Transcript Quantum Statistics

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
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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=1
*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
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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 partlces 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
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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

P r ob( E ) 
N W
Boson: 10 states, N  5,  W  10
Ferm ion: 3 states, N  5,  W  3
Classical : 10 states, N  5,  W  210
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
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Fermion
.33
.33
.20
.13
0
0
0
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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 e E / kT  1
1
 ( E  E F ) / kT
e
1
P460 - Quan. Stats.
Boson
Fermion
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“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
 E / kT
 n( E )  Ae
• kT comes from determining the average energy
(and is sort of the definition of T)
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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 )
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Inhibition/Enhancement Factors
• For Fermions, Pauili Ex. If particle in a state a
second particle can not be in that state. As dealing
with averages obtain:
I  InhibitionFactor  (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 ----->
boson
n
 n! Pn  n !( P1 )
boson
n 1
n 1
P
P
 (n  1)!( P1 )
n
 (1  n) P P
boson
1 n
• (1+n) is Boson enhancement factor
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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 equilibrium : 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
so n1 R12B  n2 R21B
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 
e e E / kT  1
n( E ) 
1
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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
so n1R12F  n2 R21F
C
 n1 (1  n2 ) R12C  n2 (1  n1 ) R21
n1
(1 n1 )
•
e E1 / kT 
n2
(1 n2 )
gives Fermi-Dirac
e E2 / kT  f (T )  e 
e e E / kT  1
n( E ) 
1
• The e term in both the FD and DE distribution
functions is related to the normalization
• we’ll see =1 for massless photons as the total
number of photons isn’t fixed
• for Fermions usually given by the Fermi energy
which usually varies only slowly with T.
e ( E  EF ) / kT  1 2
nFD ( E ) 
 when E  EF
1
1
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