LECTURE 3 Maxwell Boltzmann, Femi, and Bose Statistics Suppose we have a gas of N identical point paticles in a box of volume V. When we say gas, we mean that the paticles ae not inteacting with one anothe. Suppose we know the single paticle states in this gas. We would like to know what ae the possible states of the system as a whole. Thee ae 3 possible cases. Which one is appopiate depends on whethe we use Maxwell Boltzmann, Femi o Bose statistics. Let s conside a vey simple case in which we have 2 paticles in the box and the box has 2 single paticle states. How many distinct ways can we put the paticles into the 2 states? Maxwell Boltzmann Statistics: This is sometimes called the classical case. In this case the paticles ae distinguishable so let s label them A and B. Let s call the 2 single paticle states and 2. Fo Maxwell Boltzmann statistics any numbe of paticles can be in any state. So let s enumeate the states of the system: Single Paticle State 2 ----------------------------------------------------------------------- AB AB A B B A We get a total of 4 states of the system as a whole. Half of the states have the paticles bunched in the same state and half have them in sepaate states. Bose Einstein Statistics: This is a quantum mechanical case. This means that the paticles ae indistinguishable. Both paticles ae labelled A. Recall that bosons have intege spin: 0,, 2, etc. Fo Bose statistics any numbe of paticles can be in one state. So let s again enumeate the states of the system: Single Paticle State 2 ----------------------------------------------------------------------- AA AA A A We get a total of 3 states of the system as a whole. 2/3 of the states have the paticles bunched in the same state and /3 of the states have them in sepaate states. Femi Statistics: This is anothe quantum mechanical case. Again the paticles ae indistinguishable. Both paticles ae labelled A. Recall that femions have half intege spin: /2, 3/2, etc. Accoding to the Pauli exclusion pinciple, no moe than one paticle can be in any one single paticle state. So let s again enumeate the states of the system: Single Paticle State 2 ----------------------------------------------------------------------- A A
We get a total of state of the system as a whole. None of the states have the paticles bunched up; the Pauli exclusion pinciple fobids that. 00% of the states have the paticles in sepaate states. This simple example shows how the type of statistics influences the possible states of the system. Distibution Functions We can fomalize this somewhat. We conside a gas of N identical paticles in a volume V in equilibium at the tempeatue T. We shall use the following notation: Label the possible quantum states of a single paticle by o s. Denote the enegy of a paticle in state by ε. Denote the numbe of paticles in state by n. Label the possible quantum states of the whole gas by R. Since the paticles in the gas ae not inteacting o ae inteacting weakly, we can descibe the state R of the system as having n paticles in state =,n 2 paticles in state =2, etc. The total enegy of the state is E R = n ε + n 2 ε 2 + n 3 ε 3... = n ε () Since the total numbe of paticles is N, then we must have n = N (2) The patition function is given by Z = R e βe R = R e β(n ε +n 2 ε 2 +...) (3) Hee the sum is ove all the possible states R of the whole gas, i.e., essentially ove all the vaious possible values of the numbes n, n 2, n 3,... Now we want to find the mean numbe n s of paticles in a state s. Since P R = e β(n ε +n 2 ε 2 +...) (4) Z is the pobability of finding the gas in a paticula state whee thee ae n paticles in state, n 2 paticles in state 2, etc., one can wite fo the mean numbe of paticles in a state s: R n s e β(n ε +n 2 ε 2 +...) n s P R = (5) R Z 2
We can ewite this as ( ) Z R β ε s o β e β(n ε +n 2 ε 2 +...) = Z (6) βz ε s ln Z ε s (7) So to calculate the mean numbe of paticles in a given single paticle state s, wejust have to calculate the patition function Z and take the appopiate deivative. We want to calculate n s fo Maxwell Boltzmann, Bose and Femi statistics. Maxwell Boltzmann Statistics Let us begin by consideing the classical case of Maxwell Boltzmann statistics. In this case the paticles ae distinguishable but identical, so each paticle has the same set of single paticle enegy levels. As a esult we can wite the patition function as whee the single paticle patition function is Z = ζ N (8) ζ = e βε (9) Then ( ) ln Z = N ln ζ = N ln e βε Now we can calculate the mean occupation numbe (0) o β ln Z ε s = β N βe βεs e βε () e βεs N (2) e βε This is called the Maxwell Boltzmann distibution. It is the same as ou pevious esult when we applied the canonical distibution to N independent single paticles in a classical system. The sum ove is a sum ove single paticle states. Altenative Deivation of Maxwell Boltzmann Patition Function We can wite the patition function of the gas as Z = R e β(n ε +n 2 ε 2 +...) (3) Hee we ae summing ove all possible states R of the gas, i.e., ove all values n =0,, 2, 3,... fo each (4) 3
