Ch6 Phase Transition II - Ising model.pdf

Chapter 6

Phase transition II: Ising model

6.1 Brief remarks on history

Ernst Ising solved the 1D Ising model in 1925 and found that there is no phase transition inthe 1D model. With wrong arguments, he also concluded that 2D Ising models had no phasetransitions either. In 1941, Kramers and Wannier proved that there is a duality scaling relationbetween the high- and low-temperature expansion of the free energy of Ising models and hencewere able to exactly determine the transition temperature Tc of an Ising model in 2D squarelattices to be

tanh(J) = p2 1: (6.1)

A milestone in the development of the modern Statistical Mechanics is the exact solution ofthe 2D Ising model in a square lattice by Onsager in 1944. Later, many alternative solutionsand generalizations to other kinds of 2D lattices have been published [5]. For example, Baxterpublished in 1978 a paper titled 399th solution of the Ising Model [4]. The original Onsagerssolution is very complicated in mathematics. It was said that Onsager has tried to solve the 1DIsing model for 2 chains, 3 chains, 4 chains and more before he got the sense to the solutionof the 2D model. The calculation was very lengthy and tedious. I think this is a good lessonyoung researchers could learn. An intriguing alternative solution was given by Onsager andKaufmann in 1949 based on spin algebra [1]. Here we will introduced the solution by Schultzet al [3]. Unlike the 1D Ising model, calculation of partition functions at finite external fieldand hence the susceptibility and critical exponents are non-trivial. Interestingly, Onsager wroteon a blackboard at a conference in 1949 the critical exponent = 1=8. How he got the numberremains mysterious since he did not present any follow-up publication on that. The first pub-lished calculation of was due to C. N. Yang in 1952 [6]. T. T. Wu et al calculated anothercritical exponent = 7=4 in 1967. The exact solution of the 3D Ising model still remains un-known. The Ising model has been a basis in study of integrable systems, critical phenomena instatistical mechanics, and the renormalization group theory.

75

76 CHAPTER 6. PHASE TRANSITION II: ISING MODEL

6.2 ModelThe Ising model of spins in a lattice is given by the Hamiltonian

H = JXhi ji

zizj H

Xi

zj; (6.2)

where is the coupling constant between nearest-neighbor spins (denoted as hi ji), is themagnetic moment of a spin, H is the external magnetic field, and the Pauli matrices are

x =

0 11 0

!; y =

0 ii 0

!; z =

1 00 1

!: (6.3)

This Hamiltonian is invariant by reversing all the spins. Such symmetry with two symmetryoperations (one keeping all spins unchanged and one reserving all) is called Z2 symmetry.

In the case of ferromagnetic coupling, J < 0, the spins have lower energy when the neigh-bors are polarized parallel. Otherwise in the case of anti-ferromagnetic coupling, J > 0, thespins prefer to being anti-parallel to lower their energy.

The Hamiltonian is diagonal. The eigen states are

j 1;1; : : : ;1i (6.4)

with eigen energyU = JN++ + JN JN+ H (N+ N) ; (6.5)

where N++ (N) is the number of nearest-neighbor pairs with both of the two spins at the statej+i (ji), N+; is the number of nearest-neighbor pairs with the two spins in opposite states, andN is the number of spins in the state ji. The evaluation of the partition function

ZN = TreH(N); (6.6)

however, is non-trivial.The Ising model can be mapped equivalently to some other models including the order-

disorder transition model for alloys and the lattice gas model.In the order-disorder transition model, two kinds of atoms are mixed in a lattice. To be

specific, we consider ZnCu alloy which is a bcc lattice. Each atom could be in either a centerposition or a vertex position. At high-temperature, each kind of atoms have the same probabilityin dierent kinds of positions. The alloy is in a disordered phase. When the temperature is belowa critical value Tc, a certain kind of atoms have more probability to occupy the center positionsthan the vertices, the alloy enters into an ordered phase which can be experimentally verifiedby, e.g., X-ray diraction. The energy of a certain configuration of dierent types of atoms inthe lattice is approximately determined by the interaction between nearest neighbors,

U = AANAA + BBNBB + ABNAB; (6.7)

where NAA and NBB are the numbers of nearest-neighbor pairs of A- and B-atoms, respectively,and NAB are the number of hetero-atom nearest-neighbor pairs.

6.3. MEAN-FIELD THEORY 77

In the lattice gas model, each particle can only occupy a lattice point and a lattice point canonly have at most one particle. There is interaction between particles in nearest neighbors. Theinteraction is

ui j =

8>>>>>:+1; i = ju; (i; j) are nearest neighbors0; else

: (6.8)

The partition function of N particles in the lattice is

ZN =1

(2)N

Z Yi

dpiep2i =(2m)

Z Yi

drieU =13NT

Xall configurations

eU ; (6.9)

where T is the thermal de Brogile wavelength and the potential energy for each configuration

U = uN11 + 0 N00 + 0 N10; (6.10)

with N11, N00 and N10 denoting in turn the numbers of nearest-neighbor pairs of lattice sitesoccupied by two, one, and zero atoms.

The equivalence between the Ising model, the order-disorder transition model, and the latticegas model is transparent from their interaction energies.

