Grand canonical sensemble.pdf

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20. LECTURE 20: GRAND CANONICAL ENSEMBLE AND IDEAL QUANTUM SYSTEMS239

20 Lecture 20: Grand canonical ensemble and ideal

quantum systems

SummaryGrand canonical partition function/ensemble is introduced. Noninteracting quantum systems are discussed with the aid of grand partitionfunction.Pressures of fermions and bosons are compared.P V = 2E/3 for any ideal gas.

Key wordsgrand canonical partition function, grand canonical ensemble, fermion, boson, Bose-

Einstein distribution, Fermi-Dirac distribution

What you should be able to do Be explain why P Vis directly obtained from the grand canonical partition func-tion. Intuitively understand low temperature fermi and bose systems. In particular, compare the pressures of boson and fermion systems intuitively.Be able to derive Dt(). Note Dt .

Grand canonical ensemble/partition functionWe now introduce another ensemble. We know microcanonical, canonical, and gen-eralized canonical ensembles (e.g., pressure ensembles). The fourth law tells us (canyou directly derive this following the logic used in the derivation of the Gibbs-Duhemrelation?)

A= E T S= P V + N. (20.1)Therefore,

P V = A + N=ST E+ N, (20.2)That is,

P V

T =SE

T +

TN (20.3)This is a Legendre transformation of entropy, so there must be an ensemble thatdirectly gives P V/T orP V/kBT.


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Compare the following formulas:

S = kBlog w(E , V , X ), (20.4)

A

T =SE

T = kBlog Z(T , V , X ). (20.5)

We know with the aid of Boltzmanns principle

Z(T , V , X ) =

dE w(E , V , X )eE/kBT =

dE e[S(E)E/T]/kB . (20.6)

Thus, we can easily mimic this to get

P V

T =SE N

T =kBlog (T , V , ) (20.7)

with

(T , V , ) =

dE

N

w(E , V , N )e(EN)/kBT =

N=0

Z(T , V , N )eN. (20.8)

is called the grand (canonical) partition function, which describes the system ther-mostatted and chemostatted with a reservoir at T and chemical potential . Recallthat is the needed work to push one molecule into the system, so by adjusting you can regulate the average number of particles.

Let us summarize:

P V

T = A

T +N

T =kBlog (T , V , ) (20.9)

with

(T , V , ) =

N=0

Z(T , V , N )eN =

microstates

e(HN), (20.10)

where the summation over microstates means all the possible microstates allowedto the system irrespective of the total number of particles.

The ensemble equivalence holds here as well: IfN log N, then you can use anyensemble you wish. For example, if you have about a few thousand particles confinedin a trap, you may use the grand canonical formalism above to describe the system.

The Gibbs relation reads

d

P V

T

= Ed 1

T +

P

TdV + N d

T. (20.11)


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This has a more convenient form:

d log = Ed+ PdV + Nd(). (20.12)

Examples: adsorptionSuppose there is a gas mixture consisting of two distinct atomic species AandB. Themixture is an ideal gas and the partial pressures ofX is PX(X=A orB). The gasis in equilibrium with an adsorbing metal surface on which there are N adsorptionsites. Atom X adsorbed at a site is with energy X(which is often negative) on theaverage relative to the one in the gas phase, where X = A or B. Each surface sitecan accommodate at most one A atom, and at most 2 B atoms. One adsorbed Aatom has 2 different states, one adsorbed B atom has 1 state, but if 2 are adsorbed,then can together have 5 states. We wish to know the surface concentration of the

atoms when the surface is in equilibrium with the gas mixture. You may assumethe gas phase is huge, so you need not worry about its composition change due toadsorption. That is, the gas phase is a chemical reservoir.

Figure 20.1: Adsorption of gas particles on a metal surface: A: green (2 internal states whenabsorbed), B: red (if singly absorbed, with single internal state; if doubly absorbed with 5 internalstates).

