Lectures Pahtria

7/30/2019 Lectures Pahtria

1/75

Statistical Mechanics

Lecture notes Baruch Horovitz and class of 2007

Contents

1. Ensemble Theory 2

1a. Thermodynamics (Review) 2

1b. Micro Canonical Ensemble (MCE) 5

1c. Ensemble theory - generalities 12

1d. Canonical Ensemble (CE) 14

1e. Grand Canonical Ensemble (GCE) 19

2. Quantum Statistical Mechanics 23

2a. Ensembles for ideal quantum gases 23

2b. Ideal Bose gas 28

2c. Ideal Fermi gas 34

2d. Non- Ideal gases 36

3. Phase Transitions 39

3a. First order transitions 39

3b. Second order phase transitions Mean Field theory 45

3c. Exact results 46

3d. Landaus Theory for second order phase transitions 49

4. Non- equilibrium 534a. Kinetic theory and Boltzmanns equation: 53

4b. Brownian motion 56

4c. Fluctuation Dissipation Theorems (FDT) 60

4d. Onsagers Relations 68

Appendix: Langevins equation from a Hamiltonian 72

1


2/75

Reference books:

R. K. Pathria, Statistical Mechanics

Landau & Lifshitz, Statistical Physics

K. Huang, Statistical Mechanics

S.-k Ma, Statistical Mechanics

F. Mohling, Statistical Mechanics Methods and Applications

G. H. Wannier, Statistical Physics

1. ENSEMBLE THEORY

1a. Thermodynamics (Review)

Macroscopic state: Set of measurable coordinates of a system with many ( ) degreesof freedom, e.g. volume V, number of particles N, energy E.

Equilibrium: A macroscopic state that is uniquely determined by a small number of external

forces, e.g. pressure P, chemical potential , temperature T.

Pairs of force f and coordinate x generate work W = f dx (W is not necessarily an exact

differential, i.e. equation may not be integrated to yield a state function W).

E.g. W = P dV, W = dNMicroscopic degrees of freedom contain heat energy Q, with S the coordinate, T the force.

Q = T dS.

Thermodynamic limit: N , V with N/V const. Macroscopic state, equilibriumetc. are defined in this limit. Define: Extensive variables that increase with N, e.g. V, E, S

and intensive variables that are constants in the thermodynamic limit, e.g. N/V, P, .

First Law: Two ways to exchange energy, work or heat: E = Q + W; heat Q is due

to microscopic degrees of freedom. There exists an adiabatic process for which Q = 0.

Entropy S is a thermodynamic coordinate - proof:

2


3/75

Adiabatic process Q = 0. Equilibrium determines a

curve E(x) from dE = f dx. Between curves change is

nonadiabatic, is an integration constant.

E = E(x, ). Curves do not cross since f = Ex

is unique in equilibrium. Also E(x) is single

valued E(x, ) is monotonic in . For constant x, Q = dE = ( E

)xd

dE = f dx + d = ( E

)x

is not unique - can choose = A() with A() monotonic. Choose which is extensive

S N, and define temperature T = ES

as intensive. Assumption in proof: only one

coordinate.

dE = T dS

P dV + dN

E = E(S , V , N) is a (single valued) state function,

dE = 0. This is the first law.

Heat is not single valued -

TdS depends on how V, N change along the path (

Q = 0).Second Law: In a closed system S(t) increases with time. Exchange dE1 = dE2 in twosubsystems:

S = S1 + S2

dS

dt=

dS1

dE1

dE1

dt+

dS2

dE2

dE2

dt= 1T1 1T2 dE1dt > 0

energy flows from high to low T.In equilibrium S is maximal T1 = T2

dS =1

TdE+

P

TdV

TdN

If volumes of subsystems exchange dV1 = dV2 with T1 = T2 (hence E terms vanish),dS = ( S1

V1)E,NdV1 + (

S2V2

)E,NdV2 = (P1T

P2T

)dV1 P1 = P2 equilibrium.Particle exchange dN1 = dN2dS = (S1/N1)E,V dN1 + (S2/N2)E,V dN2 = (1/T 2/T)dN1 1 = 2 chemical equilibrium.TdS > dE+P dVdN = Q irreversible process, i.e. S increases more than its equilibriumchange.

Adiabatic process:

A process in which the energy is changed only by slow variation of external conditions so

3


4/75

that at every instant the system is in equilibrium. Furthermore, the system is thermally

isolated - no energy transfer except the external condition.

Adiabatic process is reversible: expand in d/dt, external condition (e.g. volume)

dS

dt= a + b d

dt+ 1

2c( d

dt)2

a = 0: equilibrium condition

b = 0: since dSdt

> 0 cannot depend on sign of ddt

dSd

= 12

c ddt

+ ..... 0 when ddt

0. dS = 0 in adiabatic process.Note: A reversible process in a closed system is an adiabatic process and S is constant. A

reversible process in an open system is a process for which T dS = dE+ P dV dN, i.e.heat exchange with a reservoir is allowed and dS

= 0.

Thermodynamic Functions

F = E T S Helmholtz free energydF = SdT P dV + dN F(T , V , N )If an additional variational parameter is present F is minimized to reach dF = 0.

T , V , N fixed: dF = dE T dS 0, for < 0 the process is irreversible.G = E T S+ P V Gibbs free energydG =

SdT + V dP + dN

G(T , P , N )

T , P , N fixed: dG = dE T dS+ PdV < 0 and G is minimized. = F Nd = SdT P dV N d (T , V , )T , V , fixed: d = dE T dS dN < 0 and is minimized.Extensiveness: E = E(S,V,N)

|=1:

E = T S P V + NF = P V + NG = N

= P V (1)

4


5/75

1b. Micro Canonical Ensemble (MCE)

Basic assumption:

S = kB ln(E)

Boltzmans constant: kB = 1.38 1016 erg/deg.(E) is the number of microstates for a given N , V , E , and states have equal weight :

(E) ={r}

EH{r}

{r} are microscopic degrees of freedom of the N particles.We shall now prove extensivity and maximum property: Consider two subsystems, neglect

their interaction (justified since the number of particles near the surface is small, or surfaceenergy bulk energy),

H{r1, r2} = H1{r1} + H2{r2}(E) =

{r1,r2}

EH1{r1}H2{r2}

=

E1

{r1}

E1H1{r1}{r2}

EE1H2{r2} =

E1

exp

S1(E1) + S2(E E1)

kB

Since Si Ni are large look for maximum in E1 (equivalent to steepest descent),

E1(S1(E1) + S2(E E1)) |E1 = 0

S1E1

=S2E2

T1 = T2

Expand to 2nd order

S1(E1) + S2(E2) = S1(E1) + S2(E2) 1

2k1 (E1 E1)2

1

2k2 (E2 E2)2

1

ki=

2S

E2i

Ei=Ei

= (1T

)

Ei=

1

T2Ci

Ci =

EiT

V,N

eS(E)/kB = eS1(E1)+S2(E2)

E1

exp

12k1

(E1 E1)2 12k2

(E2 E2)2

Probability that E1

= E1 is Gaussian,

(E1

E1)

2

1/2

k1

N

5


6/75

Dominant term is E1 = E1 with E1 such that T1 = T2.

If is the spacing of E1 levels (e.g. V2/3 in ideal gas)E1

dE1

exp

( 1

2k1+

1

2k2)(E1 E1)2

=1

2

k1k2k1 + k2

1/2 S(E) = S1(E1) + S2(E2) + O(ln N)

Extensivity, ie. (E) is dominated by one term: (E) = 1(E1)2(E2)

(Note: The only function f() which is additive when = 12 is f ln )

Energy fluctuation ki = T

CV.

Similarly, separate to subsystems with V = V1 + V2 or N = N1 + N2 to obtain dominant

term at Vi (hence P1 = P2) or at Ni (hence 1 = 2).

Ideal Gas (no interactions)

Classical: r (p, q). For one particle energy (p): cl

d3Nq

i (pi)=Ed3Np VN.

Note missing dimensional prefactor.

PkB T

=

ln V

N,E

= NV

Quantum: r quantum numbers. Use periodic boundary conditions, e.g. eipxL/ = 1 px =

hL

nx , nx integer < nx < = h

2

2mL

2 (n2x + n

2y + n

2z)

(E) is the number of solutions to3N

r=1 n2r =

2mh2

EV2/3

S(N , V , E ) = S(N, V2/3E)

In an adiabatic process V2/3E = const, P = EV

N,S

= 2E3V

P V5/3 = const

6


7/75

(E) irregular - No. of points on surface of 3N sphere.

Instead (N , V , E ) =

EE (N , V , E ) has reasonable E limit.

Replace by = number of microstates with E

2


8/75

ln = ln + O(ln N)

Volume near surface dominates total volume for N 1. Value of is not important.

S(N , V , E ) = kB ln = NkB ln

V

h3

4mE

3N

3/2+

3

2N kB

T =

E

S

N,V

=2E

3NkB

E =3

2N kBT 1

2mv2 = 3

2kBT

Cv = E

TN,V =3

2

NkB

P =

E

V

N,S

=2E

3V P V = NkBT

Cp

(E+ P V)

T

N,P

=5

2NkB

P

VT

N,P

is excess work at constant P.

CpCv

=5

3

Gibbs paradox:

S is not extensive. Even mixing gases 1,2 with equal T, n leads to Stotal = S1 + S2The mixing entropy is positive:

N kB ln V N1kB ln V1 N2kB ln V2 = kB

N1 lnV

V1+ N2 ln

V

V2

> 0

But this must be reversible !

Quantum mechanics - indistinguishability. Classical limit (h

0) should give

1

N!

S(N , V , E ) = NkB ln VN

+3

2NkB

5

3+ ln

2mkBT

h2

Sackur Tetrude eq. : S is extensive.

Classical derivation ...

cl=

1

w0

d3Nq

i p2i


9/75

w0 is chosen with

h 0

cl

=

1

N!C3N(

2mE

h2)3N/2 w0 = N!h3N

Where1

N! is the Gibbs correction and h is the volume of one state in the p,q space.Ex: evaluate w0 for harmonic oscillator.

Equipartition

xi = qi or pi

xi Hxj

. . .

E 12


10/75

for such systems we clearly have:j

(pjH

pj+ qj

H

qj) = 2H H = 1

2f kBT

f=No. of degree of freedom number of quadratic terms in H.Each harmonic term in the quadratic Hemiltonian makes a contribution of 1

2kT towards the

internal energy of the system and hence a contribution of 12

k towards the specific heat Cv.

