UC berkeley stat mech (physics 112) reader

8/22/2019 UC berkeley stat mech (physics 112) reader

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Contents

1 Prelude 51.1 Heat and temperature . . . . . . . . . . . . . . . . . . . 51.2 The first law of thermodynamics . . . . . . . . . . . . . . 5

1.3 The Carnot Engine . . . . . . . . . . . . . . . . . . . . . 51.4 The second law of thermodynamics and entropy . . . . . 5

2 The formal statistical mechanics theory 72.1 Entropy, temperature and pressure of ideal gas . . . . . . 82.2 Energy exchange contact between two systems, and the

relation between entropy and temperature . . . . . . . . 112.3 Energy and volume exchange contact between two sys-

tems and the relation between entropy and pressure . . . 132.4 Energy and particle exchange contact between two sys-

tems and the relation between entropy and chemical po-

tential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.5 The first law of thermodynamics . . . . . . . . . . . . . . 152.6 The second law of thermodynamics . . . . . . . . . . . . 152.7 The third law of thermodynamics . . . . . . . . . . . . . 162.8 The canonical ensemble . . . . . . . . . . . . . . . . . . . 162.9 The partition functionZand the Helmholtz free energyF 18

2.9.1 Paramagnet . . . . . . . . . . . . . . . . . . . . . 212.9.2 Maxwell distribution . . . . . . . . . . . . . . . . 222.9.3 Equal-partition theorem . . . . . . . . . . . . . . 23

2.10 The grand canonical ensemble . . . . . . . . . . . . . . . 25

2.11 The grand partition function and thermodynamic function 262.11.1 Molecular adsorption -The hydrogen storageproblem . . . . . . . . . . . . . . . . . . . . . . . 27

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3 Black body radiation 313.1 Statistical mechanics of harmonic oscillators . . . . . . . 323.2 Quantum statistical mechanics of harmonic oscillator . . 333.3 Classical description of the E-M field . . . . . . . . . . . 343.4 Quantum statistical mechanics of the cavity E-M field . . 363.5 Acoustic phonon contribution to the specific heat . . . . 37

4 Ideal Fermi gas 414.1 Particle in a box . . . . . . . . . . . . . . . . . . . . . . 414.2 Grand canonical ensemble treatment of Fermi particle in

a box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.3 The particle number constraint and the chemical potential 424.4 The total internal energy and the specific heat . . . . . . 47

4.5 The magnetic susceptibility . . . . . . . . . . . . . . . . 484.6 The physics of while dwarfs (optional) . . . . . . . . . . 50

5 Ideal Bose gas 555.1 The grand canonical ensemble treatment . . . . . . . . . 555.2 The particle number constraint . . . . . . . . . . . . . . 56

5.2.1 zero temperature macroscopic occupation of thek= 0 state . . . . . . . . . . . . . . . . . . . . . 57

5.2.2 Non-zero temperature Bose Einstein condensation 575.3 The amazing work of indistinguishability (optional) . . . 60

6 Phase transitions 636.1 Example: critical phenomena of an easy-axis ferromagnet 656.2 Universality . . . . . . . . . . . . . . . . . . . . . . . . . 666.3 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 676.4 A simple model for easy-axis ferromagnet: the Ising model 676.5 The predictions of mean-field theory . . . . . . . . . . . 70

6.5.1 The spontaneous magnetization, the exponent . 706.5.2 The exponent . . . . . . . . . . . . . . . . . . . 726.5.3 The exponent . . . . . . . . . . . . . . . . . . . 736.5.4 The exponent . . . . . . . . . . . . . . . . . . . 736.5.5 Scaling . . . . . . . . . . . . . . . . . . . . . . . . 75

6.6 The Landau theory of phase transitions . . . . . . . . . . 766.6.1 First order phase transition . . . . . . . . . . . . 78

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Statistical mechanics aims to provide a deductive method whichleads us from the microscopic physical world to the macroscopic worldstarting from the atomic or molecular structure of matter and the fun-damental dynamical principles of the microscopic world and combiningthem with the logic of probability theory. The purpose of this course isto learn how one applies statistical mechanics to understand the macro-scopic behavior of various physical systems.

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Chapter 1

Prelude

1.1 Heat and temperature

1.2 The first law of thermodynamics

1.3 The Carnot Engine

1.4 The second law of thermodynamicsand entropy

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Chapter 2

The formal statisticalmechanics theory

Let us start with the ideal gas, but looking at it from a microscopic pointof view. A microstate of the ideal gas is specified by the coordinatesand velocities (or equivalently the momenta) of all gas particles, i.e.

Microstate {r1,..., rN, p1, ..., pN}. (2.1)The space spanned by the microstate is called the phase space, it has6N dimensions and is a continuous space. In this space the number ofstates fall within an infinitesimal volume element d3r1...d

3pNaround aparticular microstate

{r1, ..., rN, p1,..., pN

}is

d = 1

N!h3Nd3r1...d

3pN. (2.2)

Here the 1N! is to ensure that the particle are identical hence solong as the values of the 3N coordinates and 3N momenta are thesame resuffling the particle labels does not lead to new state. Forexample{r1, r2..., rN, p1, p2..., pN} is regarded as the same state as{r2, r1..., rN, p2, p1..., pN}. Note that here we have performed the12 reshuffling. The factor 1

h3N (h is the Planck constant) accounts

for the Heisenbergs uncertainty principle which says the minimum er-

rors we can have in determining,say, the x-position and x-momentumof a particle is bounded by pxx h Therefore the position and mo-mentum value of a particle has fundamental fuzziness dictated by the

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Heisenberg uncertainty principle. Hence per h3N in the phase space wecan have one state. (Note that h has the dimension of length timesmomentum, hence h3N has the right unit to be the volume element inthe phase space.

2.1 Entropy, temperature and pressure of

ideal gas

The Hamiltonian governing the ideal gas is

H(r1, ..., rN; p1,...pN) =N

i=1p2i2m

. (2.3)

We note there is no interaction between the particles, and the Hamil-tonian does not depend on the particle coordinates.1

Question: How many state are there if the total energy of the wholegas is between Eand E+ E? The answer is

W =ENi=1

p2i

2mE+E

d (2.4)

Note that H(r1,...pN) =Edefines a hyper surface in the phase spaceon which the energy is a constant. To calculate this quantity we first

note that the constraint in the integral does not affect the coordinate,so the 3N coordinate integration can be done easily. It givesVN whereV is the volume of the container in which the ideal gas is confined.What we left to do is the momentum integral

ENi=1

p2i

2mE+E

d3p1...d3pN. (2.5)

To do this integral let us review a more familiar situation

Rx2+y2+z2R+R

dxdydz. (2.6)

1An interaction term will look like 12

i=jV(ri rj) where ri is the coordinate

of the ith particle.

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This is the volume of the spherical shell between radius R and R + R.For small R the answer is 4R2R. Here 4 is the surface areaa sphere of radius 1. How about in two dimensions? The answer is2RR. Here 2 is the length of a circle of radius 1. In fact thefollowing formula applies generally to any dimension

SDRD1R (2.7)

As we have calculated in class,

SD = 2 D/2

(D/2) (2.8)

is the volume of the spherical shell in D-dimension and SDis the area of

the hypersurface with radius 1 in D dimensions. Using this formula wecan compute Eq. (2.5). This is has been done in class and the answeris

S3NE3N1

2 (2m)3N2 (2.9)

Put this together with the factor ofVN and 1N!h3N we obtain

W =

VN

h3NN!

23N/2

(3N/2)(2m)3N/2(E)3N1/2

E. (2.10)

In textbooks you will often encounter a quantity called density ofstates. Its definition is the total number of states in energy intervalEandE+ E is

W = (E)E. (2.11)

Hence according to Eq. (2.10)

(E) =

VN

h3NN!

23N/2

(3N/2)(2m)3N/2(E)3N1/2

. (2.12)

If we take the logarithm ofWwe obtain (as N

)

ln W =NC1+ ln

V

N +

3

2ln

E

N

. (2.13)

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(Here we have taken into account that ln E/N > 0 as N .HereC1=

52+

32ln

43

3 ln h + 32ln m. Now suppose we hold volume

and particle number constant and ask what change to ln W would anenergy change E E+dE induce? The answer is

dW =3N

2EdE. (2.14)

We recall from the introductory prelude that for ideal gas E= 32

NkBTwhich means

dW= 1

kBTdE, (2.15)

or equivalently

dE=T(kBdW). (2.16)

We also recall that for fixed volume according to the first law of ther-modynamics

dE= dQ = T dS (2.17)

hence we can identify dS withkBdWhence up to a reference constantwe can identify

S=kBln W. (2.18)

This last equation was written in Boltzmanns tomb stone.From Eq. (2.18) we see that the entropy is a function ofE, V and

N. Here are some properties ofSfor ideal gas

S

E =

3kBN

2E =

1

TS

V =

kBN

V =

kBNNkBTP

=P

T. (2.19)

ThereforeS

E =

1

TS

V =

P

T. (2.20)

These equation are derived for ideal gas. The question is whether theyhold in general. In order to answer this question we consider system inthermal and mechanical equilibrium.

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Figure 2.1: Ludwig Boltzmann

2.2 Energy exchange contact between two

systems, and the relation between en-tropy and temperature

When two systems I and II brought in thermal contact with each other(meaning they can exchange energy) the total energy satisfies

EI+ EII=E (2.21)

and is fixed. In this way, as a function of time, the system roamsthrough the microstates of I and II under the constraint Eq. (2.21)

According to the equal weight assumption all states of I and II satisfyingEq. (??) will be visited equally frequently.

