1.4 Equipartition Theorem

Equipartition Theorem: If the dynamics of the system is described by the Hamiltonian:

2 H = Aζ + H0 where ζ is one of the general coordinates q1,p1, ,q3N ,p3N , and H0 and A are independent of ζ, then ··· 1 Aζ2 = k T 2 B βH where is the thermal average, i.e., the average® over the Gibbs measure e− . Thishi result can be easily seen by the following calculation: Aζ2e βHd3N qd3N p Aζ = − h i e βHd3N qd3N p R − 2 βAζ2 βH RAζ e− dζe− 0 [dpdq] = 2 e βAζ dζe βH0 [dpdq] R − − Aζ2e βAζ2 dζ = R − e βAζ2 dζ R − ∂ βAζ2 = R ln e− dζ −∂β Z ∂ 1 Ax2 x = ζ β = ln β− 2 e− dx −∂β ∙ Z ¸ ³ p ´ 1 = 2β 1 = k T 2 B where dζ [dpdq]=d3N qd3N p, i.e., [dpdq] stands for the phase-space volumn element without dζ.

Note that if 2 2 H = Aipi + Biqi wheretherearetotalM terms in theseX sums andXAi and Bi are constant, independent of pi,qi , then { } 1 H = kBT M h i 2 µ ¶ 1 i.e., each quadratic component of the Hamiltonian shares 2 kBT of the total energy. For example, the Hamiltonian for interacting gas particles is p2 H = i + U (r r ) 2m i j i ij − X X 7 where U (ri rj) is the potential energy between any two particles. The equipartition theo- − 3 rem tells us that the average kinetic energy is 2 kBT for every particle (since there are three translational degrees of freedom for each particle). Furthermore, for an ideal gas, i.e. gas particlesdonotinteract,wehavetheaveragetotalenergy 3 U = k TN 2 B where N is the total number of particles. Then, the speciﬁc heat for this system is

∂U CV ≡ ∂T µ ¶V ∂ 3 = k TN ∂T 2 B µ ¶ 3 = k N 2 B

8 1TheIsingModel

For example, Fe or Ni exhibit macroscopic magnetic ﬁeld via spontaneous polarization in the same direction for temperatures below the so-called critical temperature Tc, where Tc is referred to as the Curie temperature. For T>Tc, the spins are randomly orientated, there is no macroscopic magnetic ﬁeld.

1.1 Model – Ising spins The Ising model is a caricature of spin dynamics. It teaches us many interesting aspects of spin dynamics and can be used to contrast real spin systems. Especially, two dimension Ising model is a nontrivial example of phase transitions that can be worked-out exactly.

Themodel:weconsiderann-dimensional lattice (usually with periodic boundary condi- tions). For two dimensions, for example, the lattice structure can be cubic or hexagonal and it has N sitesintotal.Ateachsitethereisaspindescribedbythevariablesi ,i=1, 2, ,N with its value: ··· +1 spin up s = i 1 spin down ½ − Then, a set of si,i=1, ,N constitutes a conﬁguration of the system. The energy in a conﬁguration is{ ··· } N

E si = Jijsisj H si { } − − i,j i=1 Xh i X where J is the exchange energy between two spins and H is a constant external magnetic ﬁeld. i, i denotes the nearest next neighbor interactions. If the exchange interaction is isotropic,h theni Jij = J and N

E si = J sisj H si { } − − i,j i=1 Xh i X For J>0, it describes ferromagnetism, in which two neighboring spins like to line up in the same direction. For J<0, it is antiferromagnetism, in which two neighboring spins tend to be in the opposite direction in order to lower the energy of the system. We consider only the case J>0 below.

The partition function for the Ising model is

βE si Q (H, T)= e− { } s s ··· s X1 X2 XN N there are total 2 termsinthesummationandeachsi, ranging over 1 is independent from another. ±

1 Thermodynamic functions can be obtained via the Helmholtz free energy:

A (H, T)= kBT log Q (H, T) − For example, the internal energy is

2 ∂ A u (H, T)= kBT − ∂T k T µ B ¶ the heat capacity is ∂ c (H, T)= u (H, T) ∂T and the magnetization is ∂A M (H, T)= −∂H Itcanbeeasilyseenthat N

M (H, T)= si * i=1 + X where istheensembleaverage. h· · · i The so-called spontaneous magnetization is

M (0,T) =0 6 for the case of Jij > 0, i.e., there is a ﬁnite magnetization in the absence of the external ﬁeld.

