to get
the probability a particle has a given energy
• Probability = P(E) = g(E)*n(E)
• g=density of states = D(E) = N(E)
• n(E) = probability per state = f(E)=distribution fcn.
nMaxwell  Boltzman  e
nBose Einstein 
nFermi Dirac 
 E / kT
Classical
1
 E / kT
e e
1
Boson
1
e e E / kT  1
1
 ( E  EF ) / kT
e
1
P461 - Intro. Quan. Stats.
Fermion
5
Distribution Functions
• Relatively: Bose-Einstein – more at low Energy
Fermi-Dirac – more at high Energy. depends on T
BE
n(E)
1
MB
FD
E
Classical
nMaxwell Boltzman  e  E / kT
nBose Einstein 
1
e e E / kT  1
1
nFermi Dirac  ( E  EF ) / kT
e
1
P461 - Intro. Quan. Stats.
Boson
Fermion
6
“Derivation” of Distribution
Functions
• Many Stat. Mech. Books derive Boltzman
distribution
• the presence of one object in a state does not
enhance or inhibit the presence of another in the
same state
• Energy is conserved. Object 1 and 2 can exchange
energy and probability stays the same  need
exponential function
p( E1 ) p( E2 )  q( E1  E2 ) 
Be  E1 / E0 Be  E2 / E0  B 2 e ( E1  E2 ) / E0
 n( E )  Ae
 E / kT
• kT comes from determining the average energy
(and is sort of the definition of T)
P461 - Intro. Quan. Stats.
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Sidenote
• For Boltzman 2 particles can share energy so:
p( E1 ) p( E2 )  q( E1  E2 )
•
for Fermions, Pauli exclusion can restrict which
energies are available and how 2 particles can share
energy. This causes some inhibitions and so:
p( E1 ) p( E2 )  q( E1  E2 )
• For Bosons, there is an enhancement for particles
to be in the same state and again:
p( E1 ) p( E2 )  q( E1  E2 )
P461 - Intro. Quan. Stats.
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Inhibition/Enhancement Factors
• For Fermions, Pauli Ex. If particle in a state a
second particle can not be in that state. As dealing
with averages obtain:
I  Inhibition Factor  (1  n) with 0  n  1
as n  1  I  0 n  0  I  1
•
for Bosons, enhancement factor as symmetric
wave functions. Start with 2 particles in same state
 symmetric 
1
2
[  (1)  (2)    (2)  (1)]
      2  (1)  (2)
2
2
• giving enhancement factor of 2. Do for 3 particles
all in same state get 6 = 3!  n gives n!
• n particles. Define P1=probability for 1 particle
• Pn = (P1)n if no enhancement 
Pnboson  n! Pn  n !( P1 ) n
boson
n 1
P
 (n  1)!( P1 )
n 1
 (1  n) P P
boson
1 n
• (1+n) is Boson enhancement factor
P461 - Intro. Quan. Stats.
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Distribution Functions
• Use detailed balance to get Fermi-Dirac and BoseEinstein distribution function. Define
Rate(1  2)  R12
Rate(2  1)  R21
at equilibriu m : n1 R12  n2 R21
• for classical particles we have (Boltzman)
C
n1
R21
e  E1 / kT
C
n2
R12
e  E2 / kT


• for Bosons, have enhancement over classical
R12B  (1  n2 ) R12C
B
so n1 R12B  n2 R21
C
 n1 (1  n2 ) R12C  n2 (1  n1 ) R21
n1 (1 n2 )
n2 (1 n1 )

C
R21
C
R12
rearrange
•
 e ( E1  E2 ) / kT
n1
(1 n1 )
e E1 / kT 
gives Bose-Einstein
n2
(1 n2 )
e E2 / kT  f (T )  e 
n( E ) 
P461 - Intro. Quan. Stats.
1
 E / kT
e e
1
10
Distribution Functions II
• for Fermions, have inhibition factor (of a particle is
in the “final” state another particle can’t be added)
R12F  (1  n2 ) R12C
F
so n1R12F  n2 R21
C
 n1 (1  n2 ) R12C  n2 (1  n1 ) R21
n1
n2
E1 / kT
e

e E2 / kT  f (T )  e 
(1  n1 )
(1  n2 )
• from reaction rate balance and inhibition factor: f
must not depend on E but can depend on T
n
e E / kT  f
(1  n )
n( e E / kT  f )  f
n
1
f 1e E / kT  1
f 1  e  EF / kT
• with this definition of Fermi energy EF gives
Fermi-Dirac distribution
n( E ) 
1
e( E  EF ) / kT  1
P461 - Intro. Quan. Stats.
11
Distribution Functions III
• The e term in both the FD and BE distribution
functions is related to the normalization
• we’ll see =1 for massless photons as the total
number of photons isn’t fixed. This is not the case
for all Bosons
• for Fermions usually redefined using the Fermi
energy which often varies only slowly with T.
• For many cases can use the value when T=0
nFD ( E ) 
1
e
( E  E F ) / kT
1
1


11 2
1
if E  E F
P461 - Intro. Quan. Stats.
12
Quantum Statistics:Applications
• Determine
P(E) = D(E) x n(E)
probability(E) = density of states x prob. per state
• electron in Hydrogen atom. What is the relative
probability to be in the n=1 vs n=2 level?
D=2 for n=1 (1S)
D=8 for n=2 (2S,2P)
as density of electrons is low can use Boltzman:
nFD 
•
1
e
( E  E F ) / kT
1
e
 E / kT
E  EF
can determine relative probability
P(n  2) 8 e  E2 / kT
 ( E2  E1 ) / kT


