statistical mech.: phase transitions

8/13/2019 Statistical Mech.: phase transitions

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Lecture Notes for the C6 Theory Option

F.H.L. EsslerThe Rudolf Peierls Centre for Theoretical Physics

Oxford University, Oxford OX1 3NP, UK

May 23, 2013

Preliminary version, not yet proofread.

Please report errors and typos to [email protected] F.H.L. Essler

Part I

Statistical Mechanics and Phase Transitions

1 Brief review of some relevant quantities

Consider a classical many-particle system coupled to a heat bath at temperature T. The partition functionis defined as

Z=

configurations C

eE(C) , = 1

kBT. (1)

Here the sum is over all possible configurations C, andE(C) is the corresponding energy. Thermal averagesof observables are given by

O = 1Z

configurations C

O(C)eE(C) , (2)

whereO(C) is the value of the observableO in configuration C. The average energy is

E= 1

Z

configurations C

E(C)eE(C) =

ln(Z). (3)

The free energy is

F = kBTln(Z). (4)The entropy is

S=E F

T =kBln(Z) kB

ln(Z) =kB

T [Tln(Z)] . (5)

2 The Ising Model

Ferromagnetism is an interesting phenomenon in solids. Some metals (like Fe or Ni) are observed to acquire afinite magnetization below a certain temperature. Ferromagnetism is a fundamentally quantum mechanicaleffect, and arises when electron spins spontaneously align along a certain direction. The Ising model is avery crude attempt to model this phenomenon. It is defined as follows. We have a lattice in D dimensions

with N sites. On each site j of this lattice sits a spin variable j, which can take the two values1.

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These are referred to as spin-up and spin-down respectively. A given set{1, 2, . . . , N} specifies aconfiguration. The corresponding energy is taken to be of the form

E({j}) = J

ijij h

N

j=1j , (6)

whereij denote nearest-neighbour bonds on our lattice and J > 0. The first term favours alignmenton neighbouring spins, while h is like an applied magnetic field. Clearly, when h = 0 the lowest energystates are obtained by choosing all spins to be either up or down. The question of interest is whether theIsing model displays a finite temperature phase transition between a ferromagnetically ordered phase at lowtemperatures, and a paramagnetic phase a high temperatures. The partition function of the model is

Z=

1=1

2=1

N=1eE({j}). (7)

The magnetization per siteis given by

m(h) = 1

N Nj=1

j = 1N

hZ. (8)

The magnetic susceptibility is defined as

(h) =m(h)

h =

1

N

2

h2ln(Z). (9)

Substituting the expression (7) for the partition function and then carrying out the derivatives it can beexpressed in the form

(h) =

N

Nl,m=1

lm lm. (10)

2.1 The One-Dimensional Ising Model

The simplest case is when our lattice is one-dimensional, and we impose periodic boundary conditions. Theenergy then reads

E=N

j=1

Jj j+1 h

2(j+ j+1)

Nj=1

E(j, j+1), (11)

where we have definedN+1= 1. (12)

The partition function can be expressed in the form

Z=

1,...,N

Nj=1

eE(j ,j+1). (13)

It can be evaluated exactly by means of the transfer matrix method.

2.1.1 Transfer matrix approach

The general idea is to rewrite Z as a product of matrices. The transfer matrix T is taken to be a 2 2matrix with elements

T =eE(,). (14)

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Its explicit form is

T =

T++ T+T+ T

=

e(J+h) eJ

eJ e(Jh)

. (15)

The partition function can be expressed in terms of the transfer matrix as follows

Z= 1,...,N

T12T23. . . T N1NTN1 (16)

As desired, this has the structure of a matrix multiplication

Z= Tr

TN

.(17)

The trace arises because we have imposed periodic boundary conditions. As T is a real symmetric matrix,it can be diagonalized, i.e.

UT U=

+ 00

, (18)

where U is a unitary matrix and

= eJ cosh(h)

e2J sinh2(h) + e2J. (19)

Using the cyclicity of the trace and U U = I, we have

Z= Tr

U UTN

= Tr

UTNU

= Tr

[UT U]N

= Tr

N+ 0

0 N

= N+ +

N . (20)

But as < + we have

Z=N+1 +

+

N= N+ 1 + eNln(+/) . (21)So for large N, which is the case we are interested in, we have with exponential accuracy

Z N+ .(22)

Given the partition function, we can now easily calculate the magnetization per site

m(h) = 1

N

hln(Z). (23)

In Fig. 1 we plot m(h) as a function of inverse temperature = 1/kBT for two values of magnetic field h.We see that for non-zero h, the magnetization per site takes its maximum valuem = 1 at low temperatures.At high temperatures it goes to zero. This is as expected, as at low T the spins align along the directionof the applied field. However, as we decrease the field, the temperature below which m(h) approaches unitydecreases. In the limit h 0, the magnetization per site vanishes at all finite temperatures. Hence there isno phase transition to a ferromagnetically ordered state in the one dimensional Ising model.

2.1.2 Averages of observables in the transfer matrix formalism

The average magnetization at site j is

j = 1Z

1,...,N

jeE({j}). (24)

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2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

mh0.1

2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

mh105

Figure 1: Magnetization per site as a function of inverse temperature for two values of applied magneticfield.