Example: molecule with m atoms

C.M: 3 translations + 3 rotations non co-linear molecule

3 translations + 2 rotations co-linear

(To understand the significance of colinearity, note that quantum levels

2(+1)

2I form aclassical continuum if

2

I kBT. However, ifI 0 as in a colinear case, at kBT 2I only

the single ground state is relevant)

3m coordinates no. of vibrations = 3m-6 non-collinearor = 3m-5 collinear

Translation ( p2x

2m), rotation ( L

2x

2I) - 1 quadratic term

Vibration ( p2x

2m+ m

2x2

2) - 2 quadratic terms.

H = 12

[6 + 2(3m 6)]NkBT = (3m 3)NkBT non collinear

H = 12

[5 + 2(3m 5)]NkBT = (3m 5/2)NkBT collinear

Virial Theorem (Clausius 1870) ...

The Virial of a system is by definition, the sum of the products of the coordinates of the

various particles and the representative forces acting on them:

V 3Ni=1

qi pi = 3N kBT

E.g. ideal gas in box:

10


11/75

p = 0 only from walls at qi = r

p = Pds

where: P is pressure, ds is surface element and p ismomentum hitting ds (of all particles)

V=

r

r p = P

s

r ds = P

( r)dv = 3P V P V = NkBT

Consider now classical particles i, j with 2-body interaction

u(ri,j) : rij = |ri rj|.pi =

ri ju(|ri rj|)

i

ri pi =

i,j

ri ri

u(|ri rj|), u(r)r

=r2

r

u

r2

Sum over pairs:i

ri pi =i


12/75

1c. Ensemble theory - generalities

(p, q, t)d3Nqd3Np is the probability of microstate(p,q), with normalization

E=const

(p, q, t)d3Nqd3Np = 1

Average: f = f(p, q)(p, q, t)d3Nqd3NpEquilibrium: all observables

t= 0

t= 0. In mechanics we start with a point

p(0),q(0) in the 6N dimensional phase space, follow trajectory and average on time.

Ergodic theorem: Long time average = ensemble average, i.e. probability concept (p, q, t)

is equivalent to mechanics.

Liouvills theorem

p(t), q(t) satisfy Hamiltons equation and define a velocity in phase space

v = (q, p)

Net flow of states from volume , with surface , is = t

d

(v n)d = div(v)d =

t dtrue for any . Continuity:

t+ div(v) = 0.

Consider (qi, pi, t) for a collection of states qi(t), pi(t)

t+

i

qiqi +

pipi

+

i

qiqi

+pipi

= 0

qiqi

=H

qipi= pi

pi

and therefore qiqi + pipi = 0 leads tod

dt=

t+ [, H]poisson = 0

Local density as viewed on moving points, is constant, hence ddt

= 0.

No. of states is conserved incompressible fluid.Equilibrium:

t= 0

i

qiqi +

pipi

= 0

12


13/75

solutions:

(p, q) =

const E 12

< H(p, q) < E+ 12

microcanonical

0 otherwise

In general can depend on constants of motion, e.g. H(p, q).

[H(q, p)] canonical ensemble

In microcanonic (p, q, t) = 1

S = kB ln = kB

(p, q, t) ln (p, q, t)d3Npd3Nq

defines entropy also in other ensembles, in equilibrium. Is this form of S valid at non-equilibrium? but then

dSdt

= 0

which violates the second law.

Arrow of time

Consider volume expansion III. For N = 1020, volume increases by factor 2.The increase in number of states is 210

20!! all states of I evolve into II, but only fraction 1

21020

of states in II evolve into I. Probability suggests that S(t) increases.

Just probability is not sufficient:

Consider IIIIII, increasing volumes. States in II most probably go to III, but where dostates in II come from - most probably also from III !? not I?

Time reversal invariance: a state x1 in I evolves into x2 in II. Now reverse all momenta in

x2 Rx2.Rx2 is a state in II. Its time evolution yields x1 in I, i.e. it is possible to find a state that

lowers entropy.

The difficulty with Rx2 is that it must be prepared accurately; a minute perturbation causes

instability (chaos, the butterfly effect). Nature does not allow perfect aiming with accuracy

210

20.

13


14/75

need probability + stability, i.e. information on nearby trajectories, coarse grained,then entropy can increase with time.

Hypothesis: Universe started with low entropy, low entropy radiation from the sun produces

low entropy food etc...

1d. Canonical Ensemble (CE)

Consider a system that is embedded in a larger heat

bath. Energy exchange is allowed, i.e. Er is not fixed.

r is point in phase space of the system.

Reservoir energy ER, Er


15/75

dominant term at E:

E(S(E)/kB E) = 0 S(E)

E|E = 1

T

E is such that the MCE at E has the temperature T.

S(E)/kB E = FMCE

where FMCE is the free energy of the MCE defined with variable E, V , N .

Note: S(E(T , V , N ), V , N ) defines F(T , V , N ).

Fluctuation near E (expanding S around E):

S(E)

kB E =

S(E)

kB E +

1

2kB

2S

E2 E (E E)2 + ... ,where,

2S

E2=

(1/T)

E= 1

T21

E/T= 1

T2CV,

and T, CV are for MCE.

ZN(V, T) = eFMC E

E

e(EE)2/2kB T

2CV Gaussian.

The weight of E

= E decreases rapidly with width,(E E)2

E=

kBT2CV

E 1

N 0.

To find the partition function, we replace the Sum with Integral,E

1

dE,

where is the energy level spacing (as above Na ln ln N), and

ZN(V, T) = eF

MC E

eE2/2kB T2CV dE = eF2kBT2CV

= eFCE ,

FCE = FMCE + O(ln N),

where FCE, FMCE O(N).

Insensitivity of thermodynamic results to type of ensemble due to:

1. (E)

e(...)N

e(...)E

: exponential increase.

15


16/75

2. Thermodynamic limit: E, N .

E.g. for ideal gas, (E)eE e(3/2)Nln EE with maximum at E = 32

N kBT. For E= E,(E)eE practically vanishes.

Note: The energy fluctuations can also be identified by evaluating the specific heat:

CV =

TE = 1

kBT2

r Ere

Err e

Er=

1kBT2

E2+ E2 ,where

(E E)2 = E2 E2 = kBT2CV N.

Examples:

Classical systems: the general expression for the partition function,

ZN(V, T) =1

N!h3N

eH(p,q)d3Nqd3Np,

where, N! is Gibbs normalization, and h corresponds to a volume of one state in the

q, p phase space.

Ideal gas: H = ip2i /2m,ZN(V, T) =

VN

N!h3N

0

ep2/2m4p2dp

N=

1

N!

V

3

N,

where,

h2mkBT

, the thermal wavelength.

This corresponds to a deBroiglie wavelength for momentum mkBT i.e. h/

p2 = h/mkBT.

F = kBT ln ZN = N kBT

ln N3/V 1 ,P =

F

V

N,T

=N kBT

V,

S =

F

T

N,V

= NkB

ln V/N3 + 5/2

, as in MCE,

=

F

N

T,V

= kBT ln N3/V.

16


17/75

Consider N noninteracting molecules with internal energies i, ni molecules at level iare indistinguishable.

ZN(V, T) = {ni}

1

n0!n1!...ei ini ep2/2md3pd3q

center of mass

N

=VN

N!3N

i

ei

N, using multinomial expansion

=1

N!(Z1(V, T))

N , Z1 V3

i

ei partition of one molecule.

Note: The multinomial expansion is:

i

aiN

= N!{ni}

1

n1!n2!...an11 a

n22 ... ,

where the sum on distributions {ni} is restricted by

i ni = N.

Diatomic gas:

H = 0

electron+ (+ 1/2)

vibration+ 2k(k + 1)/2I

rotation,

Z = e0

=0

e(+1/2)

k=0

(2k + 1)e2k(k+1)/2I

s

(2s + 1) nonidentical atoms

,

where s is the spin of the two atoms.

Zrot =T

0

2ke2k2/2Idk = 2I/2,

Zrotclass = 1h2e(M2x+M2y )/2IdMxdMydxdy = 2I/2,

The two atom axis is z and dxdy = d 4 is the solid angle.

If the two atoms are identical, we need QM:

There is a constraint: s+k needs to be even. [Orbital exchange has ()k, spin exchangehas ()ss1s2, hence both Fermion and Boson symmetries are obeyed for s + k even].

17


18/75

Spin k degeneracy = 2s + 1

H2 s = 1 odd 3 orthohydrogen

s = 0 even 1 parahydrogen

D2 s = 2 even 5

s = 1 odd 3

s = 0 even 1

ZH2rot = 3

k odd

(2k + 1)e2k(k+1)/2I +

k even

(2k + 1)e2k(k+1)/2I,

ZD2rot = 3

k odd

... + 6

k even

...

T 12

2I2

s

(2s + 1).

In the last equation, the 12 is the classical reduction of angular integration by factor of

2, for identical atoms.

Equipartition: xi

H

xj

=

xi

Hxj

eHd

eHd

,

numerator =

xi

H

xjeHd =

1

dxi=jxie

H

x(2)j

x(1)j

+1

eHdij.

In general, the energy H(x(1,2)j ) at the boundaries of xj

xiHxj

= ijkBT.

E.g. ideal gas: pix

pix

pi2

2m

=

1

m

p2ix

= kBT,

E = i pi

2

2m = 32N kBT.Extreme relativistic: H =

i c |pi|,

qix =H

pix=

pixc

p2ix + p2iy + p

2iz =

cpix|pi| ,

i

pi qi

=

i

c |pi|

= H = 3N kBT (> 32

NkBT),

as the power of p smaller, the energy is higher for given temperature.

18


19/75

1e. Grand Canonical Ensemble (GCE)

System in heat bath and particle reservoir.

Er

ER, Nr

NR

Er + ER = E0 = Const.

Nr + NR = N0 = Const.

The combined system is microcanonic.

r is a point in phase space of the system, which now includes all Nr ( N0). The probabilityof a point r with Er, Nr is Pr

Pr R(N0 Nr, E0 Er)

ln Pr ln R(N0, E0) ln RN

|N0Nr ln R

E|E0Er

= ln R(N0, E0) + Nr Er = 1/kBT, is imposed by the resevoir

Pr = eNrEr /L

The grand partition function is

L( , T, V ) =

r

eNrEr e(,T,V)

Identify :0 =

r

eNrEr + =

r

Nr Er +

+

Pr

= N E+ TT

The energy E( , T, V ) shows that the solution to this differential equation is the theromo-

dynamic potential, since = F N = E N T S = E N + TT

Alternatively, if

N is dominated by one term

L eN

r,Nfixed

eEr = eNF

= F N = P V

19


20/75

Define fugacity = e

L( , T , V ) =

N=0

NZN(V, T)

So that the weight of each N = NZN/

L.

N =

ln L( , T , V )|T,V

E =

ln L( , T , V )|,V etc.