subject to the estiction n = N (5) Since this is a classical system, the paticles have to be consideed distinguishable. Thus any pemutation of two paticles in diffeent states must be counted as a distinct state of the whole gas even though the numbes n,n 2,n 3,... ae left unchanged. This was the case in ou simple example. It is not enough to specify how many paticles ae in each single paticle state, but it is necessay to specify which paticula paticle is in which state. Fo a given set of values n,n 2,n 3,..., theeae N! (6) n!n 2!... possible ways in which the paticles can be put into the given single paticle states with n paticles in state, n 2 paticles in state 2, etc. This is the numbe of distinct states since the paticles ae distinguishable. Hence Z = N! n,n 2,... n!n 2!... e β(n ε +n 2 ε 2 +...) = N! ( ) e βε n ( ) e βε n2 2... (7) n,n 2,... n!n 2!... Because of (5), this last expession is just a multinomial expansion. So we can wite o Z = ( e βε + e βε 2 +... ) N ( ) ln Z = N ln e βε whee e βε is just the patition function fo a single paticle. This is what we got befoe. Bose Einstein and Photon Statistics Hee the paticles ae to be consideed as indistinguishable, so that the state of the gas can be specified by meely listing the numbe of paticles in each single paticle state: n, n 2, n 3,... Since thee is no limit to the numbe of paticles that can occupy a state, n s can equal 0,,2,3,... fo each state s. Fo photons the total numbe of paticles is not fixed since photons can eadily be emitted o absobed by the walls of the containe. Let s calculate n s fo the case of photon statistics. The patition function is given by Z = (20) R whee the summation is ove all values n =0,, 2, 3,... fo each, without any futhe estiction. We can ewite (20) as Z = e βn ε e βn 2ε 2 e βn 3ε 3... (2) n,n 2,... e β(n ε +n 2 ε 2 +...) 4 (8) (9)
o Z = e βn ε e βn 2ε 2 e βn 3ε 3... (22) n =0 n 2 =0 n 3 =0 But each sum is a geometic seies whose fist tem is and whee the atio between successive tems is exp( βε ). Thus it can be easily summed: n s=0 e βnsεs =+e βεs + e 2βεs +... = Hence eq. (22) becomes ( )( )( Z = e βε e βε 2 e βε 3 o ln Z = ln ( ) e βεs s So if we plug this into eqn. (7), we get ln Z ε s ε s ln ( e βεs ) = e βεs (23) )... (24) (25) e βεs e βεs (26) = β β o (27) e βεs This is called the Planck distibution. We ll come back to this a bit late when we talk about black body adiation. Photons ae bosons, but thei total numbe is not conseved because they can be absobed and emitted. Othe types of bosons, howeve, do have thei total numbe conseved. One example is 4 He atoms. A 4 He atom is a boson because if you add the spin of the poton, neuton, and 2 electons, you always will get an intege. If the numbe of bosons is conseved, then n s must satisfy the condition s N (28) whee N is the total numbe of bosons in the system. In ode to satisfy this condition, one slightly modifies the Planck distibution. The esult is known as the Bose Einstein distibution (29) e β(εs μ) whee μ is the chemical potential. μ is adjusted so that eq. (28) is satisfied. Physically μ is the change in the enegy of the system when one paticle is added. Eqn. (29) is called the Bose Einstein distibution function o the Bose distibution function fo shot. Often one wites this as a function of enegy: n(ε) = e β(ε μ) n(ε) is also called the Bose-Einstein distibution. 5 (30)