We can express this in terms of the transfer matrix as

j = 1Z

1,...,N

T12T23. . . T j1jj Tjj+1. . . T N1 . (25)

Using that(T z)j1j =Tj1jj , (26)

where z =

1 00 1

is the Pauli matrix, we obtain

j

=

1

Z

Tr Tj1zTNj+1= 1

Z

Tr TNz . (27)Diagonalizing Tby means of a unitary transformation as before, this becomes

j = 1Z

Tr

UTNU UzU

= 1

ZTr

N+ 0

0 N

UzU

. (28)

The matrix Uis given in terms of the normalized eigenvectors ofT

T| =| (29)

asU = (

|+

,

|). (30)

This is easily worked out, but the expression for h = 0 is messy. For h = 0 we have

U|h=0= 12

1 11 1

. (31)

This gives

j

h=0= 0. (32)

Similarly, we find

j j+r

=

1

ZTr Tj1zTrzTN+1jr= 1ZTr UzU

r+ 0

0 rUzUNr+ 0

0 Nr . (33)

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We can evaluate this for zero field h = 0

j j+r

h=0=

Nr+ r+ Nr r+

N+ + N

+

r=er/ . (34)

So in zero field the two-point function decays exponentially with correlation length

= 1

ln coth(J). (35)

2.2 The Two-Dimensional Ising Model

We now turn to the 2D Ising model on a square lattice with periodic boundary conditions. The spin variableshave now two indices corresponding to rows and columns of the square lattice respectively

j,k = 1 , j, k= 1, . . . , N . (36)

The boundary conditions are k,N+1 = k,1 and N+1,j =1,j, which correspond to the lattice living on

k,j

j

k

Figure 2: Ising model on the square lattice.

the surface of a torus. The energy in zero field is

E({k,j}) = Jj,k

k,j k,j+1+ k,j k+1,j. (37)

2.2.1 Transfer Matrix Method

The partition function is given by

Z= {j,k} eE(

{k,j

})

. (38)

The idea of the transfer matrix method is again to write this in terms of matrix multiplications. Thedifference to the one dimensional case is, that the transfer matrix will now be much larger. We start byexpressing the partition function in the form

Z={j,k}

eN

k=1E(k;k+1), (39)

where

E(k; k+ 1) =

J

N

j=1 k,j k+1,j+1

2

[k,j k,j+1+ k+1,jk+1,j+1] . (40)

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This energy depends only on the configurations of spins on rows k and k+ 1, i.e. on spins k,1, . . . , k,Nand k+1,1, . . . , k+1,N. Each configuration of spins on a given row specifies a sequence s1, s2, . . . , sN withsj = 1. Let us associate a vector

|s (41)with each such sequence. By construction there 2N such vectors. We then define a scalar product on the

space spanned by these vectors by

t|s =N

j=1

tj ,sj . (42)

With this definition, the vectors{|s} form an orthonormal basis of a 2N dimensional linear vector space.In particular we have

I=s

|ss|. (43)

Finally, we define a 2N 2N transfer matrix T by

k

|T

|k+1

=eE(k;k+1). (44)

The point of this construction is that the partition function can now be written in the form

Z=1

2

N

1|T|22|T|3 . . . N1|T|NN|T|1 (45)

We now may use (43) to carry out the sums over spins, which gives

Z= Tr

TN

, (46)

where the trace is over our basis{|

s

|sj =

1}

of our 2N dimensional vector space. Like in the 1D case,thermodynamic properties involve only the largest eigenvalues ofT. Indeed, we have

Z=2N

j=1

Nj , (47)

where j are the eigenvalues ofT. The free energy is then

F = kBTln(Z) = kBTlnNmax 2

Nj=1

jmax

N= kBT Nln(max) kBTln

j

jmax

N , (48)where max is the largest eigenvalue of T, which we assume to be unique. As|j/max| < 1, the secondcontribution in (48) is bounded bykBT N ln(2), and we see that in the thermodynamic limit the freeenergy per site is

f= limN

F

N2 = lim

NkBT

N ln(max).

(49)

Thermodynamic quantities are obtained by taking derivatives offand hence only involve the largest eigen-value ofT. The main complication we have to deal with is that T is still a very large matrix. This posesthe question, why we should bother to use a transfer matrix description anyway? Calculating Z from itsbasic definition (38) involves a sum with 2N

2terms, i.e. at least 2N

2operations on a computer. Finding the

largest eigenvalue of a M

M matrix involvesO

(M2) operations, which in our case amounts toO

(22N).For large values ofNthis amounts to an enormous simplification.

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2.2.2 Spontaneous Symmetry Breaking

Surprisingly, the transfer matrix of the 2D Ising model can be diagonalized exactly. Unfortunately wedont have the time do go through the somewhat complicated procedure here, but the upshot is that the2D Ising model can be solved exactly. Perhaps the most important result is that in the thermodynamiclimit the square lattice Ising model has a finite temperature phase transitionbetween a paramagnetic and

a ferromagnetic phase. The magnetization per site behaves as shown in Fig.3. At low temperatures T < Tc

1 T/T

m(T)

c

1

Figure 3: Phase Transition in the square lattice Ising model.

there is a non-zero magnetization per site, even though we did not apply a magnetic field. This is surprising,because our energy (37) is unchanged if we flip all spins

j,k j,k. (50)

The operation (50) is a discrete (two-fold) symmetry of the Ising model. Because we have translationalinvariance, the magnetization per site is

m= j,k. (51)Hence a non-zero value ofm signifies the spontaneous breakingof the discrete symmetry (50). In order todescribe this effect mathematically, we have to invoke a bit of trickery. Let us consider zero temperature.Then there are exactly two degenerate lowest energy states: the one with all spinsj,k = +1 and the onewith all spins j,k = 1. We now apply a very small magnetic field to the system, i.e. add a term

E= j,k

j,k (52)

to the energy. This splits the two states, which now have energies

E = JNB N , (53)

whereNB is the number of bonds. The next step is key: we now definethe thermodynamic limit of the freeenergy per site as

f(T) lim0

limN

kBTln(Z)N2

. (54)

The point is that the contributions Z = eE of the two states to Zare such that

Z

Z+=e2N/kBT . (55)

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This goes to zero when we take N to infinity! So in the above sequence of limits, only the state with allspins up contributes to the partition function, and this provides a way of describing spontaneous symmetrybreaking! The key to this procedure is that

lim0

limN

Z= limN

lim0

Z.

