Statistical Physics II

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Leonid Levitov

Statistical Physics II

MIT 2003

Massachusetts Institute of Technology Physics Department Physics 8.334

Statistical Mechanics II

Spring 03

Last modified: February 3, 2003 Lectures: We, Fr, 2-3:30, in Rm. 12-142, by Prof Leonid Levitov; Rec: Mo, 2-3:30, in Rm. 12-142, by Mr Vadim Roytershteyn Prerequisite: Introductory level Statistical Mechanics course Course outline • Long-range order, symmetry and hydrodynamical modes; • Phase transitions, mean field theory; • Continuous and discontinuous phase transitions, the role of symmetry; • Critical phenomena, scaling transformation, universality; • Renormalization group, statistical field theory; • Wilson-Fisher fix point near d = 4, calculation of critical exponents; • Exactly solvable models: Ising, Potts, duality transformation; • Topological transitions, vortices in the XY model, Kosterlitz-Thouless RG theory; • XY and solid-on-solid duality, vortex melting of 2D crystals; • Lower critical dimension, nonlinear sigma model; and, if time permits, • Critical dynamics, scaling theory; • Field-theoretic description of localization; • Conformal invariance and Conformal Field Theory in Statistical Mechanics. Recommended texts: S-K Ma, Modern Theory of Critical Phenomena, (1976, The Bejamin/Cummings Publishing Co.) D J Amit, Field Theory, Renormalization Group, and Critical Phenomena, (1984, World Scientific) C Itzykson, J-M Drouffe, Statistical Field Theory, vol.I,II, (1991, Cambridge University Press)

8.334: Statistical Mechanics II

1 1.1

Last modified: February 7, 2003

Lecture 1. Long-range order, symmetry and soft modes Long-range order

The notion of long-range order is used to describe the situation when the state of a macroscopic system has symmetry lower than that of the system Hamiltonian. A system with a long range order is described by a symmetry group that transforms one state to another equivalent state. Two different situations are possible regarding this symmetry group: it can be (i) discrete or (ii) continuous. A slightly artificial but very useful example of a discrete symmetry is an Ising magnet model. The variables are spins on a lattice, described by σ r = ±1. The Hamiltonian of spin interaction is 1X J(r − r ′ )σr σr′ (1) H=− 2 r

where J(r − r ′ ) > 0 is exchange interaction. By convention, each pair of points r, r ′ is counted twice in the sum, which accounts for the factor 12 . At zero temperature, there are two lowest energy states: ALL spins up, σr = +1, or ALL spins down, σr = −1. Note that the overall sign change σr → −σr leaves the Hamiltonian invariant, while the two ground states are not invariant. Under a sign change they transform into each other. In the group theory terminology, the symmetry group of the manifold of (in this case, just two) states is Z2 , consisting of the identity and the sign reversal transformations. Such a situation is sometimes called spontaneous symmetry breaking. Indeed, think of someone presented with a choice of choosing a ground state. This choice is perfectly symmetric, since each of the possible ground states is equivalent to other states. However, as soon as the choice is made, the situation becomes asymmetric, because the state symmetry is lower than that of the Hamiltonian. This concept, especially in the cases when the symmetry group of possible ground states is continuous, has many applications in the high energy physics, cosmology, hydrodynamics, nonlinear dynamics and, of course, in condensed matter physics. We hope to explore this a bit more later, in homeworks, course-related projects or term papers.

1.2

Soft modes, general theory

In the situation when the symmetry group of different equivalent states is continuous, Goldstone theorem predicts that there exist soft modes, i.e. one or several branches of excitations with the energy vanishing in the limit of long wavelength.

1

Consider an example from field theory: an n–component field φ i , i = 1, ..., n, with the Hamiltonian  Z  1 2 1 2 H= πi + (∇φi ) + U(|φ|) d3 x (2) 2 2 Here the potential energy U(|φ|) depends only on the vector φ i length, but not on its orientation. The first term in Eq.(2) is kinetic energy written in terms of momentum density πi of the field φi . When the potential energy U(|φ|) has a minimum at finite φ 0 , there are infinitely many ground states, all with the same energy. They are described by constant field |φ(x)| = φ 0 with restricted length, but unrestricted direction. They manifold of ground states in this case is an n − 1 dimensional sphere. Perturb near one of these states. The state φi (x) slowly varying in space have energy which is larger but can be made arbitrarily close to the ground state energy by choosing sufficiently large wavelength. Graphically, for a two component field, n = 2, the 3D plot U vs φ looks like a Mexican hat, with the manifold of ground states being a circle near the hat rim. If the field φ orientation is slowly varying in space, with the length |φ| fixed, it costs almost no energy. (0) More quantitative estimate: for φ i (x) = φi + δφi (x) with the transverse pertubation (0) δφi (x) = Ai cos k · x such that Ai φi = 0, the energy density δH =

Z

(∇δφi (x))2 d3 x ∝ k2 A2

(3)

Indeed, δU → 0 when k → 0. Excitations are described by the Hamilton equations of motion: π˙ i = −δH/δφi = ∇2 φi − ∂U/∂φi ,

φ˙ i = δH/δπi = πi

(4)

Linearizing them for the excitations of small amplitude |δφ| ≪ |φ (0) |, obtain a wave equation (∂t2 − ∇2 )δφi = 0, which gives a linear dispersion relation ω(k) = k. Alternatively, one can start with the Lagrangean L=

Z 

1 ˙2 1 φi − (∇φi )2 − U(|φ|) d3 x . 2 2 

(5)

Linearizing L(φ) in small perturbation amplitude δφ, obtain L=

Z 

1 ˙ 2 1 (δ φi ) − (∇δφi )2 ) d3 x 2 2 

(6)

which yields the same wave equation. The mode with a “soft” dispersion relation, such that ω(k) → 0 when k → 0, is called a Goldstone mode. To summarize the discussion of the example (2), spontaneous symmetry breaking with a continuous group of ground state transformations (described also as continuous order parameter degeneracy), gives rise to Goldstone modes with soft dispersion relation. 2

It will be crucial for us that the statistical properties of such a system at low enough temperature are entirely determined by these low energy excitations. They are the thermodynamically active degrees of freedom that describe thermal energy excitations and thermodynamic fluctuations in equilibrium. Typically, by analyzing the system and characterizing the low energy excitations (soft modes), one can reduce a relatively complicated problem to a much simpler problem. Moreover, the initial microscopic problem is usually case-sensitive, while the problem obtained by isolating the low energy modes is usually more universal, with the basic features determined only by symmetry (and perhaps by several “material constants”). Comment 1: A problem very similar to Eq.(2), with n = 2 and U(φ) = 12 g(φ2 − n)2 , describes a superfluid (see Problem 1 in PS#1). The wavefunction ψ in this homework problem is complex-valued. Writing it as ψ = φ 1 + iφ2 , see that the only difference is the ¯ t ψ + c.c.. The conclusions about low energy excitations, form of kinetic energy: 21 φ˙ 2i → iψ∂ however, are unchanged. Comment 2: To see why is it so important for the Goldstone theorem to have a continuous manifold of degenerate states, consider the form of the energy U(φ) = 12 U0 φ2 with a single minimum at φ = 0. Then the ground state is φ = 0. Pertubing about it, obtain an equation with a mass term, (∂ t2 − ∇2 + U0 )δφi = 0, which gives a dispersion relation with a cutoff frequency: ω(k) = (k 2 + U0 )1/2 . No low energy excitations in this case.

1.3

Different systems, often the same symmetry classes

Now let us consider several other examples of interest for statistical mechanics. To illustrate the generality of the relation between continuous symmetry and soft modes, we shall consider seemingly distant systems: solid, liquid crystal, ferromagnet, incommensurate crystal. Elastic modes in solids are described by displacements of atoms in a crystal lattice relative to equilibrium positions. The hamiltonian of atoms and nuclei is invariant under spatial translations and Euclidean rotations. The crystal itself, however, is not invariant, since translating it by a constant vector incommensurate with lattice period gives a geometrically different but physically equivalent crystal. Similarly, spatial rotation changes the crystal to a different crystal. The soft modes of a crystal are described by displacement fields slowly varying in space. Indeed, constant displacement field correponds to an overall translation by a constant vector, and thus does not change energy. Weakly nonuniform translation will change energy, but thsi change will be small when the wavelength of spatial modulation is large. The long wavelength displacements are thus the Goldstone modes in this case. The Hamiltonian of elastic modes in a crystal is written in terms of gradients of the displacement field u i (x) as1 E(u) =

Z

!

λ 2 u + µu2ij d3 x 2 ii

3

(7)

with uij = 12 (∂i uj + ∂j ui ) the deformation tensor and Lame constants λ, µ. We stress that the form of the elastic energy can be deduced solely from a symmetry argument. Indeed, the energy E(u) cannot depend on ui , but only on its gradients ∂i uj , since an explicit dependence on u renders the energy noninvariant under uniform spatial translations ui (x) → ui (x) + ai , where a is a constant vector. Furthermore, to the lowest order in gradients, the energy has to be quadratic in the tensor ∂ i uj . One might expect that, since there are 3 × 3 = 9 different components in ∂ i uj , the energy dependence on the displacement field will be complicated. Remarkably, this is not the case. The condition of rotational symmetry leaves one with three invariants ∂ i uj ∂i uj , ∂i uj ∂j ui and ∂i ui ∂j uj . However, there is no dependence of the energy on the the antisymmetric part ∂ i uj −∂j ui of the tensor ∂i uj , since it describes an infinitesimal rotation which does not change system energy. Thus there are only two independent terms in the elastic energy (7). R 1 2 3 The kinetic energy is 2 ρu˙ i d x with ρ the crystal density. The dynamics is described R 1 2 by the Lagrangean L = ρu˙ i − λ2 u2ii − µu2ij d3 x, or (same thing!) by Newton’s equa2 tions ρ¨ ui = δE(u)/δui (8) One can look for a plane wave solution u j (x) = uj eik·x−iωt , which gives three coupled linear equations for the vibration amplitude ω 2ui = λki kj uj + µkj (ki uj + kj ui )

(9)

There are three normal modes, one longitudinal with u k k and two transverse, with u ⊥ k. All three have linear dispersion ω ∝ k. the velocities of the longitudianl and transverse sound are different. For the longitudinal mode, ω(k) =qc l k with the velocity q cl = (λ + 2µ)/ρ. For the transverse mode, ω(k) = ct k with cl = µ/ρ. Thermal excitation of long wavelength vibrational modes controls thermodynamical properties of this system. The elementary excitations associated with the vibrational modes, also called phonons, are Bose particles. The occupation number of a mode with wavevector k is given by Planck distribution: n k = 1/(eβω(k) − 1) with β = 1/T . This means that only modes with sufficiently long wavelength, ω(k) = cl,t (2π/λ) ≤ T

(10)

are thermally excited 2 . The thermodynamic potential of this system is Ω=T

Z

ln(1 − e−ωl (k)/T )

Z d3 k d3 k −ωt (k)/T + 2T ln(1 − e ) (2π)3 (2π)3

1

(11)

This simple form of elastic energy is correct for an isotropic solid, such R as a glass or a polycrystal. In a real crystal, elastic energy has more complicated form: E(u) = 12 λijkl uij ukl d3 x with λijkl an anisotropic elasticity tensor. 2 Unless stated otherwise, we use absolute temperature units with the Boltzmann constant kB = 1.