Let (E) be the density of states of the combined the system , andI(EI), II(EII) be the density of states of system I, II respectively.The following equality holds

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and EI+ dEI is given by

I(EI)II(E

EI)dEIE

(E)E . (2.23)

The function I(EI) increases very rapidly with EI, and II(E EI)decreases very rapidly withEI, so the above probability is very sharplypeaked at certainEIThe energy partition (E

I, EEI) is to be expected

almost certainly in the equilibrium state of I and II. If two systems Iand II initially have E0I and E

0II, after they are brought together the

combine system explore all possible energy partitions consistent with afixed sum. As time elapses the most probable partition will be realized,and after that the energy of the two subsystems no longer changes.Thus we say a equilibrium is established between I and II.

The condition

I(EI)II(E EI)dEIE= max (2.24)

is equivalent to

SI(EI) +SII(E EI) =max. (2.25)

Thus the as the systems moves toward equilibrium, the entropy alwaysincreases. Another consequence of Eq. (2.25) is

SIEI

|EI

= SIIEII

|EII

=EEI

. (2.26)

Suppose system I is an ideal gas then the left hand side will be equalto 1

TI. The question is what should the right hand side be equal to. To

answer that question we go back to daily experience, when two systemsare in thermal contact, they become equilibrated when the temperatureof the two system become the same. Because of this we conclude

SIIEII

|EII

=EEI

= 1

TII. (2.27)

Since system II is arbitrary (it does not have to be an ideal gas) weconclude that 1T =

SEholds for any system.

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2.3 Energy and volume exchange contactbetween two systems and the relation

between entropy and pressure

If the contact between two system I and II allows an exchange of energyand volume in such a way that the combined system visit all states witha fixed EI+ EII, VI+VIIwith equal frequency.

To find the most probable partition of E and V, we need to maximize

SI(EI, VI) +SII(E EI, V VI) (2.28)The most probable EI andVIsatisfy

EISI(EI, VI) +SII(E EI, V VI) = 0

VISI(EI, VI) +SII(E EI, V VI) = 0 (2.29)

We already know the first equation implies TI = TII. How about thesecond equation it reduces to

VISI(EI, VI) =

VIISII(EII, VII). (2.30)

Suppose system I is an ideal gas, then we know the left hand side isequal to pI/TI. Because when two systems are in mechanical contact

they become equilibrated when the pressure the is same. Hence weconclude the left hand side is equal to pII/TII(recall that TII = TI).Thus SV =p/T is also true in general.

2.4 Energy and particle exchange contactbetween two systems and the relation

between entropy and chemical poten-tial

If the contact between two system I and II allows an exchange of parti-cles as well as energy in such a way that the combined system visit all

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product states with a fixedEI+EIIandNI+NIIwith equal frequency,then the combined system satisfy the dynamics of microcanonical en-semble.

The principle of equal weight says that each one of the

(E, N)E= EN

NI=0

EIE

dEII(EI, NI)II(E EI, N NI)

(2.31)

is equally likely to be realized. The probability for system I to haveenergy between EI and EI+ dEIand particle number NIgiven by

I(EI, NI)II(E

EI, N

NII)dEIE

(E)E (2.32)

The most probable value ofEI andNImaximize the numerator of theabove equation or equivalently maximize

SI(EI, NI) +SII(E EI, N NI). (2.33)

The solution EI andNI satisfy

TI(EI, N

I) =TII(E

II, N

II)

NISI(EI, NI) =

NIISII(EII, NII)

where EI+EII=E and N

I +N

II=N. (2.34)

If system I is an ideal gas it turns out the left hand side of the secondequation isI/TI. (For the ideal gas

T = kB

32

lnE

N + ln

V

N +C1 5

2

. (2.35)

.) Because from empirical facts we know when to system can exchange

particle, equilibrium will be reached when the chemical potential be-come equal. Hence we conclude S/N = /T is generally true.

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2.5 The first law of thermodynamics

From previous discussion we have

S

EV,N=

1

TS

V E,N=

P

TS

NE,V=

T. (2.36)

Now we are ready to ask what change will E E+dE, V V +dVandN

N+dNinduce upon S. The answer is

dS= S

EV,NdE+

S

VE,NdV +

S

NE,VdN. (2.37)

Use Eq. (2.36) we can rewrite the above equation as

dE=T dSpdV +dN. (2.38)

This is the first law of thermodynamics. (This is slightly more generalthan you used to see where dN is assumed to be zero. In that case

dE=T dSpdV.)

2.6 The second law of thermodynamics

This law is interpreted by a probabilistic concept in statistical mechan-ics. When a certain inhibition (for example a dividing wall between twogas chambers) is removed the entropy of the whole system will almostcertainly increase. This is because the whole system is an isolated sys-tem and the principle of equal weight apply. Increasing entropy means

increasing the probability of realizing certain macrostate. When thenumber of particles is large, the most probable macrostate has a dom-inant probability to be realized then others.

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2.7 The third law of thermodynamics

Entropy is given by

S= kBln W (2.39)

whereWis the number of microstate consistent with the macroscopiccondition of the system. Because W 1 the entropy is always non-negative. In thermodynamics when we speak of zero entropy what wereally mean is that entropy is not of order N or V, i.e., not extensive.2

For a quantum system this requires the ground state degeneracy D benegligible compared with #N or #V. Under that condition as T 0the system goes into its ground states and S 0.

However, in nature they are systems which do show zero-point en-

tropy (at least to the lowest temperature we have studied). A famousexample is ice. Ice is a molecular crystal made ofH2O molecules. Inan ice crystal the oxygen form a diamond lattice, and the hydrogenatoms sit between the oxygen atoms on the link of the diamond lat-tice. However they do not sit at the center of the link. Because of thestoichiometry,H2O, for the four links stemming from each oxygen, twoof them have a hydrogen atom sitting closer to the oxygen in questionand the hydrogen on the other two links sit closer to other oxygens. Itturns out that for such a lattice ofNoxygens there are approximately1.56N distinct hydrogen distribution patterns. For each pattern, eachoxygen has two closer hydrogens and two farther hydrogens. This gives

rise to a residual entropy N kBln(1.56). Interestingly down to the low-est temperature, this entropy persists indicating that the degeneracy isnot lifted.

2.8 The canonical ensemble

Come back to the situation where two systems are allowed to exchangeenergy only. When system II in the previous subsection is much big-

2At non-zero temperature the entropy is extensive. The zero-point entropy canbe determined by extrapolating the finite temperature result to T = 0. Because

of the necessity of extrapolation, a non-extensive zero point entropy would requiresthe determination of finite temperature entropy to accuracy O(1/N), which is ofcourse almost impossible.

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A system emersed in a heat bath.

Figure 2.2:

ger than system I we call it an energy reservoir, or in short heatbath(Fig. (4.1)).

Let l be a microstate of system I and II(E) be the density ofstates of the heat bath. According to the principle of equal weight theprobability of realizing state l,Pl, is proportional to

II(Et El)E=eSII(EtEl)

kB eSII(Et)

kB eElkB

SIIE

Et =e

SII(Et)

kB e ElkBT .

(2.40)

In the above Et is the total energy of the combined system, T =(ES)

|1Et , and SIIis the entropy of the heat bath. Since the heat bath

is very large compared to the system under consideration Et >> El.As the result we can expand SII(Et El).

Therefore

Pl eEl/kBT, or Pl = constant eEl/kBT. (2.41)Because the probability must sum to unity when all possibilities areaccounted for, we have

l

constant eEl/kBT = 1, (2.42)

which implies

constant1 Z= l

eEl/kBT. (2.43)

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For the ideal gas

Z = 1

N!h3N d3r1...d3pNe(

p212m

+...+p2N2m

)

= VN

N!h3N

2m

3N(2.44)

From now on, we abbreviate

1

kBT . (2.45)

The normalization constant Zof the probability distribution is calledthe partition function. The ensemble defined by the probability dis-tribution Eqs (2.41 and 2.43) is called the canonical ensemble.

2.9 The partition function Z and theHelmholtz free energy F

The canonical ensemble partition function can be rewritten in terms ofthe density of states

Z(T , V , N ) =

0dE(E , V , N )eE. (2.46)

By the definition of statistical entropy we can write the above equationas

Z(T , V , N ) =

0dEeS(E,N,V)/kBeE =

0

dEeETS(E,N,V)kBT

eETS(E,N,V)kBT

0

dEe2ES(E,N,V)(EE)2

2kB

=eE

TS(E,N,V)kBT

0

dEeK(EE)2

2kB

=eETS(E,N,V)kBT

2kB/K. (2.47)

HereK=2ES(E, N , V ). In the above E is the solution of

ES(E , N , V )|E = 1T

. (2.48)

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Thus to leading order in N

kBTln Z(T , V , N ) =E

T S(E, V , N )

F(T , V , N ). (2.49)

The function F is obtained by performing the following transform onthe entropy function

F(T , V , N ) [E T S(E , N , V )]E the solution of 1T

=ES (2.50)

In thermodynamics this is precisely how we get the Helmholtz freeenergy Ffrom the entropy. Again what we do is to first solve for

1

T =

S(E , V , N )

E . (2.51)

The right hand side is a function of E,V,N. Solve this equation implicitlyfor Ewe will obtain E as a function of T,V,N. Once this function isdetermined we replace every E in E-TS(E,V,N) by this function. Asthe result we have

F(T , V , N ) = E(T , V , N ) T S(E(T , V , N ), V , N ). (2.52)Let us practice doing what describe above for the ideal gas. Here

we know, from Eq. (2.13) that

S(E , V , N ) =kBN

C1+ lnV

N +

3

2ln

E

N

. (2.53)

On the other hand from Eq. (2.44) we have

kBTln Z= kBTln VN

N!h3N

2m

3N

NkBT[C2+ ln VN

+3

2ln (kBT)]. (2.54)

HereC2+3/2 =C1+32ln

32 (recall thatC1=

52 +

32ln

43

3 ln h+ 32ln m.)