Ideal gas: with internal energies i for each particle,

Z1 =V

3

i

ei =V

3a(T)

ZN =1

N!ZN1 L =

N

N1

N!ZN1 = e

Z1

P V

kBT= Z1 = e

V

3a(T) = kBT ln P

3

kBT a(T)

N =

ln L( , V , T ) = Z1 = P V

kBT = kBT ln( n

3

a(T))

n31

Particle (N) fluctuations:

Weight W(N) = eNF(N,V,T)/LMaximum at N such that = F

N|N, i.e. can = FN equals ifN is chosen at this maximum,

N = N.

For maximum need 2F

N2|N > 0:

F(N , V , T ) = Nf(v) v =V

N

For fixed V, N

= vN

v

, hence

F

N= f(v) v f

v

2F

N2=

v2

N

2f

v 2

Since P(v) = fv

2FN2

|N = v2N Pv (or = N) we need ( Pv )|T < 0 for W(N) beinga maximum ( this is Van Hoves theorem, Huang first edition p.321 - general proof; Huang

second edition p.206 - for the special case of hardcore interaction).

Physically obvious: if P > Pext the net force increases V; now ifPV

> 0 P increases even

more and equilibrium (P = Pext) is further away.

20


21/75

CE and GCE are equivalent by this thermodynamic stability criterion ( Pv

)|T < 0.Note however, that at a 1st order phase transition, e.g. the gas-liquid transition, P/v = 0

and GCE is not equivalent to CE, hence large N fluctuations are expected.

W(N) W(N)e 12(NN)2 N2 12 = kBT Nv2(P/v)

N

N2

12 /N 1/

N 0

Energy fluctuations:

E = r

ErPr = ln L ,VE2 =

1

L

2L2

,V

E2 = E2 E2 =

2

2ln L

,V

=

E

,V

= kBT2 E

T|,V

E(N( , T , V ) , T , V )

T=

E

T

N,V

+

E

N

T,V

N

T

,V

= CV +

E

N

T,V

N

T

,V

dE = T dS

P dV + dN

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

N T,V = + TS

NT,V= T

N

F

T

N,V

T,V

= T

T

N,V

(*)

Chain rule ( xy

)z(yz

)x(zx

)y = 1prove by dx = ( x

y)zdy + (

xz

)ydz = 0 which yields (yz

)x.

N((T, ), T , V )

T ,V = N

T,V+N

T,V

T = N

T,V

TN,V +N

T,V

T

=1

T

N

T,V

T

T

N,V

=

1

T

N

T,V

E

N

T,V

from (*)

E2 = kBT2CV + kBT

N

T,V

E

N

T,V

2

= (E2)can + N2

E

N

T,V

2> (E2)can

21


22/75

(E2)12 /E 1/N 0 except at Phase transitions.

Summary: Thermodynamics:

E(S , V , N) dE = T dS P dV + dN

E T S = F(T , V , N ) dF = SdT P dV + dN

F N = (T , V , ) d = SdT P dV NdStatistical Mechanics:

= r (E,Nfixed)1 = eS(E,V,N)/kB M CE

Z =

r (Nfixed)

eEr = eF(T,V,N)/kB T CE

L =

r

eNrEr = e(T,V,)/kB T GCE

22


23/75

2. QUANTUM STATISTICAL MECHANICS

A system ofN particles is described by a wave function (r1, r2,...). Since the particles are

indistinguishable, upon exchange of particles ri

rj the wave function acquires a phase e

i

( , ri, , rj, ) = ei( , rj, , ri, ).

This particle exchange is equivalent to a rotation by in the relative coordinate ri rj.Exchanging the particles i, j twice is equivalent to a 2 rotation, which in 3-dimensions is

equivalent to the identity operator [a circle on a sphere can be smoothly deformed into a

point]. Hence e2i = 1 and there are two types of particles in nature:

= 0 symmetric for bosons (Bose Einstein statistics (BE))

= antisymmetric for fermions (Fermi Dirac statistics (FD)).

In particular fermions obey paulis exclusion principle [antisymmetric wavefunction at ri = rj

vanishes]. Note also the spin-statistics connection, i.e. integer spin particles are bosons, half

integer particles are fermions. Note also that in 2-dimensions other statistics are allowed;

since the 2 rotation is now a circle with a singular point ri = rj at its center, the circle

cannot be deformed into an identity operation. In 3-dimension one escapes in the 3rd

dimension avoiding the ri = rj singularity.

We also define a Boltzman statistics (MB) where only the particles at the same energy

level are indistinguishable, i.e. for each energy level we add a factor of 1n!

.

2a. Ensembles for ideal quantum gases

Micro Canonical Ensemble

We group the different states into energy levels

i with degeneracy ofgi. Each level is occupied

with ni particles. For the statistical treatment

we assume gi, ni >> 1.

(N , V , E ) =

{ni}

W({ni})whereW({ni}) =

iW(i) for a given distribution{ni}

23


24/75

The prime on the sum denotes the following constraints:i

ni = N and

i

nii = E

For bosons, each energy level can be occupied with ni particles within gi folders with gi 1divisions. Hence we choose ni particles from ni + gi 1 objects, W(i) = (ni+gi1)!ni!(gi1)! and forall energy levels we have

WB.E. =

i

(ni + gi 1)!ni!(gi 1)!

For fermions we choose ni sites from the gi available ones, so

WF.D. =

igi!

ni!(gi ni)! gi ni >> 1

For the Boltzman statistics any particle can occupy any of the states up to the Gibbs

correction ( 1ni!

) for the indistinguishability of the particles in each energy level

WM.B. =

i

(gi)ni

ni!

The Gibbs factor correctly accounts for permutations of particles in differen quantum states,

but also unnecessarily corrects for particles that are in the same quantum state where correc-

tion is not necessary (e.g. one symmetric state only). Therefore ni! is an over-correctionand is valid when the density is low ni


25/75

We now find the distribution ni that maximizes S. Using the method of lagrange multipliers

we demand

ln W({ni})

ini

inii

= 0.

with , to be determined below. Rewrite our 3 cases as

ln (W({ni})) =

i

ni ln

gini

a

gia

ln

1 a ni

gi

where a is defined by a = 1 (BE), a = +1 (FD) and a 0 (MB).Performing the variation yields

i ln

gini

a

i

ni = 0

so that S is maximized by the distribution

nigi

=1

e+i + a

The value of S at its maximum is therefore

S/kB = ln W({ni }) =

i

ni ( + i) +gia

ln

1 + aei

The coefficients and are determined by the constraints on N and E:

kBT

=1

kB

S

N

E,V

= N

N

E,V

+ + E

N

E,V

+ gi

a

aei1 + aei

N

E,V

+ i

N

E,V

=

since the last sum is N N

E,V

E N

E,V

. The coefficient is identified by

1

kBT =1

kB SEN,V = N EN,V + E EN,V + +

i

gia

aei1 + aei

E

N,V

+ i

E

N,V

=

The equation of state is obtained by

1

a

i

gi ln(1 + aei ) =

S

kB+

N EkBT

=P V

kBT

25


26/75

so that finally

P V =

kBT

gi ln

1 e(i)/kB T a = 1kBT giei = kBTni = kBT N a = 0

.

Canonical and Grand-Canonical ensembles

In the GCE there are no constraints on N, E. Hence we label by i all quantum numbers

that completely specify a state so that gi = 1 and

WBE{ni} = 1WF D{ni} = 1 (if all ni = 0,1) or 0 otherwise

WMB{ni} = i

1ni!

notice that WF D < WMB < WBE. E.g. with N = 2 particles in two states

BE: |aa, |bb, 12

(|ab + |ba) 3 states

FD:1

2(|ab |ba) 1 state

MB: 12 |aa, 12 |bb, |ab 2 states.

Consider briefly the CE constrained with i ni = N,ZN =

{ni}

W({ni})e

i nii

and for the MB distribution

ZN ={ni}

i

N!

ni!eini

1

N!= using multinomial expansion =

i

ei

N1

N!=

1

N!(Z1(V, T))

N

[Note: with i

0 we have ni WMB{ni} = NN/N! for N particles in N states]. For theBE and FD statistics one has to proceed to the GCE, to avoid the ni = N constraint.

L( , V , T ) =

N=0

NZN(V.T) =

N=0

{ni}

i

ei

niwith = e

here the first sum determines the constraint N on the second sum. However, since anyway

we some on all occupations, we can sum them on each level independently,

L =

n0,n1,... e0

n0

e1

n1 =

ini

ei

ni

26


27/75

now using n xn = 11x we get the result for each of the distributions

L =

i

11ei

BE

i

1 + ei

F D

ln L = Z1 =

i ei MB using

xn

n!= ex

.

The GCE relation isln L = P V

kBT=

1

a

i

ln

1 + aei

and recall a = 1 for BE, a = 1 for FD and a 0 for MB. For N, E we have :

N =

(ln L)

V,T

=

i

11

ei + a

E =

ln L

,V

=

ii

1

ei + a

The mean occupation number is given by

ni = 1L

N

{nj}

niNe

j nj j

= 1

ln L

i

,T,j=i

=1

1

ei + a=

1

e(i) + a

The figure illustrates the various cases. For FD, if > and low temperatures the mean

occupation tends to 1. For the BE case, the condition < must be valid for all to

27


28/75

maintain ni 0; this leads (see below) to BE condensation. The MB distribution is validonly when ni > 1 for all i, hence < 0 and || kBT or 1.The condition on the density is N =

i e

i = V3

which implies high temperature or

low density limit. Note that both BE and FD approach MB at this limit of

1.

2b. Ideal Bose gas

Here i = p2/2m and the summation index is i = p.

A note on momentum summations: The sum

p actually stands for a sum on integers

nx, ny, nz that define px, py, pz via periodic boundary conditions. E.g. eipxLx/ = 1 with Lx

the length in the x direction, px =2Lx

nx, hence for any function f(p)nx,ny,nz

f(p) =

LxLyLz

h3f(p)d3p =

V

h3

f(p)d3p

In radial coordinates we then have

P

kBT= 4

h3

0

p2 ln(1 ep2/2m)dp 1V

ln(1 )

N

V =

4

h3 0 p2 1ep2/2m 11

dp +

1

V

1 The p = 0 term in the original sum is singled out. It is the number of particles N0 =

1

in

the single state p = 0. [Equivalently, need ei < 1 to allow convergence for L.]The crucial property of bosons is 1 so as to keep all state occupations np 0. Fora given density the integral term decreases with T for a fixed ; to keep N/V fixed must

increase, but it is bound by = 1. The integral is bound at = 1, hence below some Tc the

integral becomes < N/V and a macroscopic part of N must populate the p = 0 state. This

is Bose Einstein condensation.