(56)

The procedure we have outlined above, i.e. introducing a symmetry breaking field, then taking the infinitevolume limit, and finally removing the field, is very general and applies to all instances where spontaneoussymmetry breaking occurs.

2.2.3 Peierls Argument

Consider the Ising model on the square lattice with boundary conditions such that all spins on the boundaryare up, i.e. take the value +1. You can think of these boundary conditions as a symmetry breaking field.The bulk magnetic field is taken to be zero. Configurations look like the one shown in Fig. 4, and can becharacterized by domains walls. These are lines separating + andspins such that

1. The(+) spins lie always to the left (right) of the wall.2. Where ambiguities remain, the wall is taken to bend to the right.

3. The lengthof the wall is defined as the number of lattice spacings it traverses.

Figure 4: A configuration of spins and the corresponding domain walls.

A wall of length b encloses at most b2/16 spins. The total number of domain walls of length b, m(b), isbounded by

m(b) 2Nt3b1,(57)

where Nt is the total number of sites. This can be seens as follows:

the first link can go into less than 2 Nt positions (up or to the right and any site, and ignoringboundaries).

subsequent links have at most 3 possible directions each.Let us denote the ith domain wall of length b by (b, i). Next consider a particular configuration = {j,k}of spins on the lattice, and define

X(b, i) =1 if (b,i) occurs in

0 else(58)

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Then the total number ofspins in is bounded by

N

b

b2

16

m(b)i=1

X(b, i), (59)

because each spin is enclosed by at least one domain wall due to our choice of boundary conditions. Takingthermal averages, we have

N

b

b2

16

m(b)i=1

X(b, i). (60)

Here the thermal average ofX(b, i) can be written as

X(b, i) = 1Z

eE(), (61)

where the sum is only over configurations, in which (b, i) occurs. Now consider the configuration obtainedfrom by reversing the spins inside the domain wall (b, i). Clearly the energies of the two configurations

Figure 5: Configurations and related by reversing all spins inside the domain wall (b, i). Shown are allthe bonds whose energies have been changed fromJ to J.

are related byE() =E() + 2bJ. (62)

This can be used to obtain a bound on Z

Z eE(

) eE()e2bJ , (63)

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where the first sum is only over configurations in which (b, i) occurs, and where we then have flipped allspins inside the domain wall. This gives us a bound on

X(b, i) = 1Z

eE() e2bJ . (64)

Now we put everything together

N

b

b2

16

m(b)i=1

e2J b

b

b2

162Nt3

b1e2J b = Nt24

b=4,6,8,...

b2

3e2Jb

. (65)

The sum overb can now be easily carried out, and the results at small T (large ) is

N 54Nte8J.(66)

So, at low temperatures we have

N

Nt 1

2 .(67)

This proves the existence of a spontaneous magnetization at low temperatures.

3 Mean Field Theory

Consider the Ising model on a D-dimensional lattice with coordination number (number of nearest neighboursites)z

E= Jij

ij h

j

j. (68)

Hereij denotes nearest neighbour bonds, and each bond is counted once. The magnetization per site is

m= 1

N

Nj=1

j. (69)

We now rewrite the energy usingi= m + (i m). (70)

In particular we have

ij =m2 + m(j m) + m(i m) + (i m)(j m). (71)

The idea of themean-field approximation

is to assume that the deviations j m of the spins from theiraverage values are small, and to neglect the terms quadratic in these fluctuations. This gives

EMF= Jij

m2 + m(i+ j ) h

j

j .

(72)

Physically, what we have done is to replace the interaction of a given spin with its neighbours by an averagemagnetic field. We can simplify (72) further by noting that

Jij

m2 = Jm2 N z2

,

ij

i+ j = z

j

j. (73)

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The mean-field energy then becomes

EMF=JN z

2 m2 (Jzm + h)

Nj=1

j .

(74)

The partition function in the mean-field approximation is

ZMF={j}

eEMF = eNJzm2

2

1

N

Nj=1

e(Jzm+h)j

= eNJzm2

2

1

e(Jzm+h)1

. . .

N

e(Jzm+h)N

= eNJzm2

2 [2 cosh(Jzm+ h)]N . (75)

The magnetization per site is

m=

1

N

Nj=1

j = N, (76)where we have used translational invariance in the last step. In mean field theory we have

m= j = 1ZMF

eNJzm2

2

1

N

N

Nj=1

e(Jzm+h)j = tanh(Jzm+ h).

(77)

This is a self-consistency equation for m.