Now let see whether we can reach Eq. (2.54) by doing the transfor-mation discussed earlier. First we set up the equation 1T =

SE.Simple

computation show this equation is

1

T =

3

2

NkBE

. (2.55)

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ThusE = 32

NkBT.Now we replaceEevery where inET S(E , V , N )by 3N

2 kBTthe result is

E T S(E , V , N ) 3N2

kBT NkBTC1+ ln VN

+32

ln32

kBT

=3N

2 kBT NkBT

C1+

3

2ln

3

2+ ln

V

N +

3

2ln(kBT)

=3N

2 kBT NkBT

C2+

3

2+ ln

V

N +

3

2ln(kBT)

= NkBT[C2+ ln VN

+3

2ln (kBT)] (2.56)

which coincide with the expression for F.Now an interesting question arise. What change will T t +

dT,V V +dV and N N + dN induce upon F? Let dE

=E(T+dT, v+dV, N+ dN) E(T , V , N ) we havedF =F(T+dT, v+dV, N+ dN) F(T , V , N )=dE dT S(E(T , V , N ), V , N ) T S

EdE TS

VdV TS

NdN

(2.57)

The first and third terms on the right hand side cancel on account ofSE

= 1/T. Thus we have

dF = SdTpdV +dN. (2.58)So

S= FTV,N

p= FV T,N

= F

NT,V. (2.59)

However you might ask wouldnt the entropy so obtained be a func-tion of T , N , V rather than a function ofE , N , V ? To get back theoriginal Edependence of the entropy we note that

ln Z= E (2.60)

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We can use this equation to solve Tas a function ofEand substitutethe result into the entropy expression obtained by differentiating F.

2.9.1 Paramagnet

The paramagnet is a disordered collection of magnetic moments. Thesimplest model describing paramagnet is

H= hm0ni=1

i. (2.61)

Here h is the applied magnet field, and i =1 describing whetherthe moment align or anti-align with the external magnetic field. Thekind of magnet in which the magnetic moments can only point along

or opposite to a fixed direction is called an easy-axis magnet. Thepartition function is given by

Z=i

i=1

ehm0i = [2 cosh(hm0)]N. (2.62)

The averaged magnetic moment is given by

M = {i}

(jj)e

hm0

Z =

j

j=1 je

m0hjj=1 e

m0hj

=N m0tanh(m0h). (2.63)

The magnetization is defined as

m=M

N =m0tanh(m0h). (2.64)

The magnetic susceptibility is given by

(h) =hm = m20

kBTsech(hm0/kBT). (2.65)

The zero-field susceptibility is therefore

0 = m20

kBT

. (2.66)

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2.9.2 Maxwell distribution

Consider a gas of interacting molecules:

H=Ni=1

p2i2m

+1

2

i=j

v(ri rj). (2.67)

An important property of the classical partition function is that

Z = 1

N!h3N

d3p1...

d3pNe

ip2i/2m

d3r1...d

3rNe(1/2)

i=j

V(rij)

= 1

N!h3NZpZr. (2.68)

In the above Zp and Zr represents the momentum and space integral

respectively. The probability that the molecule has a velocity v1,..., vNis given by

1N!h3Ne

(mv21+...+mv2N

)/2Zr1

N!h3NZpZr

=e(mv

21+...+mv

2N

)/2

Zp. (2.69)

The probability that the first molecule will have velocityv is given by

d

3

p2...d3

pNe(mv

2+...+mv2N

)/2

Zp =

emv2/2

d3p1ep21/2m

= emv

2/2

(2mkBT)3/2 (2.70)

The same is true for molecule two,three...N. Thus the probability thatany molecule has velocity v is given by

P(v) = emv

2/2

(2mkBT)3/2. (2.71)

The probability distribution in Eq. (2.71) satistisfies d3pP(v) = m3

d3vP(v) = 1. (2.72)

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Therefore it does not have the proper normalization for a velocity dis-tribution function. Instead

P(v) =m3P(v) = m3

emv2/2

(2mkBT)3/2 (2.73)

has the right normalization and is the velocity distribution function.Eq. (2.73) is called the Maxwell velocity distribution. The Maxwelldistribution satisfies

d3vP(v) = 1 (2.74)

The left hand side of the above equation can be expressed in terms ofpolar coordinate as

0 4v2P(v) = 1 the integrand

4v2 m3emv2

/2

(2mkBT)3/2 (2.75)

is called the Maxwell speed distribution.Note that the fact that molecular distribution obeys the Maxwell

distribution is a consequence of classical statistical mechanics and isindependent of whether the molecules interact or not !

2.9.3 Equal-partition theorem

The equal-partition theorem is the consequence of the following mathe-matical statement. Suppose a variablex obey the Gaussian probabilitydistribution, i.e.,

P(x) = 1

2a

ea2x2 (2.76)

The expectation value ofx2 is1

2a

x2ea2x2 = 2

2a

a

ea2x2

= 22a

a

2a

= 2a1/2 a

a1/2 =1a

(2.77)

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This is the reason why the average energy of the ideal gas is 22

NkBT.This is because the momentum distribution of any gas particle is theMaxwell distribution

P(px, py, pz) =ep

2x/(2mkBT)

2mkBT

ep2y/(2mkBT)

2mkBT

ep2z/(2mkBT)

2mkBT . (2.78)

If we calculate p2x2m

we can use the math formula derived above andreplaceaby 1mkBT. Thus

p2x

2m = 1

2mmkBT = 1/2kBT. (2.79)

Therefore

p2

2m = p

2x

2m + p

2y

2m + p

2z

2m =3

2kBT. (2.80)

In fact the equal partition theorem state more generally that ev-ery quadratic, independent degrees of freedom in the Hamiltonian con-tribute

1

2kBT (2.81)

to the average energy. The classical partition function read

Z= 1

N

dq1...dqM

dp1...dpMe

H(q,p), (2.82)

where 1N is an overall factor that is independent of temperature. Anindependent, quadratic freedom appear in the Hamiltonian as

H=C

22 +W(rest of the 2M 1 variable). (2.83)

The partition function is therefore

Z= 1

N

2kBT

C

drest of the variableseW. (2.84)

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The average energy is given by

E=

1N d

C2

2 d...eHZ =

dC2

2eC2/2

d eC2/2 =

kBT

2 . (2.85)

The importance of the result is that it is independent of the springconstant C, and for that matter any other details of the Hamiltonian!

Here is an application of the equal-partition theorem. When thetemperature is very high so that classical statistical mechanics applies,the equal partition theorem predicts that, to the extent the atoms ina solid can be treated as an elastic network, the contribution to theaverage energy of a crystal due to the vibration of its constituent atomsis equal to

3nNkBT. (2.86)

(Recall that each atom has 6 degrees of freedom, 3 from the momentumand 3 from the position.) In Eq. (2.86) N is the number of unit cell,andnis the number of atoms per unit cell. According to this predictionthe specific heat due to lattice vibration is equal to

3nkB. (2.87)

This called the Dulong-Petit law.

2.10 The grand canonical ensemble

Next we consider the situation where energy and particle exchange areallowed. When the system II is much bigger than system I we say it is aparticle and energy reservoir. Let II(Er, Nr) be the density of state ofthe reservoir, andEl, Nl are the energy and particle number associatedwith a particular microstate of system I. According to the principle ofequal weight the probability of realizing state l, Pl, is proportional to

II

(Et

El, N

t Nl)E

II(Et El, Nt Nl)E

II(Et, Nt)E

=eSII(EtEl,NtNl)SII(Et,Nt)

kB . (2.88)

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In the above Et, Nt are the total energy and total particle number ofthe combined system and SII is the entropy of the reservoir. Sincethe reservoir is very large compared to the system under considerationEt >> El andNt >> Nl. As the result we can expand the exponent as

SII(Et El, Nt Nl) SII(Et, Nt) ElESII NlNSII= El

T +

NlT

. (2.89)

Here we have used (NSII) = /T. As the result

PN,l = 1

e(ElN), (2.90)

where

l,N

e(ElN). (2.91)

The ensemble defined by the probability distribution Eqs (2.90 and2.91) is called the grand canonical ensemble.

2.11 The grand partition function andthermodynamic function

The grand partition function is given by

(T , V , ) =N

eN

dE(E)eE

=N

eN

dEe(ETS(E,V,N))

e(ETS(E,V,N)N). (2.92)HereE and N are the solutions of

ES(E , V , N ) = 1

T and = NF(T , V , N ). (2.93)

Thus

J(T , V , ) = kBTln (2.94)

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is the thermodynamic function obtained from the Legendre transform

J(T , V , ) F(T , V , N ) N|replace N by the soution of =NF(2.95)

2.11.1 Molecular adsorption -The hydrogen stor-age problem

The usage of hydrogen as future fuel has been a subject of much dis-cussion in recent years. One of the key road block for is the storageproblem. The DOE target hydrogen storage for year 2015 is 9% in

weight. Since hydrogen is the lightest element, in order to achieve thistarget it is necessary for the storage media to have the following prop-erties: (1) high surface area, (2) appropriate adsorbing binding energyso that for each explored surface the coverage 1.