1.1.1 Transfer-Matrix Method Here we discuss only the 1D Ising model. The method can be generalized to 2D with considerably diﬃculty. Consider a chain of N spins, i.e.,

s1 sN+1 ≡

For the conﬁguration s1, sN ,sk = 1, the energy of the system is { ··· } ± N N

E = J sksk+1 H sk − − k=1 k=1 X X 2 and the partition function is

β N (Js s +Hs ) Q (H, T)= e k=1 k k+1 k s s ··· s S X1 X2 XN Now we can evaluate this explicitly using a matrix formulation. First, using the periodic boundary condition, we rewrite Q in a symmetric form:

β N Js s + 1 H(s +s ) Q (H, T)= e k=1[ k k+1 2 k k+1 ] s s ··· s S X1 X2 XN It turns out that there is a natural matrix formulation underlying this expression. Deﬁne a 2 2 matrix P with its elements × 1 β[Jss0+ 2 H(s+s0)] Pss0 = e where s, s0 takesthevalueof 1 independently. More explicitly, we have ± β(J+H) P++ = e β(J H) P+ = e − − βJ P+ = P + = e− − − i.e., β(J+H) βJ e e− P = βJ β(J H) e− e − µ ¶ Therefore,

Q (H, T)= Ps1s2 Ps2s3 PsN s1 s s ··· s ··· X1 X2 XN = P N s1s1 s1 X ¡ ¢ = TrPN (a consequence of periodic BCs) N N = λ+ + λ − where λ+,λ are the eigenvalue of P. − βJ 2 4βJ 1/2 λ = e cosh (βH) sinh (βH)+e− ± ± h i λ+ λ for all values of H. P is referred to¡ as the transfer matrix.¢ ≥ −

As N , only λ+ is relevant since →∞ 1 λ N log Q (H, T)=logλ +log 1+ + N + λ Ã µ − ¶ ! log λ+ as N → →∞ Physical results:

3 1. The Helmholtz free energy per spin is

1 2 4βJ 1/2 A (H, T)= J kBT log cosh (βH)+ sinh (βH)+e− N − − h ¡ ¢ i 2. The magnetization per spin is 1 sinh (βH) M (H, T)= N 2 4βJ 1/2 sinh (βH)+e− £ ¤

For T>0, ∀ 1 M (H =0,T)=0 N therefore is no spontaneous magnetization for 1-D Ising model.

1.1.2 Symmetry Breaking

A more general model for spins is, for example, that the energy of the system is

E = J Si Sj · i,j Xh i 2 where Si is a vector with S S =S , where S is a ﬁxed number. Then the total magnetic moment is determined by · βE Tr e− M M = Tr(e βE) h i ¡ − ¢ In general, E is invariant under rotation in the absence of external ﬁeld. Therefore, M =0 h i since M and M occur with equal probability. A deep question then is how come we have ferromagnetism?− This answer lies the fundamental concept of spontaneous symmetry breaking.

4 Spontaneous Symmetry Breaking Despite the fact that a Hamiltonian is invariant under certain symmetry groups, the ground state of the system does not have to possess the same symmetry of the Hamiltonian. That is, a ground state can be degenerate. For example, a ferromagnet ground state is not rotationally invariant – any one of the degenerate ground state is a physical solution. Once a system magnetizes along a certain direction, it cannot make a transition to another direction (Although, for this transition, there is no energy required, it has almost zero probability since all spins have to rotate simultaneously exactly thesameamount). How can we mathematically capture the spontaneous symmetry breaking? In the case of spontaneous magnetizatoin, we can add an exteranl field H,thenletH 0 in the end: → β(E MH) M 1 Tr Me− − h i lim lim β(E MH) V H 0 V V ³ − ´ ≡ → →∞ Tr e− where it is important to have the correct order of¡ limiting¢ processes, which reflects the underlying thermodynamic limit. If the limit H 0 is taken before the thermodynamic limit, then we have → β(E MH) 1 Tr Me− − lim lim β(E MH) =0 V H 0 V ³ − ´ →∞ → Tr e− i.e., there would be no spontaneous magnetization.¡ ¢ For the Ising model, the Hamiltonian is invariant under the discrete, reflectional symmetry: Si Si →− For a finite external field H, we have the following phase diagram:

1.2 Curie-Weiss Model (Mean-ﬁeld model)

The energy of the Curie-Weiss model is given by

J N N E S = S S H S N i N i j j { } − i,j=1 − j=1 X X 5 which can also be expressed as

2 1 N N E S = JN S H S N i N i j { } − Ã i ! − j=1 X X or 1 N N E S = J S S H S (1) N i i N j j { } − i Ã j ! − j=1 X X X where H is the external magnetic ﬁeld. Note that

1. due to the long-rang nature of the interaction (i.e., everying spin interacts with every other spin with an equal strength of coupling in this model), we can equivallently view the model as a spin interacting with its N neighbors in an N-dimensional lattice.

2. The bracked term in Eq. (1) canbeviewedasaneﬀective magnetic ﬁeld induced by all the spins. The following approximation is based on this observation.

The partition function of this system is

βJ N N QN (T,H)= exp SiSj + βH Si ··· N S S Ã i,j=1 i=1 ! Xi XN X X We use the mean-field approximation to evaluate this partition as follows. First, we view 1 N N j Sj as if it were external effective magnetic field, and let

P N 1 m = S N j j X then

QN = exp (βJm + βH) Si ··· S S Ã i=1 ! Xi XN X N

= exp [β (Jm+ H) Si] ··· S S i Xi XN Y N

= exp [β (Jm+ H) Si] i Ã S ! Y Xi N β(Jm+H) β(Jm+H) = e + e− i Y ¡ ¢ =2N coshN [β (Jm+ H)] .

6 The corresponding Helmholz free energy under this mean-ﬁeld approximation is

AMF = kBT log QN − N N = kBT log 2 cosh [β (Jm+ H)] − 1 N £ ¤ in general, the thermal average of N j Sj is determined by

1 PN ∂ A S = N j ∂H N * j + − X µ ¶ a self-consistency requires that 1 N m = S N j * j + X i.e.,

∂ N N m = kBT log 2 cosh [β (Jm+ H)] −∂H − =tanh[β©(Jm+ H)] . ª

Phase Transition If H =0, i.e., there is no external ﬁeld, m =tanhβJm

if βJ > 1, i.e., kBT

Note that

7 1. It turns out that this mean-ﬁeld result is exact. We will demonstrate this later.

2. The Curie-Weiss model has spontaneous magnetization for T

3. Tc corresponds to the energy scale kBTc J. ∼ 1 N 4. Note that m = N j Sj should have differentvaluesfordifferent configurations. The mean-field approximation assumes that in the large-N limit, the sum converges to a mean-value withoutP fluctuations. Philosophically speaking, there is a law of large number or central-limit theorem lurking behind this mean-field approximation for the contribution from fluctuations to be neglected.

1.2.1 Large Deviation Principle for Curie-Weiss model – the exact result

n 1 1 Ωn = 1, +1 ,ρ= δ 1 + δ+1 {− } 2 − 2

Let Pn betheproductmeasureonΩn with one-dimensional marginals ρ. Clearly, for any ω = ωi,i=1, ,n Ωn, { ··· } ∈ 1 n Pn ω = { } 2 µ ¶ The energy of the system is

2 1 n n 1 n H (ω)= ω ω = ω n 2n i j 2 n i − i,j=1 − Ã j=1 ! X X The Gibbs measure is 1 βHn(ω) Pnβ ω = e− { } Zn (β) where