4

Here the integral over k adds up the contributions of modes with different wavelengths, while the factor 2 accounts for two transverse modes. This expression can be used to obtain the entropy, specific heat, and other thermodynamic quantities. Liquid crystal There is a strange state of matter intermediate between liquids and solids, called liquid crystals. These are fluids with one or another form of orientational order. Microscopically, they consist of relatively large anisotropic molecules, such as long rods, which have orientations correlated with each other throughout the entire volume of the liquid crystal. For a more detailed but still quite general discussion of a nematic state we refer to a very good book “Physics of liquid crystals,” by de Gennes. there are many interesting phases of liquid crystals. In the simplest case of a nematic ordering, the system is translationally invariant, while the rotational invariance is broken. Average molecule direction is described by a “director” vector ±n. The order parameter is a symmetric traceless tensor Q ij = Q(T )(ni nj − 31 δij ). The system has rotational degeneracy described by all possible orientations of n, – a 2D sphere with opposite points identified (a projective plane, in topolgical terms). Thus the energy cannot depend on n only. The dependence on the gradients partiali nj can be deduced from the rotational invariance of the problem. The logic is similar to that used above in the discussion of elasticity. The energy of a liquid crystal in a uniform state with constant n does not depend on the orientation of n (the ground state degeneracy). Hence the energy has to depend on the gradients ∇n. (i) E(n) must be even in n, as explained above, the states (n) and (−n) are undistinguishable; (ii) No terms linear in n. The only terms of this form invariant under rotation are div n (ruled out by (i)) and n · curln (changes sign by the transformation (x, y, z) → (−x, −y, −z)); Thus E(n) should be quadratic in ∇. The simples way to list all possible rotationally invariant terms of this form is to consider three quatities, div n,

n · curl n,

n × curl n,

(12)

and take a sum of squares. One can show (this is tedious but elementary 3 ) that this accounts for all possible terms of second order in gradients. Thus the elastic energy of a liquid crystal is  1Z  E(n) = K1 (div n)2 + K2 (n · curl n)2 + K3 (n × curl n)2 d3 x (13) 2 The constants K1,2,3 account for the energy cost of conformations with nonzero div n, n·curl n and n×curl n. They are known as the splay, twist and bend modulus, respectively. The dynamics is of relaxational form: n˙ = −γδE(n)/δn. Two relaxational soft modes with complex disperion relation ω ∝ ik 2 . Thermal excitations are fully described by the elastic energy (13). To analyze thermal excitations, one can consider a weakly modulated state, n(x) = n (0) + δn(x) with δn 3

see: Landau & Lifshits, Statistical Physics, Part I, § 140

5

transverse to n(0)) . Plugging this into (13) and expanding to the lowest nonvanishing R order in δn(x), one obtains energy of the form E(δn) = 12 K(∇δn)2 . (This expression is schematic, in fact K is a tensor with some dependence on n (0)) , but this is good enough for R the moment.) Expanding δn(x) in Fourier harmonics, δn(x) = δnk eik·x d3 /(2π)3 , obtain 1 E(δn) = K(k · δnk )2 d3 /(2π)3 (14) 2 Using the equipartition theorem, one can estimate thermal fluctuations in each mode |δnk |2 ∝ T /Kk 2 and consider thermodynamics of the system pretty much like for vibrational modes above. (To be discussed later in more detail.) Heisenberg ferromagnet: Magnetism arises due to exchange interaction of electron spins. Typically, the spins involved in magnetic ordering come from the electrons localized on crystal lattice sites, so that one can ignore orbital electron dynamics and focus on the spins and their interactions. The simplest model describing ferromagnetism, known P as the Heisenberg model, involves Pauli spin operators: H = − 12 r6=r′ J(r − r ′ ) ˆsr · ˆsr′ , where ˆsr = σ1 , σ2 , σ3 and J(r − r ′ ) > 0 is spin exchange interaction. Instead of this full quantum Hamiltonian we shall consider a classical Heisenberg model Z

H=−

1X J(r − r ′ ) Sr · Sr′ 2 r6=r′

(15)

Here Sr is a unit vector describing average spin polarization in the system at the point r. This model describes a large spin limit of the quantum problem. The classical and quantum models have identical symmetry properties, the same set of low energy modes, and hence the same thermodynamical properties. The Hamiltonian (15) is invariant under spin rotations. This is a bit more tricky than in previous examples, since this is not a physical space rotation, but rather a rotation in an “internal space.” The ground state is described by uniform magnetization with arbitrary orientation. Thus there are infinitely many ground states, and we have the same sort of continuous degeneracy as before. To find the enrgy of the soft modes, we consider a weakly inhomogeneous spin field Sr = S(0) + δSr , plug it in the energy (15) and expand in δS. This calculation is most easily carried out by rewriting Eq.(15) as H=

1X 1X J(r − r ′ ) (Sr − Sr′ )2 = J(r − r ′ ) (δSr − δSr′ )2 4 r6=r′ 4 r6=r′

(16)

(we used that S2r = 1 and added a constant term r J(0)S2r ). As long as J(r − r ′ ) falls rapidly, the points r and r ′ are close, and for slowly varying field configuration one can expand Sr − Sr′ = (rk − rk′ )∇k S + O((r − r ′ )2 ). This gives P

E(S) =

1Z 1X Jkk′ (∇k Sj )(∇k′ Sj ) = J0 (∇S)2 d3 r 2 r 2 r

(17)

with Jkk′ = 12 r J(r)rk rk′ , J0 = 16 r J(r)r 2 . Ther last expression in Eq.(17), rotationally invariant under independent spin and space rotations, applies to sufficiently symmetric P

P

6

situations, such as spins in a cubic lattice with exchange interaction being a function of distance only, when Jkk′ = J0 δkk′ . This example illustartes how a classical field description emerges from a microscopic model. We start from functions of a dicrete variable, spins on a lattice, with interaction J(r − r ′ ) that typically takes place on short length scale, |r − r ′ | or the order of lattice spacing. By considering slowly varying perturbation field configurations which locally look like one of (continuously degenerate) ground states, we arrive at a macroscopic description involving functions S r of a continuous variable r ∈ R 3 with an effective Hamiltonian (17). It is characteristic for the situations with continuous symmetry of the order parameter that the Hamiltonian is written in terms of gradients. Alternatively, one can derive (17) using Fourier representation. We expand S r = R Sk eik·r d3 k/(2π)3 , and rewrite (15) as E(S) =

1 2

Z

J(k) Sk · S−k

d3 k (2π)3

(18)

where J(k) is a Fourier transform of J(r). Note that since the spins sit on lattice sites, the Fourier transform is defined as for functions of a discrete variable. Because of that, the integral over k in Eq.(18) runs not over the entire k space but only over a domain −π/a < ki < π/a with a the lattice period. Expanding Eq.(18) in δS, we obtain E(δS) =

1Z d3 k (J(k) − J(0)) δSk · δS−k 2 (2π)3

(19)

For long wavelength spin waves, we can expand J(k) − J(0) =R J 0 k 2 + O(k 4 ). Keeping only the term J0 k 2 and going back to the real space, we have 12 J0 (∇δS)2 d3 r. However, since the background state is constant, ∇S(0) = 0, this can be brought to the form 1 E(S) = 2

Z

J0 (∇S)2 d3 r

(20)

This expression gives the energy of a weakly inhomogeneous spin-polarized state. Let us comment on the analogy with the liquid crystal energy (13). There seems to be a similarity, since the order parametr in both cases is a unit vector. The crucial difference is that in the liquid crystal case this vector defines orientation in the real physical space, while in the magnet problem it is a vector in the inner spin space. This results in a different form of the energy: a single term in (20) versus three independent terms in (13). Indeed, the only exprssion quadratic in the graduients ∇S and invariant under independent real and spin rotations is (∂ i Sj )2 , which is precisely the energy density (20). The analysis of thermodynamics is quite analogous to the liquid crystal case (see also the homework problem 2, PS#1). We shall come back to the problem (20) again later. Incommensurate crystals is an example of even more exotic symmetry. Density modulation with incommensurate periods. In 1D case, say, n(x) = A cos(k1 x) + B cos(k2 x + θ) 7

(21)

Invariance of the system energy with respect to changes of the phase θ by an overall constant. The low energy states are described by slowly varying phase θ(x) in Eq.(21). The energy is given by a gradient expansion of the form 1 E(θ) = 2

Z

K(∂x θ)2 dx

(22)

These soft modes are called “phasons,” since they describe phase fluctuations. Phyasically, they correspond to one density wave sliding freely relative to the other one. Talk about mercury chains in the material Hg 3−δ AsF6 , the “alchemist gold.”

1.4

Now, let us summarize

• Emerging classical fields: The macroscopic description of systems with broken symmetry is provided by a classical field, one or several functions in Euclidean space. This description is usually quite simple, while the microscopic origin of the ordering can be rather complicated. • Universality: Many systems seemingly very difrerent microscopically have the same or very similar description in the macroscopic limit. Different universality classes differ mainly by the symmetry of the ordering and (usually in a less important way) by interactions. • Goldstone modes: In the case of spontaneously broken continuous symmetry, the low energy excitations are described by weakly inhomogeneous order paramer field. The fluctuations of these modes govern the thermodynamic properties. • The phenomenolgical and microscopic approaches: The form of the energy describing slowly varying fields can be deduced on the basis of symmetry by listing all possible invariants in the gradient expansion and carefully analyzing symmetry requirements. The constants in such phenomenolgical hamiltonians should be calculated from a microscopic approach. The relation of this energy functional to the microscopic Hamiltonian can be simple, as in the case of lattice vibrations and Heisenberg ferromagnet model, or very complicated, as in the case of liquid crystals or incommensurate crystals.

