5.62 Physical Chemistry II Spring 2008

Transcription

1 MIT OpenCourseWare Physcal Chemstry II Sprng 2008 For nformaton about ctng these materals or our Terms of Use, vst:

2 5.62 Lecture #8: Boltzmann, Ferm-Drac, and Bose Ensten Statstcs THE DIRECT APPROACH TO THE PROBLEM OF INDISTINGUISHABILITY We could have approached the problem of ndstngushablty by treatng partcles as ndstngushable fermons or bosons at the outset. QM tells us 1. All partcles are ndstngushable 2. All partcles are ether fermons or bosons. The odd/even symmetry of a partcle's wavefuncton wth respect to exchange s determned by whether the partcle s a fermon or boson. FERMION a partcle whch obeys Ferm-Drac statstcs; many-partcle wavefunctos antsymmetrc (changes sgn) wth respect to exchange of any par of dentcal partcles: P 12 ψ = ψ 1/2 nteger spn e, proton, 3 He Sngle state occupaton number: = 0 or = 1, no other possbltes! BOSON a partcle whch obeys Bose-Ensten statstcs; many-partcle wavefunctos symmetrc (does not change sgn) wth respect to exchange of any par of dentcal partcles 4 He, H 2, photons nteger spn = any number, wthout restrcton What knd of partcle s 6 L, 7 L, D, D +? We are gong to fgure out how to wrte ({ } t ω (,g ) Ω ) = N! =1 where we are consderng level wth energy ε ι and degeneracy g. Prevously we had consdered Ω({ }) for non-degenerate states rather than g -fold degenerate ε levels.

3 5.62 Sprng 2008 Lecture 8, Page 2 Let us play wth some smple examples before generalzng results for each type of system. 3 degenerate states: A, B, C (states could be x, y, z drectons for partcle n cube) 2 partcles: 1, 2 A B C ,2 1,2 1,2 If partcles are dstngushable and there are no restrctons on occupancy, total # of dstngushable arrangements s 3 2. Note that ths s dfferent from the degeneracy of a partcular set of occupaton numbers for non-degenerate states, N!!. For each degenerate level occuped by partcles, we have a factor: partcles degeneracy of atomc state ω(n,g) = g n to correct for partcle ndstngushablty. We dvde by N! t ω(,g ) g =1 Ω ({ }) = = =1 B N! N! t 3 2 For our case 2! = 4.5 whch s not anteger so g n! to the correct total # of ways. n can be only an approxmaton Now go to F D system occupaton # s 0 or 1, ndstngushable partcles, therefore g n. revsed 2/25/08 4:35 PM

4 5.62 Sprng 2008 Lecture 8, Page 3 ω FD (n,g) = put exctaton frst n any of g states, 2nd n any of g 1, then g! 1 dvde by n! for ndstngushablty of partcles. Fnally, dvde (g n)! by (g n)! for ndstngushablty of holes. n! Ω FD ({ } t t ω FD (g, ) = 1 = 1 ) = = N! g! (g )!n! N! Now for B E A B C X X g = 3, n = 2 X X 3! X X ω FD = 2!1! = 3 what s the combnatoral factor? n partcles g dstngushable states g 1 ndstngushable parttons arrange ndst. partcles and g 1 ndst. parttons n all possble orders (n +g 1)! ω BE (N,g) = n!(g 1)! ({ } Ω BE ) = t t (n + g 1)! ω BE (,g ) =1 =1!(g 1)! = N! N! (2+3 1)! 4! for our case 2!(3 1)! = 2!2! = 6 A B C XX X X XX X X XX X X for our example ω FD < 3 ω B < 4.5 ω BE 6 revsed 2/25/08 4:35 PM