The correction in the pressure equation is negligible:

1 = 11+ < N0 >

1V

ln(1 ) = 1V

ln(< N0 > +1)always 0

The boson thermodynamics are then given by

P

kBT=

1

3g5/2()

28


29/75

N

V=

1

3g3/2() +

< N0 >

V

gn() =1

(n)

0

xn1dx1

ex 1 =

=1

n

For small we use the expansion to eliminate :

P VNkT

=

=1

a(n3)1

(a = virial coefficient)

a1 = 1, a2 = 14

2= 0.176, a3 = 0.0033 .....

For specific heat:

E = (

ln L),V = kBT2

T

V g5/2()

3

,V

=3

2kBT

P V

kBT=

3

2P V

since the virial expansion for P V has T(3)1 T5/23/2 terms.CVNk

=1

Nk

E

T

N,V

=3

2

T

P V

N k

N,V

=3

2

=1

5 32

a(n3)1

=3

2(1 + 0.0884n3 + 0.0066(n3)2 + ...)

CV = T

S

T

N,V

T0 0

hence CV looks roughly as:T

Cv

If n3 < g3/2(1) = (32

) = 2.612

0 < < 1 and< N0 >

V= 0

If n3 > g3/2(1)

V= n g 32 (1)

3> 0

1

2.612

g 32

29


30/75

with = 1 and finite density at the single

= 0 level.

Transition at n3 = g32

(1) kTc =22

m[g3/2(1)]23

n2/3

When = 1, < N1 > is negligible since 1 h2/V2/3

< N1 >

V=

1

V

1

e11 1

V

V2/3

h2V 0

At T Tc, N = 1 g3/2(1)

3n

= 1

T

Tc(n)

3/2= 1 nc(T)

n

1TTc

1

N0N

Two fluid concept: < N0 > condensate,g32

(1)

3normal component.

P

kT=

1

3g5/2() n < nc (normal)

13

g5/2(1) n > nc (condensed phase)

Pc T52 n 53

all excess particles beyond nc occupy p = 0,

< N0 >= N Nc(T), even at n withno contribution to P. Along AB P

n= 0, two

phases coexist with phase A that has n .1n

A

P

B

Pc n53

30


31/75

This corresponds to a first order phase tran-

sition, 1/n jumps by (1/n) = 1/nc as pres-

sures increases.

T

P

No

realization

Normal

Condensate on line

PcT5

2

Note that there is no realization ofP > Pc(T). The equation of state defines a 2-dimensional

surface in P , T , n; the figure is a projection of this surface on the P, T plane. There are no

points on the surface that project onto the region P > Pc(T). Consider a fixed n and an

initial high temperature where P = nkBT and P Pc(T). Now as T decreases P approachesPc(T) at Tc, and upon further decrease of T, P stays on the critical curve P(T) = Pc(T) for

all T < Tc. This feature is an artifact of the ideal gas; once hard core repulsion is added, at

sufficiently high density (in a constant T trajectory) the pressure will start to increase.

Clausius Clapeyron: dPcdT

=S/N

( 1n

)

S is the jump in entropy. We will evaluate S directly:

S

N kB=

P V

N kBT+

E

NkBT

kT=

5

2

P V

NkBT

kBT=

52

g5/2()

n3 ln T > T c

52

g5/2(1)

n3= 5

2

g5/2(1)

g3/2(1)N

NT Tc

At T < Tc, S N < N0 >= Nnorm, no entropy for the condensate. Above the Pc(T)line S/N 0, while just below this line

N =V

3g3/2(1) S

N=

5kB2

g5/2(1)

g3/2(1), (1/n) = 1/nc =

3/g3/2(1)

Consider now the derivative

dPcdT

=5

2k

g5/2(1)

3obtained from Pc =

kT

3g5/2(1)

This proves the Clausius Clapeyron relation.

31


32/75

Note that S implies a Latent heat TS 1st order transition.

Specific heat (T < Tc):

CvNk

= 3V2N

g5/2(1)d

dT( T

3) = 15

4( 52 )n3

T3/2

At T > Tc,dCvdT

dg3/2()d

=g1/2()

1 Tc

T

1.5

1.925

CvNk

The shape of CV(T) is similar to the transition of4He, which looks like the letter .

For 4He by this theory Tc = 3.13K, by experimental results Tc = 2.19

K.

Note: In a real superfluid there is a finite critical velocity vc below which there is no dissi-

pation to the flow. This implies that at momentum k and frequency the current states

have excitations = vk (note the transformation to the moving frame evtd/dx) which cannot

decay into the excitations (k) of the system at rest, hence vc = min{(k)/k}. The idealgas is therefore not a real superfluid since excitation energies are k =

2k2/2m, so that

vc = 0 at k 0.

With repulsive weak interaction g

one has at k 0 k = k

NgV m

1/2,

and critical current is finite either

from this limit or from a roton

minimum (see figure) so that

vc = 0/k0. k0k

0

k

Black body radiationPhotons in thermal equilibrium correspond to bosons with = 0 since there is no conserva-

tion law for photons, hence N minimizes F, i.e. F/N = 0 = .

Photon spectrum is k = ck with |k| = k. Photons have two polarization states, = 1.

Z ={nk,}

e

n, knk, =k,

n=0

ekn =

k

2

1 ek

32


33/75

ln Z = 2k ln(1 ek ), same as the GCE L with = 0. (Note F = when = 0).nk = 1

(k)ln Z =

2

exp(k) 1n() is the number of photons with the frequency .

k

nk = V(2)3

4k2nkdk = 4V

(2)3

(

c)2d(

c)nk

0

n()d

u() = n() =3

exp() 1

2c3

This is the Plancks formula for energy density.

The total energy

E

V=

0

n()d =2k4B

153c3T4

cV T3 is unbounded as can con-tribute.R is the rate of radiation through a small hole in a cavity. We need to average only velocities

with vz > 0

vzvz>0 = c/2

0

cos d(cos )/

0

d(cos ) =c

4

R = EV

c4

= 2

k4

B603c2

T4 T4

where is Stefans constant (1879) = 5.670 108 Watt/(meter)2(Kelvin)4.We use periodic boundary quantization where n has integer entries,

k = c|k| = c2|n|V1/3

kV

= 13

kV

Therefore

P =1

Vln Z =

1

3Vkknk P V = 1

3E P = 4

3cT4

(for nonreletavistic bosons k V2/3 kV = 23 kV P V = 23 E).

Note that (P

V)T = 0 n2

If the photon had a mass m ,then at kBT mc2 Stefans constant would change by afactor 3/2, inconsistent with experiment photon mass = 0 (i.e. less than the experimentalkBT /c

2.)

Phonons

33


34/75

Consider N atoms in a lattice that have 3N vibration modes, labeled by phonons with

wavevector k and 3 polarizations. In the Debye model: = c|k|, < m

3N = k3 = 3V

(2)3 4k2dk

m

0

f()d, f () =32

22c3V

Integrating f() yields the cutoff frequency m = c(62n)1/3; the Debye temprature is TD :

kBTD = m.

Z =

ni

e(

i ini) =3Ni=1

1

1 ei

where i = {k, polarization}.

ni = 1

(i)ln Z =

1

exp(i)

1

E =

ln Z =

i

ini =m

0

f()

exp() 1 d =

E =

3N kBT(1 38 TDT + ...) T TD3N kBT

4

5

( TTD

)3 + O(exp(

TD

T

)) T

TD

Experiments on specific heat confirm that noninteracting phonons are normal modes of solids

at low T.

2c. Ideal Fermi gas

P V

kT= ln L =

p,spin

ln (1 + exp(p2/2m))

Where = exp () and

np = 11

ep2/2m + 1

so that N = p,spinnp.34


35/75

P

kBT=

4

h3g

0

dpp2 ln (1 + exp(p2/2m)) = g3

f5/2()

where g = 2s + 1 is the number of spin states

n = 43

g 0

dpp2 11

ep2/2m + 1)= g

3f3/2()

fn() =1

(n)

0

xn1

1

exp(x) + 1dx

=

l=1

()l+1 l

ln

0 < < (unlike bosons!)For n3 1 or 1:

n3/g = 2/23/2 + ..., [np 1g

n3 exp(p) is MB form]

P

nkBT=

g

n3( 2/25/2 + ...) = 1 + 1

g25/2n3 + ...

This is the virial expansion. Consider next n3 11

gn3 =

4

3[(ln )3/2 +

2

8(ln )1/2 + ...] + O(1/)

Where eF, so that by comparing T 0 terms

F =2

2m(

62n

g)2/3

To confirm the T = 0 result np 1

exp ((pF))1so that all the states

with p < F are occupied. Therefore

N = gV

h3

p


36/75

E =3

5NF[1 +

5

12(

kBT

EF)2 + ...] =

3

2P V.

The first term is g

|p|


37/75

For low density, each pair contributes independently:

Z = Z0[N(N1)

2 ( d3r1d3r2

V2eU(r1r2)

1) + 1]

0

11e

U(r)

r

F = F0 12

kBTN2

V2

(eU(r1r2) 1)d3r1d3r2

B(T) 12

(1 eU(r))d3r

F = F0 + kBTN2

VB(T)

P = F

V =N k

BT

V (1 +N

V B(T))

assuming convergence (U = 1r

).

B

>0 b00

... +


38/75

Evaluate now Z1, Z2: For Z1 there are no effects of interactions or of quantum statistics, i.e.

Z1 =

p

ep2/2m =

V

3 =

h2mkBT

Consider next Z2 = Tr eH for N = 2, first the classical problem:

Z2 =1

2!

d3pep

2/2m

2 1

h6

d3r1d

3r2eU(r1r2)

=V

26

d3reU(r)

From the expression above for B we obtain B(T) = 12

(eU(r1r2) 1)d3r1d3r2 as in the

previous derivation.

Consider next the quantum problem with the 2-particle Hamiltonian

H = 2

2m

2r1 + 2r2+ U(r1 r2) = 24m2R 2m 2r + U(r)where r = r1 r2, R = 12 (r1 + r2). The eigenvalues ofH are P

2

4m+ En where P is the center

of mass momentum, hence

Z2 =

P

eP2/4m

n

eEn =23/2

3V

n

eEn

Note that En contain information on statistics, i.e. En is restricted to eigenfunctions that

are symmetric in r for bosons, or to antisymmetric ones for fermions.