8

8.334: Statistical Mechanics II

Problem Set # 1

Due: 2/12/03

Long-range order, symmetry and soft modes 1. Superfluid dynamics Interacting Bose gas in a superfluid state is described by the Hamiltonian H=

Z

g ¯ h ¯2 |∇ψ|2 + (ψψ − n)2 d3 r 2m 2 !

(1)

with ψ(r) the Bose condensate wavefunction. Here m, g and n are particle mass, interaction, and density. a) Find the ground state. Compare the symmetry of the Hamiltonian and the symmetry of the ground state. How many soft modes do you expect to have in this system? b) The dynamics is described by the Gross-Pitaevsky equation 1 i¯h∂t ψ =

δH h ¯2 2 ¯ − n)ψ ∇ ψ + g(ψψ = − 2m δ ψ¯

(2)

The equation for ψ¯ is a complex conjugate of Eq.(2), −i¯h∂t ψ¯ = δH/δψ. Consider a perturbed ground state ψ(r, t) = ψ (0) + δψ(r, t) and linearize in δψ(r, t) the pair of equations for ψ and ψ¯ (Eq.(2) and its conjugate). Find a plane wave solution and characterize soft modes in a superfluid. How does the dispersion relation ω(k) look at small k? c) Treating the excitations found in part b) as Bose particles, find the entropy and specific heat of the superfluid (1) at low temperature T ≪ gn. 2. Spin waves in a Heisenberg ferromagnet Consider a classical spin model of a ferromagnet 1X H=− J(r − r ′ ) Sr · Sr′ 2 r6=r′

(3)

Here Sr is a unit vector and J(r − r ′ ) > 0 is spin exchange interaction. This model describes a large spin limit of the quantum Heisenberg problem. a) Find the ground state. Compare the symmetry of the Hamiltonian (3) and the symmetry of the ground state. How many soft modes do you expect to have in this system? b) The dynamics of each spin is precession in the magnetic field of neighboring spins, ∂t Sr = Sr × Br ,

Br ≡ −δH/δ Sr =

X

J(r − r ′ ) Sr′

(4)

r′

Consider a long wavelength spin wave, S r (t) = S(0) r + δ Sr (t), linearize Eqs.(4) in δ S, and find a plane wave solution for spin waves. How does the dispersion relation ω(k) look at small k? c) Treating spin waves as Bose particles, find the entropy and specific heat of the ferromagnet (3) at low temperature T ≪ J. P d) Spin system in external magnetic field is described by H → H − r µ Sr · Bext with µ the spin magnetic moment (Bohr’s magneton for electron spin). Consider once more the above questions a), b) and c). Comment on the difference with the B ext = 0 case. 1

This equation is also known as the nonlinear Schr¨odinger equation and the time-dependent Ginzburg-Landau equation.

8.334: Statistical Mechanics II

Problem Set # 1

Due: 2/12/03

Long-range order, symmetry and soft modes 1. Superfluid dynamics Interacting Bose gas in a superfluid state is described by the Hamiltonian H=

Z

h ¯2 g ¯ |∇ψ|2 + (ψψ − n)2 d3 r 2m 2 !

(1)

with ψ(r) the Bose condensate wavefunction. Here m, g and n are particle mass, interaction, and density. a) Find the ground state. Compare the symmetry of the Hamiltonian and the symmetry of the ground state. How many soft modes do you expect to have in this system? The ground state is √ |ψ| = neiθ (2) where the phase θ is arbitrary. The Hamiltonian (1) is invariant under gauge transformation ψ → eiα ψ. The grounds states do not have this symmetry. b) The dynamics is described by the Gross-Pitaevsky equation1 i¯h∂t ψ =

δH h ¯2 2 ¯ − n)ψ = − ∇ ψ + g(ψψ δ ψ¯ 2m

(3)

The equation for ψ¯ is a complex conjugate of Eq.(3), −i¯h∂t ψ¯ = δH/δψ. Consider a perturbed ground state ψ(r, t) = ψ (0) + δψ(r, t) and linearize in δψ(r, t) the pair of equations for ψ and ψ¯ (Eq.(3) and its conjugate). Find a plane wave solution and characterize soft modes in a superfluid. How does the dispersion relation ω(k) look at small k? Consider perturbation δψ of the ground state ψ = ψ (0) + δψ. Then the linearized equation for δψ and δ ψ¯ are   h ¯2 2 ∇ δψ + g ψ 2 δ ψ¯ + nδψ i¯h∂t δψ = − 2m   h ¯2 2 ¯ −i¯h∂t δ ψ¯ = − ∇ δ ψ + g ψ¯2 δψ + nδ ψ¯ 2m

(4) (5)

Seek the plain wave solution δψ ∝ exp(−iωt + kr). Then the dispersion relation is 2 2

det

h ¯ ω − k2m¯h − gn −gne2iθ 2¯ h2 −gne−2iθ −¯hω − k2m − gn

The solution is ω(k) = 1

v u u t

k2 2m

!2

h ¯2 +

gn 2 k m

!

=0

(6)

(7)

This equation is also known as the nonlinear Schr¨odinger equation and the time-dependent Ginzburg-Landau equation.

q

For small k we obtain ω = ck where c = gn/m. c) Treating the excitations found in part b) as Bose particles, find the entropy and specific heat of the superfluid (1) at low temperature T ≪ gn. The thermodynamic potential is Ω=

1 β

  d3 k −β¯ hω ln 1 − e (2π)3

Z

(8)

We are concerned with the low-k modes described by the dispersion relation ω = ck. In this case all the dimensional quantities can be scaled out of the integral: 4π Ω= 4 β (¯hc)3 (2π)3

Z

∞

0



−x

ln 1 − e

The entropy is



T4 x dx = − 2 6π (¯hc)3 2

Z

∞ 0

x3 dx π2T 4 = − ex − 1 90(¯hc)3

(9)

2π 2 T 3 ∂Ω = S=− ∂T 45(¯hc)3

(10)

The specific heat is

2π 2 T 3 ∂S = ∂T 15(¯hc)3 2. Spin waves in a Heisenberg ferromagnet c=T

(11)

Consider a classical spin model of a ferromagnet H=−

1X J(r − r ′ ) Sr · Sr′ 2 r6=r′

(12)

Here Sr is a unit vector and J(r − r ′ ) > 0 is spin exchange interaction. This model describes a large spin limit of the quantum Heisenberg problem. a) Find the ground state. Compare the symmetry of the Hamiltonian (12) and the symmetry of the ground state. How many soft modes do you expect to have in this system? The ground state is S = S0 where S(0) is uniform. The Hamiltonian is invariant under uniform rotations in the spin space, i.e. the degeneracy space is a 2D sphere. It looks like there should be two soft modes, but actually there is only one. b) The dynamics of each spin is precession in the magnetic field of neighboring spins, ∂t Sr = Sr × Br ,

Br ≡ −δH/δ Sr =

X

J(r − r ′ ) Sr′

(13)

r′

Consider a long wavelength spin wave, Sr (t) = S(0) r + δ Sr (t), linearize Eqs.(13) in δ S, and find a plane wave solution for spin waves. How does the dispersion relation ω(k) look at small k? The linearized equation is (0) ∂t δS = S(0) (14) r × δB + δSr × B The perturbation of the magnetic field is δB =

X r′

J(r − r ′ )δSr′ =

X r′

1 J(r − r ′ ) δSr + (r − r′ ) · ∇δSr + ((r − r′ ) · ∇)2 δSr + . . . 2 



(15)

Since X

J(r − r′ )(r − r′ ) = 0

(16)

J(r − r′ )(r − r′ )2 = 6J0

(17)

r

X r

(see the note after equation (17) in Lecture 1), we obtain the so-called Landau-Lifshits equation ∂t δS = J0 S(0) × ∇2 δS

(18)

Choose S(0) = zˆ, δS = δSx xˆ + δSy yˆ and seek a plane wave solution δS ∝ exp(−iωt + kr). Then the dispersion relation is ! iω J0 k 2 det =0 (19) −J0 k 2 iω or ω(k) = J0 k 2 (20) c) Treating spin waves as Bose particles, find the entropy and specific heat of the ferromagnet (12) at low temperature T ≪ J. Z

Ω=T The integral is Z

0

Therefore

∞



−β¯ hω

ln 1 − e

2



 d3 k

4πT = 3 (2π) (β¯hJ0 )3/2 (2π)3

−x2

dx x ln 1 − e



1 =− 3

Z

0

∞

Z

∞

0



dx x2 ln 1 − e−x

y 3/2 dy 1 = − Γ(5/2)ζ(5/2) y e −1 3

Γ(5/2)ζ(5/2) 5/2 T 6π 2 (¯hJ0 )3/2 5Γ(5/2)ζ(5/2) 3/2 S= T 12π 2 (¯hJ0 )3/2 5Γ(5/2)ζ(5/2) 3/2 c= T 8π 2 (¯hJ0 )3/2 Ω=−

2



(21)

(22)

(23) (24) (25)

d) Spin system in external magnetic field is described by H → H − r µ Sr · Bext with µ the spin magnetic moment (Bohr’s magneton for electron spin). Consider once more the above questions a), b) and c). Comment on the difference with the Bext = 0 case. a’) In magnetic field the ground state is S r = S(0) where S(0) k B. There is no degeneracy. b’) The linearized equation of motion now is P

(0) ∂t δS = S(0) × ∇2 δS − Bext × δS r × δB + δSr × Bext = J0 S

(26)

Choose Bext , S(0) k zˆ and look for plane wave solutions. Then det

iω J0 k 2 + |Bext | 2 −(J0 k + |Bext |) iω

!