βHn(ω) Zn (β)= e− Pn (dω) ZΩn 1 = e βHn(ω) − 2n n Ωn X∈ We discuss the magnetization S (ω) 1 n n = ω n n i i=1 X We are looking for the following property if there is a phase transition as the mean-ﬁeld approximation indicates: S (ω) n 0 for β 1 n → ≤ m (β) for β>1 → ± 8 where m (β) (0, 1). That is, we want to study the question: ∈

Sn (ω) Pnβ dx =? n ∈ ½ ¾

Sn i.e., is there a large deviation principle for n with respect to Pnβ?Notethat S n [ 1, +1] n ∈ − In order to demonstrate the equivalence between Laplace principle and the large-deviation principle, we need to ﬁnd a rate function Iβ on [ 1, +1] such that for any continuous function − f :[ 1, +1] R, − →

1 Sn nf( n ) lim log EPnβ e =sup f (x) Iβ (x) n n x [ 1,+1] { − } →∞ n o ∈ − Note that

2 Sn 1 Sn β Sn nf( ) nf( n )+n 2 ( n ) P e n = e Pn (dω) E nβ k l Zn (β) Ωn n o Z 1 nf(x)+n β x2 Sn = e 2 Pn dx (2) Zn (β) [ 1,+1] n ∈ Z − ½ ¾

β Sn 2 n ( ) βHn(ω) Note that e 2 n is simply e− . 2.

2 n β ( Sn ) Zn (β)= e 2 n Pn (dω) ZΩn n β x2 Sn = e 2 Pn dx [ 1,+1] n ∈ Z − ½ ¾ 3. Cramier Theorem (which is the special case of what we discussed before):

1 1 n Sn(ω) Let Λ = 1, +1 ,ρ = δ 1 + δ+1,ω Ωn,Sn (ω) j=1 ωj,then ,n N {− } 2 − 2 ∈ ≡ n ∈ satisﬁes the large-deviation principle on [ 1, +1] with rate function: n o − P 1 1 I (x)= (1 x)log(1 x)+ (1 + x)log(1+x) 2 − − 2

Proof: For x [ 1, +1] , ∈ − S (ω) n x n ∼

9 n if and only if there are approximately 2 (1 x) components such that ωj =1and n − 2 (1 + x) components such that ωj =+1. Therefore,

Sn (ω) 1 1 Pn x Pn Ln ( 1) = (1 x) ,Ln (+1) = (1 + x) n ∼ ≈ − 2 − 2 ½ ¾ ½ 1 1 ¾ nIρ( (1 x), (1+x) ) e− 2 − 2 ( Large-deviation principle) ≈ { } ∵

where Ln ( 1) is the empirical frequency of 1 in ω. Note that − − 1 1 1 1 1 2 (1 x) 1 2 (1 + x) Iρ (1 x) , (1 + x) = (1 x)log − + (1 + x)log 2 − 2 2 − 1 2 1 µ½ ¾¶ 2 2 = I (x)

QED

Applying Cramier theorem to the denominator and numerator of Eq. (2),weobtain

1. 1 β 2 lim log Zn (β)= sup x I (x) n n x [ 1,+1] 2 − →∞ ∈ − ½ ¾ 2.

1 nf(x)+n β x2 Sn β 2 lim log e 2 Pn dx =sup f (x)+ x I (x) n n [ 1,+1] n ∈ x [ 1,+1] 2 − →∞ Z − ½ ¾ ∈ − ½ ¾ with 1 1 I (x)= (1 x)log(1 x)+ (1 + x)log(1+x) 2 − − 2 Now define β 2 β 2 Iβ (x) I (x) x inf I (y) y ≡ − 2 − y [ 1,+1] − 2 ∈ − ½ ¾ therefore 1 Sn nf( n ) lim log EPnβ e =sup f (x) Iβ (x) n n x [ 1,+1] { − } →∞ n o ∈ − that is Sn satisfies the Laplace principle on [ 1, +1] , therefore, Sn satisfies large-deviation n − n principle with the rate function Iβ (x) , i.e.,