38


39/75

3. PHASE TRANSITIONS

Order Parameter Example Tc(K)

Liquid-gas density H2O 647

Ferromagnetic magnetization F e 1044

Antiferromagnetic sublattice magnetization F eF2 78

Bose condensation superfluid amplitude 4He 2

Superconductivity electron pair amplitude Y Ba2Ca3O7 90

Binary fluid concentration of one fluid CCl4 C7F14 302Binary alloy density of one kind on a sublattice Cu Zn 739Ferroelectric polarization Triglycine sulfate 322

Ferroelastic q = 0 distortion Ni T icharge density wave q= 0 distortion NbSe3 59

Metal-Insulator

Percolation fraction of sites in percolating cluster

Definition: n-th order phase transition has a discontinuity in the n-th derivative of a free

energy. Hence a 1st order transition has a jump in entropy (F/T) which is the latentheat. A 2nd order transition has a jump in Cv (contains

2F/T2).

3a. First order transitions

T

nnl

ng

mid point

(assymetry)

T

M

coexisting domains

Tc

Infinitesimal change in P (liquid-gas on left figures) or in H (magnetic field in a ferromagnet

39


40/75

on right figures) leads to a jump in density n (on left) or in magnetization (on right). In

P T plane (left) or H T plane (right) first order transition ends at a critical point Tc.

T

P

liquid

gas

solid critical point

triple point

T

H

Consider the liquid-gas phase transition as a prototype of a 1st order transition

Liquid-gas: 1st transition, no symmetry change.

Solid-liquid: 1st order transition, change in translation symmetry.

Fixing V below Tc leads to liquid/gas coexistence.

Fixing P leads to a jump in V from fully gas phase to fully liquid phase.

40


41/75

Clapeyrons equation

g(P, T) =1

NG(P , T , N )

where G(P , T , N ) is the Gibbs free energy.

g(P, T) = g2(P, T) g1(P, T)

where gi(T, P) is formally continued across Tc.

Define s = S/N, v = V /N

s = s2 s1; v = v2 v1g

T

P

= s;

g

P

T

= v

41


42/75

By the chain rule g

T

P

T

P

g

P

g

T

= 1

we get(g/T)P(g/P)T = PTg = sv

Along the transition line P(T) we get g = 0 =const.

dP

dT=

s

v=

latent heat

Tv.

Van-der Waals equation

A rough argument:

Vef f = V bP = Pkin a

V2

where b N is the excluded volume and a N2 is a measure of attractive forces amongthe molecules of the system; a N2 since there are N2/V2 pairs. Hence

PkinVeff = (V b)

P +a

V2

= N KBT

At T > Tc we get one solution for

the equation, and for T < Tc we get

three.

The three roots merge to one

root at inflection point Pc, Vc, Tc

The parameters a, b are sample specific. To identify them in term of Pc, Vc, and Tc rewritethe equation in the form

(V Vc)3 = 0 =

(V b)

Pc +a

V2

NkBTc

V2Pc

By identifying each power of V in this form we write a, b in terms of Pc, Vc, Tc, and with

P = P/Pc, T = T /Tc, V = V /Vc we obtain

P +

3

V2V 1

3=

8

3T

42


43/75

which is the law of corresponding states all substances have the same equation of state

when their P, V , T are measured in units of the critical point.

Note the region with PV

> 0 is unstable so that (N)2

0.Alternative derivation (S.K.Ma p.470)G0(T , P , N ) = E T S+ P V

where

E =3

2NKBT a1NN

V

attraction from neighboring N/V atoms, and

S = kB ln 1N!

(V b1N)N

3N

where b1N is the excluded volume and is the thermal wavelength.

Proper G(T , P , N ) is obtained at minimum with respect to V which is a redundant variable

for G.

G0(T , P , N ; V) =3

2NKBT a1 N

2

V kBTNln

V b1NN3 + N+ P V

If G0V

= 0, we get the Van der Waals equation with a = a1N2 and b = b1N .

G0 shows (see figure) that with 3 solutions 1 is metastable and 1 is unstable (max of G0)

For P1 < P < P2 gas is stable, liquid is metastable.

For P2 < P < P3 liquid is stable, gas is metastable.

43


44/75

P2 is when gas and liquid are degenerate G0(V1) = G0(V2).

Maxwells construction

In general G0 = F(V , T , N ) + P V where P(V) = F/V. At P2:

0 = G0(V2) G0(V1) =V2

V1

[P(V) + P]dV .

Hence the area between P = P2 line and the P(V) curve vanishes this is Maxwells

construction.

An alternative derivation: In terms of F(V) choose V1 and V2 such that parallel tangents

generate one line with FV1

= FV2

.

The line isF2 F1V2 V1 =

F

V1

P2(V2 V1) = (F2 F1) =V2

V1

P(V)dV

which is the same Maxwells construction. The line is

realized by coexistence of the two phases with fractions

x and 1 x, respectively (0 < x < 1),

V = xV1 + (1 x)V2

Fline = xF1+(1x)F2 = (V V2)F1 + (V1 V)F2V1 V2 < F

Phase separation at P2 since V1 < V < V2 has lowerFline then F on the continuous curve.

When V is changed carefully to avoid strong fluctua-tion, metastable phases can persist until they become

unstable; this leads to hysteresis.

44


45/75

3b. Second order phase transitions Mean Field theory

Consider ferromagnetism as a prototype second order phase transition. Localized indepen-

dent moments

in a magnetic field B lead to magnetization

M = NeB eB

eB + eB= N tanh(B) .

Mean field theory for interactions: neighbors with magnetization M/V induce an addi-tional field, B B + M

V, interaction strength.

= M = N tanh B + MV

has M = 0 solutions even if B = 0.

If T = Tc, 1 =NV

2/kBTc

Expand near Tc:

M =TcT

M 13

M

V

3N = M (Tc T)1/2 .

Susceptibility at T > Tc : B 0, M 0

M = N

B +

M

V

=

M

B

B=0

=N 2

1 N 2/V 1

T Tc .

Microscopic Model

Consider two neighbors of spin 1/2, total spin is S = 0, 1 with energy difference which defines

the exchange energy J. The requirement for an antisymmetric wavefunction requires a

symmetric orbital for the singlet S = 0 and antisymmetric one for the triplet S = 1, hence

a large energy difference from the difference Coulomb interactions. Hence J is much larger

45


46/75

then that of interacting magnetic dipoles. Consider

si sj = 12

(si + sj)2 s2i s2j

= 1

2S(S+ 1) 1

2 3

2=

3/4 S = 01/4 S = 1

.

= Interaction energy = Jn.n. sisj +const (n.n. is summation on nearest neighbors). means that each pair is summed once.If the crystal has an easy axis for the spin, si (sz)i, hence two models:

H = J

i,j

si sj s spin operator: Heisenberg model

H = J

i,jij i = 1: Ising model.

Solution of the Ising model by mean field:

HMF = 12

J

i

i no. of nearest neighbors.

12

is needed so that

HMF

= 12

N J2 , 12

N no. of bonds. (e.g. N/2 odd sites each

generates bonds.)

= i = i=ie Ji/2

i=e Ji/2

{j=i} eJ

j=i

j /2

{j=i}

eJ

j=i

j /2= tanh

J 1

2

= kTc = 12 J.

3c. Exact results

Mapping between systems

1. Lattice gas model to Ising

Na atoms occupy sites in lattice with N cells; Naa number of nearest neighbors, each with

energy 0, g(Na, Naa) is the number of configurations with a given Na, Naa (a non-trivial

function). Grand partition

LG =

NaNa

Naag (Na, Naa) e

0Naa.

46


47/75

Canonical partition of Ising has the same form:

N+ no. of + spins, N++ no. of ++ neighbors. Draw one line from all + sites to all their

neighbors:

Number of lines = N+ = 2N++ + N+

same for N N = 2N + N+, N+ + N = N

=i,j

ij = N++ + N N+ = 4N++ 2N+ + 12 Ni

i = N+ N = 2N+ N

EIsing = 4JN++ + 2(J B)N+ 12 J BNZ = e

N

12

JB

N+

e2(JB)N+N++

g (N+, N++) e4J N++

Since configuration counting is the same as in lattice gas we get the same function g(Na, Naa),

hence the two problems are equivalent as in the following table:

Ising Lattice gas

N+ Na

4J 0

e2(JB)

1N

FIsing +12

J B PM = N+ N, N+N = 12

MN

+ 1

1v

= NaN

order parameter

2. Binary alloy to lattice gas

N11, N22, N12 no. of pairs of each type

EA = 1N11 + 2N22 + 12N12

As above

N1 = 2N11 + N12

N2 = 2N22 + N12

=

N12 = N1 2N11N22 =

12

N + N11

N1

47


48/75

N1 + N2 = N

EA = (1 + 2 212) N11 +

(12 2) N1 + 12 2N

, Z(N1, T) =N11

g (N11) eEA

Lattice gas Binary alloyNa N1

0 1 + 2 212F F + (12 2) N1 + 12 2N

Ising Model in 1D: Exact Solution

Z =1=

2=

...

N=

e

JN

k=1

kk+1 +12

B k

(k + k+1)

; 1 = N+1

Consider the bond (k, k + 1) with the elements

e J kk+1+12B(k+k+1)

This element can have 4 values for k =

and k+1 =

. Call these elements

P1,1 , P1,1 , P1.1 , P1,1 and define a matrix

P =

P1,1 P1,1P1,1 P1,1

= e(J+B) eJ

eJ e(JB)

Note that P is symmetric by the choice of k + k+1 for the B term in the Hamiltonian.

P is defined in a spinor space |, e.g. +|P| = P1,1.E.g. for N = 3 the partition Z has 23 terms, one of them is

= P1,1P1,1P1,1 = +|P|++|P||P|+

In general we need

Z =

all |i=|

1|P|2 . . . k|P|k + 1k + 1|P|k + 2 . . . N|P|N + 1

=

1=N=(PN)1,N = T r(P

N) = N1 + N2

48


49/75

1, 2 are the eigenvalues of P: 2 2eJ cosh(B) + 2 sinh (2J) = 0

= 1,2 = eJ cosh (B ) [e2J + e2J sinh2 (B)] 12

2 < 1, N , 1 dominates, hence1

N ln Z ln 1 and is an analytic function (exceptat T = 0). For the magnetization we have

M = 1

B

ln Z = N sinh (B)[e4J+sinh2 (J)]

12

= M(B = 0, T > 0) = 0, M(T = 0) = N sign(B), hence an ordered phase is only atT = 0. The susceptibilty:

= MB

|B=0 = N2kB Te2J/kB T|T0 indicates ordering at T = 0.

No phase tradition in 1-dimension general argument:

Low energy excitations are walls |, energy is 2J (relative to the ground state), whileentropy is S = kB ln N since the wall can be positioned at any bond.

= F = E T S = 2J kBT ln N < 0 no long range order in 1-dimension at any finitetemperature.