=0

(27)

or ω(k) = ω0 + J0 k 2

(28)

where ω0 = |Bext |. The dispersion relation has a cut-off at ω = ω 0 . There are no low-frequency modes. c’) Z   d3 k Γ(5/2) =− 2 T 5/2 f5/2 (e−λ ) (29) Ω = T ln 1 − e−β¯hω 3 (2π) 6π (¯hJ0 )3/2 where λ=h ¯ βω0 fm (z) =

(30)

1 Γ(m)

Z

∞ 0

m−1

∞ X zj

dx x = z −1 ex − 1 j=1 j m

(31)

At high temperatures, T ≫ h ¯ ω0 the thermodynamic Al properties are the same as at B ext = 0, while at low temperatures, T ≪ h ¯ ω0 , there are almost no thermally excited modes and the thermodynamic quantities are exponentially small Ω ∝ T 5/2 e−¯hω0 /T S ∝ ω0 T 1/2 e−¯hω0 /T c ∝ ω02T −1/2 e−¯hω0 /T

(32) (33) (34) (35)

8.334: Statistical Mechanics II

1 1.1

Last modified: February 22, 2003

Lecture 2. Phase transitions, the mean field approach Phase transition types

The degree of ordering in every system is negotiated between interactions that enforce order and thermal motion acting to enhance disorder. As a result of the competition between the tendencies to minimize energy and to maximize entropy, many systems can form two (or more) different macroscopic phases, depending on parameters, such as temperature, pressure, etc. The phases are in equilibrium at certain temperature and pressure values, usually forming a transition line. Because of different thermodynamic properties of the phases, away from the transition line one of them is lower in energy than the other one. For the phases coexisting in thermodynamic equilibrium, certain conditions must be satisfied. Thermodynamic variables such as pressure, chemical potential, etc., have to be equal. However, for distinct phases the quantities such as entropy and density are typically not equal, — they experience a jump at the transition line (thus a finite latent heat and volume change). Such discontinuous transitions are known as phase transitions of first order, or type I transitions. The canonical example is a liquid–gas transition. Remind of the phase diagram, coexistence line, Clausius-Clapeyron equation for the slope of this line, the van der Waals equation of state, and describe the critical point at which the phase transition line terminates. The characteristic feature here, from the symmetry point of view that we are developing, is that the two phases are not different in symmetry in any way. Indeed, the liquid and gas states differ only by density, but have the same order (rather, disorder!) characteristics. Many other examples of this nature exist. In fact, most of the phase transitions, especially in material science, metallurgy, chemistry, etc. are of type I. Briefly mention binary alloys, a solid state version of the liquid–gas transition. A lattice gas problem: two kinds of atoms occupying sites of the same lattice at random. Due to atom interaction, ordering takes place at low temperature. Depending on the interaction details, the ordering can be different. If, for example, like atoms attract each other, while unlike atoms repel each other, the system phase separates in the phases rich in atoms of one particular sort. For more detail see PS2, Problem #2. Besides the transitions in which system state and properties experience a jump, there is a totally different kind of phase transitions, known as transitions of second order, or type II. In such transitions, as first pointed out by Landau, there is an abrupt symmetry change, while the state of a system may evolve continuously through the transition. The symmetry change may occur, e.g. via a small displacement of atoms in a crystal lattice (breaking lattice periodicity), ordering of spins in a magnet (breaking the time reversal symmetry), 1

Bose condensate formation (breaking the gauge invariance), etc. More generally, the second order transitions are related with some kind spontaneous symmetry breaking, that takes place below certain temperature. Let us illustrate the difference in symmetry change between the type I and II transitions by an example. Consider a binary alloy with cubic lattice structure, in which like atoms repel, while unlike atoms attract. When the density of the two atoms is equal, there are two ground states at T = 0, with atoms of each type occupying one of the two sublattices of the cubic lattice: a b a b b a b a a b a b b a b a

or

b a b a a b a b b a b a a b a b

(1)

These states have lower symmetry than the disordered state at high temperature, since they are not invariant under translation by lattice period. Such a translation maps one of the two states onto the other one, and reverse. This means that there is an abrupt symmetry change when an ordering like this takes place. Note that the overall density of atoms, however, does not change. This is in contrast with the above example of a type I transition, the phase separation in a binary alloy, where the low temperature phases have the same symmetry as the high temperature phase, while there is a density jump at the transition. Although the quantities like entropy, density, etc. are continuous at the transition (hence the name “type II”), the abrupt symmetry change leads to a number of very striking physical phenomena, most notably, divergences and singularities in physical quantities of a power law form, strong fluctuations and slow relaxation, the appearance of soft modes. The physics of these transitions is conceptually extremely rich, and many areas of physics besides statistical mechanics, have benefited from the theory of second order phase transitions. Which is an excuse for us to spend a fair fraction of the course studying the properties of such transitions.

2

Mean field theory

There is a very simple and useful method that starts with a microscopic model and enables to estimate the temperature at which the transition occurs, and to relate the quantities below and above the transition. It is highly popular because of broad applicability and relatively high accuracy, not to mention simplicity. Let us illustrate it for the Ising ferromagnet model H=−

1 X J(r − r ′ )σr σr′ 2 r6=r′

(2)

with σr = ±1. The exchange interaction J(r − r ′ ) > 0, in principle, can be arbitrary. However, the approximation to be made can be justified rigorously only for exchange 2

interaction of large radius, such as ′

J(r − r ) =



U, |r − r ′ | < R; 0, |r − r ′ | > R,

with R ≫ 1

(3)

The partition function of the system in magnetic field H is 



X 1 X Z = tr exp  β J(r − r ′ )σr σr′ + σr βH  2 r6=r′ r

(4)

where tr stands for a sum over all spin variables σ r = ±. Without the couplings J(r − r ′ ) one would have independent spins, each polarized by magnetic field as P σr eσr βH (σr ) = Pσr =± σr βH = tanh(βH) (5) σr =± e But with the couplings the problem becomes more complicated. :–( The central idea of the mean field approach is to approximate the interacting problem (4) by a simpler interacting partition function. Usually this is the point when insight enters the analysis. In this case, since the interaction radius is large, each spin experiences an average field of many spins, and thus we can replace spin–spin interaction by an interaction of each spin with an average magnetization of spins around it. :–) There are many ways to do it, some of which are illustrated in the homework (PS#2). One particularly nice way is to rewrite the product σr σr′ in the partition function (4) as σr σr′ = (M + (σr − M))(M + (σr′ − M))

(6)

where M = (σr ). Expanding this to first order in the deviations δσ r = σr − M, M 2 + M((σr − M) + M(σr′ − M) + O(δσr2 ),

(7)

we approximate the exponent in (4) by X X 1 1 X β J(r − r ′ )(M(σr + σr′ ) − M 2 ) + βH σr = − NβJM 2 + β(JM + H) σr (8) 2 r6=r′ 2 r r

where J = r′ 6=r J(r − r ′ ), and N is the total number of sites. The terms ignored in making this approximation correspond to correlations of the spins. This may be too drastic and even questionable when interactions are nearest neighbor. For interactions of large radius, due to a large number of interacting spins and the central limit theorem, the replacement of a sum of fluctuating spin variables by its average value is a reasonable approximation. This yields an answer for Z, P

1

2

Z = e− 2 N βJM [2 cosh(β(JM + H))]N , 3

(9)

Figure 1: Mean field free energy of Ising ferromagnet −0.4

Free energy F(m)

−0.5

T = 0.8 Tc

−0.6 T=T

c

−0.7

T = 1.2 Tc −0.8

−0.9 −1.5

−1

−0.5

0 0.5 Magnetization m=〈 σ〉

1

1.5

and for the free energy per site 1 F (M) = −T ln Z/N = JM 2 − β −1 ln [2 cosh(β(JM + H))] 2

(10)

with irrelevant constant subtracted. We have not yet specified M. The value M is determined by minimizing F (M), since the most probable value corresponds to the lowest free energy. Evaluating the derivative dF/dM and setting it to zero, obtain M = tanh(β(JM + H))

(11)

Interpretation: the field polarizing each spin is enhanced by JM due to a feedback from other polarized spins. The corresponding contribution, JM, is called the molecular field. In the theory of magnetism it was introduced by Weiss. There are, in general, several different solutions to Eq.(11). To decide which of them should be chosen, it is instructive to consider the system in the absence of external field. For H = 0 and small M, the free energy Taylor expansion is 1 F (M) = J(1 − βJ)M 2 + O(M 4 ) 2

(12)

Depending on the sign of the M 2 term, there are two situations. For large T > J, there is single minimum of F (M) at M = 0. At low T < J, the point M = 0 is a local maximum, with two degenerate minima M = ±M 0 . The system has to choose one of them and then the σr → −σr symmetry is spontaneously broken. The value T c = J is the mean field theory result for the critical temperature. 4

At finite H, the function F (M) is asymmetric, the two minima are not degenerate. In this case, there is no symmetry loss, or bifurcation, upon lowering temperature. The spins are polarized along H even at high temperature and this polarization gradually increases as T goes below Tc . An interpretation of the difference from H = 0, from symmetry point of view, is that the Hamiltonian in the presence of a finite H does not possess the σr → −σr symmetry. Hence no symmetry breaking occurs. Another example is provided by the Heisenberg model H=−

1X J(r − r ′ ) Sr · Sr′ 2 r6=r′

(13)

with Sr unit classical vectors. The partition function of the system in magnetic field H has the form   X X 1 Z = tr exp  β J(r − r ′ )Sr · Sr′ + βSr · H (14) 2 r6=r′ r

where tr stands for an integral over the configuration space (N independent integrals over unit sphere |Sr | = 1 for N spins). Introducing the mean polarization M = hS r i and repeating the above steps (6), (7), (8), we have  Z 1 F (M) = JM2 − β −1 ln (15) exp(β(JM + H) · S)dS 2 with the integral taken over the sphere |S| = 1. The integral over dS is 2π

Z

1

−1

eβ(JM +H)x dx = 4π

sinh(β(JM + H)) β(JM + H)

(16)

Minimizing F (M), obtain an equation for magnetization: M = coth(β(JM + H)) − 1/(β(JM + H))

(17)

Despite looking very different from the above result (11) for the Ising model, this equation has the same qualitative properties. The equation for critical temperature T c = 13 J is obtained by setting H = 0 and expanding the right hand side of Eq.(17) in powers of M. Comment 1: The mean field estimate for T c is almost always an overestimate. This is because thermal fluctuations, partly ignored in the derivation, reduce the molecular field. Comment 2: The large radius of interaction condition, required for the mean field approach to be accurate, is often replaced within a nearest neighbor interaction model by a requirement of a large number of neighboring spins. Since in a d-dimensional cubic lattice the number of nearest neighbors of each site is 2d, this is equivalent to taking the limit of a high space dimension.

5

3

Summarize • Type I phase transitions are discontinuous, accompanied by jumps in physical quantities, such as density or entropy, latent heat and volume change. The symmetry is typically not changed in such transitions. • Type II phase transitions are very different. In such transitions, physical quantities vary continuously, while symmetry changes abruptly. The phase transformation in a type II transition is described by spontaneous symmetry breaking. • Mean field theory: a very successful approximation technique that captures correctly all qualitative features of the type I and II transitions. It gives numerically accurate results for the phase transitions in systems with large interaction radius or large number of interacting neighbors.