5 5.62 Sprng 2008 Lecture 8, Page 4 always true compare ω(,g ) factors n Ω({ }) term by term. Ths means that for a degenerate level s always largest for BE and smallest for FD. Why? Before startng the general, rgorously dervable result for for BE, corrected Boltzmann, and FD, we need to derve a relatonshp between µ and q for corrected Boltzmann statstcs. A = kt ln Q = kt ln (q N /N!) = NkT ln q + kt ln N! for large N, ln N! = N ln N N. Ths s Strlng s approxmaton. Very, very useful. A = NkTln q + NkTln N NkT µ A N T,V = ktln q + ktln N = ktln q / N µ kt = ln(q / N) = ktln q + ktln N + NkT(1 N) kt ( ) e µ kt = q / N q = Ne µ/kt whch s a very convenent form. The probablty of fndng one partcle out of N n level ε s P = n = e ε N q kt whch are the standard defnton and statstcal mechancal values of P = Ne ε kt q replace q = Ne ε kt 1 ( Ne µ kt ) = e ( ε µ ) kt Ths s the corrected Boltzmann result for. Notce that when ε < µ, > 1 whch volates the assumpton upon whch the valdty of corrected Boltzmann depends. Note also, that when T 0, the only occuped levels are those where ε µ. When ε > µ and T = 0, = 0. Note further, from the derved T dependence of µ µ = kt ln (q/n) lm µ = 0 T 0 revsed 2/25/08 4:35 PM

6 5.62 Sprng 2008 Lecture 8, Page 5 Thus, as T 0, the only occuped level n ε = 0 whch has occupancy n ε= 0 = 1. It s clear that we need to replace corrected Boltzmann statstcs by BE or FD as T 0 and whenever B becomes comparable to 1. Assert (derved for Grand Canoncal Ensemble, where µ, V, and T are held constant, pp of Hll) +1 s FD 1 = 1 s BE e ε µ ± 1 no 1 s B Ths equaton obeys the expectaton FD B BE < < In the lmt of T 0 ε = µ ε < µ ε > µ BE < 0 (mpossble) 0 FD B 1 (llegal) 0 Frst, let's make sure exact result for s correctly normalzed to total number of partcles. N = F 1 = F e ( ε µ )/kt ± 1 normalzaton correcton factor so F = N 1 e ( ε µ )/kt ± 1 N = Fn = ( (e ε µ ) kt 1 ± 1) (ε j µ) kt j e ± 1 normalzed exact result normalzaton factor approxmate result beng checked for normalzaton ( Now make the Boltzmann approxmaton e ε j µ ) kt 1 for all j revsed 2/25/08 4:35 PM

7 5.62 Sprng 2008 Lecture 8, Page 6 The factors of ±1 are small and can be neglected, thus: N = ( e ε µ ) kt e ( ε j µ ) = kt j Ne ε kt = = j e ε j kt Ne ε kt q CORRECTLY REDUCES TO BOLTZMANN STATISTICS RESULT! PLOT the FD and BE dstrbuton functons for vs. ε µ. kt Note that: * cannot be larger than 1 for FD * goes to when ε = µ for BE * when 1, we are no longer allowed to use corrected Boltzmann. (ε µ)/kt For large ε µ and consequently large e or << 1, revsed 2/25/08 4:35 PM

8 5.62 Sprng 2008 Lecture 8, Page 7 FD = BE and when these occupaton numbers are equal << 1! Most atoms and molecules at "ordnary" temperatures are n Boltzmann regme where << 1 or q >> N. In ths case, there s no dfference between FD and BE statstcs. Doesn't matter f the molecule s a fermon or boson. So the Boltzmann statstcs we have developed s vald over a wde range of molecules and condtons. EXCEPTIONS: 4 He and H 2 are BOSONS. Must be treated as such at T close to 0K. 3 He s FERMION. Must be treated as such at T close to 0K. Notce that the exceptons are lghter atoms and molecules at low T. That's because as you make partcle less massve, the spacngs n energy between the partcle's states get larger, leadng to fewer avalable states. If fewer states are avalable, goes up, and eventually the dfference between FD and BE statstcs becomes dscernble. ε = h 2 8ma 2 n 2 + n 2 + n 2 ( x ) y z ε3 ε7 ε5 ε ε2 ε3 ε ε1 ε1 LIGHT PARTICLE * n s larger * may have to use FD or BE statstcs HEAVY PARTICLE * n s smaller At "normal" temperatures > ~20K, can treat 3 He, 4 He, H2 wth Boltzmann statstcs. revsed 2/25/08 4:35 PM

9 5.62 Sprng 2008 Lecture 8, Page 8 So temperature plays a role here too. We'll talk about ths next tme. BUT ELECTRONS always have to be treated as FERMIONS for all "normal" temperatures (<3000K), because ther s 1. The valence electrons of the Au atoms whch make up a gold crystal are delocalzed throughout the crystal. These electrons can be thought of as an electron gas contaned wthn the crystal. Ths s called a free electron model where the energy levels of electrons are partcle-n-box energy levels. The average number of electrons n each electron state s gven by the Ferm-Drac dstrbuton functon = 1 ( ) kt + 1 e ε µ F.D. At T = 0 We fll each state wth 1 e n order of ncreasng energy. The energy at whch we run out of e s s ε = µ. In Sold State Physcs language µ ε F FERMI ENERGY whch s the maxmum energy that an e can have at T = 0. AT T > 0K revsed 2/25/08 4:35 PM