6.3 Mean-field theoryThe exact evaluation of the Ising model is known for one- and two-dimensional lattices. To gainsome insights before getting into the mathematical details, let us take a look of the mean-fieldapproximation.

6.3.1 Mean-field approximationSuppose there are totally N spin-1/2s. The number of spins pointing up/down is N+=. Themagnetization of the lattice is

M =N+ NN+ + N

: (6.11)

The probability of having a spin pointing up or down is

p = N=N =1 M2

: (6.12)

On average, the probability of having two spins pointing both up or down is

p; = p2: (6.13)

Suppose each spin has n nearest neighbors. Neglecting the boundary eects for a large system,the total number of nearest neighbor pairs is

N = nN=2: (6.14)


So the numbers of nearest neighbor pairs pointing parallel are

N; =n2Np2 =

n8N (1 M)2 : (6.15)

The number of anti-parallel nearest neighbor pairs is

N+; =n4N1 M2

: (6.16)

Notice that in the above derivation, the distribution of the spins pointing up or down is assumeduniform, which is the core assumption of the mean-field approximation. We can imagine thatif the system is large enough, the deviation of a real distribution from the uniform one (i.e.,fluctuation) should be negligible. Under some special circumstances, the fluctuation could belarge and the mean-field theory would be inadequate.

Now with the uniform distribution assumption, the energy of the spins for a particular mag-netization M is

U(M) = JN++ + JN JN+ H (N+ N) = NnJ2 M2 NHM: (6.17)

The average energy per spin is

u U=N = HM + 12nJM2: (6.18)

The dierential is

1

@u@M

= H nJM He ; (6.19)

which gives an eective magnetic field. The microscopic interaction between spins now is takenas a uniform internal macroscopic field nJ

M, or a Bragg-William mean field.

6.3.2 Phase transition and symmetry breakingFor a ferromagnetic coupling (J < 0), the energy has too minima at M = 1 when the externalmagnetic field H is vanishing. The two ground states corresponding to the states in whichall the spins point up or down. The two-fold degeneracy is due to the Z2 symmetry of themodel. The two degenerate states are macroscopically dierent (or classically distinguishable).A superposition of the two

+j + + + +i + j i;is a Schrodinger cat state, which would not survive in a noisy environment. So at zero temper-ature, the system has to decide between the two fully polarized states. Collapsing into eitherstate breaks the Z2 symmetry.

Now let us consider the state at finite temperatures. The partition function is

Z =X

all configurations

eU : (6.20)


In the mean-field approximation, we takes the interaction energy all the same for the samemagnetization (which is of course not the reality but the fluctuation would be small for a largesystem). For the same number of spins pointing up N+, we still have many dierent microscopicstates with the number

N!N+!(N N+)! :

So the partition function is

Z =NX

N+=0

eU(N+)N!

N+!(N N+)! NX

N+=0

ZN+ : (6.21)

For a large system, N+ and N N+ are much greater than one, the partition function is over-whelmingly dominated by the maximum term ZN+ for which N+ = NM. To determine themaximum term, we use

@ lnZN+@N+

= 0: (6.22)

With the Sterlings formula,

ln ZN+ = U(N+) N+ lnN+N

(N N+) ln N N+N= N u(N+) TkB (p+ ln p+ + p ln p) : (6.23)

The free energy per spin as a function of magnetization is

F = kBT lnZN+ = HM + 1

2nJM2

!+ kBT

1 + M2

ln1 + M2

+1 M2

ln1 M2

!(6.24a)

=u T s; (6.24b)

where the entropy per spin is

s = kB (p+ ln p+ + p ln p) :

Eq. (6.22) amounts to finding the magnetization with minimum free energy. At zero tempera-ture, the free-energy minima are the same as the interaction energy minima, at which the spinswith ferromagnetic coupling are fully polarized. When the temperature is raised, the role of en-tropy (randomness) becomes important. The fully polarized state has zero entropy. Decreasingmagnetization would increase the entropy. When the temperature is very high, the reduction offree energy by increasing entropy would be much more significant than by reducing the interac-tion energy (or the interaction is irrelevant as in a non-interacting system). The system at hightemperature would be of no magnetization in absence of external magnetic field H. The phasetransition between the ordered ferromagnetic and the disorder states is due to the competitionbetween the interaction energy and the entropy.

The minimum free energy equation gives

H + nJM + kBT2

ln1 + M1 M = 0; (6.25)


(a)

T=1.6 Tc

0.4 Tc

(b)0.85 Tc

1.15 Tc

Figure 6.1: (a) The intersections between the two curves y = M and y = tanh(TcM=T ) deter-mine the free energy minima. (b) Free energy as a function of magnetization M for varioustemperatures.

orH nJM = kBT tanh1 M: (6.26)

This is the Bragg-William equation.Consider the ferromagnetic case (J < 0). For spontaneous magnetization (symmetry break-

ing), we set H = 0. The B-W equation becomes

M = tanhTcMT

; (6.27)

whereTc nJkB (6.28)

is a positive temperature. The tanh curve is the steepest at M = 0 with slope Tc=T . So, as shownin Fig. 6.1 (a), when T > Tc, the tahn curve and the curve y = M has only one intersection atM = 0 and the free energy has one minimum [Fig. 6.1 (b)]. When T < Tc, the two curves havethree intersections and the free energy at M = 0 becomes a local maximum, i.e., the systemis unstable at M = 0. The two intersections at M0 correspond to two minima of the freeenergy. The system has to decide to take one. The spins are spontaneously polarized and the Z2symmetry is broken. Tc is the critical temperature for the ferromagnetic phase transition.