2D Ising model: Onsagers solution (1944)

KTc = 2.269J

M (Tc T)1/8 near TcCv ln(1 TTc )

3d. Landaus Theory for second order phase transitions

Partition sum is done first locally to define a coarse grained order parameter, slowly

varying. F{M(r)} is determined by symmetry for small M(r), i.e near a 2nd order phasetransition. Coarse graining involves a finite cluster F{M(r)} has an analytic expansion.For ferromagnet, M M, symmetry+analiticity

F{M(r)} = 12

a(T)M2(T) +1

4bM4(T) +

1

2c(M)2

where a(T) = a(T Tc) so that in Mean Field the transition occur at Tc.

49


50/75

Mean Field: M(r) = M is homogenous, no fluctuations.

FM

= 0 M =

a

b(Tc T)

F = 14 a2

b (Tc T)2

T < Tc0 T > Tc

S = FT

continuous (S < 0 at T < Tc, needs correction by fluctuations, see below.)

cV = TST

has a jump, and is (Tc T)0 at T < Tc.

Consider next B = 0 and add M B to F:

T > Tc, a(T Tc)M B = 0 = MB |0 1TTc

T = Tc, bM3 B = 0 M B1/3

Consider fluctuations at T > Tc: neglect M4, M(r) = V1/2

k Mke

ikr

F{M(r)}d3r = 12 k (a + ck2)|Mk|2 since eikr+ikrd3r = V k,k. Note that Mk = Mkfor a real M(r).

The weight ofk modes is e(a+ck2)|Mk|2. The correlation function is then

< M(r)M(0) >= V1

k < MkMk > eikr = V1

k e

ikr

k

d(ReMk)d(ImMk)|Mk|

2e(a+ck2)|Mk|

2

d(ReMk)d(ImMk)e

(a+ck2)|Mk|2

= V1

keikr

d(|Mk|2)|Mk|2e(a+ck2)|Mk|2

d(|Mk|2)e(a+ck2)|Mk|2

= V1

keikr

((a + ck2))ln

0

dxe(a+ck2)x

< M(r)M(0) >= V1

k

eikr

(a + ck2)=

kBT

4rcer/

where =

ca

(T Tc) 12 is the correlation length. Consider next the free energy

Z =

k

d|Mk|2e(a+ck2)|Mk|2

k

[(a + ck2)]1

50


51/75

F = kBT[

k

ln (a + ck2) + const.]

Therefore

CV = T2F

T2

1/a0 ddk

(a + ck2)2+ less singular terms.

a0 is a lattice constant, d=dimensions. Using k = k

CV 1a2

a0 ddk

(1 + k2)2d 4d (T Tc) d42

Validity of mean field:M2(0) Tc (similar result at T < Tc) :

M2(0) ddk(a+ck2)

kd1dk(a+ck2)

(a)d2 |T Tc|(d2)/2

Comparing with M2 (Tc T) shows that mean field is valid at T Tc if d > 4.At d < 4 mean field breaks down at T Tc

Critical exponents:

t = TTcTc

, T > Tc; for exponents t = TcTTc

, T < Tc

Exponent Definition Mean Field Fe

, Specific heat cV t 2 d/2 0.12 Order parameter t 1/2 0.34

, Susceptibility x

t 1 1.33

Order parameters at Tc B1/ 3 4.2 (Ni), correlation length t 1/2

at Tc M(r)M(0) 1/rd2+ 0 0.07

Universality: exponents depend only on symmetry of order parameter and dimensionality.

51


52/75

Scaling assumption:

is the single length responsible for singular behavior.

obtain relations between exponents.

e.g. M(r)M(0) 1rd2+

g(r/) at Tc need g g(0) = 0In terms of

M

, integration on all configurations M(r), and the Hamiltonian H we have forthe susceptibility

= MB

|B0 = B

M M(0)e(H

M(r)Bddr)M e(H

M(r)Bddr)

|B=0 =

M

M(0)M(r)ddreH

M eH=M(0)M(r) ddr

since at T > Tc the MB=0 = 0 term vanishes (from /B of denominator.)

rd+2g(r/)ddr d+2 d (t)2 = (2 )

Renormalization Group:

Consider the 1-dimensional Ising model Z = k= eJk kk+1Sum on all even sites to obtain a new form for Z

2e(J12 + J23) = e(a

+ J13)

Identify a, J by two cases:13 = + case: e

2J + e2J = e(a+J)

13 = case: 2 = e (aJ)

Z(k) odd i i=

e

J odd k kk+2with the renormalized coupling J given by e 2J

= 2e 2J+e 2J

At Tc H is scale invariant, i.e. J = J Tc = 0.

52


53/75

4. NON- EQUILIBRIUM

4a. Kinetic theory and Boltzmanns equation:

f(r, v, t) is particle density in phase space d3rd3v. In absence of collisions

r, v r + vt , v + Fm

t

so that volume A moves to an infinitesimally close volume B with f changed only by collisions,

f( r + vt, v +F

mt , t + t) f( r, v, t) ( f

t)collt

Effect of collisions: If one initial particle is in A the result is certainly outside A (up to the

differential volume of A) this is a loss term; conversely with one final particle in B,

Rtd3

rd3

v = no. of collisions with one initial particle in ARtd3rd3v = no. of collisions with one final particle in B [= A + O(t)]

( ft

)collt = (R R)t

Assume binary collisions (dilute gas) v1 , v2 v1 , v2

v1 + v2 = v1 + v2

v12 + v2

2 = v1 2 + v2

2

Define V = 12

( v1 + v2) , u = v2 v1 ; V = 12 ( v1 + v2) , u = v2 v1

V = V , |u| = |u|A collision is defined by an angle in the center of mass, i.e. between u, u. V , u, are

independent parameters which determine V, u.

Consider u u + du , as is fixed (i.e. d is independent ofd3u).For u u + du the triangle (u, u + du) is similar to (u, u + du), hence |du| = |du|Choosing an orthogonal basis set du

1, du

2, du

3it transforms to a rotated orthogonal basis

with axes equal in magnitude, hence d3u = d3u

Since d3V = d3V d3v1d3v2 = d3v1d3v2.

Assumption of molecular chaos: This is the central assumption of the Boltzman equa-

tion.

The number of pairs with velocities ( v1, v2) about to collide is the product

53


54/75

f(r, v1, t)d3rd3v1 f(r, v2, t)d3rd3v2

i.e. both particles do not know about each other until they collide, hence no correlation

before the collision.The scattered flux of v2 hitting v1 is

I =

d3v2

d() |v1 v2| f(r, v2, t)

where () is the scattering cross section.

f(r, v1, t)d3v1 is the number of target particles, therefore Rd

3v1 = f(r, v1, t)d3v1 I

For R define a collision by variables v1,v2 v1, v2 so that v1 A.

The scattered flux ofv

2 onv

1 is

I =

d3v2

d()

v1 v2 f(r, v2, t) = d3v2 d() |v1 v2| f(r, v2, t)using microscopic time reversal () = () (spin and internal quantum states are ne-

glected).

f(r, v1, t)d3v1 is the number of target particles (no integral since ( v1,

v2, ) determinev1).

Therefore, Rd3v1 = f(r, v1, t)d3v1 I.

Since d3v1d3v2 = d3v1d3v2 the probability for the reverse collision process is

R =

d3v2

d() |v1 v2| f(r, v1, t)f(r, v2, t)

Hence Boltzmans transport equation:

(

t+ v1 r + 1

mF v1)f(r, v1, t) =

()dd3v2 |v1 v2| [f(r, v2, t)f(r, v1, t) f(r, v2, t)f(r, v1, t)]

Time reversal would imply that f(r, v1, t) also solves this equation. However, changingt t, and all v v in this equation, the left side changes sign (acting on f(r, v1, t))while the right hand side does not change sign (acting on f(r, v, t) with the various v).Hence time reversal, although valid on the microscopic level, is broken on the macrospcopic

54


55/75

level. [If t t and also vi vi the equation is invariant, i. e. reversing the order ofcolliding particles; assumption of molecular chaos looses this symmetry.]

Consider the case with no external force F = 0 and no r dependence (e.g. no density waves).In equilibrium f

t= 0, therefore the right side is equal to 0 for all v1, which requires

f(v2)f(v1) = f(v2)f(v

1)

hence ln f(v) is conserved in all collisions. The conserved quantities are momentum and

energy, therefore

ln f(v) = A + B

v + c

(v)2 =

b(v

v0)

2 + ln a

f(v) = aeb(v v0)2

which is the Maxwell distribution (up to a center of mass velocity).

Boltzmans H theorem

As a candidate for entropy, Boltzman defined the function H(t):

H(t) f(v, t)ln[f(v, t)]d3v

where f(v, t) is a solution of Boltzmanns equation.

We show now that dH(t)dt

0 so that H(t) decreases with time. Define as a shorthand

f1 = f( v1, t) f2 = f( v2, t)

f1 = f(v1, t) f

2 = f(

v2, t)

(2)

Thus we get: f1t

=

d3v2

d() |v1 v2| (f2f1 f2f1)

dH

dt=

d3v2

d3v1

d() |v1 v2| (f2f1 f2f1)(1 + ln f1)

by changing v1 v2, using eq. for f2t and taking 12 of both, we get

dH

dt=

1

2 d3v1

d3v2 d() |v2 v1| (f2f1 f2f1)[2 + ln(f1 f2)]

55


56/75

Now, using equations forf1t

,f2t

, i.e. fi fi we get

dH

dt=

1

2

d3v1

d3v2

d()

v2 v1

(f2f1 f2f1)[2 + ln(f1 f2)]

And finally, adding the last two equations

dH

dt=

1

4

d3v1

d3v2

d() |v2 v1| (f2f1 f2f1)[ln(f1 f2) ln(f1 f2)]

Since (x y) ln xy

> 0 for all x, y we have

dH

dt 0 (3)

In equilibrium f2f1 = f2f1 and

dHdt

= 0.

Therefore, a candidate for entropy is (H) which increases with time and is maximal inequilibrium. [Note that f(r,v,t) is coarse grained by the assumption of molecular chaos,

therefore initial velocities vi and final velocities vi are distinguished].

4b. Brownian motion

Brown (1828) - the random motion of pollen particles in a solution.

Einstein (1905) - random walk problem.

Consider one dimension: Pn(m) is the probability of arriving at coordinate m after n steps,

i.e. one needs 12 (n + m) steps to the right and12 (n m) steps to the left,

Pn(m) = 12n n![ 12

(n + m)]![ 12

(n m)]!where 12 is the equal probability of going right or left. Normalization:

nm=n Pn(m) = 1

Averages: m = 0, m2 = n. For m n (since

m2 =

n n)) and using Stirlings limitwe get:

Pn(m) 22n

em2/2n

Defining x = ml (l is a step size in space) and t = n ( is the time for each step)), we get

56


57/75

x2 =l2

t = 2Dt

so that D = l2

2is the diffusion coefficient.

