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 P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

“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) P461 - Intro. Quan. Stats.

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: P461 - Intro. Quan. Stats.

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 P1=probability for 1 particle Pn = (P1)n if no enhancement -----> (1+n) is Boson enhancement factor P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

Distribution Functions II 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 P461 - Intro. Quan. Stats.

Distribution Functions III 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 P461 - Intro. Quan. Stats.

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) P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

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 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.

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 0 L P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.

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 = P461 - Intro. Quan. Stats.

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 P461 - Intro. Quan. Stats.