6.3.3 Critical exponentsTo consider the critical phenomena, we expand the free energy near the critical point at Tc =nJ=kB, Mc = 0, and Hc = 0.

1. Critical magnetizationWe define a dimensionless temperature

t = (Tc T )=Tc: (6.29)The Bragg-Willian equation can be expanded as (at H = 0)

M(1 + t + t2 + ) = tanh1 M = M + 13M3 + ; (6.30)


or1 + t + t2 + = 1 + 1

3M3 + (6.31)

For T ! Tc 0+, the scaling isM t; (6.32)

with critical component = 1=2.

2. Critical susceptibilityThe susceptibility is

=

@M@H

!H=0

: (6.33)

We shall check how it scales as the temperature approaches to the critical value from below orfrom above. For this purpose, we expand the BW equation as

HkBT

+TcTM = M +

13M3 +

15M5 + : (6.34)

By dierential we have

kBT+TcTM0 = M0 + M2M0 + M4M0 + : (6.35)

As T ! Tc + 0+, we have M ! 0 at H ! 0, so!

kBTcjtj+ ; (6.36)

with the critical exponent + = 1.For T ! Tc 0+, we use M !

p3jtj at H = 0 and get!

2kBTct ; (6.37)

with the critical exponent = 1.

3. H-M scalingAt the critical temperature, we see how the magnetization scales with the external magneticfield. The BW equation becomes

HkBTc

+ M = M +13M3 + : (6.38)

Thus we haveH M (6.39)


4. heat capacityThe specific heat capacity is

C =@u@T

= (H + nJM) @M@T

: (6.40)


For T > Tc, M = 0 at H = 0, soC = 0 t+ ; (6.41)

with the critical exponent + = 0. For T ! T 0+, M jtj1=2 at H = 0, thus

C ! 32kB t ; (6.42)


5. Comparing with van der Waals gasThe critical exponents can be summarized as

+ = = 0; (6.43a) = 1=2; (6.43b)

+ = = 1; (6.43c) = 3: (6.43d)

These critical exponents are exactly the same as van der Waals gases if we make the mapping

P$ H; v vc c $ M:

The two phase transitions look very dierent at the first sight. But they are deeply related toeach other by their symmetry breaking classes and order parameters. The liquid-gas transitionbreaks the reflection symmetry of the system (along the liquid-gas surface, the system is notinvariant under reflection) and the order parameter, being the dierence between the densitiesof the two sides of the surface, is a real number (scalar). The ferromagnetic transition breaks thesymmetry between the pointing up and pointing down configurations and the order parameter,being the dierence between the probability of spins pointing up and that of spins pointingdown, is also a real number (scalar). In both cases, a Z2 symmetry is broken.

6.4 Exact solution of one-dimensional Ising modelThe one-dimensional Ising model has Hamiltonian

H = hNXi=1

zj + JN1Xi=1

zizi+1: (6.44)

Here we have assumed the periodic boundary condition N+i i. The eigen states are

jsi js1; s2; : : : ; sNi; (6.45)

with eigen energy

Es = hNXi=1

si + JN1Xi=1

sisi+1; (6.46)

6.4. EXACT SOLUTION OF ONE-DIMENSIONAL ISING MODEL 83

where si = +1, or 1. The partition function isZ =

Xs

eEs =Xs1=1

Xs2=1

XsN=1

ehs1Js1s2es2Js2s3 esN1JsN1sNehsNJsN s1 : (6.47)

An intriguing method of evaluation of the summation above is to write it into a matrix product

Z =Xs1=1

Xs2=1

XsN=1

hs1jDCjs2ihs2jDCjs3ihj isN1jDCjsNiisN jDCjs1i; (6.48)

with the definition

hsjDjs0i ss0 exp(hs); (6.49a)hsjCjs0i exp Jss0 ; (6.49b)

or in the matrix form in the basis of fj + 1i; j 1ig

D "exp(+h) 0

0 exp(h)#= cosh(h) + Z sinh(h); (6.50a)

C "exp(J) exp(+J)exp(+J) exp(J)

#= exp(J) + X exp(J); (6.50b)

where X=Y=Z are short-hand notations of Pauli matrices x=y=z. The partition function becomes

Z =Xs1=1

hs1j (DC)N js1i = Trh(DC)N

i= Tr

D1=2CD1=2

N: (6.51)

Now the problem is reduced to the solution of the eigen values of a 2 2 matrix D1=2CD1=2.Suppose its eigen values are + and with + > , the partition function would be

Z = N+ + N ; (6.52)

and the free energy per spin is

G = kBTN

lnN+ +

N

= kBT ln + kBT 1N ln2666641 +

+

!N377775! kBT ln +; (6.53)

at the thermodynamic limit (N ! 1). The partition function of a one-dimensional Ising mod-el thus is related to the ground state (which has largest partition probability +) of a zero-dimensional model. In general, the partition function of a d-dimensional classical system isrelated to the ground state of a (d 1)-dimensional quantum system. This feature will later beused to calculate the partition function of a two-dimensional Ising model.

