1. 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:
but if wavefunctions overlap, can’t tell “1” from “2” from “3” and so the same state

2. 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= E=5 + 1 E=1 + 3 E= E=4 + 1 E=2 + 3E= * 1 E=4 + 2E=1 + 2E= *1E=3 + 1E=2 + 1E=1 + 2E= E=3 + 3E= E=3 + 3E=1 + 1E= E=2 + 2E= *2E=2 + 2E=1 + 1E= E=2 + 4E=1
can eliminate some for (b. Fermion) as do not obey Pauli exclusion. Assume s=1/2 and so two particles are allowed to share an energy state. Only those * are allowed in that case

3. 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 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

4. 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)
Energy Probability: Classical Boson Fermion

5. 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.
Classical
Boson
Fermion

6. Distribution Functions
Relatively: Bose-Einstein – more at low Energy Fermi-Dirac – more at high Energy. depends on T BE MB
n(E) 1 FD E
Classical
Boson
Fermion

7. “Derivation” of Distribution Functions SKIP
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
kT comes from determining the average energy (and is sort of the definition of T)

8. Sidenote SKIP For Boltzman 2 particles can share energy so:
for Fermions, Pauli exclusion can restrict which energies are available and how 2 particles can share energy. This causes some inhibitions and so:
For Bosons, there is an enhancement for particles to be in the same state and again:

9. Inhibition/Enhancement Factors SKIP
For Fermions, Pauli Ex. If particle in a state a second particle can not be in that state. As dealing with averages obtain:
for Bosons, enhancement factor as symmetric wave functions. Start with 2 particles in same state
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 
(1+n) is Boson enhancement factor

10. Distribution Functions SKIP
Use detailed balance to get Fermi-Dirac and Bose-Einstein distribution function. Define
for classical particles we have (Boltzman)
for Bosons, have enhancement over classical
gives Bose-Einstein

11. Distribution Functions II SKIP
for Fermions, have inhibition factor (of a particle is in the “final” state another particle can’t be added)
from reaction rate balance and inhibition factor: f must not depend on E but can depend on T
with this definition of Fermi energy EF gives Fermi-Dirac distribution

12. Distribution Functions III DO
The ea 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

13. 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:
can determine relative probability
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)

14. 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 = total number of “objects” (particles) gives normalization factor A for n(E). For classical
note dependence on N and T

15. 1D H.O. : BE and FD
Do same for Bose-Einstein and Fermi-Dirac. BE with s=0
Fermi-Dirac with s=1/2. Density extra factor of 2
“normalization” depends on N and T

16. 1D H.O. :Fermi-Dirac
“normalization” varies with T. Fermi-Dirac easier to generalize
T=0 all lower states fill up to Fermi Energy
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

17. Density of States: Boson or Fermi Gas
any system determine density of states D=dN/dE
can do for a gas of uninteracting but overlapping identical particles
density of states the same for Bosons or Fermions but how they are filled (the probability) and so average energy, etc will be different
for Fermions (i.e. electrons), Pauli exclusion holds and so particles fill up lower states
at T=0 fill up states up to Fermi Energy EF. Fermi energy depends on density as gives total number of particles available for filling up states

18. 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
go to 3D. ni>0 and look at 1/8 of sphere L

19. 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
convert to energy depends on kinematics relativistic
non-relativistic

20. 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
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
energy per volume per energy interval =