This problem can be analyzed using what we have already learned.Consider an adsorbent surface having Nsites each of which can adsorbone gas molecule. In the following we shall model the H2gas as an idealgas.3 We treat the H2 gas as the energy and particle reservoir that setthe chemical potential and the temperature for the adsorbates. Bydefinition the reservoirs energy and particle number is far greater than

the substrate. The temperature pressure and the chemical potentialof the ideal gas exerts on to the absorbed molecules can be computedfrom 1T =

SE,

pT =

SV, T = SN. (The pressure of the ideal gas is

controlled by a piston as shown in Fig. (2.3)). These differentiationsgives 1T =

3NkB2E ,

pT =

NkbV , and Eq. (2.35), respectively. Assuming that

an adsorbed molecule has energy 0 compared to one in the free state,the goal is to determine the coverage as a function ofP.According to Eq. (2.35) the chemical potential exerted by an ideal gasis given by

T

=

kB

3

2

lnE

N

+ lnV

N

+C1

5

2. (2.96)

3H2 is not an ideal gas because the molecule has internal rotation and vibrationdegrees of freedom. Lets ignore such complications.

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Figure 2.3:

If we replace Eby 32 NKBT, VN by

kBTP and insert the value for C1 and

combine the sum of log into the log of the product we obtain

kBT = ln

PkBT

h22mkBT

3/2.

(2.97)

This is the chemical potential the ideal gas exerted onto the moleculesthat are absorbed on the substrate.

The statistical mechanics of the adsorbed molecules is describedby the grand canonical ensemble with chemical potential given byEq. (2.97). For n adsorbed molecules distributed among N adsorb-

ing sites there areN!

n!(N n)! (2.98)

different adsorbing patterns. Each of them is a microstate of theadsorbed system. The grand partition function of the substrate is

=Nn=0

N!

n!(N n)!en(0+) =

1 +e(0+)

N. (2.99)

The probability ofn molecules being adsorbed is

en(+0) N!n!(Nn)!

. (2.100)

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Thus the average number of adsorbed molecule is

n = 1

Nn=0

nen(+0)

N!

n!(N n)! = 1

ln =

N

e(0+) + 1 .(2.101)

Since is the chemical potential of the particle reservoir which, in thiscase, is the ideal gas, we have, according to Eq. (2.97)

e = p

kBT

h22mkBT

3/2. (2.102)

Substitute this result into Eq. (2.101) we obtain

=

n

N =

p

p+p0(T) (2.103)

where

p0(T) =2mkBT

h2

3/2e0/kBTkBT. (2.104)

When p = p0(T) in the above equation the coverage is = 1/2. Putin the number for the H2 mass, planck constant, Boltzmann constant..etc we obtain

p0(T) = 115091 atm TTroom e0/kBT. (2.105)The material science question is what material we use for the sub-

strate. In our language this translate into what kind of 0 we need.There are two constraints. (1) We would like to achieve considerablecoverage at around room temperature and around ambient pressure.(2) We would like the substrate to release a substantial fraction of theadsorbate when the temperature is heated from room temperature tosay 100o C. Lets say we would like

(Troom, 2atm) =f(1.4Troom, 2atm) =

2

3f (2.106)

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0.0 0.2 0.4 0.6 0.8 1.0Release Fraction0.0

0.2

0.4

0.6

0.8

1.0

Coverage

0.0 0.2 0.4 0.6 0.8 1.0Release Fraction0.0

0.2

0.4

0.6

0.8

1.0

0eV

Figure 2.4: 0 and fas a function of the release fraction.

These conditions translate into2

2 + 115091e0/26 =f

2

2 + 115091 1.4 e0/(1.426) =2

3f. (2.107)

Here I have used the fact thatkBTroom 26meVand have expressed0in unit of meV. We would like to solve the above equations for fand0.It turns out that0 = 0.4361eV andf= 0.997 solve the above equation.Thus we need to search for a material which can bind H2 molecule atthe binding energy of 0.436 eV. The above result was obtained for the

release fraction 1 2/3 = 1/3. For different release fractions the resultfor 0 and f are slightly different. In Fig. (2.4) I plot 0 and f as afunction of the release fraction.

Thus a high surface area material which binds the hydrogenmolecule at binding energy in the range of 0.3-0.5 eV is what we arelooking for.

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Chapter 3

Black body radiation

Quantum statistical mechanics works exactly the same as its classi-cal counterpart. The biggest difference is the specification of the mi-crostate. In most of the cases we shall encounter, each microstate isthe energy eigenstate of a Hamiltonian. For example the canonicalensemble partition function of a quantum system is given by

Z=k

eEk (3.1)

Here k labels the eigenstate and Ek is the eigen energy of the state.The grand partition function is given by

=N

k

e(Ek(N)N) (3.2)

where Ek(N) is the eigen energy of state k when the particle numberis N. The relation between the partition function and the free energyare exactly the same as we talked about before.

If you have not taken introduction to quantum mechanics it is theright time for you to read the suggested materials in bspace.

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3.1 Statistical mechanics of harmonic os-cillators

A harmonic oscillator is governed by the following Hamiltonian

H= p2

2m+

K

2x2. (3.3)

Classically the oscillation frequency is given by

K/m. The classical

equation of motion (the Newton equation) is given by

2x

t2 +2

0x= 0 (3.4)

where0=

K/m is the oscillation frequency. According to the equalpartition theorem the average kinetic and potential energy each con-tributeskBT /2 to the total energy, hence

H =kBT. (3.5)

For a collection of independent harmonic oscillators (each with a dif-ferent frequency) the hamiltonian is given by

H=Ni=1

p2i2mi

+Ki

2 x2i . (3.6)

The important thing about the equal partition theorem is that regard-less of the values ofmi and Ki each modes contribute kBT to thetotal internal energy, as the result the total energy ofNmodes of har-monic oscillator is NkBT. The specific heat due to these oscillatorsis

C=E

T =N kB. (3.7)

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3.2 Quantum statistical mechanics of har-monic oscillator

According to the quantum theory of harmonic oscillator, each mi-crostate is labeled by a non-negative integer n.1 The energy associatedwith such microstate is given by

En= (n +1

2)h0 (3.8)

where 0 =

K/m is the classical oscillation frequency. Physicallynis the number of oscillation quanta present in the nth excited state.Each oscillation quantum carries energy h0 hence nquanta carry en-ergy nh0. The ground state energy h0/2 is the energy due to the

zero point oscillation. The partition function of a quantum harmonicoscillator is given by

Z=n=0

e(n+1/2)h0 =eh0/2n=0

enh0 =eh0/2 1

1 eh0 .(3.9)

In arriving at the final result we have used the formula for geometricseries

n=0

xn = 1

1 x (3.10)

for

|x

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For kBT >>h the above result is approximate by

E

h0/2 +kBT. (3.13)

Aside from the zero point energy the second term, namely the energydue to the thermal excitation, is the same as the equal partition theoremprediction. For a collection of N harmonic oscillators Eq. (3.12) ismodified to

E=i

hi/2 +

hiehi 1

, (3.14)

wherei is the oscillation frequency of the ith oscillator.

3.3 Classical description of the E-M field

In the absence of charge and current density, the magnetic field of theEM oscillation satisfies Maxwells wave equation

2B

t2 c22B= 0. (3.15)

In a cavity with linear dimension L L Lthe magnetic field satisfiesthe boundary condition B = 0 on the walls. In that case B can beexpanded in terms of normal modes

B=

n1,n2,n3

B(n1,n2,n3)sin(n1x/L) sin(n2y/L)sin(n3z/L) (3.16)

where n1,2,3 are positive integers. (The negation ofn1,2,3 reverses thesign of the sine function hence does not give rise to a new mode).

In the following we shall introduce a vector

k=

L(n1, n2, n3) (3.17)

to replace the triplet (n1, n2, n3). After so doing Eq. (3.16) becomes

B=k

Bksin(kxx) sin(kyy) sin(kzz) (3.18)

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where kx, ky, kz denotes the three component ofk. In the three dimen-sional space spanned by kx, ky, kz the allowedk lies in the first octant.They form a simple cubic grid with lattice constant /L. Note thatthe origin in this k-space is excluded because the corresponding Bkis zero. Substituting Eq. (3.18) into Eq. (3.15) we obtain

2Bkt2

+c2k2Bk= 0. (3.19)

Eq. (3.19) looks exactly the same as Eq. (3.4), namely, the equationof motion of a simple harmonic oscillator. Moreover since there is anormal mode for each k we have a collection of independent simpleharmonic oscillators. An important difference between Eq. (3.19) andEq. (3.4) is that the coordinates of the normal mode in Eq. (3.19)

is a vector not a scalar. However we could write down Eq. (3.19) foreach component ofBk. In doping so we convert Eq. (3.19) into threeequations, one for each component ofBk. However do Bk really havethree independent components? The answer is no. To see that we recallthe magnetic field obeys B = 0 (there is no magnetic monopole).Substitute Eq. (3.16) into this equation we obtain

k Bk= 0. (3.20)

hence for each k the only allowed Bk components are transverse tok. Therefore for each k there are two independent components ofBk.

Another thing to note is the oscillation frequency implied by Eq. (3.19)is

k= c|k|. (3.21)

According to the equal partition theorem the total thermal energy ofthe cavity EM field is

E= 2k

kBT (3.22)

here the prefactor 2 is due to the two independent direction ofBk foreach k. Since there are infinitely many allowed k, the above resultsuggests the total internal energy of an cavity EM field is infinite!

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3.4 Quantum statistical mechanics of thecavity E-M field

Quantum mechanically the total energy (omitting the zero point oscil-lation energy) associated with the cavity EM field is given by

E= 2k

nkh(k) (3.23)

where

nk = 1

eck 1 . (3.24)

When L is very large, the allowed k points form a very dense grid.