Now let us give the final result for the 1D Ising model. The matrix

D1=2CD1=2 ="exp(J + h) exp(+J)

exp(+J) exp(J h)#

(6.54)


has eigen values

= eJ cosh (h) qe2J + e2J sinh2 (h): (6.55)

The free energy per spin at the thermodynamic limit is

G = kBT ln"eJ cosh (h) +

qe2J + e2J sinh2 (h)

#: (6.56)

The free energy is analytical for all real and h. The magnetization is

m = @G@h

=kBT+

@+@h

=sinh(h)q

e4J + sinh2(h): (6.57)

For h = 0, m = 0, i.e., there is no spontaneous magnetization at any finite temperature. Forferromagnetic coupling (J < 0), when the temperature approaches zero (! 1),

m! 1; (6.58)for arbitrary small external field h (as long as sinh(h) exp(2J) which would be the casewhen T J=kB and h > kBT ). In this sense, we can take the absolute zero temperature as thecritical point.

The exact result is obviously dierent from the mean-field theory. The discrepancy is due tothe fluctuation. To show this, let us calculate the fluctuation of the magnetization. The secondorder correlation function of the fluctuation is defined as

g(n) hzjzj+ni hzjihzj+ni: (6.59)This correlation quantifies how much the spins are in a long-ranged order. In the Bragg-Williammean-field theory discussed before, this correlation function is zero. In the exact solution (weconsider the case for h = 0), the correlation is

g(n) hzjzj+nih=0 m2 = hzjzj+nih=0: (6.60)For h = 0, the partition function is

Z =Xs

eJP

i sisi+1 : (6.61)

The correlation function can be calculated from the partition function by varying the couplingsbetween dierent spins,

g(n) =hs js j+ni = hs js j+1s j+1s j+2 s j+n1s j+ni=1Z

Xs

s js j+1s j+1s j+2 s j+n1s j+neJP

i sisi+1

=

0BBBBB@ 1()nZ Xs

@

@J j

@

@J j+1 @

@J j+n1e

Pi Jisisi+1

1CCCCCAJ j=J

: (6.62)

6.5. 2D ISING MODEL: HIGH-TEMPERATURE EXPANSION 85

Noticing that the matrices

C = exp(Ji) + X exp(Ji) (6.63)for dierent Ji commute with each other, we haveX

s

eP

i Jisisi+1 =Yi

+(Ji) +Yi

(Ji); (6.64)

with(Ji) eJi e+Ji : (6.65)

The correlation function becomes

g(n) =

0BBBBB@ 1()nQi +(Ji) @@J j @@J j+1 @@J j+n1Yi

+(Ji)

1CCCCCAJ j=J

= tanhn (J) : (6.66)

Defining the correlation length

1ln (tanh jJj) > 0; (6.67)

the correlation function has the form

g(n) = (1)n exp(n=); (6.68)where the 1 is for ferromagnetic and anti-ferromagnetic couplings, respectively. As T ! 0,the correlation length

! 2 exp(2jJj); (6.69)which diverges at the critical point (Tc = 0).

6.5 2D Ising model: High-temperature expansionWe consider the Ising model in a 2D lattice

H = JXhi ji

i j; (6.70)

in which hi ji denotes a pair of spins in the nearest neighbors and i take values from 1. Thepartition function is

ZH = Tr exp

0BBBBBB@JXhi ji

i j

1CCCCCCA = 2 coshq=2 KN 2NTrYhi ji

1 + vi j

; (6.71)

where K J; v tanhK, and q is the number of nearest neighbors of each site. The productterm

I 2NTrYhi ji

1 + vi j

;


(a) (b)

Figure 6.2: (a) Closed loops of 4 bonds (solid), 6 bonds (dotted), and 8 bonds (dashed). (b) Asquare lattice (solid) and its duality lattice (dotted).

can be expanded with respect to v for jKj 1. The expansion can be done similarly to thecluster expansion using the diagram representation.

Notice that a term likeTri1 j1i2 j2 ik jk ;

is non-zero if and only if the sites connected by the bonds

i1 j1i2 j2 ik jk;form a loop (not necessarily connected). The first order term consists of the smallest loops, i.e.,those loops around a unit cell which contains four bonds, called unit loops. The trace of the firstorder terms yields

2N (coshK)2N [Nvq] :Similarly, the next order would contain loops of 6 bonds, and then 8 bonds [see Fig. 6.2(a)].Counting all such loops, we can group diagrams according to how many unit cells are enclosedby the loops,

ZH =

21 v2

!N "1 + Nvq +

qN2v2(q1) +

N(N q 1)2

v2q + #; (6.72)

in which the second term within the brackets is contribution of terms with unit loops, the thirdterm is contribution of terms with two annexed unit loops, the fourth term is contribution ofterms with two separated unit loops, etc.

