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08_562ln08

# 08_562ln08 - MIT OpenCourseWare http/ocw.mit.edu 5.62...

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MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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5.62 Lecture #8: Boltzmann, Fermi-Dirac, and Bose–Einstein Statistics THE DIRECT APPROACH TO THE PROBLEM OF INDISTINGUISHABILITY We could have approached the problem of indistinguishability by treating particles as indistinguishable fermions or bosons at the outset. QM tells us 1. All particles are indistinguishable 2. All particles are either fermions or bosons. The odd/even symmetry of a particle's wavefunction with respect to exchange is determined by whether the particle is a fermion or boson. FERMION — a particle which obeys Fermi-Dirac statistics; many-particle wavefunction is antisymmetric (changes sign) with respect to exchange of any pair of identical particles : P 12 ψ = – ψ 1/2 integer spin e , proton, 3 He Single state occupation number: n i = 0 or n i = 1, no other possibilities! BOSON — a particle which obeys Bose-Einstein statistics; many-particle wavefunction is symmetric (does not change sign) with respect to exchange of any pair of identical particles 4 He, H 2 , photons integer spin n i = any number, without restriction What kind of particle is 6 Li, 7 Li, D, D + ? We are going to figure out how to write t ( { } ω ( n i ,g i ) Ω n i ) = N! i = 1 where we are considering level i with energy ε ι and degeneracy g i . Previously we had considered ({n i }) for non-degenerate states rather than g i -fold degenerate ε i levels.
5.62 Spring 2008 Lecture 8, Page 2 Let us play with some simple examples before generalizing results for each type of system. 3 degenerate states: A, B, C (states could be x, y, z directions for particle in cube) 2 particles: 1, 2 A B C 1 2 1 2 2 1 2 1 1 2 2 1 1,2 1,2 1,2 If particles are distinguishable and there are no restrictions on occupancy, total # of distinguishable arrangements is 3 2 . Note that this is different from the degeneracy of a particular set of occupation numbers for non-degenerate states , N! ! . n i For each degenerate level occupied by particles, we have a factor : degeneracy of atomic state to correct for particle indistinguishability. We divide by N! t t n i ω (n i ,g i ) g i ω ( n ,g) = g n particles i = 1 = i = 1 ( { } ) = Ω B n i N! N! 3 2 For our case 2! = 4.5 which is not an integer so g n n ! can be only an approximation to the correct total # of ways. Now go to F–D system occupation # is 0 or 1, indistinguishable particles, therefore g n. revised 2/25/08 4:35 PM

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5.62 Spring 2008 Lecture 8, Page 3 ω FD ( n ,g) = put excitation first in any of g states, 2nd in any of g–1, then g! 1 divide by n ! for indistinguishability of particles. Finally, divide (g n )! n ! by (g – n )! for indistinguishability of “holes”. g i ! ( g i n i ) !n! N!
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