In this continuoum formP(x) =

14Dt

ex2/4Dt

In general, D is defined by the linear response:

j(r, t) = D n(r, t)

If there is a gradient, then there is a current which responds to reduce this gradient.

The continuity equation is j + t

n(r, t) = 0. Taking a gradient yields

2n(r, t) 1D

n(r, t)t

= 0

This is the diffusion equation. The solution with an initial value n(r, t = 0) = N 3(r) is

n(r, t) =N

(4Dt)3/2er

2/4Dt

with

0n(r, t)4r2dr = N. This solution shows < r(t) >= 0 and

< (r(t))2

>=

1

N4 0 n(r, t)r4dr = 6DtLangevins equation

A particle of mass M is immersed in a medium which produces a random fluctuating force

MA(t). A(t) has time correlation

A(t + ) A(t)

which decays fast with .

The correlation time col is the time between collisions of medium particles and the particle

of mass M. The medium leads also to irreversibility, i.e. friction.

.v= v + A(t)

1/ is the time scale for approaching equilibrium (see below).

We assume col


58/75

r v = 12

d

dtr2

r.

v=1

2

d2

dt2r2 v2 = r (v + A)

average : d2

dt2

r2

+ d

dt

r2

= 2

v2

= 6kT/M

r2 = 6kTM

t + c1 + c2et

Initial value r(t = 0) = 0 ddt

r2 = 2r v = 0 at t = 0.

r2 = 6kTM

[t 1

(1 et)]

t > 1/

r2 6kT

Mt diffusive

Friction is related to diffusion D, D = kTM

. is also related to mobility - responce to

constant force , e.g electric field E for particle with charge e.

.

v= v + A(t) + eE

M

In steady state.v= 0 v = eE/M E

D = kT/e Einsteins relation

So far the random force A was not explicit. Consider next the velocity fluctuations. We

assume now E = 0; if E= 0 one needs just to shift v v + e E/M.

v(t) = v(0)et + et t

0

eu A(u)du

v(t) = v(0)etv2(t)

= v2(0)e2t + e2t

t0

t0

e(u1+u2) A(u1)A(u2)du1du2 = v2(0)e2t + C2

(1e2t )

t should have equilibrium 12

Mv2 = 32

kT C = 6kT/M.Strength of fluctuating force , both fluctuations and friction are due to the medium.

58


59/75

v2(t)

= v2(0) + (

3kBT

M v2(0))(1 e2t )

General solution for

r2

(v2(0)

= 3kBT /M)

:

d

dt2

r2

+ d

dt

r2

= 2

v2

r2 = 12

v2(0)(1 et )2 3kBTM2

(1 et)(3 et ) + 6kBTM

t =

= v2(0)t2 + O(t3) t > 1/

Velocity correlations: Kv() v(t)v(t + )= v2(0)e(2t+) + e(2t+)

t0

t+0

e(u+u)C(u u)dudu =

=

v2(0)e(2t+) + C2

e(2t+)(e2t 1) > 0

v2(0)e(2t+) + C2e(2t+)(e2(t+) 1) < 0

Kv() = v2(0)e|| + ( 3kTM

v2(0))(e|| e(2t+))

=3kBT

Me|| at long times t + , t >> 1/ .

Note that replacing v2(0) by its equilibrium average yields Kv() = (3kBT /M)e|| at all

times t.

Reaching equilibrium at large times Kv() becomes t independent

stationary medium.

Note that for correlations of just one component, e.g. vx, we have

Kvx () = vx(0)vx() = kB TM e||.Below the Fourier transform is needed:

vx () =

Kvx ()eid =

2kBT

M

2 + 2

59


60/75

4c. Fluctuation Dissipation Theorems (FDT)

General properties of averages of stationary medium

Consider a variable x(t) such that Kx() =

x(t)

x(t + )

is independent of the time t.

Kx(0) > 0; [x(t) x(t + )]2 = 2[Kx(0) Kx()] 0

it follows that |Kx()| Kx(0) is a decreasing function, and usually Kx() 0. AlsoKx() = x(t )x(t) = Kx(), with shift t t .Consider now the Fourier transform x() :

x(t) =

x()eit d

2

For a precise derivation one needs x(t) to be finite only in the interval [T2 , T2 ]. The spacing between distinct values with cos( T

2) = 0 or sin( T

2) = 0 is = 2

T. (See Wannier

p. 481). For T considerKx() =

d

deiti(t+) x()x() /(2)2

Since this is t independent, perform the integral

... dtT

Kx() =

d

d 2

T( +

)ei

x()x()(2)2

=

d|x()|2ei/(2)2

This is the Wiener-Khinchin theorem:

x() =x(t)x(t + ) eid =

2

|x()|2|x()|2 is called the intensity, or the power spectrum of x.As an example, consider Langevins equation:

(i + )v() = A() |v()|2

=

|A()|2

(2+2)

from the theorem above v() =A()2+2

which is consistent with previous result: v() =

2kBTM

2+2

and A() =2kBT

M, the latter corresponds to white noise.

Note also that ix() = v(), hence

|x()|2 = |v()|2

2 x() = 2kBT

M

(2 + 2)2

60


61/75

Dissipation and Response functions

An external force adds the Hamiltonian term F(t)x. The Hamiltonian becomes:

H = p2

2m+ V(x; env) F(t)x

where env stands for the coordinates and momenta of the environment. For an explicit

system + environment see the Appendix.

The equations of motion are:

x = Hp

= pm

, p = Hx

= Vx

+ F = mx

hence F(t) is indeed an external force.

The rate at which the systems energy changes is

dHdt

=

=0 H

pp +

H

xx +... + H

t= x dF

dt

where the sum of the first two terms on the right hand side vanishes by Hamiltons equations

for x, p; the ... terms are similar derivatives with respect to the coordinates and momenta

of the environment which also vanish due to their corresponding Hamiltons equations.

The dissipation rate is the rate of energy absorption from the external source, averaged

on time. This includes an average on the enviorment x(t) (e.g. on the random force inLangevins equation) and on the explicit time dependence in F(t), hence

dE

dt= xdF

dt

The dissipation rate can be expressed in terms of a response function x() where the linear

response at small forces F is defined by

x() = x()F()

It is more compact to consider a single frequency F(t) = 12

f0eit + 1

2f0 e

it (although a

Fourier sum can be added)

x(t) = 12

x()f0eit + 1

2x()f0e

it with x() = x()

The dissipation is then:

61


62/75

dE

dt=

x

F

t

=

i

4[x()f0eit + x()f

0 e

it][f0eit f0 eit] = i

4|f0|2[x()x()]

dEdt

= 12

|f0|2Imx()

Consider now the specific case of Langevins equation, where the environment leads to

friction and random force terms. Averaging on the random force,

M(2 i) x() = F(), hence the response has the same frequency as the source andthe response function is

x() = x()F()

=1

M(2 + i)

The dissipation rate is therefore proportional to Imx() =

M(2+2)2

Comparing with x() we get the Fluctuation - Dissipation theorem:

x() =2kB T

Imx()

Note: same result holds for v(), v() with an external source term in the Hamiltonian

F(t)p/M (exercise).

In conclusion, x() are fluctuations in absence(!) of external force.x() linear response and energy dissipation due to external force.

Applications

1. Kapplers experiment (1931) : fluctuations in angle of a mirror suspended on a fine wire

with a restoring force

12

C2

= 1

2kBT determine kB. Note : Variance is independent of

gas density. Equilibrium is achieved after time col, 1/.

62


63/75

Low density: rare strong deflections High density: frequent weak deflections

2. A square vane of area 1cm2, painted white

on one side and black on the other, is at-

tached to a vertical axis and can rotate freely

about it as shown. Assume that the vane can

support a temperature difference between its

black and white sides. Suppose the arrange-

ments is placed in He gas at room tempera-

ture and sunlight is allowed to shine on the

vane. Explain qualitatively why : (a) At ex-

tremely small densities the vane rotates. (b)

At some intermediate (very low) density the

vane rotates in a sense opposite to that in (a);

estimate this intermediate density and the cor-

responding pressure. (c) At a higher (but still

low) density the vane stops.

Answers :

(a) Radiation pressure on white > black.

(b) Black heats up and nearby hotter atoms exert higher pressure. Gas is in local equilibrium

with density n 1cm3.(c) Viscosity is effective at col


64/75

3. Electrical circuit:

L dIdt

=

RI+ V(t)

V(t) - voltage fluctuations

Langevin analogy: dvxdt

= vx + Ax (t) => R/L

Correlation of A is determined by 12

Mv2x = 12 kBTML

= 12

L I2

vx(w) =2kB T

M

2+2

I () =

2kB TL

R/L

2+(R/L)2


65/75

Fluctuation Dissipation Theorem (FDT): quantum version

Fluctuations

Consider a system in equilibrium, no external force. For x (t) (Heisenberg representation)

define a symmetrized correlation

K() =1

2x (t) x (t + ) + x (t + ) x (t) .

where a subscript x is omitted here, for convenience. The symmetrized correlation involves

a hermitian operator and has an obvious classical limit. note, however, that there are FDTs

for non-symmetrized correlations.

Note that K() = K(

) and K() is real. The average is defined by

... =

n

n| ... |n eEn/Z.

where Z =

n eEn is the partition sum.

Introduce a unit operatorm

|m m| = 1 and define (time independent) matrix elementsxmn by

n| x (t) |m = ei(EnEm)t/xnm.

Therefore

K() = 12Z

n,m e

i(EnEm)t/xnmxmnei(EmEn)(t+)/eEn + (t t + ) =

= 12Z

n,m e

inm|xnm|2eEn + c.c. where nm = EnEm () =

K() eid =

Z

n,m

eEn |xnm|2 [( nm) + ( + nm)]

Interchange n m in the first ( nm),

() = Z n,m eEm + eEn |xnm|

2( + nm) =

= Z

n,m

eEn

e(EmEn) + 1 |xnm|2( + nm) =

= Z

1 + e

n,m

eEn |xnm|2( + nm)

Dissipation

Add coupling to an external force, F(t):

H = H0 + V(t) where V(t) =

F(t)x .

65


66/75

We use here Shrdingers representation to avoid the form HH(t) = H(t). The responsefunction () gives x(t) in terms of F(t ) at all previous times, i.e. > 0:

x(t)

=

0

()F(t

)d.

Assume, without loss of generality,

F(t) =1

2f0e

it +1

2f0 e

it.

Therefore

x(t) = 12

0

()

f0e

i(t) + f0 ei(t)

d.