It may also be possible to have a low-temperature expansion for T ! 0, if the excitationfrom the ground state can be classified in some way. Particularly, for the Ising model withferromagnetic coupling, the ground state is known, and the excited states are spin flips from theordered state. There will be cases of one spin flipped, two spins in the nearest neighbors flipped,two spins not connected flipped, and so on. Thus the partition function can be expanded as

ZL = eE0"1 + Nuq +

N(N q 1)2

u2q +qN2u2(q1) +

#; (6.73)

6.6. EXACT SOLUTION 2D ISING MODEL* 87

with the expansion parameter u e2J=(kBT ). There is a one-to-one mapping between a flippedspin in this low-temperature expansion and a unit loop in the high-temperature expansion. Or aunit loop can be viewed as a spin in a duality lattice [see Fig. 6.2(b)]. Thus we have

ZH(v) =

2u1 v2

!NZL(u): (6.74)

Thus the high- and the low-temperature expansions are related to each other by the followingmapping

tanh e2K u, v e2 tanhK: (6.75)The solution of

u = v;

istanhK = tanh =

p2 1;

for which the factor2u

1 v2 = 1;and the two expansion results would have exactly the same value. The two functions, however,are expanded from dierent orders (with dierent symmetries) of the system and have dier-ent analytical properties. At low-temperature, the high-temperature expansion (if converges)contains entropy contribution from two equivalent ordered phases, but the low-temperature isobtained from only one local minimum. Thus we expect an order-disorder phase transitionoccur at

K = ;

which is the result derived from the exact solution of the 2D Ising model.

6.6 Exact solution 2D Ising model*

6.6.1 Exact solution by transfer matrix methodWe consider a 2D Ising model on a NN square lattice. The Hamiltonian in absence of externalfield is

H = JXi; j

zi; jzi+1; j J

Xi; j

zi; jzi; j+1; (6.76)

with periodic boundary conditionsi; j = i+N; j = i; j+N . Here for convenience we have changedthe sign of J so that the ferromagnetic coupling is positive. The eigen states are

jsi js11; s12; : : : ; s1N; s21; s22; : : : ; s2N; : : : ; sN1; sN2; : : : ; sNNi; (6.77)with eigen energy

Es = JXi; j

si jsi+1; j JXi; j

si jsi; j+1; (6.78)


where si j 2 f+1;1g. The partition function is

Z =Xs

eEs

=Xsi j

eKP

j s1 js1; j+1eKP

j s1 js2; jeKP

j s2 js2; j+1eKP

j s2 js3; j eKP j sN jsN; j+1eKP j sN jsN+1; j ; (6.79)

where K J=(kBT ). We can denote the states of a row of spins by a state vector asjsi;1ijs1;2i jsi;Ni jii; (6.80)

with the jth component of i being i( j) si; j. Each row has 2N possible states. Similarly tothe solution of the 1D Ising model, we define 2N 2N transfer matrices as

hjDj0i exp0BBBBBB@K NX

j=1

( j)0( j + 1)

1CCCCCCA ;0 ; (6.81a)hjCj0i exp

0BBBBBB@K NXj=1

( j)0( j)

1CCCCCCA : (6.81b)Obviously, D is a diagonal matrix in the basis of ji. The partition function function can bewritten as

Z =X

1;01;2;

02;:::;

0N

h1jDj01ih01jCj2ih2jDj02i h0N jCj1i

= TrC1=2DC1=2

N: (6.82)

The matrices C and D, being 2N 2N matrices, can be expressed as functions of 3N Pauli-matrices X=Y=Z. Let us first consider C. We notice that the matrix elements of C can befactorized as functions of (( j); 0( j) for dierent js, namely,

hjCj0i =Yj

eK( j)0( j): (6.83)

Thus C can be written as the outer product of N 2 2 matrices as

C =NOj=1

C j; (6.84)

withh( j)jC jj0( j)i eK( j)0( j); (6.85)

or

C j ="eK eK

eK eK

#= eK + eKX j =

p2 sinh(2K) exp(X j); (6.86)

wheretanh() exp(2K) = e2J=(kBT ): (6.87)


Thus

C =NOj=1

C j = [2 sinh(2K)]N=2 eP

j X j : (6.88)

Note that would be an imaginary number if J < 0 (anti-ferromagnetic coupling). Throughoutthis chapter, we shall consider the ferromagnetic case. Now let us consider D. It should be notedthat here the Pauli matrices (X=Y=Z) j are not for real spins but introduced for mathematicalconvenience. The matrix elements of D would involve coupling between neighboring indices( j)0( j + 1). In terms of the Pauli matrices introduced for expressing C, it is obvious that

D = exp

0BBBBBB@KXj

Z jZ j+1

1CCCCCCA : (6.89)The problem is reduced to finding the maximum eigen value of the matrix

D C1=2DC1=2 = [2 sinh(2K)]N=2 exp0BBBBBB@2 X

j

X j

1CCCCCCA exp0BBBBBB@KX

j

Z jZ j+1

1CCCCCCA exp0BBBBBB@2 X

j

X j

1CCCCCCA : (6.90)This amounts to finding the ground state eigen value of an 1-dimensional XY model in anexternal field, which can be solved by the Jordan-Wigner transformation.