3.1 Solution of the self-consistency equation for h= 0

For zero field the self-consistency equation reads

m= tanh(Jzm). (78)

This can be solved graphically by looking for intersections of the functions g1(m) = tanh(Jzm) andg2(m) =m. There are either one or three solutions

m=

0 ifJz 1. (79)

We still have to check whether these solutions correspond to mimima of the free energy per site. The latteris

fMF=

1

N

ln(ZMF) = Jzm2

2 1

ln[2 cosh(Jzm)]. (80)

We have2fMF

m2 =J z

1 Jz

cosh(Jzm)

. (81)

This is negative for m = 0 and Jz > 1, and hence this solution corresponds to a maximum of the freeenergy and hence must be discarded. This leaves us with

m=

0 ifT > Tc

m0 ifT < Tc,

(82)

where the transition temperature is

Tc= JzkB

. (83)

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3.2 Vicinity of the Phase Transition

Let us define a dimensionless variable, that measures the distance in temperature to the phase transition

t= T Tc

Tc. (84)

For|t| 1 we obtain the following results1. Magnetization per site

m

h=0

0 ifT > Tc,3t ifT < Tc.(85)

2. Magnetic susceptibility in zero field

=m

h

h=0

1kBTc

t1 ifT > Tc,1

2kBTc(t)1 ifT < Tc,

(86)

3. Free energy per site and heat capacity per volume

fMF

h=0

kBTln2 +

0 ifT > Tc

3kBTc4 t2 ifT < Tc.(87)

C

V = T

2fMFT2

h=0

0 ifT > Tc3kB

2 ifT < Tc.(88)

4 Landau Theory of Phase Transitions

4.1 Phase Transitions

Physically a phase transition is a point in parameter space, where the physical properties of a many-particlesystem undergo a sudden change. An example is the paramagnet to ferromagnet transition in Fe or Nishown in Fig. 6.

Mathematically a phase transition is a point in parameter space, where the free energy F = kBTln(Z)becomes a nonanalytic function of one of its parameters (i.e. For some of its derivatives becomes singularor discontinuous) in the thermodynamic limit.

For a finite system this can never happen, because

Z= configurations C

eE(C)/kBT (89)

is a finite sum over finite, positive terms. Hence all derivatives are finite and well defined as well.Phase transitions are usually divided into two categories:

1. First Order Phase Transitions.

Here the free energy is continuous, but a first derivative is discontinuous. At the transition there isphase coexistence. The magnetization per site is a first order derivative of the free energy with respectto the magnetic field h. Therefore the phase transition at h = 0 and T < Tc in Fig. 6 is first order.

2. Second Order Phase Transitions.

These are characterized by a divergence in one of the higher order derivatives (susceptibilities) ofthe free energy. The phase transition as a function ofT for h = 0 in Fig. 6 is second order.

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h

T

Tc

c

h

m(T,h)

T Tc|t| ifT < Tc t= T TcTc .(91)

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is a critical exponent.

2. Susceptibilities

At the critical point the system is very sensitive to external perturbations. The singularity in theresponse of the order parameter to a field conjugate to it is characterized by critical exponents .

For our magnet(T) =

h

h=0

1

VM(h, T) |t| .

(92)

Figure 8: Critical behaviour of the magnetic susceptibility.

3. Heat Capacity

A third critical exponent is associated with the heat capacity

C(T) = T2F

T2

A+|t|+ ifT > TcA|t| ifT < Tc.

(93)

Depending on the signs of this may or may not be singular, see Fig. 9.

Figure 9: Critical behaviour of the heat capacity.

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4.2.1 Universality

The critical exponents are insensitive to microscopic details of the system under consideration and arecharacteristic of the critical point. A consequence of this is that completely different systems can exhibitthe same critical behaviour!

4.3 Landau Theory

Landau Theory is a general approach to phase transitions that

is phenomenological in nature and deals only with macroscopic quantities; applies only to the neighbourhood of a critical point, where the order parameter is small.

Landau theory is constructed as follows.

1. Identify the order parameter(s) characterizing the phase transition.

2. Form a coarse-grained order parameter density (r). Think of this as a magnetic moment density,

averaged over atomic distances. This is a continuum field.

3. Consider the free energy density to be a functionalof the order parameter field (r). The free energyis then

F =

dDrf[(r)].

(94)

The partition function is the path integral

Z=

D(r)eF. (95)

4. By construction of the order parameter, the latter is small close to our critical point. This allows usto expand f[(r)] is a power series around = 0. Assuming the order parameter to be a real scalar,we have

f[] const h + 22 +12||2 + 33 + 44 + . . .

(96)

Here the coefficient of the gradient term is fixed by convention to be 1/2. This makes in generaldimensionful

dim[(r)] = (length)1D/2. (97)

The only linear term that is not a total derivative is

h, whereh is an external field (a source) cou-pling to the order parameter. Total derivative terms can be dropped, because the only give boundarycontributions to F. The coefficientsn are a priori all functions of temperature.

5. Finally, we use symmetries and the fact that we are interested in the vicinity of a critial point toconstrain thej.

If we truncate our expansion at order 4, then thermodynamic stability requires

4> 0. (98)

If4 < 0 the free energy density would be unbounded from below and could become infinitely

negative, which is forbidden.

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If we know that the system is invariant under certain symmetry operations, e.g.

, (99)

then the free energy must respect this symmetry. A ferromagnet has the symmetry (99) in absenceof a magnetic field because oftime-reversal invariance. Hence we must have 3= 0 in this case.