6

8.334: Statistical Mechanics II

Problem Set # 2

Due: 2/19/03

Phase transitions, the mean field approach, Landau theory 1. Ising ferromagnet In an Ising model of a ferromagnet the spin degrees of freedom are discrete, σ r = ±1, and the Hamiltonian of spin interaction is 1 H=−

1X J(r − r ′ ) σr σr′ 2 r6=r′

(1)

Here J(r − r ′ ) > 0 is spin exchange interaction. For spin interaction of large radius, the effect of spin interaction can be analyzed using the mean field approach. For that, the interaction of a spin σ r with all other spins is replaced by interaction with an effective field H(σr ) = −σr Br ,

Br ≡ −δH/δσr =

X

J(r − r ′ ) hσr′ i ,

(2)

r′

with hσr′ i the statistical average values of the spins. For the interaction of large radius, replacing spins by their average values in the sum over r ′ can be justified by the central limit theorem. a) Find spin magnetization m = hσ r i for the problem (2) and, using the relation between B r and hσr′ i, obtain an equation for the magnetization. b) Show that there is a temperature Tc , called Curie temperature, below which this equation has a nonzero solution. Sketch m(T ). How does it behave at T ≪ Tc and |T − Tc | ≪ Tc ? c) Near T = Tc , find the free energy F , entropy S and specific heat C above and below Tc . Sketch the temperature dependence S(T ) and C(T ). 2. The Binary Alloy In a model of a binary alloy, each cite of the lattice can be occupied by a particle of type A or one of type B. The interactions between the different types of particles are given by J AA (r − r ′ ), JBB (r − r ′ ) and JAB (r − r ′ ). Consider an alloy consisting of N A particles of type A, and NB particles of type B. Assume attractive an interaction energy J AA , JBB < 0 between like neighbors A − A and B − B, but a repulsive energy JAB > 0 for an A − B pair. In a generic alloy, J AA 6= JBB . a) What is the gound state of the system at zero temperature, the particle densities n A = NA /(NA + NB ) and nB = NB /(NA + NB ) being fixed? b) At finite tmperature, estimate the total interaction energy assuming that the atoms are randomly distributed among lattice cites, i.e. each cite occupied independently with probabilities pA = nA , pB = nB . c) Estimate the mixing entropy of the alloy within the same approximation. Assume N A,B ≫ 1. d) Using the above, obtain a free energy function F (x), where x = (nA − nB ). Show that the requirement of convexity of F (x) breaks down below a critical temperature T c . 1

This model describes a situation when due to a spin-orbital crystal field spins are aligned with some particular crystal axis.

e) Sketch F (x) for T > Tc , T = Tc and T < Tc . For T < Tc there is a range of compositions x1 < x < x2 where F (x) is not convex and thus the composition is locally unstable. Plot the so-called spinodal lines2 x1,2 (T ) in the (T, x) plane. f) Describe what happens with an alloy cooled from higher to lower temperature, depending on the composition. In the (T, x) plane sketch the phase separation boundary, dividing the (T, x) plane into the stability regions of the A rich and B rich phases and the unstable region. 3. Ising ferromagnet, a different approach Let us solve Problem 1 by using the method of Problem 2. To map the spin problem (1) on the P binary alloy problem, consider a microcanonical ensemble with fixed total magnetization M = r σr . In this case, out of total N spins, N+ = (N + M)/2 is up and N− = (N − M)/2 is down. a) Assuming the occupation of different sites to be statistically independent, and using the occupation probabilities p ± = (N ± M)/2N, find the energy of spin interaction as a function of M. b) Find the entropy and the free energy as a function of M. Sketch F (M) at different temperatures. Show that F (M) does not satisfy the convexity requirement at T < T c . Find the critical temperature Tc . c) Below Tc find the values of M for which the uniform state is stable. d) Evaluate the free energy and compare the result with that of Problem 1. 4. Symmetry and transition type Consider an Ising ferromagnet (1) with the nearest neighbor interaction: J(r − r ′ ) 6= 0 only when |r − r ′ | = 1. Map this problem onto the binary alloy problem. Now, note that the ferromagnetic ordering emerges via a second order phase transition, whereas the ordering transition in a generic binary alloy is first order. Can you interpret/explain this by a symmetry argument in the Landau theory spirit? 5. Liquid crystal ordering transition In some liquid crystals, the order parameter is the local electric quadrupole moment, which can be represented by a symmetric traceless tensor Q ij , tr Q = 0. The specific form of the tensor Qij depends on the ordering type. For instance, in nematic liquid crystals discussed in class, the relation between Qij and nematic director vector n is Q ij = Q(T )(ni nj − 31 δij ). Consider a spatially uniform state (no gradients!). Near the phase transition from liquid into liquid crystal, consider the Landau free energy F (Q) as power series in Q. What are the various lower order terms that are allowed, by rotational symmetry, to appear in the free energy? Show that this suggests that the transition into the ordered phase should be the first order for this system.

2

The spinodal line indicates onset of metastability and hysteresis effects.

8.334: Statistical Mechanics II Problem Set # 2 Phase transitions, the mean field approach, Landau theory 1. Ising ferromagnet In an Ising model of a ferromagnet the spin degrees of freedom are discrete, σr = ±1, and the Hamiltonian of spin interaction is1 H=−

1X J(r − r ′ ) σr σr′ 2 r6=r′

(1)

Here J(r − r ′ ) > 0 is spin exchange interaction. For spin interaction of large radius, the effect of spin interaction can be analyzed using the mean field approach. For that, the interaction of a spin σr with all other spins is replaced by interaction with an effective field H(σr ) = −σr Br ,

Br ≡ −δH/δσr =

X

J(r − r ′ ) hσr′ i ,

(2)

r′

with hσr′ i the statistical average values of the spins. For the interaction of large radius, replacing spins by their average values in the sum over r ′ can be justified by the central limit theorem. a) Find spin magnetization m = hσr i for the problem (2) and, using the relation between Br and hσr′ i, obtain an equation for the magnetization. m=

1 Tr σe−βσBef f = tanh βBef f Z

(3)

b) Show that there is a temperature Tc , called Curie temperature, below which this equation has a nonzero solution. Sketch m(T ). How does it behave at T ≪ Tc and |T − Tc | ≪ Tc ? Since tanh x < x for x > 0, for βJ ≥ 1 the only solution is m = 0. At βJ > 1 there is a none-zero solution. The magnetization as a function of temperature is shown on Fig. 1. Suppose T is close to Tc = J. Then we expect magnetization to be small and can expand the RHS of the equation (3):  3 Tc 1 Tc m + O(m5 ) (4) m= m− T 3 T and solve for m: m ≈ (3τ )1/2 , τ ≡ (Tc − T )/T (5) For small temperatures, T ≪ T c , i.e. large βJ, tanh x → 1 − 2e−2x + O(e−4x ) and we obtain m(T ) ≈ 1 − 2e−2Tc /T

(6)

c) Near T = Tc , find the free energy F , entropy S and specific heat C above and below Tc . Sketch the temperature dependence S(T ) and C(T ). 1

This model describes a situation when due to a spin-orbital crystal field spins are aligned with some particular crystal axis.

Figure 1: Mean field magnetization vs temperature, Ising ferromagnet 1

Magnetization m=〈 σ〉

0.8

0.6

0.4

0.2

0 0

0.2

0.4

0.6 T/T

0.8

1

1.2

c

The expression for the free energy is given in Lecture 2: 1 F = Jm2 − T ln(2 cosh βJm) 2

(7)

For T > Tc F (T ) = −T ln 2. Hence S = ln 2 and c = 0. For T < Tc  T  1 T 1 F (T ) = Jm2 + ln 1 − tanh2 (βJm) − T ln 2 = Jm2 + ln(1 − m2 ) − T ln 2 2 2 2 2

(8)

Expand this expression in m 1 T F (T ) = (Tc − T )m2 − m4 + O(m6 ) − T ln 2 2 4

(9)

Near Tc the first two terms are of the same order and the higher-order terms can be dropped: F (T )T 0 0, τ 0 0, τ 0 for an A − B pair. In a generic alloy, JAA 6= JBB . a) What is the ground state of the system at zero temperature, the particle densities nA = NA /(NA + NB ) and nB = NB /(NA + NB ) being fixed? If the concentrations were not fixed, the system would have two ground states — all sites occupied by particles of type A or all sites occupied by particles of type B. For fixed densities the system will phase separate into A–rich and B–rich regions. b) At finite temperature, estimate the total interaction energy assuming that the atoms are randomly distributed among lattice cites, i.e. each cite occupied independently with probabilities pA = nA , pB = nB . The energy per site is 1 1 E = JAA n2A + JBB n2B + JAB nA nB (13) 2 2 c) Estimate the mixing entropy of the alloy within the same approximation. Assume NA,B ≫ 1. The entropy is S = ln Γ where Γ=



M NA



=

N! = eN ln N −NA ln NA −NB ln NB NA !NB !

(14)

(we used the Sterling formula, n! = (2πn) 1/2 (n/e)n ). Therefore the entropy is S = −nA ln nA − nB ln nB

(15)

d) Using the above, obtain a free energy function F (x), where x = (nA − nB ). Show that the requirement of convexity of F (x) breaks down below a critical temperature Tc . The free energy is F = E − T S. Introduce x = nA − nB so that nA,B = (1 ± x)/2. Then E(x) = Const + and S(x) = −

1 1 (JAA + JBB − 2JAB ) x2 + (JAA − JBB )x 8 4

1 1+x 1+x (1 + x) ln − (1 − x) ln 2 2 2 









(16)

(17)

Now we compute ∂ 2 F/∂x2 : ∂2F 1 1 = (JAA + JBB − 2JAB ) + T 2 ∂x 4 1 − x2

(18)

The first term is negative while the last term is positive . The lowest value of the second term, as a function of x, is T /2 (at x = 0). Hence ∂ 2 F/∂x2 > 0 for T > Tc = (2JAB − JAA − JBB )/4 and becomes negative at T < Tc , first near x = 0. e) Sketch F (x) for T > Tc , T = Tc and T < Tc For T < Tc there is a range of compositions x1 < x < x2 where F (x) is not convex and thus the composition is locally unstable. Plot the so-called spinodal lines2 x1,2 (T ) in the (T, x) plane.