4
e
P(n  1) 2 e  E1 / kT
 4e 10.2 / .026  10 171 (!)
for E  10.2eV
T  3000 K
• If want the ratio of number in 2S+2P to 1S to be .1
you need T = 32,000 degrees. (measuring the
relative intensity of absorption lines in a star’s
atmosphere or a interstellar gas cloud gives T)
P461 - Intro. Quan. Stats.
13
1D Harmonic Oscillator
• Equally spaced energy levels. Number of states at
each is 2s+1. Assume s=0 (Boson) and so 1
state/energy level
• Density of states
N
1
D( E ) 

E 
• N = total number of “objects” (particles) gives
normalization factor A for n(E). For classical

N   D ( E )n( E )dE
0

A  E / kT
AkT
N 
e
dE 
0 

N
 A
kT
• note dependence on N and T
P461 - Intro. Quan. Stats.
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1D H.O. : BE and FD
• Do same for Bose-Einstein and Fermi-Dirac. BE
with s=0
1
nBE 
Be E / kT  1


1
1
1
N    n( E )dE   E / kT
dE
0
0
Be
1
1
 BBE 
1  e  N / kT
•
Fermi-Dirac with s=1/2. Density extra factor of 2
nFD 
N 
1
Be E / kT  1

2
0 
 BFD 
n( E )dE  

0
1
e N / 2 kT  1
2
1
dE
E / kT
 Be
1
(  e  EF / kT )
• “normalization” depends on N and T
P461 - Intro. Quan. Stats.
15
1D H.O. :Fermi-Dirac
• “normalization” varies with T. Fermi-Dirac easier
to generalize
• T=0 all lower states fill up to Fermi Energy
1
1
n( E  EF )  1 ( E  EF ) / kT

(T  0)
e
1 0 1
1
1
n( E  EF )  0

(T  0)
( E  E F ) / kT
e
1  1
EF 2
2 EF
N
N 
 1 dE 
or EF 
0 

2
• could have obtained by inspection (top of well)
• In materials, EF tends to vary slowly with energy
(see BFD for terms). Determining at T=0 often
“easy” and is often used. The Fermi energy is
always where n(E)=1/2
P461 - Intro. Quan. Stats.
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Density of States “Gases”
• # of available states (“nodes”) for any wavelength
• wavelength  momentum  energy
• “standing wave” counting often holds:often called
“gas” but can be solid/liquid. Solve Scrd. Eq. In 1D
d2
dx 2
  a  0
n 
L
 ( x  0)   ( x  L)  0 0
n 2
L
2 L
 kn 

 n  k 
L n

2
2L
n
• go to 3D. ni>0 and look at 1/8 of sphere
kx 
n x
L
ky 
n y
L
k z  Lnz
1 4n 3
# nodes 
 Degeneracy
8 3
1 4
( 2 L)3

 Deg  3
take derivative
8 3

4V
 N ( )d  4  Degeneracy  d

P461 - Intro. Quan. Stats.
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Density of States II
• The degeneracy is usually 2s+1 where s=spin. But
photons have only 2 polarization states (as m=0)
• convert to momentum
p
h


hn
2L
# nodes 
1 4
8 3
(2 s  1)(
2 Lp 3
h
)
D( p )dp  2 (2 s  1)( 2hL ) 3 p 2 dp
• convert to energy depends on kinematics
relativistic E  pc dE  cdp
D( E )dE  2 (2s  1)( 2hcL )3 E 2 dE
• non-relativistic
8V 2
 3 3 E dE
ch
for 
p2
p
E
 dE  dp
2m
m

2L 3
m
D( E )dE  (2s  1)( ) 2mE
dE
2
h
2mE

2L 3
 (2 s  1)( ) 2m 3 / 2 E 1/ 2 dE
2
h
P461 - Intro. Quan. Stats.
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Plank Blackbody Radiation
• Photon gas - spin 1 Bosons - derived from just stat.
Mech. (and not for a particular case) by S.N. Bose
in 1924
• Probability(E)=no. photons(E) = P(E) = D(E)*n(E)
• density of state = D(E) = # quantum states per
energy interval
2
8 V E dE
 3
3
c
h
• n(E) = probability per quantum state.
Normalization: number of photons isn’t fixed and
so a single higher E can convert to many lower E
n
1
 E / kT
e e
1

1
e
E / kT
1
  if E  0
• energy per volume per energy interval =
8 E 3
T ( E )  E D( E )n( E )  3 3 E / kT
c h (e
 1)
P461 - Intro. Quan. Stats.
19
Phonon Gas and Heat Capacity
• Heat capacity of a solid depends on vibrational
modes of atoms
• electron’s energy levels forced high by Pauli Ex.
And so do not contribute
• most naturally explained using phonons
- spin 1 psuedoparticles
- correspond to each vibrational node
- velocity depends on material
• acoustical wave <---> EM wave
phonon
<---> photon
• almost identical statistical treatment as photons in
Plank distribution. Use Bose statistics
• done in E&R Sect 11-5, determines
dE
cV 
dT
P461 - Intro. Quan. Stats.
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