10 5.62 Sprng 2008 Lecture 8, Page 9 Electrons move from occuped to unoccuped states as T s ncreased. Must move to unoccuped states because > 1 not allowed. Ths s the orgn of conductvty n metals. [More on ths n second half of course.] revsed 2/25/08 4:35 PM

11 5.62 Sprng 2008 Lecture 8, Page 10 Non-Lecture Alternatve Dervaton of for B-E and F-D Partcles (ths s actually a sum over sets of )e β ε states, ) Canoncal p. f. Q = Ω({ } occupaton numbers, { }, and over { } = ω { } ( )e β ε ω( ) s the number of dstngushable ways of arrangng partcles n the ε sngle- partcle energy level. The form of ω(n) depends on the partcle statstcs. If t were possble to evaluate ths sum, then we could determne from kt lnq kt Q ε = Q ε V,T, N kt ({ })e β ε = kt Ω Q n { } Ω )e β ε ({ } { } =. Q Now we wll evaluate Q approxmately by fndng the sngle set of occupaton numbers { } that gves the maxmum term n the sum over occupaton numbers that defnes Q. The approxmatos to set Q equal to the value of ths maxmum term n the sum. There remans a sum over states,. Ths approxmaton can only be vald f the maxmum term n the sum s vastly larger than any term correspondng to a dfferent set of occupaton numbers. Ths s a common and useful approxmaton statstcal mechancs. Fnd the maxmum term n the sum, call t Q M and assume Q M Q A = kt lnq M βa = lnq M = {ln ω ( ) βε }, we have kept only the sngle set of occupaton numbers that gves the maxmum term n the sum that defnes Q. revsed 2/25/08 4:35 PM

12 5.62 Sprng 2008 Lecture 8, Page 11 The prme on the mples the constrant N = Use Lagrange multplers to mpose ths constrant so that the constraned sum can be replaced by an unconstraned sum, ( ) βa = lnq M = {ln ω βε } + λ N A β = λ N V,T unconstraned =0 sum value to be chosen but A µ V = µ Thus λ = βµ =. N,T kt Insert the derved specfc value for λ, rearrange βa = {ln ω ( ) β ( ε µ ) } Nβµ, Nβµ βa = { ln ω( ) β ( ε µ ) } and use the thermodynamc dentty G = Nµ = A + pv β(nµ A) = βpv = lnq + Nβµ = {ln ω ( ) β ( ε µ) } Now we choose partcular forms for ω( ) and nsert them nto the above equaton for βpv. Note that we have not yet addressed the approxmaton of Q by Q M. A. Corrected Boltzmann g s the degeneracy of the sngle-partcle ε energy level, and s the number of partcles n the assembly n the ε energy level. revsed 2/25/08 4:35 PM

13 5.62 Sprng 2008 Lecture 8, Page 12 n ω B = g n! lnω B = n ln g n ln n + n βpv = { ln g ln + β ( ε µ) } to obtan the maxmum term n ths sum, we take a dervatve wrt each. For each (βpv ) = {ln( g ) ln 1+1 β ( ε µ )} = 0, thus V,T, N B = g e βε eβµ (βpv ) (note 2 2 = (1/ ) < 0 whch assures that the extremum s a maxmum and not a mnmum). Snce q = Ne βµ, we get the standard corrected Boltzmann result when we replace e βµ by N/q, B = g e βε N q, moreover, when we replace the orgnal sum over sets of { } that defnes βpv by the specfc set { } that gves the maxmum term n the sum, we get Thus βpv = ln g + β ( ε µ ) g = e β ε µ) ( ln (g ) = β ε µ). ( whch s the deal gas law. βpv = = N revsed 2/25/08 4:35 PM