6.6.2 Solution of 1D XY modelJordon-Wigner transformation

To write the matrix in a standard XY model, we make a unitary transformation (a rotation aboutY-axis)

X j ! Z j; Y j ! Y j; Z j ! X j: (6.91)One can verify that the commuting properties of the Pauli matrices are kept. In this coordinates,

D = [2 sinh(2K)]N=2 exp

0BBBBBB@2 Xj

Z j

1CCCCCCA exp0BBBBBB@KX

j

X jX j+1

1CCCCCCA exp0BBBBBB@2 X

j

Z j

1CCCCCCA : (6.92)The Jordan-Wigner transformation transforms the spin operators into Fermi operators. It isdefined by

+j = cyj exp

0BBBBBB@i j1Xl=1

cyl cl

1CCCCCCA ; (6.93)where the raiser/lower operator are defined by j (X j iY j)=2. The other Pauli matrices inthe representation of the fermion operators are

Z j = iX jY j = 2cyjc j 1: (6.94)The intuitive understanding of this transformation is obtained by noticing that a fermion opera-tor change the number of particles in a site between 0 and 1 while a spin raiser/lower operator


changes the spin quantum number between the two values 1 and +1. The anti-commutingproperties of fermion operators are produced by the phase factor defined as the product of achain of spin operators. The inverse transformation is

cyj +j exp0BBBBBB@i j1X

l=1

Zl + 12

1CCCCCCA : (6.95)With the relation

exp

0BBBBBB@i j1Xl=1

Zl + 12

1CCCCCCAn = n exp0BBBBBB@i j1X

l=1

Zl + 12

1CCCCCCA ; (6.96)for j < n and j > n, respectively, it is easy to verify the fermionic commutators

fcm; cyng = n;m; fcn; cmg = fcyn; cymg = 0: (6.97)Note that the fermion operators are collective in the sense that it contains a chain of spin opera-tors.

In the representation of the fermion operators,

C = [2 sinh(2K)]N=2 exp

0BBBBBB@2Xj

cyjc j N1CCCCCCA = h2 sinh(2K)e2iN=2 exp

0BBBBBB@2Xj

cyjc j

1CCCCCCA : (6.98)For the expression of D, we find

X jX j+1 =+j +

j

+j+1 +

j+1

=cyj c j

cyj+1 + c j+1

; (6.99)

if j < N. For the special case j = N,

XNX1 =+N +

N +1 +

1

= exp

0BBBBBB@i N1Xj=1

cyjc j

1CCCCCCA cyN + cN cy1 + c1= exp

0BBBBBB@i NXj=1

cyjc j

1CCCCCCA cyN + cN cy1 + c1 (1)P cyN + cN cy1 + c1 : (6.100)where

P Xj

cyjc j; (6.101)

denotes the number operator. We notice that the operator C conserves the number of fermionsand the operator D changes the number of fermions always by 0 or 2. So all the states can beclassified into two invariant subspaces of C and D, with an even or odd number of particles, inwhich (1)P = 1, respectively. Thus, D can be written in the form as

D = exp

0BBBBBB@K NXj=1

cyj c j

cyj+1 + c j+1

1CCCCCCA ; (6.102)


with the boundary conditions

cN+1 = c1; even particles; (6.103a)cN+1 = + c1; odd particles: (6.103b)

Note that both boundary conditions keep the periodic boundary conditions for the physical spinssince the fermion operators are defined up to a phase factor (gauge freedom). Eventually,

D =h2 sinh(2K)e2

iN=2exp

0BBBBBB@Xj

cyjc j

1CCCCCCA exp0BBBBBB@K NX

j=1

cyj c j

cyj+1 + c j+1

1CCCCCCA exp0BBBBBB@X

j

cyjc j

1CCCCCCA ;(6.104)

corresponding to a 1D fermion model with bilinear coupling, and periodic or anti-periodic(twisted) boundary conditions for odd or even particles, respectively. Such a model can beexactly diagonalized by transforming the operators in real space into the momentum space(Fourier transformation),

bq N1=2Xj

c jeiq j=N ; (6.105)

with

q = 1;3; : : : ;(N 1); for even particles; (6.106a)q =0;2;4; : : : ;N; for odd particles; (6.106b)

where N has been assumed to be an even number without loss of generality. The inverse trans-formation is

c j N1=2Xq

bqeiq j=N : (6.107)

In the momentum representation

C =h2 sinh(2K)e2

iN=2exp

0BBBBBB@2Xq

byqbq

1CCCCCCA h2 sinh(2K)e2iN=2Yq0

Cq; (6.108a)

D = exp

0BBBBBB@KXq

"2 cos

qNbyqbq i sin

jqjN

byqb

yq + bqbq

#1CCCCCCA Yq0

Dq; (6.108b)

with

Cq exp2byqbq + 2b

yqbq

; ; (6.109a)

Dq exp2K cos

qN

byqbq + b

yqbq

i2K sin q

N

byqb

yq + bqbq

; (6.109b)

for 0 < q < N, and

C0 exp2by0b0

; (6.110a)

D0 exp2Kby0b0

; (6.110b)

CN exp2byNbN

; (6.110c)

DN exp2KayNaN

: (6.110d)

Now the 2N 2N matrix D is factorized into products of single particle operators with dierentjqj which are commutable with each other.