Bose Einstein Distibution Function μ = 0 K 5.0 5.0 n(e) 5.0 5.0 T = K T = 5 K T = 0 K T = 20 K 25.0 50.0 25.0 0.0 25.0 50.0 E [K] Notice that if ε<μ,thenn(ε) < 0 which doesn t make much sense. The Bose distibution only makes sense fo ε>μ. We can explicitly deive (29). In ode to satisfy the condition (28), one multiplies the patition function by a fudge facto exp( αn). α is then adjusted to satisfy eqn. (28). α is an example of what is called a Lagange multiplie. Z = R e β(n ε +n 2 ε 2 +...) e αn = e β(n ε +n 2 ε 2 +...) e α(n +n 2 +...) R = e (α+βε )n (α+βε 2 )n 2... n,n 2,... = e (α+βε )n e (α+βε 2)n 2... (3) n =0 n 2 =0 (32) We use Z instead of Z because we have an exta facto of exp( αn). Z is the gand patition function that we met when we discussed the gand canonical ensemble. This is just a poduct of simple geometic seies. Hence ( )( ) Z =... (33) e (α+βε ) e (α+βε 2) o ln Z = ln ( e (α+βε)) (34) Recall that when we discussed the gand canonical ensemble and the gand patition function, we set α = βμ (35) whee μ is the chemical potential. We ae basically assuming that we have a system of weakly inteacting bosons in contact with both an enegy and paticle numbe esevoi, 6
and that the tempeatue T and the chemical potential μ ae fixed by the esevoi. So we can ewite (34) to obtain the gand patition function: ln Z = ln ( e β(ε μ)) (36) We can now calculate the aveage value of N: N = Ne β(n ε +n 2 ε 2 +...) e βμn Z R = e β(n ε +n 2 ε 2 +...) e βμn β Z μ R = Z β Z μ = ln Z β μ = [ ln ( e β(ε μ))] β μ = βe β(ε μ) β e β(ε μ) = e β(ε μ) (37) μ is fixed by setting N = N whee N is the total numbe of bosons in the system. Finally we will calculate the aveage numbe of bosons in state s: n s e β(n ε +n 2 ε 2 +...) e βμ(n +n 2 +...) Z states = ( ) e β(n ε +n 2 ε 2 +...) e βμ(n +n 2 +...) Z β ε s states = ln Z β ε s = [ ln ( e β(ε μ))] β ε s = ( β)e β(εs μ) (38) β e β(εs μ) o (39) e β(εs μ) Note that we can ecove the esult fo photons by setting μ =0. Alsonotethat N = s n s (40) 7
We will etun to the Bose Einstein distibution when we discuss black body adiation. Femi Diac Statistics Recall that femions have half intege spin statistics and that at most one femion can occupy each single paticle state. This means that 0 o. We can easily get some idea of what n s is by consideing the vey simple case of a system with just one single paticle state. In this case n s n s e βnsεs (4) n s e βnsεs In this case the sums just have 2 tems. The denominato is e βnsεs =+e βεs (42) The numeato is n s=0, n s=0, n s e βnsεs =0+e βεs (43) So we have e βεs (44) +e βεs o (45) e βεs + Fo a eal system we have many single paticle states and many paticles. The expession fo n s in this case must satisfy the condition that the numbe of paticles is a constant: N (46) The coect fomula which satisfies this condition (46) is s e β(εs μ) + Often one wites this as a function of enegy: f(ε) = e β(ε μ) + (47) (48) f(ε) is called the Femi distibution function. μ is adjusted to satisfy the constaint (46). As in the Bose Einstein case, μ is called the chemical potential. This is basically the same as the Femi enegy. Notice that f(ε = μ) =/2. This is always tue of the Femi distibution. 8
f (E) Femi Distibution Function T = 0 T > 0 μ(τ=0)= E F E We can fomally deive the Femi distibution in much the same way as we deived the Bose distibution. We once again conside a system of weakly inteacting femions in contact with both an enegy and a paticle numbe esevoi. The gand patition function is given by Z = e β(e μn) = = = states n,n 2,n 3,... e β(n ε +n 2 ε 2 +...) e βμ(n +n 2 +...) e β(ε μ)n e β(ε 2 μ)n 2... n,n 2,n 3,... e β(ε μ)n e β(ε 2 μ)n 2 n =0 n 2 =0... = ( +e )( β(ε μ) +e ) β(ε 2 μ)... = ( ) +e β(ε μ) (49) and ln Z = ln ( +e β(ε μ)) (50) So the mean numbe of paticles in the system is N = ln Z β μ = βe β(ε μ) β +e β(ε μ) (5) o N = e β(ε μ) + (52) The mean numbe of femions in state s is β ln Z = ( β)e β(εs μ) (53) ε s β +e β(εs μ) 9
o Note that e β(εs μ) + N = s (54) n s (55) We will etun to this when we discuss metals. Summay Fo futue efeence the two expessions fo the aveage numbe of paticles in the sth state fo bosons and femions ae: e β(εs μ) e β(εs μ) + bosons (56) femions (57) A moe succinct way to wite ou esults fo the quantum statistics of ideal gases is (59) e β(εs μ) ± whee the uppe sign efes to Femi statistics and the lowe sign efes to Bose statistics. If the gas consists of a fixed numbe of paticles, μ is detemined by s s (58) e β(εs μ) ± = N (60) In geneal the numbe N of paticles is much smalle than the total numbe of single paticle states s. Classical Limit Let us conside 2 limiting cases. Conside the low density limit whee N is vey small. The elation (60) can then only be satisfied if each tem in the sum ove all states is sufficiently small, i.e., if n s oexp[β(ε s μ)] fo all states s. The othe case to conside is the high tempeatue limit. Since β =/k B T, the high tempeatue limit coesponds to small β. Nowifβ wee 0, we would have s ± = N (6) which is a disaste fo both the Femi Diac and Bose Einstein cases. But β =0means that T =. Let s assume that the tempeatue is high but not infinite, so that β is small but not 0. At high tempeatues, lots of high enegy states ae occupied. By high enegy, I mean that ε s μ. In ode to satisfy the fixed N constaint of eqn. (60), it is necessay to have exp[β(ε s μ)] (62) 0