S (ω) n nIβ (x) Pnβ dx e− n ∈ ∼ ½ ¾

In the limit of n , only those minima x∗ of Iβ (x) such that →∞

Iβ (x∗)=0

10 will have a nonvanishing contribution. Obviously, x∗ satisﬁes

Iβ0 (x∗)=0 i.e., 1 I0 (x∗)=βx∗ or x∗ =(I0)− (βx∗) Note that I (x)= 1 (1 x)log(1 x)+ 1 (1 + x)log(1+x) , It is easy to verifty that 2 − − 2

x∗ =tanhβx∗

Therefore, we have three solutions: 0, m (β) , i.e., ±

j ωj W. δ0 for 0 β 1 Pnβ dx = 1 1 ≤ ≤ n ∈ ⇒ δm(β) + δ m(β) for β>1 ½P ¾ ½ 2 2 − i.e., the set εβ = x [ 1, +1] ,Iβ (x)=0 { ∈ − } Sn(ω) is the support of Pnβ dx . Note that x∗ εβ. n ∈ ∈ We can also computen the freeo energy, which is simply

1 β 2 lim log Zn (β)= sup x I (x) n n x [ 1,+1] 2 − →∞ ∈ − ½ ¾

β 2 Clearly, maximizing x I (x) leads to the solution x∗. 2 −

1.3 Renormalization Groups/Phase Transitions

Critical properties reflect the nature of long wavelength fluctuation, not of short-distance details. An effective and efficient way of understanding their underlying dynamics is by coarse-graining out the irrelevant scales all the way up to the correlation length.

1.3.1 Block-spin Transformation The concept of the block spins captures the "average" spin within a block materials near the critical point, where the notion of average depends on a particular coarse-graining procedure inuseanditaccentuatesthefactthattheﬂuctuations with the block are not relevant. We illustrate this notion using the one-dimensional Ising model. Although we have already learnt that it has no phase transition at ﬁnite (i.e., T =0)temperatures, it is nevertheless 6

11 an instructive example in which the coarse-graining (block-spin transformation) procedure can be carried out exactly. The Ising model is N N

E = J sisi+1 H si − i − i=1 or X X N N

βE = j sisi+1 h si − i − i=1 X X where j = βJ, h = βH. The corresponding transfer matrix is j+h j e e− P = j j h e− e − µ ¶ 1 µ µν ν ≡ ⎛ µ ⎞ µ ⎜ ⎟ βJ βH ⎝ ⎠ where µ = e− , and ν = e− . For ferromagnetic J>0, without loss of generality, we can assume h 0, then, µ, v [0, 1]. Recall that the partition function is ≥ ∈ N QN = TrP . Now consider the block which consists of every two nearest next neighbors and rewrite this QN as if the site were block spins, i.e., 2 N/2 QN = Tr P 2 where P2 P would be the transfer matrix¡ for¢ the chain of spins which are block type. Obviously,≡ 2 1 1 µ + 2 2 ν + 2 µ ν ν P2 = P = 2 ⎛ 1 2 ν ⎞ ν + µ + 2 ⎜ ν µ ⎟ If the block spin can be viewed as effective⎝ spins, P2 should be⎠ expressed in the form: 1 µ0 µ0ν0 P2 = c ⎛ ν0 ⎞ µ0 ⎜ µ0 ⎟ Note that the over-all constant c is needed⎝ for finding⎠ such a form – the diagonal and off- diagonal terms in the matrix gives rise to three equations. Now we have three unknowns, µ0,ν0, and c. i.e.,

c 2 1 = µ + 2 2 µ0ν0 µ ν 2 cν0 2 ν = µ + 2 µ0 µ 1 cµ = ν + 0 ν 12 which have the following solutions:

1 1 1 2 1 1 − 4 µ = ν + µ4 + + ν2 + 0 ν µ4 ν2 µ ¶ 1 µ 1 ¶ 4 2 2 4 2 2 ν0 = µ + ν µ + ν − 1 1 1 2 1 1 − 4 c = ¡ ν + ¢ ¡ µ4 + ¢ + ν2 + ν µ4 ν2 µ ¶ µ ¶