We wish to know the average number of A and B atoms on the metal surface.Since we do not know how many particles are on the surface, it must be convenient

to use the grand canonical ensemble. Assuming the chemical potentials ofAandBas

Aand B, respectively, write down the partition function of the metal surface:

=

1 + 2e(AA) + e(BB) + 5e2(BB)N

. (20.13)


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The needed chemical potentials can be computed with the aid of the ideal gas statisti-cal mechanics. We have done that calculation, so let us copy the needed results:

A= kBTlog(PA/nQA), B =kBTlog(PB/nQB) (20.14)

Here,nQXis the quantum density depending on Tand the mass (see above (??)),andPQX=nQXkBTmay be called the quantum pressure. We know (we ignore thevolume (or the area) change)

d log = Ed+ NAd(A) + NBd(B), (20.15)so we obtain

NA= N 2e(AA)

1 + 2e(AA) + e(BB) + 5e2(BB) (20.16)

and

NB =N e(B

B)

+ 10e2(B

B)

1 + 2e(AA) + e(BB) + 5e2(BB). (20.17)

Microstates for non-interacting indistinguishable particle systemsLet us consider a system consisting of non-interacting particles. Suppose the statesof a single particle are numbered as i= 1, 2, . If we assume that all the particlesare indistinguishable, then to specify a microstate of a system consisting of suchparticles, we have only to count the number ni of particles in the i-th one particlestate (ni is also called the occupation numberof the ith one particle state; do NOTconfuse the microstate of the whole system and the one particle states). Or, we haveonly to make a table of the occupation numbers{n1, n2, }; we may identify thistable and the microstate.

To study the thermodynamics of such a system, we should use the grand canonicalensemble, because we have not specified the total number of particles.

Grand partition function of indistinguishable particle systemLeti be the energy of the i-th one particle state. The total energy

Eand the total

number of particles Nof the microstate{n1, n2, } can be written asE=

i=1

ini, (20.18)


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and

N =i=1

ni. (20.19)

Then, the grand canonical partition function must be

(, ) =

n1,n2,eE+N. (20.20)

Using the microscopic descriptions ofEandN((20.18) and (20.19)), we can rearrangethe summation as

=i

i, (20.21)

where

ini

exp[(i )ni]. (20.22)

This quantity may be called the grand canonical partition function for the one particlestatei.

Boson and fermionsIn the world it seems that there are only two kinds of particles:bosons: there is no upper bound for the occupation number;

fermions: the occupation number can be at most 1 (the Pauli exclusion principle).This is an empirical fact. Electrons, protons, 3He, etc., are fermions, and 4He,D,

etc., are bosons.There is the so-called spin-statistics relationthat the particles with half odd inte-

ger spins are fermions, and those with integer spins are bosons. The rule applies alsoto compound particles such as hydrogen atoms. Thus, H and T are bosons, but theirnuclei are fermions. D and 3He are fermions. 4He is a boson, and so is its nucleus.

For a neutral system consisting of + and charged particles (e.g., the usualelectron-nucleus system) it is proved that at least + or species must be all fermionsfor the system to be stable. Here, stable means that there is a positive number Bsuch that the system energy E satisfies E >NB, where N is the number of par-ticles in the system. That is, for the world to be stable, we desperately need fermions.

Ideal boson systems


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For bosons, any number of particles can occupy the same one particle state, so

i =

n=0

e(i)n = 1 e(i)

1

. (20.23)

The mean occupation number of the i-th state is given by

ni =

n=0

nie(i)n/i, (20.24)

so we conclude

ni =

logi

=kBT

logi

T

= 1

e(i) 1 . (20.25)

This distribution is called the Bose-Einstein distribution.If the ground-state energy is zero, then the ground state occupancy is

nground = 1e 1 , (20.26)

but this should not be negative, so 0. That is, notice that the chemical poten-tial must be smallerthan the ground state energy to maintain the positivity of theaverage occupation number.