Figure 2: Free energy vs atom concentration (binary alloy) −0.1 2

F=0.5J(nA−nB) −h(nA−nB)+T(nA ln nA+nB ln nB)

−0.12

J=1, h=0.03

−0.14

TT

c

−0.22 −1

−0.5

0

0.5

Atom concentration fraction x=(nA−nB)/(nA+nB)

1

The free energy is plotted on Fig. 2 To find spinodal lines we solve T < T c ∂ 2 F/∂x2 = 0: 1 Tc = 2 1−x T

(19)

x = ±(1 − T /Tc )1/2

(20)

which gives Spinodal lines are plotted on Fig. 3 f ) Describe what happens with an alloy cooled from higher to lower temperature, depending on the composition. In the (T, x) plane sketch the phase separation boundary, dividing the (T, x) plane into the stability regions of the A rich and B rich phases and the unstable region. If F ′′ (x) < 0 the uniform state is unstable. The system separates into A–rich and B–rich phases with x = xA or x = xB given by common tangents (see Fig. 4). Volume fractions of the A,B rich phases are (xB − x)/(xB − xA ) and (x − xA )/(xB − xA ) respectively. 3. Ising ferromagnet, a different approach Let us solve Problem 1 by using the method of Problem 2. To map the spin problem (1) on the P binary alloy problem, consider a microcanonical ensemble with fixed total magnetization M = r σr . In this case, out of total N spins, N+ = (N + M)/2 is up and N− = (N − M)/2 is down. a) Assuming the occupation of different sites to be statistically independent, and using the occupation probabilities p± = (N ± M)/2N, find the energy of spin interaction as a function of M. 1 E = − J(p+ − p− )2 2 2

The spinodal line indicates onset of metastability and hysteresis effects.

(21)

Figure 3: Two phase coexistence in a binary alloy (JAA=JBB) 1 0.9 0.8

Temperature, T/T

c

0.7 0.6

Stability boundary 0.5 0.4

Spinodal lines

0.3 0.2 0.1 0 −1

−0.8

−0.6 −0.4 −0.2 0 0.2 0.4 0.6 Concentration fraction x=(n −n )/n +n ) A

B

A

0.8

1

B

Figure 4: Volume fractions in a mixture of two phases −0.12 −0.125

Volume ratio VA/VB = (x’’−x)/(x−x’)

Free energy, a.u.

−0.13 −0.135 −0.14 −0.145 −0.15 −0.155 −0.16 −1

X’ −0.8

X

−0.6 −0.4 −0.2 0 0.2 0.4 0.6 Concentration fraction x=(nA−nB)/(nA+nB)

X’’ 0.8

1

where p± = (1 ± m)/2. b) Find the entropy and the free energy as a function of M. Sketch F (M) at different temperatures. Show that F (M) does not satisfy the convexity requirement at T < Tc . Find the critical temperature Tc . S = −p+ ln p+ − p− ln p−   1 T 1+M 1−m 2 F = E − T S = − Jm + (1 + m) ln + (1 − m) ln 2 2 2 2   ∂2F T 1 1 T = −J + + = −J + ∂m2 2 1+m 1−m 1 − m2

(22) (23) (24)

Note that ∂ 2 F/∂m2 becomes negative near m = 0 at T < Tc . (See the Fig. 2 in Problem 2). c) Below Tc find the values of M for which the uniform state is stable. Stable states correspond to energy minima ∂F/∂m = 0. This gives −Jm +

T 1+m ln =0 2 1−m

(25)

or,

J m = tanh m T the same equation for m as in Problem 1. d) Evaluate the free energy and compare the result with that of Problem 1. Combining equations (23) and (25) we obtain: 



1 T 1+M 1−m F = − Jm2 + (1 + m) ln + (1 − m) ln 2 2 2 2 " !# " # 2 2 1 1 T 1 − m + m 1 T 1 − tanh (Jm/T ) ln + m ln = Jm2 + ln = − Jm2 + 2 2 4 1−m 2 2 4 1 = Jm2 − T ln (2 cosh(Jβm)) 2 



(26)

(27) (28) (29)

4. Symmetry and transition type Consider an Ising ferromagnet (1) with the nearest neighbor interaction: J(r − r ′ ) 6= 0 only when |r − r ′ | = 1. Map this problem onto the binary alloy problem. Now, note that the ferromagnetic ordering emerges via a second order phase transition, whereas the ordering transition in a generic binary alloy is first order. Can you interpret/explain this by a symmetry argument in the Landau theory spirit? In the binary alloy problem we found a type I transition, while in the ferromagnet it is a type II transition. To see why this is consistent with the mapping, use the results of Problem 3: 1 = − J (p+ − p− )2 + T (p+ ln p+ p− ln p− ) ; 2 1 FBinary alloy = − (2JAB − JAA − JBB ) (pA − pB )2 + (JAA − JBB ) (pA − pB ) 2 +T (pA ln pA + pB ln pB )

FIsing

magnet

(30) (31) (32)

Thus the binary alloy problem can be mapped to the magnet with J = 2J AB −JAA −JBB in magnetic field h = −(JAA − JBB ). We observe that • Finite magnetic field removes the m → −m symmetry of the magnet problem hence there is no type II transition • Conserved particle number in the alloy problem translates into the fixed M total in the magnet problem, i.e. this is a microcanonical ensemble • Cooling a magnet with fixed M gives rise to phase separation into the regions m up and m down. In a finite external field this is a type I transition, just like in the alloy. 5. Liquid crystal ordering transition In some liquid crystals, the order parameter is the local electric quadrupole moment, which can be represented by a symmetric traceless tensor Qij , tr Q = 0. The specific form of the tensor Qij depends on the ordering type. For instance, in nematic liquid crystals discussed in class, the relation between Qij and nematic director vector n is Qij = Q(T )(ni nj − 31 δij ). Consider a spatially uniform state (no gradients!). Near the phase transition from liquid into liquid crystal, consider the Landau free energy F (Q) as power series in Q. What are the various lower order terms that are allowed, by rotational symmetry, to appear in the free energy? Show that this suggests that the transition into the ordered phase should be the first order for this system. Consider the free energy of a liquid crystal expanded in Q ij . In each order, we write all the invariant combinations F (Q) = F0 + AQij Qij + B1 (Qij Qij )2 + B2 Qij Qjp Qpq Qqi + CQij Qjp Qpi + gradient terms

(33)

There are no other invariants, since Q is traceless: Q ii = 0. In the matrix form the free energy is 

F = F0 + ATr Q2 + B1 Tr Q2

2

+ B2 Tr Q4 + CTr Q3 + gradient terms

(34)

We found a cubic invariant allowed by the symmetry. This excludes type II transition in this system.

8.334: Statistical Mechanics II

1 1.1

Last modified: February 12, 2003

Lecture 3. Symmetry change at a phase transition. Landau theory. Order parameter. Symmetry classification

Landau theory of type II phase transitions can be viewed as a generalization of of the observations made in Lecture 2, when we discussed ordering transitions in a ferromagnet and binary alloy. This theory is using powerful argument based solely on the symmetry change in the transition. Suppose one has to describe a continuous phase transiton in which the symmetry group is reduced from G 0 to its subgroup G ∈ G0 , where G0 and G are the symmetry groups of the disordered and ordered states, respectively. The ordering is characterized by a function ρ(x) that changes through the transition. In a structural transition ρ(x) can be particle density, in a magnetic transition it is spin density, and so on. In a disordered state above Tc ρ(x) is invariant under the symmetry group G0 . Below the transition, this quantity (we call it ρ 1 (x)) is invariant only under the transformations from the subgroup G1 . Hence one can, following Landau, write the (n) difference δρ = ρ1 − ρ0 in terms of the basis functions φ i of irreducible representations of the group G0 : X′ X (n) (n) δρ(x) = ci φi (x) , (1) n

i

where n labels irreducible representations of G, and i labels the basis functions of each representation. The prime sign indicates that the unit representation is excluded from the sum, since it contributes equally to ρ 1 and ρ0 . At the transition T = Tc all quantities (n) ci vanish. (n) The coefficients ci characterize the system in thermodynamic equilibrium below the transition, and thus their values provide minimum for the thermodynamic potential Φ. (n) In a type II transition just a bit below T c the values ci can be arbitrarily small. Thus (n) the potential can be expanded in powers of c i . A number of general features of this expansion can be understood by noting that, since the function δρ is invariant under G, (n) (n) the coefficients ci transform the same way as the functions φ i . Since the potential is invariant under coordinate transformations and other symmetry operations from G, each (n) order of the expansion is given by some invariant polynomials in c i . Now, each irreP (n) ducible representation has no linear invariants and just one quadratic invariant, i (ci )2 . (n) Therefore the series expansion of Φ in c i begin with Φ = Φ0 +

X′

(n)

A

X

(P, T )

n

i

(n) (ci )2

!

(n)

+ O((ci )3 )

(2)

Thermodynamic stability requires A (n) (P, T ) > 0 for all n above Tc . In order to have (n) broken symmetry, i.e. nonzero c i , below Tc , one of the coefficients A(n) (P, T ) has to 1

change sign at T = Tc . (Simultaneous sign change of two ot more A (n) can take place only at isolated points of the (P, T ) plane.) Since only one of the representations of G corresponds to thermodynamic instability of a disordered state, we can discard all the terms in the expansion (2) except the one (n) which changes sign at Tc . The quantities ci that become nonzero in the ordered state below Tc are called the order parameter. Focussing on this representation, from now on we drop the index n, always assuming that the representation is the one that corresponds to the instability at T c . It is useful to use the notations X c2 = c2i , ci = c xi (3) i

with

P

2 i xi

= 1. Let us now consider the higher order terms in the expansion (2):

Φ = Φ0 (P, T ) + c2 A(P, T ) + c3

X

Cα (P, T )fα(3)(xi ) + c4

α

X

Bα (P, T )fα(4) (xi ) + O(c5) (4)

α

Here the functions fα(3) , fα(4) , ..., are invariant polynomials of order 3, 4, etc. Each sum over α has as many terms as there are invariants of the required order. These terms determine the ordering just below the transition, and define the transition type. To summarize, there are three main ingredients in this phenomenological theory: the symmetry groups G0 and G above and below the transition, and the representation of G 0 that describes the order parameter properties. Usually, there is no ambiguity in G 0 and G, while some insight or additional experimental input is needed to choose the correct form of the order parameter (i.e. the relevant representation of G 0 ).

1.2

Transition type deduced from symmetry.