The Fourier transform is

() =

0

()eid (() = 0 for < 0),

therefore

x(t) = 12

()f0eit +

1

2()f0 eit,

since () =

() [() is real]. In terms of the density matrix (t) corresponding to thefull Hamiltonian H, the dissipation rate is

dE

dt= Tr{ d

dt(H)} = i

Tr{[, H]H} + Tr{H

t} (4)

= Tr{x Ft

} = xFt

(5)

where the equation of motion of is used, as well as the cyclic property of the trace. The

result is formally the same as in the classical case, except that x is a quantum expectationvalue.

Using the response function and averaging on time,

dE

dt=

1

2(()f0eit + ()f0 eit)

i

2(f0eit f0 eit),

dEdt

=i

4|f0|2[() + ()] =

2|f0|2Im[()].

66


67/75

Therefore Im[()] measures the dissipation rate. Since this is positive

Im[()] > 0 for > 0.

Golden Rule for the Dissipation

The rate of transition between states at time T is

Wnm =1

2

m|V|ndt2 1T ==

1

2

12

dt

ei(EmEn)t/xmn(f0eit + f0 e

it)2 1T

We note that

1

T

ei(mn)tdt2 = 1T T /2T /2 ei(mn)tdtT /2T /2

ei(mn)t

dt =

=1

T

T /2T /2

ei(mn)t[2(mn )]dt = 2(mn ).

Wnm = 22

|f0|2|xmn|2[(nm ) + (nm + )].

The energy dissipation rate can now be calculated

Q =1

Z

n,m

(Em En)WnmeEn =

=

22|f0|2 1

Z

n,m

eEn|xnm|2[( + nm) ( nm)] =

=

22|f0|2 1

Zn,m(eEn eEm)|xnm|2( + nm).

Noting that

eEn eEm = eEn 1 e(EmEn) = eEn 1 emn ,and using the function with mn = , we get

Q =

2|f0|2(1 e) 1

Zn,meEn|xnm|2(nm + )

67


68/75

Im[()] =

(1 e)n,m

1

ZeEn |xnm|2(nm + ).

() = coth 12 Im[()]The coth

12

can be interpreted as the mean fluctuation of an oscillator coordinate:

x2osc

=

2m

(a + a)2

2e 1 + 1 = coth

1

2

.

Note that in the limit 0 (i.e. for kBT >> ) the quantum FDT becomes

()

2kBT

Im[()].

which is our previous classical FDT.

4d. Onsagers Relations

Onsagers relations (1931) consider the response to deviations from equilibrium. The re-

sponses are measured by a flow processes xi in coordinates xi (e.g heat flow, electric current,

mass transfer, etc.), while the deviation from equilibrium corresponds to forces fi, e.g

gradients in temperature, potential, pressure, etc. The forces fi can be external ones, or

they can form spontaneously as a fluctuation.

If the system is close to equilibrium, we assume a linear relation between the forces and the

responding currents:

xi = ijfj ,

where summation convention is used (i.e. repeated indices are summed). ij are known as

kinetic coefficients. [In general, however, the relation is non-local in time and can be written

as xi(t) t

ij(t t)fj(t)dt.]We consider xi as coordinates of the entropy S so that the forces are fi = S/xi. For small

deviations from equilibrium values xi we expand near the entropy maximum

S = S(xi) 12

ij(xi xi)(xj xj)

fi = Sxi

= ij(xj xj)

68


69/75

xi = ijjk (xk xk) .

Consider

xifj = kB

xi

xjeij (xixi)(xjxj )/2kB

eij (xixi)(xjxj )/2kB

where the limits are allowed due to fast convergence.By partial integration xifj = kBij

xixj = ik fkxj = kBij.

This is a type of FDT the fluctuations xixj are related to the dissipation ij.Consider xi()xj(0) with microscopic time reversal (dissipation ij and irreversibilityarise after average on the environment, or on the ensemble.)

xi()xj (0) = xi()xj(0) = xi(0)xj()

where in the second form both times are shifted by , allowed in equilibrium.

|0 xi(0)xj(0) = xi(0)xj(0) ij = ji

These are Onsagers relations.

DefineF = 12 ijfifj xi =

F

fi

S =S

xixi = fixi = fi

F

fi= 2F

S > 0, hence F > 0 for all choices of fi, therefore the matrix ij is positive definite.

A few more properties of ij:

In presence of a magnetic field B time reversal involves B B, hence ij(B) = ji(B);

e.g. for the Hall conductance xy(B) = yx(B).In presence of angular velocity , similarly, ij() = ji().Furthermore, if for some coordinate xi xi, xj +xj under time-reversal, thenij = ji

69


70/75

Alternative derivation:

Consider fi as external forces with a Legendre transformed entropy S S = S (xi xi)fi.Since fi =

Sxi

dS = (xi xi)dfi, i.e. S = S(fi).Weight of a configuration is now

eS/kB = eS/kB(xixi)fi/kB

here S = S{xi} has all orders of xi.xif=0 = 0, i.e. flow is only in response to fj = 0.

xi =

xieS/kB(xjxj )fj /kB

eS/kB(xjxj )fj /kB=

xie

S/kB [1 (xj xj)fj/kB]eS/kB [1 (xj xj)fj/kB] +

= xixj0 fj/kB +O(f2)

Here linear response is explicit

ij =

xixj

0 /kB as above.

xixj0 is evaluated with fi = 0 and microscopic time reversal can be used as above ij = ji

Application: Bi-metal junctions

Using E1 + E2 =const, and Q1 + Q2 =const we have

dE12 = dE1 = dE2, dQ12 = dQ1 = dQ2

and since dE = T dS+ V dQ we obtain

dS =dE1T1

+dE2T2

V1T1

dQ1 V2T2

dQ2 = dE12 x1

1

T1+

1

T2

f1

+ dQ12 x2

V1T1

V2T2

f2

E12 = 11

1

T1+

1

T2

+ 12

V1T1

V2T2

Q12 = 21

1

T1+

1

T2+ 22

V1T1

V2T2

70


71/75

Seebeck coefficient is defined by V = T in an open circuit, i.e. Q12 = 0:

V1 V2 = 2122

T2 T1T

= 2122T

(to lowest order in gradients)

Peltier coefficient , corresponds to T1 = T2 and is defined by

E = Q = 1222

12 = 21 = T This is Kelvins relation.

71


72/75

Appendix: Langevins equation from a Hamiltonian

This appendix considers a microscopic model for an environment that produces the dissipa-

tion and the random forces (t) in Langevins equation

mq+ q+V

q= (t) .

The derivation is based on Caldeira and Legget, Ann. Phys. 149, 374(1983) and provides a

basis for Brownian motion.

Consider a linear coupling of the particle coordinate q to a set of Harmonic oscillators, with

coordinates Qi and momenta Pi; the linearity is valid when the particles trajectories are

sufficiently confined.

H = P2

2m+ V(q) +

Ni=1

iqQi Coupling of oscillators to a heat bath

+i

P2i2Mi

+ 12

Mi2i Q

2i

Baths Energy

+V(q)

Where V(q) is chosen to retain the original minimum ofV(q). Minimizing H with respect

to Qi gives Qi =i

Mi2iq, hence q dependent potential terms appear

i

2iMii

q2(1 + 12

) which

we wish to cancel by

V(q) =

i2i

2Mi2iq2 .

The equations of motion for Qi are

Qi + 2i Q =

iMi

q(t)et

where retarded response is defined by 0+, i.e. at t the coupling to the environ-ment vanishes. The solution is

Qi(t) = Q0i (t) +

iMi

dtq(t)

d

2

ei(tt)+t

2 2i

where Q

0

i is the solution of

Qi +

2

i Q = 0. Let us shift i and rewriteQi(t) = Q

0i (t) +

iMi

dtq(t)

d

2

ei(tt)

( i i) ( + i i) et

To solve this integral we need to take the upper contour for t > t (see figure, so that ei(tt)

vanishes on the upper half circle). The closed contour integration has two poles (see figure)

and therefore

Qi(t) = Q0i (t) +

iMi

t

dtq(t)i

eii(tt

)

2i e

ii(tt)

2i

et

72


73/75

Hence,

Qi(t) = Q0i (t)

iMii

t

dtq(t)sini(t t)et

= Q0i (t) i

Mi2iq(t) +

iMi2i

t

dtq(t)cosi(t t) (*)

where partial integration is done, and at t we use et 0; in the final form we set

= 0.Equation of motion for q(t):

mq+V

q+

V

q=

i

iQi

=

iQ0i +

i

2iMi2i

q(t)

i

2iMi2i

t

dtq(t)cosi(t t)

Note the cancelation of Vq

. Define the spectral density as

J() = 2

i

2iMii

( i)

so that

mq+V

q+

2

0

dJ()

tdtq(t)cos (t t) =

i

iQ0i .

To obtain linear dissipation, as in Langevins equation, we assume an ohmic bath

J() = . Using

0

d cos (t t) = 12 Re

ei(tt)d = (t t)

we get

mq+ q+V

q=

i

iQ0i

We identify now the random force (t) = i iQ0i (t). Free oscillators evolve asQ0i (t) =

1

iQ0i (0)sinit + Q

0i (0)cosit

73


74/75

With random initial conditions we have Q0i (0) = Q0i (0) = 0 so that Q0i (t) = 0.For a thermal distribution (classical oscillators)

Q0i (t)Q

0i (0) = (Q

0i (0))

2

cosit =kBT

Mi2

i

cosit

(t)(0) = kBT

i

2iMi2i

cosit =2

kBT

0

dJ()

cos t .

With J() = we obtain

(t)(0) = 2kBT (t) ()

i.e. the noise correlation relates to the dissipation .

To check FDT, we need to evaluate the response of (t) to a source external to the oscillatorsqex(t); this source may or may not be the particles position q(t). Adding a term qex(t)(t)to the Hamiltonian, where qex(t) = qex()e

it+t, the response is then defined by () =

()qex(). Since qex(t) affects the equation of motion for Qi(t) in the same way as q(t) in

Eq. (*) above,

(t) =

i

2iMi2i

qex(t)

i

2iMi2i

t

dtqex(t)cosi(t t)

= 2 0 dJ(

) qex()eit+t + 2 0 dJ(

) t dteit+t cos[(t t)]iqex()=

2

0

dJ()

2

0

dJ()

(

1

+ + i+

1

+ i )

qex()eit+t .

Therefore, for a general bath (not necessarily an ohmic one)

Im() = J()

For an Ohmic bath J() = with Eq. (**) we recover the classical FDT. This relates the

fluctuations of the bath to the response () of the bath to an external force.

Note that there is a separate FDT relating the fluctuations of the particle q(t)q(0) and itsresponse q() to an external force, i.e. adding to the Hamiltonian a term q(t)ex(t). ForV(q) = 0 this was che

Documents

Lectures Pahtria