Eigenvalues

We just need to diagonalizeDq C1=2q DqC1=2q ; (6.111)

for 0 < q < N. The basis states are

j0; 0i; j1; 0i; j0; 1i; j1; 1i: (6.112)

The one-particle states j1; 0i and j0; 1i are already eigen states with eigen values

d1;0(q) = d0;1(q) = exp2 + 2K cos

qN

: (6.113)

In the basis of fj1; 1i; j0; 0ig, the following operators can be written into Pauli matrices

byqbq + byqbq =Z + 1; (6.114a)

ibyqbyq + ibqbq =Y; (6.114b)

and thereforeDq = e2+2K cos

qN eZe2K(Z cos

qN +Y sin

qN )eZ : (6.115)

The eigen values are

d(q) = exp2 + 2K cos

qN (q)

; (6.116)

withcosh[(q)] = cosh(2K) cosh(2) + sinh(2K) sinh(2) cos

qN: (6.117)

By convention, (q) is taken to be positive. As a function of q, the minimum of (q) is reachedas q! N,

(q) > min = 2 jK j : (6.118)Also, for q! 0,

(q)! 2 (K + ) : (6.119)Obviously, the largest eigen value for each q is d+(q).

States with even number of particles

Since the largest eigen value state has an even number (2) of particles, the largest eigen valuein the subspace of even particles is just

0 =h2 sinh(2K)e2

iN=2 YN>q>0

d+(q)

= [2 sinh(2K)]N=2 exp

0BBBBBB@12 Xq

(q)

1CCCCCCA ; (6.120)where q 2 f(N 1);(N 3); : : : ; (N 1)g.


States with odd number of particles

For states in the subspace of an odd number of particles, the momentum q could be 0 or N, forwhich the eigen values are given by

C1=20 D0C1=20 jni0 = exp(+2nK + 2n)jni0 = dn(0)jni0; (6.121a)

C1=2N DNC1=2N jniN = exp(2nK + 2n)jniN = dn(N)jniN ; (6.121b)

for the number of particles in the states to be n = 0 or 1. We should consider two dierentsituations.

(i) K > .In this case, if all the largest eigen values of dierent q are taken, the numbers of particles instates with 0 < jqj < N are even, the number of particles with q = 0 is 1, and that with q = N is0. Thus the total number of particles is odd. So the largest eigenvalue in the subspace is

o = [2 sinh(2K)]N=2 e2KY0


Largest eigenvalue of the whole Hilbert space

In the thermodynamic limit (N ! 1), we should haveNXk=1

(2k N 1) =NXk=1

(2k N): (6.128)

Thus in the case K > , we have the largest eigenvalue is

max = e = o = [2 sinh(2K)]N=2 exp

0BBBBBB@12 Xq

(q)

1CCCCCCA ; (6.129)and the state is two-fold degenerate. For K < , the largest eigen value is

max = e = [2 sinh(2K)]N=2 exp

0BBBBBB@12 Xq

(q)

1CCCCCCA > o; (6.130)and the state is non-degenerate.

6.6.3 Phase transitionIf the state with the largest eigenvaluemax is non-degenerate, the systemmust have no magneti-zation. Otherwise, a reflection of the system would lead to another state of the same eigenvalue.The magnetization is expected to occur when the largest-eigenvalue state has degeneracy. Thusthe phase transition temperature is determined by

K = ; (6.131)

or with the definition tanh() exp(2K) and K = J=(kBT ),

tanh

JkBTc

!= exp

2JkBTc

!: (6.132)

The solution isTc =

2

ln(p2 + 1)

JkB

2:269 JkB

: (6.133)

6.6.4 Critical exponentsThe free energy per spin is

G = kBTN2

lnNmax = kBT2

ln [2 sinh(2K)] kBT2

1N

Xq

(q): (6.134)

Defining = q=N, we have

() =(q) = arccoshcosh(2K) cosh(2) + sinh(2K) sinh(2) cos

qN

=arccosh [cosh(2K) cosh(2) + cos ] ; (6.135)

6.7. LIMITATION OF MEAN-FIELD APPROXIMATION 95

where sinh(2K) sinh(2) = 1 has been used. The summation over q can be written as

I 1N

Xq

(q) =1

Z 0()d = ln [2 cosh(2K) cosh(2)] +

2

Z 0ln

1 +p1 A2 sin2 2

d;

(6.136)where A 2 sinh(2K) cosh2(2K). So the free energy is

G = kBT ln [2 sinh(2K)] kBT

Z 0ln

1 +p1 A2 sin2 2

d (6.137)

The internal energy per spin isu = @(G); (6.138)

from which the specific heat can be calculated. The result near the critical point is

c(T ) 2kB

2JkBTc

!2ln1 TTc

+ constant: (6.139)This logarithm divergence is strikingly dierent from the mean field theory. In the expressionof a power-law divergence,

c(T ) 1 TTc

; (6.140)the logarithm divergence (which is much slower than any power-law divergence) is a specialcase with = 0+.

The calculation of the spontaneous magnetization is non-trivial (One has to figure out therepresentation of the spin operators in fermion operators). The critical exponent first writtendown by Onsager and than calculated by C. N. Yang is = 1=8 in

m (T ! Tc 0+) (Tc T ); (6.141)which is dramatically dierent from the mean-field result = 1=2.