ConsiderZ=

D(r) e

dDrf[(r)]. (100)

Clearly, the vicinity of the minimum of f[(r)] will contribute most to the partition function.We will see that for translationally invariant systems the minimum corresponds to r-independentorder parameters (i.e.(r) = 0). In order to understand the nature of the phase transition, wetherefore can simply look at the minima of the potential V() =2

2 +44, which is done in

Fig. 10. We see that the phase transition corresponds to2 changing sign at T =Tc. So in the

Figure 10: Minima of the potentialV(). For2> 0 the minima occurs at = 0, so there is no ferromagneticorder. For 2 < 0 there are two minima at =0, corresponding to the emergence of a spontaneousmagnetization.

vicinity of the transition we have (by Taylor expanding 2 in T Tc)

2(t) =At + O(t2) , t= T TcTc

, A >0.

(101)

The parameter 4 is also temperature dependent, but this dependence is subleading

4(t) =4(0) + O(t). (102)

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If we have 3 < 0 the transition is generically first order. To see this we again use that ina translationally invariant system the minima of the free energy density will be r-independent,so that we merely need to scrutinize the potential V() to understand the nature of the phasetransition.

Figure 11: Minima of the potential V() for 3 < 0. Decreasing the value of2 leads to a discontinuousjump in the order parameter at some critical value. The transition is therefore first order.

4.3.1 Saddle Point Approximation

The Landau field theory

Z=

D(r)e

dDrf[(r)], (103)

is still difficult to analyze. For the example we have discussed, it reduces to the Euclidean space version of

the4 theory you have encountered in the field theory part of the course. In order to proceed we thereforeresort to further approximations. Thesaddle-point approximationtakes into account the thermodynamicallymost likely configuration (r), i.e. the configuration that minimizes the free energy

F =

dDr

h + 22 +1

2||2 + 44

. (104)

This is obtained by functional extremization

F

(r)= 0. (105)

The resulting nonlinear differential equation is

2(r) + 22(r) + 443(r) h= 0.(106)

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It is easy to see that the solutions to (106) that minimize the free energy are in fact r-independent (if weignore boundary conditions). The reason is that 12 ||2 0, and hence this contribution to F is minimizedby constant solutions. For constant fieldh the potential V((r)) =h+22 +44 is also minimizedby constant solutions. Hence for zero field h = 0 the saddle point approximation gives the following resultsfor the most likely configurations

(r) =

0 if2> 0 T > Tc0 =

224 if2< 0 T < Tc

.

(107)

So in the saddle point approximation we obtain a second order phase transition (because 0 vanishes at thecritical point) from a paramagnetic to a ferromagnetic phase.

4.3.2 Mean Field Exponents

Using the saddle point solution we can determine the corresponding approximation for the critical exponents.

Order parameter.Using that 2= At fort = (T Tc)/Tc< 0, we have

0=

A

24|t| 12 .

(108)

This gives the critical exponent

=1

2.

(109)

Magnetic susceptibility.Differentiating (106) with respect to h gives for r-independent solutions

22

h+ 124

h2 = 1. (110)

The zero-field susceptibility is thus

=

h

h=0

= 1

22+ 1242. (111)

Using that 2= At and

2

0= A|t|/24 this becomes

=

12At ift >0,

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The saddle point contribution to the free energy is

F

V

0 ift >0,

A2kBTct244 ift 0,A2kBT

2cV

24ift


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The first factor is merely a constant, which we will denote byN, while the second factor is rewrittenas

Z[h] = Nexp

1

2

dDrdDrh(r)G(r r)h(r)

,

G(r) = dDp(2)D eipr

p2 + 22 . (127)

Taking functional derivatives we have

(r)(0) =G(r) e|r|22

|r|D12, |r| . (128)

This gives the correlation length

= 1

22 1

t12

, (129)

and thus the critical exponent

= 12

.(130)

In the ordered phase t < 0 we expand f[(r)] around one of the minima at0. The choice ofminimum implements spontaneous symmetry breaking. We have

V() =22 + 4

4 220+ 440+ (2+ 6420)( 0)2 + . . . (131)We may drop the constant and retain only the contribution quadratic in = 0 (this is theGaussian approximation in the ordered phase), which gives

f[(r)] 1

2 ||2

+ 22

. (132)

Here 2 =22 > 0. We may now copy the calculation in the disordered phase and obtain for theconnected correlation function

(r)(0) = (r)(0) (r)(0) e|r|42

|r|D12, |r| . (133)

We see that the correlation length scales as |t|1/2, giving again = 1/2.

4.4 Other Examples of Phase Transitions

4.4.1 Isotropic-Nematic Transition in Liquid Crystals

Liquid crystals are fluids of rod-like molecules. At high temperatures their centres of mass are randomlydistributed and the rods are randomly oriented. At low temperatures, in the nematic phase the rodsspontaneously align along a common axis, see Fig. 12. What is the order parameter characterizing thistransition?

Let us associate a unit vectorn(r) with a molecule at position r. The first guess one may have is to taken(r) as the order parameter. This will not work, because the two vectors n(r) andn(r) describe thesame orientation of the molecule. Hence the order parameter must be invariant under

n(r) n(r). (134)

So how about something quadratic likeni(r)nj (r). (135)

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high temperatures low temperatures

n!

Figure 12: At low temperatures the rod-like molecules spontaneously align an axis n.