Under the condition thatnk does not change appreciably if we move kfrom a lattice point to its nearest neighbor, we can replace the sum inEq. (4.8) by an integral

E= 2

d3k

(/L)3hk

ehk 1 (3.25)

Here we need to remember that thekintegral is over the first octant andthe reason we divide by (/L)3 is because for every (/L)3 in k-spacethere is onek point. Because the integrand only depends on the modu-lus ofk we can changed3kto the radial integral 4

8k2dk(4/8 because

we only integrate over the first octant). In addition since k= /hcwecan change the radial k-integral into an integral over the frequency.

E=

0dD()

h

eh 1 . (3.26)

Here the photon density of states can be determined as follows

D()d= 2

48

c

2dc

(/L)3 . (3.27)

Therefore

D() = L3 2

c32. (3.28)

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Substitute this result into Eq. (3.29) we obtain

E=V

0 d h

3

c32 1

eh 1 , (3.29)where V =L3 is the total volume. The energy density is given by

U=E

V =

0

d h3

c32

1eh 1 . (3.30)

The integrand of Eq. (3.30) is referred to as the blackbody spectralfunction in the literature:

(, T) = h3

c32

1eh 1 . (3.31)

This universal function describes all emission spectrum of cavity EMfield. By changing the integration variable to y = h and performingthe following dimensionless integral

0

dx x3

ex 1=4

15 (3.32)

we obtain the energy density of blackbody radiation

U= 8k4B5

15c2h3

T4. (3.33)

This is called the Stefan law.

3.5 Acoustic phonon contribution to the

specific heat

The quanta of lattice vibration is called phonon. A solid is made up ofatoms arranging in a periodic array. The unit cell is a group of atomswhich, upon periodic repetition, generate the whole solid. The acousticvibration is a dynamic displacement in which atoms in the same unit

cell displace by the same amount in the same direction. The opticalvibration is the type of dynamic displacement where atoms in the sameunit cell are displaced by different amount and/or different direction.

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Like EM field, the vibration in a solid is characterized by wavevectorsatisfyingk=

L(n1, n2, n2). The associated displacement is given by

u(R, t) =k

uk(t)sin(kxRx)sin(kyRy) sin(kzRz), (3.34)

where R is the position of atoms in the periodic arrangement. TheFourier coefficient uk(t) satisfies the equation of motion of harmonicoscillators. However unlike EM field there is no no constraint on thedirection of uk (such as k uk = 0). As the result there are threeindependent directions foruk, two perpendicular tok(called transversemodes) and one parallel to k(call the longitudinal modes).

The dispersion of acoustic phonon is very similar to that of photon.

(k) =clk for longitudinal modes

=ctk for transverse modes. (3.35)

Here cl and ct are the longitudinal and transverse phonon velocities.Following Eq. (3.28) the number of phonon modes in frequency range +d is given by

D()d= V1

c3l+

2

c3t

222

d. (3.36)

Here the factor of 2(1) multiplying 1c3t

( 1c3l

) reflects the fact that for each

wavevector there are two (one) independent transverse (longitudinal)

modes. There is however a constraint on the total number of normalmodes. For a solid with volumeV there are N =V /a3 unit cells (a isthe lattice constant). In each unit cell the atoms can displace in threedifferent directions, hence there are in total N = 3N modes. Debyeincorporates this constraint by imposing a maximum frequency D sothat

D() =V1

c3l+

2

c3t

222

for hD. (3.37)

He chooses D so that D0

dD() = N V1

c3l+

2

c3t

3D62

= 3N. (3.38)

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In terms ofD Eq. (3.37) read

D() =

9N2

3D for < D

= 0 for > D. (3.39)

Using Eq. (3.39) we compute the average energy to be

E= D

0

9N2

h33D

h

eh 1 d. (3.40)

At temperature where

hDkBT

>>1 (3.41)

Eq. (3.40) gives

E 3k4B

4

5h33DNT4. (3.42)

The heat capacity derived from that is

C=TE=

12k4B4

5h33D NT3

. (3.43)

TheT3 specific heat is often taken as the hallmark of acoustic phonons.

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Chapter 4

Ideal Fermi gas

4.1 Particle in a boxWe focus on spin 1/2 Fermi particles confined in a L L L box ofvolume V =L3. This time, however, we practice a different boundarycondition - the periodic boundary condition:

(x+L, y,z) = (x, y, y), (x, y+L, z) =(x,y,y),

(x,y,z+L) =(x, y, y). (4.1)

Under this boundary condition the eigenfunction of a particle in thebox is given by

k = 1

L3eikr (4.2)

where

k=2

L(n1, n2, n3), (4.3)

where n1,2,3 can take all integer values. The associated eigen energy is

k =h2|k|2

2m . (4.4)

Thus in the three dimensional space spanned by the allowed k is asimple cubic grid with lattice constant 2/L.

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4.2 Grand canonical ensemble treatmentof Fermi particle in a box

First we focus on the grand canonical ensemble. The quantum statesare characterized by

|{nk} (4.5)wherenk is the fermion occupation number of the k, state. Paulisexclusion principle tells us that each nk is either 1 or zero.

The partition function is given by

=k,

1nk=0

e((k))nk =k,

1 +e((k))

. (4.6)

Here =1/2 is the z-component of the spin of the fermion. Theaverage number of particlenkis given by

nk = 1

{np}

nke((p))np =

1

1

nk=0e((k))nk

1

nk=0e((k))nk

= 1

1 +e((k))

1 +e((k))

= e((k))1 +e((k))

=

1

e((k)) + 1 (4.7)

Eq. (5.6) is called the Fermi Dirac distribution. Note that because theenergy in Eq. (??) does not depend on the spin, nk is independent of.

4.3 The particle number constraint andthe chemical potential

If the total number of particle is fixed (which correspond to the usualsituation of a solid) the following constraint must be satisfied

N= 2k

1

e((k)) + 1 (4.8)

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Here the k sum is performed the simple cubic lattice (n1, n2, n3)2/L.Here the overall factor of two comes form the fact that every fermionhas two spin states. The equality in Eq. (4.8) is made possible byadjusting the chemical potential . In fact Eq. (4.8) determines thechemical potential as a function of temperature. WhenL is very large,thek point is very dense. Under the condition that nk does not changeappreciably if we move k from a lattice point to its nearest neighbor,we can replace the sum in Eq. (4.8) by an integral

N= 2

d3k

(2/L)31

e((k)) + 1 (4.9)

We divide by (2/L)3 is because for every (2/L)3 in k-space there isone k point.

N = 2L3

d3k

(2)31

e((k)) + 1

=L3

2

0

k2dk 1

e((k)) + 1

=

0dD()

1

e() + 1. (4.10)

Here D() is the density of states, i.e., the number ofk state betweenenergy and+d, of the free Fermi gas:

D()d=L3

2

2mh

2m2

d

h = 2L3d

m3/22h32

. (4.11)

and

N = 2L3 m3/2

2h32

0

d

e() + 1

(4.12)

In the zero temperature limit the left hand side of the above equationbecomes

N/L3 = = 2 m3/2

2h32

0

d

(4.13)

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The solution is

F

= h2

2m(942)1/3. (4.14)

This energy is often referred to as the Fermi energy. At zero tem-perature, all free fermion states are occupied below this energy, whileempty above this energy. In terms ofFand the number of particle Nthe density of state is

D() = 3N

2F

F, (4.15)

or

D() =D(F)

F

. (4.16)

For low but non-zero temperature, the chemical potential deviatesfrom the Fermi energy. In this case the formula that determines thechemical potential is given by

= 2 m3/2

2h32

0

d

e() + 1, (4.17)

or equivalently

2

3

3/2F =

0

d

e() + 1. (4.18)

For low temperatures it is desirable to find analytic expression foras a function of temperature. In order to do that let us first considerhow to evaluate the following integral

0

df()(), where f() = 1

e() + 1, and (0) = 0. (4.19)

In the case of Eq. (??) () = 23 3/2.Using integration by parts

0

df()() =

0()f(). (4.20)

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Figure 4.1: f versus .

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The advantages of Eq. (4.20) is that f() is a sharply peaked function(about = ) at low temperatures:

f

= [e() + 1][e() + 1]

. (4.21)

This suggests that in evaluating the right hand side of Eq. (4.20)we should expand around = :

() =() + ( )() +( )2

2 () +... (4.22)

Because Eq. (4.21) is an even function of , all odd powers inEq. (4.22) drop out after integration. For even powers we have

0 d

[e() + 1][e() + 1] ( )2n

(kBT)2n

dx x2n

[ex + 1][ex + 1]= (kBT)

2n2 (2n)!(1 22n+1)(2n).(4.23)

Here we have used the fact that

dx xm

(ex + 1)(ex + 1)= 2m!(1 2

2m)(m). (4.24)

For 2n = 0, 2, 4, 6 the above resultis 1, (kBT)

22/3, (kBT)474/15, (kBT)

6316/21. Use the aboveresult in Eq. (4.20) we obtain

2

3

3/2F =

0

f() = 2

3

0

3/2f()

=2

3

3/2 +

3

8

2

3(kBT)

2 + 3

1285/274

15(kBT)

4 +...