The susceptibility has the critical behaviour (due to T. T. Wu et al)

(0;T ) jT Tcj ; (6.142)with = 7=4, again dierent from the mean-field value = 1.

6.7 Limitation of mean-field approximationAbove the Ising model is studied without specifying the lattice structure or even the dimension-ality. The phase transition exists regardless of the dimension of the lattice. By exact solution (tobe discussed later), however, Ising has shown that there would be no phase transition in a one-dimensional Ising model. Without exactly evaluating the partition function, the non-existenceof phase transition in 1D Ising models can be understood by considering the scaling propertiesof the interaction energy and the entropy.


Consider the state at zero temperature, in which all the spins pointing to the same direction.The state is a ground state and the entropy is zero. Let us first check howmuch the energy wouldbe increased if the chain contains Nd defects or domain walls, at which the polarizationdirection of the spins are switched. The interaction energy of two anti-parallel spins is greaterby (2J) than that of two parallel spins. Since the interaction is short-ranged (non-vanishingonly between nearest neighbors), the energy increased would be

E 2JNd: (6.143)The number of choices of the sites of these Nd defects is

N!Nd!(N Nd)! : (6.144)

This would increase the entropy by

S = kb lnN!

Nd!(N Nd)! kBNd lnNdN kB(N Nd) ln N NdN > kBNd ln

NNd

: (6.145)

The change of free energy isF = E TS : (6.146)

For a macroscopic system (N ! 1) at a finite temperature, no matter how low it is, the freeenergy reduction due to entropy increase always dominates the increase of the interaction ener-gy, until Nd becomes a number in the order of N. Thus the phase transition in 1D Ising modeloccurs at T = 0, or, there is no finite-temperature phase transition. The long-range order cannotsustain the fluctuation in a 1D Ising chain.

The existence of phase transition in 1D Ising models predicted by the mean-field theoryshows the inadequacy of the approximation in accounting for the fluctuation which is particu-larly important in 1D systems.

Now let us consider a 2-dimensional case. We notice that in 2D a point-like fluctuationcannot destroy the ferromagnetic order. A domain wall has to be a closed loop. To be simple,let us consider straight-line defects cut through the lattice of L L spins. The length of thedefect is L. The interaction energy increased by m such line defects is

E 2JmL: (6.147)The increase of entropy is

S kBm ln(L=m); (6.148)with a logarithm scaling with the system size, much slower than the linear scaling of the energyincrease. Thus the long-range ferromagnetic order in a 2D Ising model is stable against thefluctuation as long as the temperature is not too high.Note: The arguments above apply to the Ising model in which the phase transition breaks adiscrete symmetry. For a long-range order breaking a continuous symmetry such as the ferro-magnetic order in the Heisenberg model which has the interaction

H = JXhi ji

Si S j; (6.149)

6.7. LIMITATION OF MEAN-FIELD APPROXIMATION 97

The case is dierent. The Heisenberg model has continuous rotational symmetry. A fullypolarized ferromagnectic state which is a ground state breaks the continuous symmetry. Nowin 2D, a domain line can be formed with a transition region of thickness in the order of L. Theenergy change per line defect is

E J1 cos

L

(thickness) (surface size) J

1 cos

L

L L J: (6.150)

The entropy will still scale as m ln L, much faster than the energy scaling. Thus, the long-rangeorder breaking a continuous symmetry in a system with short-ranged interaction cannot sustainthe fluctuation in 2D space. This is a specific example of the famousMermin-Wagner theorem(also known as Mermin-Wagner-Hohenberg theorem or Coleman theorem) which states:No finite-temperature phase transition can occur to break a continuous symmetry in a 1- or2-dimensional system with short-range interaction.The rigorous proof actually follows the arguments above for the 2D Heisenberg model.

Bibliography

[1] T. D. Lee, Statistical Mechanics, (Shanghai, 2006), x 3.4-3.5.[2] R. K. Pathria, Statiscal Mechanics, 2nd ed. (Elsevier, Singapore, 2001), x 11.5-11.7.[3] M. Plischke and B. Bergersen, Equlibrium Statistical Physics, 2nd ed. (World Scientific,

Singapore, 1994), x 3.1-3.5, x 5.1.[4] R. J. Baxter, 399th solution of the Ising model, J. Phys. A: Math. Gen. 11, 2463 (1978).

[5] R. J. Baxter, Exactly Solved Models in Statistical Mechanics (Academic Press, New York,1982).

[6] C. N. Yang, Phys. Rev. 85, 809 (1952).

[7] N. D. Mermin and H. Wagner, Absence of Ferromagnetism or Antiferromagnetism in One-or Two-Dimensional Isotropic Heisenberg Models, Phys. Rev. Lett. 17, 1133 (1966).

[8] P. C. Hohenberg, Existence of Long-Range Order in One and Two Dimensions, Phys. Rev.158, 383 (1967).

[9] S. Coleman, There are no Goldstone bosons in two dimensions, Commun. Math. Phys. 31,259 (1973).

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Ch6 Phase Transition II - Ising model.pdf