The problem with this expression is that it is different from zero even for randomly oriented molecules(which is what one has at very high temperatures). Indeed, using a parametrization of the unit vector fora single molecule in terms of polar coordinates we have

n=

sin cos sin sin

cos

. (136)

Then averaging over all possible orientations gives

ninj = 14

0

d

20

d sin ninj = 1

3i,j= 0. (137)

This consideration suggests to try

Qij = ninj 13

i,j(138)

as our order parameter. At very high temperatures, when molecules have random orientations, this is zero.On the other hand, if the molecules are aligned in the z -direction, i.e. n=ez, we have

Q=13 0 00 13 0

0 0 23

. (139)So this seems to work. Given the order parameter, how do we then determine the free energy? In the hightemperature phase the free energy must be invariant under rotations of the molecules, i.e. under

Q(r) RQ(r)RT. (140)This suggests the following expansion for the free energy density

f[Q(r)] =1

2|Q|2 + 2Tr[Q2] + 3Tr[Q3] + 4

Tr[Q2]

2+ . . .

(141)

Here|Q|2 =3i,j,k=1(kQij)2. The presence of a cubic term suggests that the transition is first order,which is indeed correct.

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4.4.2 Superfluid Transition in Weakly Interacting Bosons

Let us recall the second quantized Hamiltonian for weakly repulsive bosons

H=

d4r

c(r)

2

2m

c(r) +

U

2c(r)c(r)c(r)c(r)

. (142)

In the superfluid phase we have macroscopic occupation of the zero momentum single-particle state

c(p= 0) =

N0. (143)

This implies that(r) = c(r)= 0 , T < T c. (144)

At high temperatures the expectation value vanishes, so that we may take (r) as a complex valued orderparameter. The Landau free energy describing the transition can then be constructed by noting that itshould be invariant under U(1) transformations

(r)

ei(r). (145)

This gives

F[] =

dDr

1

2m||2 + 2|(r)|2 + 4||4

.

(146)

Part I I

Random Systems and Stochastic Processes

In nature there are many phenomena, in which some quantity varies in a random way. An example isBrowninan motion, which refers to the motion of a small particle suspended in a fluid. The motion, observedunder a microscope, looksrandom. It is hopeless to try to compute the position in detail, but certainaverage

Figure 13: Cartoon of a path of a particle undergoing Brownian motion.

featuresobey simple laws. Averaging over a suitable time interval is difficult and one therefore replaces timeaveraging of a single irregularly varying function of time by averaging over an ensemble of functions. Thelatter must of course be chosen in such a way that the two results agree

time average ensemble average. (147)

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5 Random Variables

A random variable is an object X defined by

1. a set{xj} of possible values (either discrete or continuous);

2. a probability distributionPX(xj) over this set

PX(xj) 0 ,

j

PX(xj) = 1 (this becomes an integral in the continuous case.). (148)

Example: Let Xbe the number of points obtained by casting a die

{xj} = {1, 2, 3, 4, 5, 6} , PX(xj) =16

. (149)

5.1 Some Definitions

The probability distribution PX(x) can be characterized by the moments ofX

Xn

dx xn PX(x).

(150)

The average value ofXis the first momentX, while the variance is 2 = X2 X2.The Fourier transform ofPX(x) is called the characteristic function

X(k) =

dx PX(x)e

ikx =

n=0(ik)n

n! Xn eikX.

(151)

The last equality shows that the characteristic function is the generating function for the moments

Xn = (i)n dn

dkn

k=0

X(k). (152)

The cumulants ofPX(x) are defined as

Cn(X) = (i)n dn

dkn

k=0

ln

X(k)

.(153)

The first few cumulants are

C1(X) = X , C2(X) = X2 X2 , C3(X) = X3 3X2X + 2X3. (154)

The Gaussian distribution is defined by

PX(x) = 1

22e

(xx0)2

22 . (155)

Its characteristic function isX(k) =e

ikx0 12k22 . (156)

The cumulants are C1

(X) =x0

, C2

(X) =2, Cn>2

(X) = 0. Hence the Gaussian distribution is completelydetermined by its first two cumulants.

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5.2 Discrete-time random walk

To see the above definitions in action we consider a tutor on (without loss of generality) his way back toSummertown after a long night out (needless to say it must be a humanities tutor). He moves along Banburyroad, by making each second a step forward/backward with equal probability. Modelling Banbury road bya line, his possible positions are all integers

< n t2> t1. (177)

Evolution equation

P1(y2, t2) =

dy1 P1|1(y2, t2|y1, t1)P1(y1, t1) , t2 > t1.

(178)

A MP is called stationary if P1(y, t) is time-independent and P1|1(y2, t2|y1, t1) depends only on the timedifferencet2 t1 (and y1,2).

7.1 Examples of Markov Processes

The discrete-time random walk is a MP. The Wiener process, defined by

P1|1(y2, t2|y1, t1) = 12(t2 t1)

e (y2y1)

2

2(t2t1) ,

P1(y, t) = 1

2te

y2

2t , (179)

is a non-stationary Markov process. It was originally invented for describing the stochastic behaviour

of the position of a Brownian particle.

7.2 Markov Chains

A Markov chain is a MP, in which the random variable only takes a finite number of values and in-volves discrete time steps. The probability P1(y, t) can then be represented as a N-component vector,and P1|1(y2, t2|y1, t1) Tas an N Nmatrix. The normalization condition (173) becomes

Nj=1

Tjj = 1,

(180)

which is often referred to as probability conservation.

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Example: Two state process with Y= 1 or Y= 2, with conditional probabilities

P1|1(1, t + 1; 1, t) = 1 q ,P1|1(2, t + 1; 1, t) = q ,P1|1(1, t + 1; 2, t) = r ,

P1|1(2, t + 1; 2, t) = 1 r . (181)Introducing a two-component vector

p(t) =

P1(1, t)P1(2, t)

, (182)

the evolution equation for the process can be expressed as a vector equation

p(t + 1) =

1 q r

q 1 r

T

p(t).