Therefore, P2 can be written as

j +h +c j +c e 0 0 0 e− 0 0 P2 = j +c j h +c e− 0 0 e 0− 0 0 µ ¶ where c exp (c0) . To carry out the coarse-graining procedure, we now ask the following ≡ question: Is there an Ising model which has exactly the same transfer matrix as P2? Clearly, theanswerisaﬀermative and the corresponding Hamiltonian is

βE0 = (j0Si Si +1 + h0Si + c0) . − 0 0 0 i X0 where i0 =1, 2, ,N/2.The last term only gives rise to a shift in the total energy and is ··· independent of the spin configuration Si . Hence, no effect on thermodynamics. { 0 } The computation above has a simple block spin coarse-graining interpretation: If we group spin si and spin si+1 toformablockwiththerulethatthevalueofsi is used to represent the effective spin of the block, we perform the sum over all possible configuration

13 of si+1 to obtain a new Hamiltonian of block spins. Mathematically,

N N j0 sisi+1+h0 si QN = e i=1 i=1 s s ··· s S S X1 X2 XN N j s s + h0 (s +s ) = e i=1 0 i i+1 2 i i+1 s s ··· s S X1 X2 XN N j s s + h0 (s +s ) = e 0 i i+1 2 i i+1 s s ··· s i=1 X1 X2 XN Y

= Ps1s2 Ps2s3 PsN 1sN − s s ··· s ··· X1 X2 XN

= Ps s Ps s Ps s Ps s ··· ··· 1 2 2 3 3 4 4 5 ··· s1 s3 s2i 1 sN 1 sN Ã s2 !Ã s4 ! X X X− X− X X X

Ps2i 1s2i Ps2is2i+1 PsN 2sN 1 PsN 1sN ×··· − ···⎛ − − − ⎞ Ã s2i ! sN 1 X X− = P 2 ⎝ P 2 P 2 ⎠ s1s3 s3s5 sN 2sN ··· ··· ··· − s1 s3 s2i 1 sN 2 sN X X X− X− X ¡ ¢ ¡ ¢ ¡ ¢ h0 i j0s2i 1s2i+1+ (s2i 1+s2i+1)+c0 = e 0 − 2 − ··· ··· S s1 s3 s2i 1 sN 2 sN X X X− X− X βE = C e− 0 ··· ··· s1 s3 s2i 1 sN 2 sN X X X− X− X which involves only eﬀective spins of blocks. This illustrates that the corase-graining procedure is realized using partial sums. In the case of the Ising model, the renormalization group constitutes the follwoing two steps:

1. Coarse-graining via the block-spin transformation:

(j, h) (j0,h0) → with the length-scale change from 2 to 1 (since every two spins form a single new, eﬀective block spin). After the block-spin transformation, we have regained the form of the Hamiltonian exactly modullo an unimporant constant. This transformation forms a semi-group map.

2. Seek ﬁxed points of this semi-group map. Physically, some of these ﬁxed points correspond to critical points of phase transitions. This is determined by the fact that the only relevant length scale is the longwave for critical phenomena. The relevant scale is measured by the correlation length ξ. Therefore, at a critical point ξ in the thermodynamic limit N . →∞ →∞

14 For the one dimensional Ising model, the fix points of the block-spin transformation is u =0, u =1, and v =1 v is any positive value ½ ½ u =0corresponds to j = ,theeffectiveexchangeenergydiverges,orT 0.; v = 1 corresponds to zero external∞field. It can be easily shown that (u, v)=(0, 1) is an→ unstable fixed point under block-transformation. The second fixed point corresponds to the effective j =0, (i.e., no interaction or T ) with an arbitrary field. This is a stable fixed point under the block transformation.→∞ Note that since spins do not interact or T at this fixed point, then all spins are independent from one another and they have equal→∞ probability of lining up or down.

the invariant length scale under the semi-group map is ξ =0 or ξ = ∞ the latter of which correspond to a phase transtion at T =0. For the one-dimensional Ising spins, there is no T =0ﬁxxed point for ξ = , therefore, there is no phase transition at ﬁnite temperature –6 this is what we already learned∞ in previous lectures.

1.4 Momentum-Space Renormalization

Contrast between coarse-graining in real space and coarse-graining in momenum space: