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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 probaility to be in the n=1 vs n=2 level? D=2 for n=1 D=8 for n=2 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) P460 - Quan. Stats. II 1 1D Harmonic Oscillator • Equally spaced energy levels. Number of states at each is 2s+1. Assume s=0 and so 1 state/energy level • Density of states D( E ) N E 1 • N = total number of “objects” (particles) gives normalization factor for n(E) N D ( E )n( E )dE 0 N A E / kT 0 e dE AkT N A kT • note dependence on N and T P460 - Quan. Stats. II 2 1D H.O. : BE and FD • Do same for Bose-Einstein and Fermi-Dirac 1 1 nBE n FD Be E / kT 1 Be E / kT 1 N 1 0 n( E )dE 1 BBE BFD 1 ( e EF / kT ) 1 e N / kT e N / kT 1 • “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 E F ) 0 (T 0) ( E E F ) / kT e 1 1 N EF 0 1 1 dE EF or EF N • In materials, EF tends to vary slowly with energy (see BFD for terms). Determining at T=0 often “easy” and is often used. Always where n(E)=1/2 P460 - Quan. Stats. II 3 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 4n 3 # nodes Degeneracy 8 3 1 4 ( 2 L) 3 Deg 3 take derivative 8 3 4V N ( )d 4 Degd P460 - Quan. Stats. II 4 Density of States II • The degenracy 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 (2s 1)( 2 Lp 3 h ) D( p)dp 2 (2s 1)( 2hL )3 p 2 dp • convert to energy depends on kinematics E pc dE cdp relativistic D( E )dE 2 (2s 1)( 2hcL )3 E 2 dE 8V 2 E dE 3 3 ch for • non-realtivistic 2 p p E dE dp 2m m 2L 3 m D( E )dE (2 s 1)( ) 2mE dE 2 h 2mE 2L 3 (2 s 1)( ) 2m 3 / 2 E 1/ 2 dE 2 h P460 - Quan. Stats. II 5 Plank Blackbody Radiation • Photon gas - spin 1 Bosons - dervied 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 c h3 • 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 e E / kT 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) P460 - Quan. Stats. II 6 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 P460 - Quan. Stats. II 7