14 5.62 Sprng 2008 Lecture 8, Page 13 B. Ferm-Drac g! ω FD (g,n) = n!(g n)! βpv = {ln ω ( ) β ( ε µ ) } βpv = {g ln g g ln + (g )ln(g ) + ( g ) β ( ε µ) } g + + (g ) = 0 The extremum term n the sum over sets of { } s obtaned from takng dn result = 0. and settng ( pv ) = {ln( g ) ln β ( ε µ)} = 0 for all. β V,T, N ln g = β( ε µ) g g 1 = e β ( ε µ) = g = e β ( ε µ) +1 Thus FD 1 = g e β ε µ) +1, ( replacng the sum n Q by ts maxmum term, Q M βpv = {g ln g ln g ln (g ) + ln (g ) β ( ε µ) } = g ln g g ln (g ) + ln g β ( ε µ ) but, for the term n the sum over sets of { } that has the maxmum value ln g = β( ε µ ) and the last two terms n { } cancel. We obtan revsed 2/25/08 4:35 PM

15 5.62 Sprng 2008 Lecture 8, Page 14 βpv = g ln g = g ln 1 n g g = g ln 1 e β ε 1 ( µ) +1 e β ( ε µ) = g ln +1 1 e β ( ε µ) +1 ( = g ln e β ε µ) +1 ( = g ln 1 e β ε µ) whch s not the deal gas law. However, at hgh ε, [ ln 1+ e β(µ ε ) ] e β(µ ε ) because, for x 1 ln ( 1 1+ x) = ln x + x and, for x = e β(µ ε ) when ε µ Thus e β(µ ε) 1. βpv = g e β(µ ε ) = g e βµ e βε e βµ = (q N ) [ + e β(µ ε )] whch s the deal gas law! βpv = Ng e βε = N q = N q q C. Bose-Ensten Thus ( ) = (g + n 1)! ω BE!(g 1)! lnω BE ( ) = {(g + 1)ln (g + 1) (g + 1) ln + (g 1)ln (g 1) + g 1} = {(g + 1)ln (g + 1) l (g 1)ln (g 1)} βpv = lnq + Nβµ = {lnω BE ( ) β ( ε µ ) }. { } The set of occupaton numbers that gves the largest contrbuton to βpv s obtaned from revsed 2/25/08 4:35 PM

16 5.62 Sprng 2008 Lecture 8, Page 15 thus (βpv ) g + 1 ln n = 0 = ln (g + 1) + β ( ε µ) g + 1 V,T,N ln g + 1 = β ( ε µ) g 1 +1 = e β( ε µ) BE 1 = (g 1) e β ε ( µ) 1 Snce g 1, we obtan the usual result BE g = e β ( ε µ ) 1 and, when ε = µ, the denomnator vanshes and n BE can be very large. Ths s the Bose- Ensten condensaton. If ε < µ, the nonsense result of a negatve occupaton number s obtaned. D. Accuracy of the Maxmum Term Approxmaton (see Hll, Appendx II, pp ). Q ( )e βε t ({ } ω ) { } { } where t({ }) s a typcal term n the sum and mposes the mplct constrant = N. Expand the value of the typcal term t({ }) n the sum as a power seres n devatons from the specal set of occupaton numbers { nˆ } that gve the maxmum value of ω( ) e βε, t M = t ({ nˆ }). We have already used the requrement that 0 = value of t M and the set { nˆ }. Thus ω( ) e βε for all to fnd the ( ˆ ) 1 n lnt 2 n ) = lnt ) lnω ({ } ({ ˆ } δ. 2 n The frst nonzero term n the expansonvolves the second dervatves because all of the frst dervatves were requred to be zero (condton for the maxmum term) revsed 2/25/08 4:35 PM

17 5.62 Sprng 2008 Lecture 8, Page 16 lnt ({ nˆ }) = 0 and the e βε term s used up n the extremum condton. Thus the true value of Q s gven terms of the value of the maxmum term n the sum over { } as 1 2 lnω ( nˆ ) 2 Q = Q M 1+ exp δ + 2 {δ } 2 n but, snce we showed that the set { nˆ } maxmzes t({ }), all of the second dervatves must be negatve. Ths means the factor multplyng Q M s (1 + e x ) where x > 0 hence Q Q M. To make ths argument stronger we need to compute these second dervatves and also realze that the exp[ ] contans many addtve negatve terms, thus e x 0. The dervaton of the second dervatves lsted below s left as an exercse for you: Statstcs Boltzmann Ferm-Drac Bose-Ensten 2 lnω ( nˆ ) 2-1/ g (g ) g 1 n (g + 1) revsed 2/25/08 4:35 PM