(183)

T is a square matrix with non-negative entries, that is in general not symmetric. The matrix elementsTijare the rates for transitions from state j to state i. It is useful to rewrite the equation in components

pn(t + 1) =

j

Tnjpj (t). (184)

Thenpn(t + 1) pn(t) =

j

Tnjpj(t)

j

Tjn

1

pn(t) =

j

Tnjpj(t) Tjnpn(t). (185)

Now consider our time interval to be small instead of 1. Then (185) turns into a differential equation, called

a Master equationdpn(t)

dt =

j

Tnjpj(t) Tjnpn(t).(186)

This has a nice physical interpretation as a loss/gain equation for probabilities: the first term on theright-hand side is the rate of transitions from state j to state n times the probability ofj being realized, i.e.the total gain of probability for state n. The second term on the right-hand side is the rate of transitionsfrom staten to state j times the probability ofn being realized, i.e. the total loss of probability for state n.Let us now return to the discrete form (183), which can be iterated to give

p(t + 1) =Tt+1p(0). (187)

While T is generally not symmetric, it is nevertheless often diagonalizable. Then there exist left and righteigenvectors such that

T|R =|R , L|T =L| , L|R =, , (188)and Tcan be represented in the form

T =

|RL|. (189)

Using the orthonormality of left and right eigenvectors we have

Tt+1

=

t+1 |RL|.

(190)

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In our example

1 = 1 , L1| = (1, 1) , |R1 = 11 + r/q

r/q

1

. (191)

2= 1 q r , L2| = (qr

, 1) , |R2 = 11 + r/q

11

. (192)

So for large t we have

Tt+1 |R1L1| = 1r+ q

r rq q

, (193)

and hence

p() = 1r+ q

r rq q

p(0). (194)

8 Brownian Motion

We now want to think of Brownian motion as a Markov process. Let v1, v2, . . . be the velocities of theparticle at different time steps. Then vk+1 depends only on vk, but not on v1, . . . , vk

1.

8.1 Langevin Equation

One approach to Brownian motion is via a stochastic differential equation, the Langevin equation for thevelocityv(t) (more precisely the velocity in D = 1 or a component of the velocity in D >1)

dv(t)

dt = v(t) + (t).

(195)

Here the first term on the right hand side is damping term linear in v, while the second term represents theremaining random forcewith zero average (t) = 0. This is often referred to as noise. For simplicity wewill assume collisions to be instantaneous, so that forces at different times are uncorrelated

(t)(t) = (t t). (196)Given our assumptions about the noise, we can calculate noise-averaged quantities quite easily. We have

d

dt

v(t)et

=

dv

dt+ v

et

=(t)et

, (197)

where in the last step we used the Langevin equation (195). Integrating both sides of this equation betweent= 0 and t = t, we obtain

v(t) =v(0)et +

t

0dt (t)e(tt

).

(198)Averaging this over the noise, we find the average velocity

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

0dt(t)

=0

e(tt) =v(0)et .

(199)

Similarly we have

v2(t) =

v(0)et + t

0dt (t)e(tt

)

v(0)et +

t0

dt (t)e(tt)

= v2(0)e2t + e2t t0

dtdt(t)(t) (tt)

e(t

+t

) (200)

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Carrying out the time integrals this becomes

v2(t) =v2(0)e2t + 2

(1 e2t).(201)

The displacement of the particle is

x(t) x(0) = t

0dt v(t) =

v(0)

(1 et) +

t0

dt t

0dt (t)e(t

t). (202)

Assumingx(0) = 0, the average position of the particle is

x(t) = v(0)

(1 et).(203)

Finally, we want to determine the particles mean square deviation

x(t) x(t)2 = x2(t) x(t)2. (204)Substituting (202) and using again that x(0) = 0 this becomes t

0dt1

t10

dt2e(t1t2)

t0

dt1

t10

dt2e(t1t2)(t2)(t2)

(t1t2)e(t1t2)

=

t0

dt1

t10

dt2e(t12t2)

tt2

dt1et1

1(et2et)

.(205)

Carrying out the remaining two integrals we find

x(t) x(t)2 = 2

t 3

(1 et) 23

(1 et )2.(206)

So at very late times we have

x2(t) = 2

t + . . .

(207)

The displacement grows like

t, which is characteristic of diffusion. Finally, we may relate / to thetemperature of the fluid by noting that

v2(t

)

=

2

. (208)

On the other hand, by equipartition we have

m

2v2 kBT

2 (209)

Combining these two equations, we arrive at

2 =

kBT

m .

(210)

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8.2 Fokker-Planck Equation

We now want to derive a differential equation for the probability P1(v, t) of our particle having velocity vat time t from the Langevin equation

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

0

dt e(tt)(t). (211)

Our starting point is the general evolution equation

P1(v, t + t) =

duP1|1(v, t + t|u, t) P1(u, t). (212)

It is convenient to consider the integral

=

dv [P1(v, t + t) P1(v, t)] h(v), (213)

whereh(v) is is test function (infinitely many time differentiable,h(v) and all of its derivatives going to zeroat infinity etc). On the one hand, we have to linear order in t

dv [P1(v, t + t) P1(v, t)] h(v) =

dv P1(v, t)

t t h(v). (214)

On the other hand, using (212) we have

=

du dv h(v) P1|1(v, t + t|u, t) P1(u, t)

dv h(v)P1(v, t). (215)

Relabelling the integration variable from v tou in the second term, and using that normalization conditiondv P1|1(v, t + t|u, t) = 1, we obtain

= du P1(u, t) dv P1|1(v, t + t|u, t) [h(v) h(u)] . (216)Expandingh(v) around u in a Taylor series gives

=

du P1(u, t)

n=1

h(n)(u)

dv P1|1(v, t + t|u, t)

(v u)nn!