(4.25)

Solve the above equation for we obtain

= F

2

12

(kBT)2

F

4

80

(kBT)4

3

F

+... (4.26)

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4.4 The total internal energy and the spe-cific heat

The average energy is given by

E =

0dD()

e() + 1. (4.27)

The specific heat of the free electron gas at constant volume and con-stant particle number is defined as

CV,N =TE(T , V , N ) = 1

kBT2

0

dD() ( +T )

[e() + 1][e() + 1]

= 1kBT2D(F)F

0

d 5/2

( +T

)[e() + 1][e() + 1] (4.28)

We can change the integration variable to z= ( )

CV,N = 1

T

D(F)F

dz(kBT z+)

5/2(kBT z+T )

[ez + 1][ez + 1]

1T

D(F)F

dz(kBT z+)

5/2(kBT z+T )

[ez + 1][ez + 1]

1

T

D(F)

F

dz[5/2 + 5

23/2kBT z+...](kBT z+T

)

[ez + 1][ez + 1]

1T

D(F)F

dzT

5/2F

+ 52

3/2F (kBT)

2z2

[ez + 1][ez + 1] (4.29)

Recall that = 26k2BT

Fand

dx

[ez + 1][ez + 1]= 1,

dx x2

[ez + 1][ez + 1]=2/3 (4.30)

Eq. (4.29)reduces to

CV,N=2

3k2B

2D[F]T. (4.31)

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Using the result in Eq. (4.11) we obtain

CV,N=N kBkBT

2F . (4.32)

Eq. (4.31) is an important formula for experimental condensed matterphysics, because by measuring the heat capacity we can deduce thedensity of states at the Fermi energy via

CV,NT

=2

3k2B

2D[F]. (4.33)

4.5 The magnetic susceptibility

The magnetic susceptibility of the free Fermi gas is also an impor-tant quantity to study experimentally. In the presence of an appliedmagnetic field, the spin up and spin down energy undergoes oppositeZeeman shift, namely,

For spin up (k) (k) BhFor spin down (k)

(k) +Bh (4.34)

In the aboveB is the magnetic moment of each electron. In this case,the average number of spin up and spin down electrons are no longeridentical.

N+ =12

0

dD() 1

e(Bh) + 1

N =12

0

dD() 1

e(+Bh) + 1. (4.35)

The new constraint on the chemical potential is given by

N=1

2

0

dD()

1

e(Bh) + 1+

1

e(+Bh) + 1

(4.36)

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When the magnetic field is very weak, to linear order in h is the sameas Eq. (4.26). The magnetic moment induced by the applied field isgiven by

M=B(N+ N) = B2

0

dD() 1

e(Bh) + 1 1

e(+Bh) + 1

.

(4.37)

For weak applied field, we can expand the difference in the integrandin power of the applied field

M = BBh

0dD()f()

= 2Bh

0dD()f() =

2Bh

0dD()

[e() + 1][e() + 1].

(4.38)

As usual we expand D() around:

D() =D(F)

F=D(F)

F

+

F

2 ( )

F

82( )2 +..

.

(4.39)

Therefore

M =2

Bh

0 dD(F)

F F

82 ( )2

[e() + 1][e() + 1]

=2BhD(F)

dx

F

F

82(kBT)

2x2 1

[ex + 1][ex + 1]

=2BhD(F)

F

F

82(kBT)

2 2

3

hD(F)2B1

2

24

(kBT)2

2F

. (4.40)

The magnetic susceptibility is defined as

=M

h|h=0= D(F)2B

1

2

24

(kBT)2

2F

. (4.41)

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Figure 4.2: Subrahmanyan Chandrasekhar

Combining Eq. (4.33) and Eq. (6.6) we see that as T 0 the ratioCN,V/T

=

23

k2B2

2B(4.42)

approaches a universal constant.

4.6 The physics of while dwarfs (optional)

The white dwarfs are cold stars stabilized against gravitational collapse

by the quantum pressure of their electrons. Let the number of nucleons(protons and neutrons) be N,and the number of electron per nucleonbe q (q 1/2). If R is the radius of the star, the total gravitationalenergy is given by the following estimate.

The total mass of the star is MN where M is the nucleon mass.(We have omitted the electron mass.) The mass density of the star isgiven byM=

MN4R3/3 . The gravitation energy is given by

Eg = G R

0M

4r3

3

M4r2dr

r = 3GM

2N2

5R . (4.43)

The total kinetic energy of the electron is

EK=

0dD()

e() + 1. (4.44)

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Let us assume that we can treat the electron as non-relativistic andregard them as at T= 0 (we will come back to see whether that is agood approximation). Under that approximation = F where

F = h2

2m

94

qN4R3/3

21/3=

3h2

32 N

22q21/3

4mR2 , (4.45)

and

EK = F

0D()= 2

F0

3qN

4F

F=

3NqF5

=9( 32 )

1/32/3h2N5/3q5/3

20mR2 . (4.46)

Therefore the total energy of the star is given by

Etot(R) = 3GM2N2

5R +

9( 32

)1/32/3h2N5/3q5/3

20mR2 . (4.47)

The optimal radius satisfies

REtot(R) = 0. (4.48)

The solution is

R=9

4

2/3 h2q5/3GmM2N1/3

(4.49)

Put in the numbers (G= 6.67 1011m3s2kg1) we obtain

R=7.29 1025m

N1/3 (4.50)

The solar mass is 1.99 1030kg which amounts to N = 1.19 1057.This means

R= 6881

(M/M)1/3Km. (4.51)

A white dwarf whose mass is one solar mass has electron density

4.4 1029/cm3 The Fermi energy associated with that isF 2.1 105eV. (4.52)

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The electron rest mass is mc2 5.1105eVtherefore the Fermi energyis about 2/5 of the electron rest mass, hence the electrons in the whitedwarf are quite relativistic. In general a white dwarf of mass Ms hasFermi energy

F 2mc2

5

MsM

4/3. (4.53)

The Fermi temperature of the solar-mass white dwarf is TF 2.4 109K. The temperature of the white dwarf is about 105K hencecompared to its Fermi temperature it is justifiably almost at absolutezero.

For star whose mass is much bigger than the solar mass, the Fermienergy given by Eq. (4.53) can become much larger than the rest mass

of the electron. In that case, we should use the ultra-relativistic limitto compute the average kinetic energy as a function of radius. Therelativistic kinetic energy

(k) hck (4.54)which give the density of states

D() = 2V 2

2c3h32, (4.55)

the same as that for the photon gas. With the density of states given

by Eq. (4.55) we recompute the Fermi energy in the ultra-relativisticlimit and find it to be

Ne = V 3F3c3h32

F = hc(32e)1/3. (4.56)

The average kinetic energy is given by

EK= F

0dD() =

3Ne4

hc(32e)1/3. (4.57)

Substitute Ne = qN and e = qN/(4R3/3) into the above equation

we obtain

EK=3hc

4

9N4q4

4

1/3

R . (4.58)

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Combined with the gravitational energy it yields

Etot= 3GM2N2

5R +3hc

49N

4q4

4 1/3

R . (4.59)

We note that the two terms in the above equation have the same radiusdependence. Thus if

3GM2N2

5R >

3hc

4

9N4q4

4

1/3R

, (4.60)

or

N >2.0 1057 Ms> 1.7M (4.61)the gravitational attraction will overwhelm the electron quantum pres-sure, and the star will gravitationally collapse despite the electron quan-tum pressure.

When the density of the star get so high that inverse beta decay

e +P+ n + (4.62)occurs, at that stage almost all the protons and electrons are convertedinto neutrons and neutrinos. The neutrino will escape and carry energyaway. Eventually the degeneracy pressure of the neutron fight againstthe collapse. In that case we can do sa similar calculation as in the

non-relativistic limit of the electron pressure to get

R=9

4

2/3 h2GM3N1/3

. (4.63)

(note that compared with Eq. (4.49) we see that m M and q 1.)This yield, for a star twice the solar mass a radius

R 9.7Km. (4.64)Using Eq. (4.45) (with q 1 andm M) we can compute the Fermienergy of such star. The answer is

F 141MeV 19

Mc2. (4.65)

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Chapter 5

Ideal Bose gas

5.1 The grand canonical ensemble treat-ment

We focus on spin zero Bose particles in a box of volume V =L3, againunder periodic boundary condition. Like the Fermi case the eigenfunc-tion of any single particle is given by

k = 1

L3eikr (5.1)

where

k=2

L(n1, n2, n3), (5.2)

where n1,2,3 take all integer values. The associated eigen energy is

k =h2|k|2

2m . (5.3)

All of the above eigen information are the same as the Fermi case.Again the quantum states are characterized by

|{nk} (5.4)where nk is the boson occupation number of the k state.

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Figure 5.1: Satyendra Nath Bose

We focus on the grand canonical ensemble. The partition function is given by

=k

nk=0

e(k)nk+nk =k

1 e((k))

1. (5.5)

The average number of particlenk is given by

nk = 1

{np}

nke((p))np =

1

nk=0e((k))nk

nk=0e((k))nk

= 1

1 +e((k))1

1 +e((k))

1 = 1e((k)) 1 (5.6)Eq. (5.6) is called the Bose Einstein distribution. Because the averageparticle number is positive definite for all nk therefore

0. (5.7)

5.2 The particle number constraint

The particle number constraint requires

N=k

nk=k

1e(k)1

. (5.8)

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5.2.1 zero temperature macroscopic occupation ofthe k= 0 state

Let us start with the zero temperature limit. In this case all the bosonswill sit in the lowest energy ,.i.e.,the k = 0 state. In other wordsnk=0=N. In order for this to be consistent with the Bose distributionfunction we must have

lim

1

e|| 1=N. (5.9)

Here we have used the fact that 0. Eq. (5.9) implies that e|| =1 + 1N or || = ln(1 + 1N) = 1N. In the last step we used the fact thatwe are interested in the thermodynamic limit where N

. Solving

the above equation we obtain

|| = 1N

. (5.10)

Clearly as and Nboth approach infinity 0. Thus at zero tem-perature = 0 (this result holds even when Nis finite.)

