1. P460 - Quan. Stats.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. P460 - Quan. Stats.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=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

3. P460 - Quan. Stats.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 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

4. P460 - Quan. Stats.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 0.4.42.33 1.27.26.33 2.17.16.20 3.095.08.13 4.048.04 0 5.019.02 0 6.005.02 0

5. P460 - Quan. Stats.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. P460 - Quan. Stats.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 kT comes from determining the average energy (and is sort of the definition of T)

7. P460 - Quan. Stats.7 Sidenote 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:

8. P460 - Quan. Stats.8 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: 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 P 1 =probability for 1 particle P n = (P 1 ) n if no enhancement -----> (1+n) is Boson enhancement factor

9. P460 - Quan. Stats.9 Distribution Functions 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

10. P460 - Quan. Stats.10 Distribution Functions II for Fermions, have inhibition factor (of a particle is in the “final” state another particle can’t be added) gives Fermi-Dirac 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.