In a continuous phase transition (type II) the variables c i found by minimizing Φ should become small near T c (as in the prototype case of Φ(c) = A(T − Tc )c2 + Bc4 ). One can ask when does the potential (4) has this propoperty. This mathematical question has a simple answer: 1 If the transition in a system described by the potential (4) is of type II, no cubic invariants are allowed. To prove this statement, one simply has to consider Φ(c) at T = Tc , when A = 0, and note that if C 6= 0 there exist other mimima c = c ∗ with Φ(c∗ ) < Φ(0). This simple but very basic observation allows to predict which symmetry changes can take place as a type II transition, and which cannot. One has to look for cubic invariants of the order parameter representation that describes the transition and if they exist, the transition is guaranteed to be type I, whereas if there are no cubic invariants, the transition is most likely of type II. There are powerful goup theoretic methods for searching for cubic invariants. Mathematically speaking, this is a question of whether the tensor cube of the representation invloved in the transition, when split into irrreducible components, contains 1

The absence of cubic invariants is a necessary condition. Another necessary condition is that the 4th order terms in (4) are positive and can stabilize the ordered phase at small c.

2

a unit representation. To find out, one can compute the sum n 3 = g∈G χ3 (g), where χ(g) are representation characters. This sum gives the dimension of the invariant subspace in the tensor cube, and therefore, if it is zero, there no cubic invariants, while if it is nonzero, there are n3 invariants. Although this general methodology is quite helpful in analyzing some complex situations 2, in simple cases the absence or presence of cubic invariants can be figured out by a direct inspection. Two examples illustrating this approach are provided by ferromagnetic phase transition in a spin system and by crystallization of a liquid into a solid. In a ferromagnetic transition, the order parametr is magnetization m, a scalar in the Ising model, or a vector in the Heisenberg model. The disordered state is time-reversal invariant: the transformation m → −m leaves Hamiltonian invariant, and thus it is part of the symmetry group G0 . Therefore, there are no cubic invariants and Landau free energy in this case has the form P

Φ(m) = A(P, T )m2 + B(P, T )m4 ,

A(P, T ) = a(T − Tc ) near Tc

(5)

with a > 0, B > 0. Thus the ferromagnetic transition is second order. In crystallization transition, ordering is described by the appearance of a density modulation δρ(x) in a uniform liquid state. Choosing the order parameter to be the P amplitude δρk of density harmonics in the crystal phase, δρ(x) = k δρk eik·x one can write the thermodynamic potential as Φ(δρ) =

X k

A(k)δρk δρ−k +

X

B(k, k ′ )δρk δρk′ δρ−k−k′ + O(δρ4 )

(6)

k,k ′

Here the cubic terms arize from the series expansion of Φ in density variations in a liquid. Note that they are not eliminated by any symmetry of the liquid phase hamiltonian. To see how the cubic terms affect the transition, we need to analyze the relations between different harmonics. Let us first discuss quadratic terms. In a simple model, the function A(k) is isotropic (due to rotational invariance of liquid state) and has a single minimum in |k| = k 0 . For example A(k) = A1 + A2 (|k| − k0 )2 . Suppose that, as a function of temperature, the minimal value of A(k) becomes negative (i.e. A 1 + A2 < 0). As soon as this happens, all harmonics with |k| ≈ k 0 become unstable. Which combination of harmonics gives the lower energy is decided by the 4th and higher order terms, but without going into this discussion, let us assume that there is a star of vectors k i that defines the density harmonics in the ordered state. The symmetry of the set of vectors k i reflects the symmetry of the crystal state It is very often the case that some of the vectors from the star added together give another vector from the same star. Simple examples: ordering of a 2D liquid in a triangular crystal lattice, with the hexagonal star of harmonics k i , or face-centered cubic ordering 2

The group theory is especially useful in solids, where there are 240 different space groups even before the magnetic ordering is taken into account. To understand how these symmetries can be broken at a phase transition, and which transition types are possible, often requires using heavy group-theretic tools.

3

in 3D (the star is defined by a tetrahedron, 6 vectors pointing along the edges plus their opposites). In fact, it has been shown by Landau that for any crystal symmetry, one can find triplets of harmonics that contribute to the cubic term. As a result, on very general symmetry grounds, crystrallization is a first order transition. This analysis shows that, despite symmetry being spontaneously broken in crystallization, the transition is of type I, i.e. discontinuous. Another case, where on pure symmetry grounds one can reject type II transition, is formation of liquid crystals (see homework problem 5, PS#2)

1.3

Thermodynanmic quantities

Let briefly list the results derived for the type II transition with a scalar order parameter, such as the Ising model or a binary alloy. For concretenens, we discuss transition in a magnetic system with free energy of the form Φ = −aτ m2 + bm4 + mh ,

τ = (Tc − T )/Tc

(7)

where h is the magnetic field. More generally, the field that couples linearly to the order parametr is called ordering field. At h = 0, the order parameter equlibrium value is m=



0, (aτ /2b)1/2

at τ < 0 at τ > 0

(8)

From that, the free energy is 0 at τ < 0, and F ∝ −τ 2 at τ > 0. The specific heat CM = −T (∂ 2 F/∂T 2 )M

(9)

has a jump at the transition. The zero field susceptibility χ = (∂M/∂h)T ∝ τ −1

at τ < 0

(10)

These results illustrate that thermodynamic quantities have singularities at the type II transition point. The specific form of temperature dependence at singularity can be modified by the effects of fluctuations, if those are strong.

1.4

Isomorphic phase transitions.

Another useful consequence of the symmetry approach is that it allows to identify isomorphic phase transitions in which the symmetry change is the same. Isomorphic transitions may take place in very different systems, and despite that, they have the same macroscopic charactersitics. Such transitions are saiud to belong to the same universality class. One example, discussed already in Lecture 2, is the relation between the Ising phase transition, the liquid-gas critical point, and the binary alloy. In all three cases, the ordering is describes by a scalar order parameter. The symmetry is exact in the case of the Ising 4

problem (m → −m), and in the case of a binary alloy (a ↔ b), but only approximate near the liquid-gas critical point (however, it becomes exacvt asymptotically right at this point). For convenience, we list analogous quantities in a table: System described by Ising model Density-like variable Field-like variable

Thermodynamic potential for the density variable Thermodynamic potential for the field variable Order parameter φ Ordering field h Susceptibility (∂φ/∂h)T Specific heat Cφ

1.5

Curie point ferromagnet

of

a

Magnetization M = −(∂Φ/∂H)T Magnetic field H = −(∂F/∂µ1 )T

Liquidgas critical point of a one component liquid Density ρ = −(∂P/∂µ)T Chemical potential µ = F + P V = [∂(ρF )/∂ρ]T

Critical point of a binary mixture

F (T, M) = Φ + Mh, dF = −SdT + hdM

Concentration x = −(∂µ1 /∂µ∗ )T The difference of the chemical potentials of the components, µ∗ = µ2 − µ1 = (∂F/∂x)T ρF (T, ρ) = −P + ρµ , F (T, x) = µ1 + µ∗ x , d(ρF ) = −ρSdT +µdρ dF = −SdT + µ∗ dx

Φ(T, h) = F − Mh , dΦ = −SdT − Mdh

−P (T, ρ) , −dP ρSdT − ρdµ

M

(ρ − ρc )/ρc

x − xc

h (∂M/∂h)T

(µ − µ(ρc , T ))/Tc (Tc /ρc )(∂ρ/∂µ)T

(µ∗ − µ∗ (ρc , T ))/Tc Tc (∂x/∂µ∗ )T

= µ1 (T, µ∗ ) = F − µ∗ x , dµ1 = −SdT − xdµ∗

CM = −T (∂ 2 F/∂T 2 )M ρCV = −T (∂ 2 ρF/∂T 2 )V Cx = −T (∂ 2 ρF/∂T 2 )x

Summarize

• Symmetry change determines many qualitative features of a phase transition, such as its type (I or II) and universality class. • Spontaneous symmetry breaking occurs discontinuously, via a type I transition, when the free energy expansion in powers of order parametr contains cubic terms. • Fluctuations are not accounted for in the Landau theory. Many conclusions, including the predicted temperature dependence of thermodynamic quantities, hold only when fluctuations are small.

5

8.334: Statistical Mechanics II

Problem Set # 3

Due: 2/26/03

Landau theory, susceptibility & fluctuations 1. Ising ferromagnet. a) Consider the Landau functional for an Ising ferromagnet near transition. Evaluate the free energy by minimizing the functional and find entropy and specific heat near the transition. Compare with the result of Problem #1, PS2. b) Using Landau theory of a ferromagnet in an external field, consider magnetization as a function of temperature. Make a plot showing M(T ) for different values of magnetic field, positive and negative. c) Find magnetic susceptibility χ = dM/dH in zero field separately above and below the transition. Compare the two results. 2. Ising antiferromagnet, Landau theory. Antiferromagnet on a cubic lattice is described by spin Hamiltonian with negative exchange interaction between nearest neighbors: X 1 σr σr′ , H= J 2 |r−r′|=1

J>0

(1)

with σr = ±1 Ising spins. This interaction favors opposite spins on neighboring lattice sites, which results in a ground state with spins opposite on each of the two sublattices. In fact, there are two equivalent ground states related by an overall spin sign change. (Compare with the two ground states in a binary alloy pictured in figure (1), Lecture 2.) a) Develop Landau theory for the phase transition in Ising antiferromagnet. Phenomenologically, choose the order parameter to be staggered magnetization, i.e. the difference of magnetization on two sublattices. Derive the form of Landau free energy from a symmetry argument, show that it is the same as for the ferromagnet. b) Now, describe an antiferromagnet near the phase transition in the presence of an external magnetic field. Using symmetry argument, show that there is no coupling of uniform field to the staggered magnetization linear in the order parameter. c) Generalize this approach by including the average magnetization in the Landau functional. Write the most general free energy form, allowed by symmetry, in the presence of an external field. Find magnetic susceptibility χ above and below the transition. Determine the form of singularity in χ(T ) at T = Tc . 3. Curie and Neel susceptibilities. Here we consider ferromagnet and antiferromagnet in magnetic field, starting with the microscopic hamiltonian and solving the problem using the mean field approach. a) From the mean field equation for the Ising and Heisenberg ferromagnet magnetization derived in class (Eqs.(11),(17), Lecture 2), derive susceptibility in zero field at T > T c . Express the resulting χ(T ) in terms of T /Tc (Curie law). b) From the mean field approach, find the zero field susceptibility of an Ising antiferromagnet at T > Tc (Neel law).

c) Using the mean field approach, find the zero field susceptibility χ of an Ising antiferromagnet at T < Tc . Does the behavior near Tc agree with the Landau theory result of Problem 2? 4. Ornstein-Zernicke formula. a) Use Landau theory to study the correlation function hm(x)m(x′ )i in an Ising ferromagnet near R −ik·x Tc . For the order parameter Fourier harmonics m k = e m(x)d3 x, derive the formula hmk m−k i =