5.2.2 Non-zero temperature Bose Einstein con-densation

Because the chemical potential can be zero the Bose distribution func-tion

nk= 1

e(k) 1 . (5.11)

can be singular at k = 0. The fact that the summand of Eq. (5.8)can be singular for k = 0 call for a more careful look at the previousprocession from

k to

d3k(2/L)3

. In fact this replacement is good for allk except k= 0 Thus we write

k

nk= n0+

k,|k|>0

nk= n0+|k|>0

d3k(2/L)3

nk. (5.12)

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For the second term of the above equation we can proceed as usual andconvert

|k|>0

d3k(2/L)3

nk to

0+dD() 1

e() 1 . (5.13)

As a result Eq. (5.8) read

N=n0+

0+dD()

1

e() 1 . (5.14)

HereD() is the density of state, identical to that of the free Fermi gas

D() =L3m3/2

2h32. (5.15)

The maximum number of particle that can be accommodated bythe second term of Eq. (5.14) is obtained when we set = 0 in thatterm

N(T) =

0+dD()

1

e 1=D0

0+d

e 1=D0(kBT)

3/2

0+dx

x

ex 1

=D0(kBT)

3/2

2 (3/2). (5.16)

In the above

D0 = L3 m

3/2

2h32

, (5.17)

and 0

x

ex 1=

2 (3/2) 2.31516. (5.18)

The density corresponds toN is

(T) = m3/2

2h32(kBT)

3/2

2 (3/2). (5.19)

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For a system with a fixed number of particle, hence an fixed density, there exists a temperature Tc such

(Tc) =. (5.20)

For temperature lower Tc the total number of particles accommodatedby the k= 0 states can no longer account for all the particles. Whenthat happens a fixed fraction of the particles have to be accommodatedby the k= 0 state, or n0. The phenomenon that at low temperaturesa infinite number of particles all sits in the same state is called Bose-Einstein condensation. Substitute Eq. (5.19) into Eq. (5.20) we obtain

m3/2

2h32

(kBTc)3/2 2

(3/2) = , (5.21)

which implies

Tc = 2h22/3

kBm(3/2)2/3. (5.22)

The thermal De Broglie wavelength is defined as

(T) = h2mkBT

. (5.23)

At the critical temperature

3(Tc) =(3/2) 2.61248. (5.24)If we define interquartile distance das

d3 = 1, (5.25)

then Eq. (5.24) implies

(Tc)

d =(3/2)1/3 1.38. (5.26)

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For T < Tc the condensate fraction

f =n0

N =

(T)

=

(Tc)

(T)

(Tc) = 1

T3/2

T3/2C. (5.27)

AsT Tc the condensate fraction vanishes as

f(T) 32

Tc TTc

. (5.28)

For temperature below Tc the chemical potential is pinned at zero.The internal energy is given by

E =

0dD()

e 1=D0

0d

3/2

e 1=D0(kBT)

5/2

0dx

x3/2

ex 1=D0(kBT)5/2 3

4 (5/2)

. (5.29)

The specific heat is given by

CV,N=15D0

(5/2)

8 k

5/2B T

3/2. (5.30)

The value of(5/2) is approximately 1.34.

5.3 The amazing work of indistinguisha-

bility (optional)

In the above discussion we have seen that for temperature below Tc thenumber of bosons in the ground states become of order N. Howeverdespite the minute energy difference E = h

2

2m(2L)

2, the number ofatom in the first excited state is proportional to

1

eE 1kBT

E =

kBTh2

2m(2L)

2=

kBTh2

2m(2d)

2N2/3. (5.31)

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As the result the ratio between the number of ground state versus thefirst excited state particle is proportional to

N2/3

N 0 asN . (5.32)

This is pretty amazing given the fact that asL the thermal energyfar exceed E.

To better appreciate the above discussion, let consider a simperproblem with only two energy states, one has energy zero and the otherenergy > 0. First, let consider Ndistinguishable particle case. Thepartition function is

Z=N

n=0N!

n!(N

n)!

en. (5.33)

Here n is the number of particle in the upper energy states. The aboveis a binomial series and can be easily summed to give

Z=

1 +eN

. (5.34)

Next, we compute the average number of particle in the upper state

n =Nn=0 n

N!n!(Nn)!

en

Z = 1

ln Z. (5.35)

The result is

n =N 1e + 1

. (5.36)

As the result the ratio between the number of atoms in the two statesare given by

NN0

=N 1

e+1

N N 1e+1

=e, (5.37)

as expected from classical statistical mechanics.Now let us consider the particles to be indistinguishable bosons. In

this case the partition function is given by

Z=Nn=0

en. (5.38)

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Note that the factor N!n!(Nn)!

in Eq. (5.33) is missing. This is becausethe particle are indistinguishable, i.e., we can not count different per-mutation of the particle (with the same total number of particle in theupper and lower states) as different state. Of course Eq. (5.38) can alsobe summed easily to give

Z=1 e(N+1)

1 e 1

1 e as N . (5.39)

The total number of particle in the upper state is then

n =Nn=0 ne

n

Z = 1

ln Z=

1

e 1 . (5.40)

Note that for all finitetemperature T > 0 the number of particles inthe excited state is of order 1 not N. Hence

NN0

=1

e1

N 1e1 0 as N . (5.41)

This striking contrast with the distinguishable particle result is theact of indistinguishability. More precisely, in the distinguishable par-ticle case the Boltzmann weight unfavor the occupation of the excitedstate. However the N!n!(Nn)! favors it. As the balance between the en-ergy and entropy factors, the number of particles in the excited state

remains of order N. For bosons, the entropy factor is entirely removed!

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Chapter 6

Phase transitions

This chapter is about phase transitions, in particular, continuous phasetransitions. The behaviors of various physical measurable quantitiesnear the critical point of a phase transition are called critical phenom-ena. The task we are facing is to understand critical phenomena withinthe framework of statistical mechanics.

The reason for being interested in phase transitions is because theconcepts we develop in understanding them are widely applicable innumerous other areas of physics. Among them the concept ofrenor-malization group is particularly important.

Renormalization group is not a mathematical group, rather it is astrategy. In specific, it is a strategy that is useful when we face a system

with infinite (or a huge number) degrees of freedom. Huge number ofdegrees of freedom does not necessarily makes a problem difficult. Thedifficulty only arises when these degrees of freedom interactwith oneanother. For concreteness, let us imagine each degree of freedom is abit, or rather, a spin that can point in either the up or down directions.

The renormalization group strategy can be simply stated as follows.We first solve the problem involves a few spins. We realize that dueto the spin-spin interaction, the low energy states often involve thespins in question act collectivelytogether. To be more precise, we seekfor a description of the low energy states of these few spins in terms

of the behavior of a single effective degree of freedom. We then usethe original microscopic Hamiltonian to figure out how do the effectivespins interact. Next, we solve a few effective spin problem and describes

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the low energy states in terms of a new effective spins. Clearly thisprocedure can be continued forever. Some times, if we are lucky, theinteraction between the effective spins are the same as that between theoriginal spins, except the strengthof the interaction is modified. Undersuch condition, the iterated spin effective spin procedure describedabove traces out a trajectory in the space panned by the parametersdescribing various types of interaction in the Hamiltonian. Tremendousinsight can be gained by just analyzing the trajectory as a function ofiterations. For instance by knowing the limiting behavior after a largenumber of iterations, we often can make statement about the state ofthe system. Furthermore by finding fixed pointsat which the trajectorybifurcate, and analyze the flow near them we can extract informationabout phase transitions and critical phenomena.

There are numerous examples of phase transitions. Some hap-pen at finite temperature (e.g., ferromagnetic and antiferromagnetictransition, superconducting phase transition, liquid-gas critical point,order-disorder transition of adsorbed atoms, etc.) and some happenat zero temperature (e.g. the metal-insulator transition, superconduc-tor or superfluid-insulator transition, quantum magnetic phase transi-tion, etc). The latter class of transitions are called quantum phasetransitions. In this chapter we shall concentrate on classical phasetransitions where the transition happens at non-zero temperature.

Let me now give an example of continuous phase transition and

discuss its associated critical phenomena. Consider the easy-axis ferro-magnet Y F eO3. This material has two magnetic phases: the param-agnetic phase and the ferromagnetic phase. In the paramagnetic phasethe local magnetic moments arrange in a random fashion such that if weaverage over them there is no net magnetization. In the ferromagneticphase, on the other hand, the local moments align with one another sothat a net magnetization results.

When we study a magnetic system experimentally, temperatureT isan important control parameter. Experimentally it is found that thereexists a critical temperatureTc, above which the system is paramagnetic

and below which it is ferromagnetic. The behavior of the system in theneighborhood of Tc is called the critical phenomena, which we shalldiscuss in the following.

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6.1 Example: critical phenomena of aneasy-axis ferromagnet

The specific heat Ch diverges asT Tc. The precise fashion by whichsuch divergence occurs is

Ch A+( T TcTc

); T > Tc

A( Tc TTc

); T < Tc. (6.1)

Experimentally0.11. The number is called a critical exponent.In the following we shall define

t T

Tc

Tc . (6.2)

The magnetization m is defined as the average magnetic momentper unit cell. Above Tc,m = 0. BelowTc the system develops non-zeromagnetization. However, the direction of such spontaneous magneti-zation is undetermined until an infinitesimal external magnetic field isapplied. If we define the spontaneous magnetization m0 as

m0(T) = limh0

m(T, h), (6.3)

we find

m0(T) sign(h)B(t) ; T < Tcm0(T) = 0; T > Tc. (6.4)

For easy-axis ferromagnet 0.32.The magnetic susceptibility records the response of the system to

an externally applied magnetic field:

(T) m(T, h)h

. (6.5)

Experimentally it is found that as T Tc the zero-field susceptibilitydiverges according to

(T) C+t; T > Tc C(t); T < Tc. (6.6)

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The value of is 1.24.From Eq. (6.6) we know that the derivative ofm(Tc, h) is infinite at

h= 0. In fact experimentally it is found

m(Tc, h) D+h1/ : h >0 D(h)1/ : h 1. The experimentalvalue for is 4.8.