such that n s (63) fo all states s. (Remembe that thee ae many moe states s than paticles N.) This is the same condition that came up in the low density case. We call the limit of sufficiently low concentation o sufficiently high tempeatue whee (62) o (63) ae satisfied the classical limit. In this limit n s educes to e β(εs μ) (64) Plugging this into (60), we get s s o Thus e β(εs μ) = e βμ s e βμ = e βεs = N (65) N s e βεs (66) e βεs N (67) s e βεs Hence we see that in the classical limit of sufficiently low density o sufficiently high tempeatue, the Femi Diac and Bose Einstein distibution laws educe to the Maxwell Boltzmann distibution. Relation of Z and Z In deiving the Bose Einstein and Femi Diac distibutions, we used the gand canonical patition function. We can use Z to obtain an excellent appoximation to the canonical patition function. We can wite Z = = = = e β(e μn) all states e βe e βμn all states e βμn e βe N =0 all states with N paticles N =0 e βμn Z(N ) (68) whee Z(N ) is the canonical patition function fo N paticles. Since Z(N )isaapidly inceasing function of N and e βμn is a apidly deceasing function of N (fo μ<0), the gand patition function is shaply peaked at N = N. Sowecanwite Z = N =0 e βμn Z(N ) = Z(N)e βμn (Δ N ) (69)
whee the width of the peak is Δ N.Thus since ln (Δ N ) is negligible. O ln Z = lnz(n)+βμn +ln(δ N ) = ln Z(N)+βμN (70) ln Z = βμn +lnz (7) This is the elation between the patition function Z and the gand patition function Z. Chemical Potential Since e βμn Z(N ) is a shaply peaked function at N = N, wecanusethistodeive an expession fo the chemical potential μ. Consideln [ e βμn Z(N ) ]. By definition the maximum of this is given by N [ln Z(N )+βμn ]= ln Z(N ) + βμ = 0 (72) N o μ = ln Z(N) ln Z(N) = k B T β N N This is useful fo calculating the chemical potential μ. = F N Othe Conventions fo the Gand Canonical Ensemble In some books, the fugacity y is defined by (73) y = e βμ (74) (Some books use z o λ to denote the fugacity.) The gand patition function is given in tems of the fugacity by Z(V,T,y) N =0 y N Z N (V,T) (75) with Z 0. In the gand canonical ensemble, pessue is defined by PV = k B T ln Z(V,T,y) (76) whee V is volume. Some books define a themodynamic potential Ω(V,T,y) by Ω(V,T,y) = PV = k B T ln Z(V,T,y) (77) (Do not confuse the themodynamic potential with the numbe of micostates of a system, even though both ae sometimes denoted by Ω.) In tems of the gand patition function, the mean numbe of paticles N is [ ] N = y ln Z(V,T,y) (78) y V,T 2
and the mean enegy E is [ ] E = ln Z(V,T,y) β y,v (One can compae this to Eq. (70) in Lectue 9 [ ] E = μn β ln Z to see the impotance of noting what is kept constant and what isn t in taking deivatives.) The fist law of themodynamics then becomes μ,v (79) (80) ΔE = Q W + μδn (8) o, in diffeential fom, de = TdS pdv + μdn (82) Gibbs Paadox Revisited Now back to eq. (7). Plugging in Z fo bosons fom (34), we have ln Z = βμn ln ( e β(ε μ)) (83) Similaly fo femions eq. (50) yields ln Z = βμn + ln ( +e β(ε μ)) (84) We can combine these two expessions: ln Z = βμn ± ln ( ± e β(ε μ)) (85) At high tempeatues e β(ε μ) is small and we can expand the logaithm to obtain ln Z = βμn ± ( ) ±e β(ε μ) = βμn + N (86) whee in the last step we used the high tempeatue limit of the Bose and Femi distibutions: n = N e β(ε μ) ± = N e β(ε μ) = N at high tempeatue e βμ e βε = N (87) 3
Taking the logaithm of both sides of (87) yields ( ) βμ +ln e βε =lnn (88) o ( ) βμ = ln N +ln e βε (89) Plugging this into (86), we get ln Z = βμn + N ( ) = N ln N + N + N ln e βε (90) The fist two tems ae Stiling s appoximation to N!: ln N! = N ln N N (9) The last tem in (90) is the Maxwell Boltzmann patition function. ( ) ln Z MB = N ln e βε (92) So (90) becomes ln Z = ln N!+lnZ MB (93) o Z = Z MB (94) N! This was the esolution to the Gibbs paadox. Without quantum mechanics we had to put in the facto N! by hand. But now we see that the indistinguishability of the paticles comes out natually. 4