Ideal fermion systemsFor fermions, at most one particles can occupy the same one particle state, so

i =1

n=0

e(i)n = 1 + e(i). (20.27)

The mean occupation number of the i-th state is given by

ni =1

n=0

nie(i)n/i, (20.28)

so we conclude

ni =

logi

=kBT

logi

T

= 1

e(i) + 1. (20.29)


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This distribution is called the Fermi-Dirac distribution.Notice that

nground = 1e + 1

. (20.30)

It is important to recognize the qualitative features of this Fermi-Dirac distributionfunction (see Fig. 20.2). The distribution has a cliff of width of orderkBT. In theT 0 limit, it has a vertical cliff at = , which is called the Fermi energy(Fermilevel).201 Notice that >0 is required, if the temperature is low enough.

0

1/2

1

expected

occupation

number

k TB

symmetric around this point

Figure 20.2: The cliff has a width of orderkBT. is called the Fermi potential. The symmetrynoted in the figure is the so-called particle-hole symmetry

Classical limitIn order to obtain the classical limit, we must take the occupation number 0 limit toavoid quantum interference among particles. The chemical potential is a measureof the strength of the chemostat to push particles into the system. Thus, we must

make the chemical potential extremely small: .In this limit both Bose-Einstein (20.26) and Fermi-Dirac distributions (20.30)

reduce to the Maxwell-Boltzmann distribution as expected:

ni Nei , (20.31)

whereN = e is the normalization constant determined by the total number ofparticles in the system.

Notice that is far away from the situation where quantum effects areimportant 0 for bosons and >0 for fermions.

201Do not forget that for a fixed N is temperature dependent. at T= 0 is called the Fermienergy.


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Intuitive picturesBefore going to the equations of state, let us to try to build our intuition. Supposethere are only three one-particle states with energies 0, and 3, and there are three

particles. Make a table of all the microstates of the system for bosons and fermions.

3

0

4 0 3 2 4 3 5 7 9total energy 6

Figure 20.3: identical particles in three states; fermion (leftmost) and boson cases.

Fermions first:microstate 0 3 total energy1 1 1 1 4

Bosons:microstate 0 3 total energy1 3 0 0 02 2 1 0 3 2 0 1 34 1 2 0 25 1 1 1 46 1 0 2 67 0 3 0 38 0 2 1 59 0 1 2 710 0 0 3 9

Suppose there are 100 identical spinless202 bosons or fermions whose s-th one-particle state has an energy s = s (s N). These particles do not interact. Forthe boson case, at T= 0 all the particles are in the lowest energy one-particle state(see Fig. 20.4). For fermions, all the low-lying one particle states are completelyfilled up to some energy level (called the Fermi level) that corresponds to . Noticethat the ground state energy of the fermion and boson systems are quite differentcompared with the single particle ground state energy.

The low-lying excited microstates are also in Fig.20.4(right). For the boson caseall the particles have equal chance to be excited, but in the case of fermions, onlythe particles near the Fermi level can be excited (and excited particles leave holes).

202Spinless implies that these particles do not have any internal degrees of freedom.


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This should tell you something about the specific heat of these systems (later).

100

0

100particles

100

0almost 100particles

ground microstate

Fermi level holes

fermion

case

boson

case

boson

casefermion

case

example of low energy excited microstate

Figure 20.4: Ground states and low-lying excited levels for fermions (left in each panel) andbosons (right). Do not confuse one particle states and microstates. Here, the lowest energy (groundstate) microstate is described on the left, and one example of low-energy excited microstate isillustrated on the right.

Pressure of ideal systemsThe distinction between fermions and bosons show up clearly in pressure:

P V

kBT= log =

i

log

1 e(i) . (20.32)IfT , V , are the same, then the pressure of the system consisting of the particleswith the same single-particle energy states (i.e., the the same density of states)shows the following ordering (BE = Bose-Einstein, MB = Maxwell-Boltzmann, FD= Fermi-Dirac):

PBE> PMB > PFD. (20.33)

To seePBE> PFD we use (20.32). Letx = e(i) and we compare log(1 x)

and log(1 +x) (see Fig.20.5). This figure must be enough. This is also easily seenfrom 1/(1 x) = 1 + x + x2 + >1 + xfor x (0, 1).