Experimentally using neutral diffraction we can determine the equaltime correlation function between two local moments

G(r; T) m(0)m(r). (6.8)

It is found that

G(r; T) = E+rd2+

er/(T); T > Tc

=m0(T)2 +

Erd2+

er/(T) : T < Tc. (6.9)

This define the critical exponent . The value of is 0.05.Finally the correlation length defined in Eq. (6.9) also diverges at

Tc. The singular behavior follows

(T)

F+t; T > Tc

F+(t); T < Tc. (6.10)The value of is 0.65.

So we have

0.11 0.32 1.24 4.80 0.65 0.05

6.2 Universality

The reason that we are interested in the above critical exponents isbecause they are universal. In other words the numerical value for

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... for any easy-axis ferromagnet that exhibit a Curie transition isthesame. What is even more surprising is that the same set of numbersdescribe the critical behavior near the liquid-gas critical point of, say,CO2, as long as the following translation is made:

liquid gas critical point : P T cferromagnet paramagnet critical point : h T m m0.

(6.11)

Apparently the critical exponents for the easy-plane and isotropicferromagnets differ from those of the easy-axis ones. Therefore the no-tion of universality only exists in the weak sense. The purpose of oursubsequent lectures is to discuss the origin of such universality and ex-

plore different universality classes.

6.3 Scaling

Experimentally it is found that

Ch(T, h) = |t|Sc,(h|t|y)m(T, h) = |t|Sm,(h|t|y)(T, h) = |t|S,(h|t|y)G(r, T , h) = 1

rd2+SG,(r|t|; h|t|y), (6.12)

where y = , and SO, are scaling functions whose form depend onwhether t >0 or t


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whereSi= 1,J >0,i, jrun through the sites of a simple cubic lattice,andij denotes the nearest neighboring lattice sites. In subsequentdiscussions we shall be interested in understanding the predictions ofthe statistical mechanics of the above model (the Ising model).

In statistical mechanics we study the expectation values of variousphysical quantities. To be more precise, a physical quantityO is afunction of the Sis, i.e. O[Si]. In the above and for the rest of thecourse we shall us the symbolO[Si] to denote thatO depends on allthe Sis.

The expectation value ofOis defined as

O

[Si] O[Si]eHZ , (6.14)

where= 1/kB

T, and

Z [Si]

eH. (6.15)

For example when we want to compute the average magnetization wehave

O = 1N

i

Si, (6.16)

where N is the total number of lattice sites (of course we shall beinterested in the thermodynamic limit where N ). As anotherexample when we want to compute the average spin-spin correlation

function we have

O(r) = 1N

i

SiSi+r. (6.17)

In the presence of external magnetic field Eq. (6.13) becomes

H Jij

SiSj hi

Si. (6.18)

Thus

m = 1

Ni

Si = 1N

T r[SieH]

Z(T, h)

= 1

N

Zh

1

Z=

1

N

ln Zh

= fh

. (6.19)

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In the above T r is short for

[Si], and

f(t, h) kBT

N ln Z (6.20)is the Helmholtz free energy density. Applying one more h-derivativeto Eq. (6.19) we obtain

=m

h =

2f

h2

= 1

N{

2Z/h2Z (

Z/hZ )

2}

=

Ni,j

{T r[SiSjeH]

Z(T, h)

T r[SieH]

Z(T, h)

T r[SjeH]

Z(T, h)

}

=

N

i,j

[SiSj SiSj]. (6.21)

Eq. (6.21) is the famous fluctuation-dissipation theorem.Now let us get some feeling about the the Ising model. Clearly as

T 0 the spins Si will align to minimize the energy. The groundstate has two-fold degeneracy, i.e., spins all up or spins all down. Wenotice that for Eq. (6.13)Si Si(global spin reversal) is a symmetryof the Hamiltonian. However, global spin reversal does not leave theground state invariant. When such happens, the physical state has

less symmetry than the Hamiltonian we say that there is spontaneoussymmetry breaking.

Clearly the spins will become disordered as we raise the tempera-ture. In particular when kBT >> Jthe interaction between neighbor-ing spins can not compete with the entropic effect of disordering. Asthe result we expect a paramagnetic to ferromagnetic phase transitionas the temperature is lowered.

Clearly it is extremely naive to expect a simple model like Eq. (6.13)to describe the behavior of a real material (say Y F eO3). The reasonthat such model description has any value at all is because the univer-

sality, i.e., it is plausible that the critical behavior of Eq. (6.13) (i.e.the predicted values of critical exponents) will be the same as that ofreal systems.

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Unfortunately it is very difficult to compute quantities such asEq. (6.15), Eq. (6.19) and Eq. (6.21). The root of the difficulty liesin the fact that in Eq. (6.18) spins on different sites interact with eachother. Were it for such interactionZ and f can be computed exactly.Indeed setting J= 0 in Eq. (6.18) simple calculations give

Z= [2 cosh h]Nf= kBT ln[2 cosh h]. (6.22)

Indeed, the interaction between infinite degrees of freedom is at theheart of many-body problems.

6.5 The predictions of mean-field theory6.5.1 The spontaneous magnetization, the expo-

nent

In the absence of exact solution for Eq. (6.13) ind= 3 (or in d = 2 inthe presence of non-zeroh), we have to resort to approximate methods.Among various such methods the most important and the most physicalone is themean-field theory. The key intuition behind mean-field theoryis the following. Through the spin-spin interaction, the neighbors of agiven spin exert an effective magnetic field on it. This effective magnetic

field polarize the spin so that it points in the same direction as theaverage magnetizationof its neighbors. Formally we replace Eq. (6.13)by the following Hamiltonian

Hmf= i

heff,iSi, (6.23)

where

heff,i= h +Jj

CijSj. (6.24)

In the above Cij = 1 ifi, j are nearest neighbor and = 0 otherwise. TodetermineSj we require the following self-consistentcondition to be

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met:

Si =T r[Sie

Hmf]

T r[eHmf]= tanh (heff,i)

= tanh (h +Jj

CijSj). (6.25)

In the case where the external magnetic field is uniform, we expect Sito be independent ofi. In that case Eq. (6.25) becomes quite simple:

S = tanh (h +JzS), (6.26)

wherezis the number of nearest neighbors of a site, or the coordination

number. For the hypercubic lattice in d dimension z= 2d.A simple way to solve Eq. (6.26) is by graphical means. Let us

first concentrate on the case where h= 0. If we plot the straight liney = x and the curve y = tanh (Jzx), it is simple to see that whenJz < 1 the two graphs intersect each other only when x = 0, whilewhen Jz >1 they intersect at three points. The temperature

TcJzkB

, (6.27)

at which the transition between the previous two cases occurs, marks

the mean-field phase transition. For T > Tc the only solution ofEq. (6.26) isS = 0, while for T < Tc two extra solutionsS = m0become allowed. We shall later prove that under the latter conditionthe free energy is lower when S = m0. Since asT Tc m0 0, wecan expand tanh J zS on the right hand side of Eq. (6.26) in powerseries of JzS whenT Tc

S J zS 13

(J zS)3 +.... (6.28)

Thus m0 satisfies

1 =TcT1

3(

TcT

)3m20+O(m40) +..., (6.29)

71


72/80

or

m0 3T3

T3c (TcT 1). (6.30)

Thus as T Tcm0 |t|1/2, (6.31)

which implies

=1

2. (6.32)

6.5.2 The exponent

WhenT =Tc buth >0 Eq. (6.26) read

S = tanh [S +ch], (6.33)

wherec 1kBTc . If we again expand the right hand side of the aboveequation in power series we obtain

S = S +ch 1

3(S +ch)3

+..., (6.34)

or

ch=1

3(S +ch)3. (6.35)

Eq. (6.35) implies that ash 0

S h1/3, (6.36)

hence

= 3. (6.37)

72


73/80

6.5.3 The exponent

For the more general case whereTis slightly higher thanTcand h


74/80

It is important to realize that Eq. (6.26) is equivalent to

f

S= 0, (6.44)

i.e. the self-consistent mean-field solution is the one that minimizesthe free energy in Eq. (6.43). First, we use Eq. (6.43) to prove thatfor T < Tc and h= 0 the free energy associated with theS=m0solution of Eq. (6.26) is lower than that of the S = 0 solution. To dothat we plot the function

fmf=Jz

2S2 kBTln {2cosh[(JzS)]} (6.45)

versusS. For smallS we can expand the ln and the cosh inEq. (6.45). The first three terms read

fmf=kB{Tln 2 + Tc2T

(T Tc)S2 + T12

(TcT

)4S4}. (6.46)First we notice that fconsists of even terms only, thus f(S) =

f(S). Second, it is clear that forT < Tc the global minima offoccurs atS = 0. By finding the solution to f/S|S=m = 0, andsubstitute the solutionS = m back into Eq. (6.46), we obtain forT Tc:

fmf = kBTln 2 ; T > Tc

= kB{Tln 2 +3

4

T

T2c (T Tc)2

} ; T < Tc. (6.47)As the result we find

Ch=0= T2f

T2 = 0 ; T > Tc

=kBT

T2c[9

2T 3Tc] ; T < Tc. (6.48)

Therefore at T = Tc the value of Ch=0 is 3kB

2 , thus the specific heat

is discontinuous across the transition in mean field theory. If we in-sist on putting this behavior in the framework described by algebraic

singularity, we conclude

= 0. (6.49)

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