D(n)(u)

. (217)

Integrating the nth term in the sum n times by parts, and using the nice properties of the function h(u),then leads to the following expression

= du h(u)

n=1 u

n

P1(u, t)D(n)

(u). (218)

Using that (214) and (218) have to be equal for any test function h(u), we conclude that

P1(v, t)

t t=

n=1

v

nP1(v, t)D

(n)(v).

(219)

This starts looking like our desired differential equation. What remains is to determine the quantities D(n)(v)

D(n)(v) = dw P1|1(w, t + t|v, t)

(w v)nn!

= dz P1|1(v+ z, t + t|v, t)

zn

n!

= 1

n![v(t + t) v(t)]n. (220)

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We see that D(n)(v) are related to the moments of the velocity difference distribution! We can use theLangevin equation to determine them, and the result is

v(t + t) v(t) = v(0)ett ,v(t + t) v(t)

2 = t + O(t)2

,v(t + t) v(t)n = O(t)2 , n 3. (221)Substituting these into (219) and then taking the limit t 0, we arrive at the Fokker-Planck equation

tP1(v, t) =

vvP1(v, t) +

2

2

v 2P1(v, t).

(222)

This is a second order linear PDE for P1(v, t) and can be solved by standard methods. For initial conditionsP1(v, 0) =(v v0) we find

P1(v, t) = 122(t)

expv v(t)222(t) ,(223)

where

2(t) =

2(1 e2t ), v(t) =v0et . (224)

In the limit t this turns into the Maxwell distribution

P1(v, t) =

e

v2

=

m

2kBTe mv22kBT . (225)

8.2.1 Moments of the Velocity Difference Probability Distribution

Let us see how to derive (221), starting from the Langevin equation

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

0dt e(tt

)(t). (226)

For a very small time interval t we have

v(t + t) = v(0)e(t+t) + t+t

0dt e(t+tt

)(t)

= v(0)et(1 t) + t0

dt e(tt)(t)(1 t) + t+tt

dt e(tt)(t) + O(t2)= (1 t) v(t) +

t+tt

dt e(tt)(t) + O(t2). (227)

Hence v(t) =v(t + t) v(t) is given by

v= vt + t+t

tdt e(tt

)(t) + O(t2). (228)To derive (221) we rewrite (228) in the form

v(t) =v0(t)t +

dtK(t, t)(t), (229)

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where

v0(t) = v(0)et ,K(t, t) = (t t)(t)e(tt)t + (t t)(t + t t)e(tt). (230)

Let us now assume that (t) is Gaussian distributed. Then the probability distribution for the noise is the

functionalP[(t)] =e

12

dt 2(t). (231)

Averages are then given by the path integral

(t1) . . . (tn) =D(t) (t1) . . . (tn) e 12

dt 2(t). (232)

As this is Gaussian, we may use Wicks theorem to calculate averages

(t) = 0 ,(t)(t) = (t t) ,

(t1)(t2)(t3) = (t1)(t2)(t3) + (t3)(t1)(t2) + (t2)(t3)(t1) = 0 ,(t1)(t2)(t3)(t4) = (t1)(t2)(t3)(t4) + . . . (233)

The probability distribution for v(t) can be obtained from the generating function

Z() = ev(t) =ev0(t)tD(t) e 12

dt[2(t)2K(t,t)(t)]. (234)

Changing variables to(t) =(t) K(t, t), (235)

the generating function becomes

Z() =ev0(t)t+

2

2

dt K2(t,t)

. (236)It is then a straightforward matter to calculate the moments

v = dZd

=0

=v0(t)t ,

v2 = d2Zd2

=0

=

dt K2(t, t) =

t+tt

dt e(tt) + O(t)2

= t + O(t)2 ,vn = dnZ

dn

=0= O(t)2 , n 3. (237)

8.3 Diffusion Equation

Finally, we would like to obtain a differential equation for P1(x, t). The difficulty is thatx(t) is nota Markovprocess, becausex(t + t) depends on both x(t) andv(t). On can treat the combined evolution ofx(t) andv(t), but we will follow a simpler route. Recalling that

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

0dt e(tt

)(t), (238)

we see that v(t) becomes a random variable for t 1 with

v(t) = 0 , v(t)v(t) = 2

e|tt|. (239)

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Now let us imagine that we observe the Brownian particle only at sufficiently long time intervals t, t 1,and describe only these coarse grained positions. Then we may replace

v(t)v(t) = 2

e|tt|

2 (t t). (240)

This is because 2e|tt

| is substantially different from zero only if|t t| < 1, and

dt

2e|tt

| =

2. (241)

The differential equation for the position

dx(t)

dt =v(t), (242)

then turns into a special case (of no damping) of the Langevin equation we solved for the velocity. Wetherefore can use our previous results to conclude that

tP1(x, t) =

222

x2P1(x, t).

(243)

This is the diffusion equation with diffusion coefficient

D=

22. (244)

Its solution for initial conditions P1(x, 0) =(x x0) is

P1(x, t) = 12D|t t0|

exp

x x024D|t t0|

.

(245)

statistical mech.: phase transitions

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