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-log(1-x)

log(1+x)

x

4 2 2 4

4

2

2

4

1 2 3 4

O

1

2

3

4

1234

1

2

3

4

Figure 20.5: log(1 + x) is always below x. Now, log(1 x) is obtained from log(1 + x) bymaking mirror images x x and than y y. Obviously, x < log(1 x).

In contrast, ifT , V , N are the same (the usually more interesting case than theabove case), then

PFD > PMB > PBE. (20.34)

To show this requires some trick, so a formal demonstration is below with fine letters,but intuitively this can be understood by the extent of effective particle-particleattraction as is illustrated in Fig.20.6. The figure not only suggests the pressures,but it also suggests the extent of particle number fluctuations. The particle densityfluctuations in a boson system are larger than those in a fermion system.

BE

FD

MB

2/3

2/4=1/2

0

Figure 20.6: Two-particle two box illustration of statistics. The numbers in the right denote the

relative weights of the states for which effective attraction can be seen. (BE = Bose-Einstein, MB= Maxwell-Boltzmann, FD = Fermi-Dirac)

As seen from the effective attraction weights in the above figure, the fermion system


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exhibits the largest pressure; fermions avoid each other (Paulis exclusion principle),so they hit the wall more often.

may be demonstrated as follows: Classically, P V =N kBT, so we wish to demonstrate (Nand Nneed not be distinguished, since we consider macrosystems)logFD > N >log BE. (20.35)

Let us see the first inequality:203 Writingxj = e(j), we have

logFD N =j

log(1 +xj) xj

1 +xj

, (20.36)

where xj =e(j). We are done, because for x >0204

log(1 +x) x1 +x

>0. (20.37)

Similarly, we can prove the second inequality in (20.35).

Universal P-E relation: introductionLetD() denote the number of single particle states whose energy is between and+d. This quantity is called the one-particle state density (or density of statesofone-particle system). If we know this, the pressure (20.32) can be rewritten as

P V = kBT

d D()log

1 e() . (20.38)We know for a classical ideal gas

P V =23

E, (20.39)

whereEis the internal energy. In this caseD() is the density of state for a particleconfined in a 3D box of volume V, which we will denote by Dt(), where the sufficet implies the translational degrees of freedom. Miraculously, this is true for non-interacting fermions and bosons (and their mixtures as well). The formulas for P VandEare quite different from the classical case. Here,

E=

d Dt()n() =

d Dt()

e() 1 . (20.40)203The reader might wonder why we cannot use MB to demonstrate the formula; the reason

is that in this grand partition function and that in FD or BE are distinct. Remember thatwe keep N; inevitably depends on statistics, so we cannot easily compare the Boltzmann factore() in each term.

204Consider the derivatives.


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Density of statesTo demonstrate this, we need Dt(). I explain a quick way to derive the formulaappropriate in a statistical mechanics course. We count the number of microscopic

states for a single particle up to some energy . To do this we use the classicalquan-tum mechanics correspondence: the number of quantum state in the phase volumeelement dpdq is given by dpdq/h3. Then, the total number of quantum states for asingle particle whose energy is less than or equal to must be give by

1

h3

qV

dq

|p|


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Performing an integration by parts, we get

P V = kBT

0

dDt()log 1 e()

0

kBT

d

0

dDt()

d

dlog 1 e

()

.(20.46)

The first term vanishes (you must check this205), so

P V = kBT

d

0

dDt()

d

dlog

1 e() (20.47)

= 23

kBT

dDt()

d

dlog

1 e() , (20.48)

= 23

kBT

dDt()

e()1 e() (20.49)

=

2

3

d Dt()

e() 1=2

3E. (20.50)

Now, a bomb making question. There is an isolated metal container of volume V andenergyEwithN fermions whose chemical potential is = 10 eV. These bosons areadiabatically converted into bosons. Can we use this mechanism to make a bomb?Notice that since EandVare unchanged, the pressure does not change. [Hint: Youmust know 1 eV is roughly equivalent to 105 K.]

205 0 for the boson case is only slightly tricky, but 1/2 log 0 saves the day.

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