A +1

(2)

ξ 2 k2

Find the correlation length ξ and the constant A at T > T c . b) What is the relationship of this correlator with the susceptibility χ? c) Consider the correlation fucntion C 2 (x−x′ ) = hδm(x)δm(x′ )i in real space. Find its functional form in D = 3 using the above formula. d) Consider the correlation fucntion C 2 (x − x′ ) in the ordered state, at T < Tc . Use equipartition theorem to show that the structure factor form (2) does not change (up to numerical factors in A and ξ). From that, derive a real space form of C 2 . 5. Goldstone mode fluctuation suppressed by an external field. a) Scattering of light in a dielectric medium takes place due to spatial fluctuations of refraction index, i.e. the dielectric permeability δǫ ij . The angular distribution of scattered light is determined by the correlator hδǫij δǫi′ j ′ i of the fluctuating part of permeability. This quantity is called the scattering tensor1 . Show that when an electromagnetic plane wave with wavevector q and polarization E is scattered on the fluctuations in the medium, the scattered field E ′ is determined by the correlator of permeability δǫ Fourier harmonics hEi′ (Ej′ )∗ i ∝ V hδǫim (k)δǫ∗jn (−k)iEm En∗ ,

k = q′ − q

(3)

where V is the system volume, δǫ ij (k) = δǫij (x)e−ik·x d3 x and q′ is the wavevector of a scattered wave. b) It is interesting to apply this result to scattering in a nematic liquid crystal 2 caused by fluctuations of the nematic order parameter n. In this case, the fluctuating part of dielectric permeability δǫij is proportional to the traceless tensor n i (x)nj (x) − 13 δij . Consider a liquid crystal in the presence of a uniform static electric field. The polarizability of anisotropic rod-like molecules is much higher along the molecule axis than in the transverse direction, αk ≫ α⊥ . Hence the field is trying to align the molecules, which is described by adding a term − α2 (E · n)2 to the liquid crystal free energy (Eq.(13), lecture 1). (Here α = α k − α⊥ is a positive constant.) Show that with the field applied, the lowest energy state is indeed n k E. c) Consider the fluctuations n = n(0) + δn, with δn ⊥ n(0) , by expanding the free energy to the lowest nonvanishing order in δn. Find the correlator hδn k δn−k i and show that applied electric field suppresses scattering. The possibility to change the medium from opaque to transparent by an external field has various practical applications (such as the liquid crystal displays). R

1

Definition and discussion of the scattering tensor can be found in: Landau & Lifshits, Electrodynamics of Continuous Media, §117 2 Scattering in nematic liquid crystals is discussed in: Landau & Lifshits, Electrodynamics of Continuous Media, §122

8.334: Statistical Mechanics II Problem Set # 3 Landau theory, susceptibility & fluctuations 1. Ising ferromagnet. a) Consider the Landau functional for an Ising ferromagnet near transition. Evaluate the free energy by minimizing the functional and find entropy and specific heat near the transition. Compare with the result of Problem #1, PS2. Starting from 1 (1) F = − aτ m2 + bm4 2 and minimizing F , obtain    ± 1 paτ /b τ > 0  − 1 a2 τ 2 τ > 0 2 16b m∗ = F (m∗ ) = (2)   0 τ Tc . Express the resulting χ(T ) in terms of T /Tc (Curie law). For Ising ferromagnet M = tanh(βJM + βh). For small h and M (above Tc = βJ), this yields M = βJM + βh or M = βh/(1 − βJ). Hence we obtain Curie law χ=

1 T − Tc

(16)

For Heisenberg ferromagnet above Tc = βJ/3 M = coth(βJM + βh) −

1 βJM + βh

(17)

Linearizing, we obtain M = (βJM + βh)/3 and χ=

1 β/3 = 1 − βJ/3 3(T − Tc )

(18)

b) From the mean field approach, find the zero field susceptibility of an Ising antiferromagnet at T > Tc (Neel law). The mean field result for magnetization of Ising antiferromagnet is the same as for ferromagnetic case, but with J < 0. For Ising antiferromagnet this leads to Neel law χ=

1 T + Tc

(19)

with Tc = |βJ|. c) Using the mean field approach, find the zero field susceptibility χ of an Ising antiferromagnet at T < Tc . Does the behavior near Tc agree with the Landau theory result of Problem 2? To apply the mean field theory to AFM, consider an ordered state with magnetization different on two sublattices. The free energy of the sublattices is 1 F1 = Jm1 m2 − T ln [2 cosh β(Jm2 + h)] 2 1 F2 = Jm1 m2 − T ln [2 cosh β(Jm1 + h)] 2

(20) (21)

The total free energy per spin is F = (F1 + F2 )/2. Minimizing it with respect to m1 and m2 , we obtain m1 = tanh β(Jm2 + h) m2 = tanh β(Jm1 + h) Since J is negative, the ground state for h = 0 is m0 = m1 = −m2 . The susceptibility is   1 β ∂(m1 + m2 ) 1 ∂(m1 + m2 ) χ= h→0 = 2 cosh2 β(Jm0 ) 2 + J h→0 2 ∂h ∂h Hence

χ(T < Tc ) =

1 1 − m2 (τ ) = T + Tc (1 − m2 (τ )) T cosh2 (βJm) − J

(22) (23)

(24)

(25)

where m(τ ) is the solution to m = tanh(β|J|m). The function χ(τ ) is continuous at Tc , since χ(Tc ) = 1/2Tc is equal to the value of (19) at Tc . In the ordered state m2 (τ ) ∝ τ near Tc , hence χ(τ ) has a cusp at Tc (see Fig. 2).

Antiferromagnet susceptibility vs temperature

Susceptibility χ=dm/dh

0.5

0.4

0.3

0.2

0.1

0 0

0.5

1

1.5

2

2.5

3

Temperature T/T

c

Figure 2: Susceptibility of Ising antiferromagnet 4. Ornstein-Zernicke formula. ′ a) Use Landau theory to study the correlation function )i in an Ising ferromagnet near R hm(x)m(x −ik·x Tc . For the order parameter Fourier harmonics mk = e m(x)d3 x, derive the formula hmk m−k i =

A +1

ξ 2 k2

Find the correlation length ξ and the constant A at T > Tc .  Z  1 1 2 2 4 F = − aτ m + bm + K(∇m) d3 x 2 2 Above Tc F has minimum at m ¯ = 0. Expanding F in fluctuations δm we obtain  Z  X a|τ | + Kq 2 1 1 2 2 F (δm) = a|τ |(δm) + K(∇ δm) d3 x = |δmq |2 2 2 2 q Equipartition theorem gives h|δmq |2 i =

T A = 2 2 2 a|τ | + Kq ξ q +1

(26)

(27)

(28)

(29)

with A = T /a|τ | and ξ 2 = K/a|τ | b) What is the relationship of this correlator with the susceptibility χ? The susceptibility is χ = dm/dh = 1/a|τ |. Therefore A = T χ. This agrees with the general theorem about fluctuations and linear response. c) Consider the correlation function C2 (x − x′ ) = hδm(x)δm(x′ )i in real space. Find its functional form in D = 3 using the above formula.

Z ∞ Z π iq(x−x′ ) cos θ ′ e eiq(x−x ) A d3 q A C2 (x − x ) = = 2 q 2 d q d cos θ = 2 2 3 1 + ξ q (2π) 4π 0 1 + ξ 2q2 0 Z ∞ Z ∞ 1 sin q(x − x′ ) A ∂ cos q(x − x′ ) A = 2 q dq = − dq = 2π x − x′ 0 1 + ξ 2q 2 2π 2 (x − x′ ) ∂x 0 1 + ξ 2q2   ′ A ∂ π −|x−x′ |/ξ A e−|x−x |/ξ =− 2 e = (30) 2π (x − x′ ) ∂x 2ξ 4πξ 2 |x − x′ | ′

Z

d) Consider the correlation function C2 (x − x′ ) in the ordered state, at T < Tc . Use equipartition theorem to show that the structure factor form (26) does not change (up to numerical factors in A and ξ). From that, derive a real space form of C2 . In the ordered state (τ > 0) m ¯ = (aτ /4b)1/2 and for fluctuations m = m ¯ + δm the free energy is    Z  Z  1 1 1 1 2 2 3 2 2 2 2aτ δm + K(∇ δm) d3 x (31) F (δm) = 6bm ¯ − aτ δm + K(∇ δm) d x = 2 2 2 2 √ Therefore C2 has the same form as in part c) with a → 2a, ξ → ξ/ 2, A → A/2.

5. Goldstone mode fluctuation suppressed by an external field. a) Scattering of light in a dielectric medium takes place due to spatial fluctuations of refraction index, i.e. the dielectric permeability δǫij . The angular distribution of scattered light is determined by the correlator hδǫij δǫi′ j ′ i of the fluctuating part of permeability. This quantity is called the scattering tensor1 . Show that when an electromagnetic plane wave with wavevector q and polarization E is scattered on the fluctuations in the medium, the scattered field E′ is determined by the correlator of permeability δǫ Fourier harmonics

hEi′ (Ej′ )∗ i ∝ V hδǫim (k)δǫ∗jn (−k)iEm En∗ , k = q′ − q (32) R where V is the system volume, δǫij (k) = δǫij (x)e−ik·x d3 x and q′ is the wavevector of a scattered wave. In the scattering medium Di = (ǫij + δǫij )Ej . Starting with the wave equation ∇ × ∇ × E = ω 2 /c2 D, obtain ω2 ω2 ∇ × ∇ × Ei − 2 ǫij Ej = 2 δǫij Ej (33) c c The radiation field with wave vector k ′ at a point far away from the scattering region is Z ′ ω 2 eik R ′ E = (δǫij Ej (x))d3 x (34) 2 4πc R x∈V For plane wave E(x) = E exp(ikx), the radiation field Z ′ eik R ′ E ∝ δǫij e−iqx d3 x R x∈V 1

(35)

Definition and discussion of the scattering tensor can be found in: Landau & Lifshits, Electrodynamics of Continuous Media, §117

where q = k ′ − k. The integral over scattering region of size much larger than EM wavelength is the Fourier harmonic of ǫij . The field intensity is XZ Z ′ ′ ∗ hEi (Ej ) i ∝ hδǫin (x)δǫjm (x′ )ie−iqx+iq1 x1 d3 xd3 x1 (36) k ′ k1′

where q = k ′ − k, q1 = k1′ − k1 . Change the variables to X = (x + x1 )/2, y = x − x1 . Then XZ Z XZ Z ′ −iqx+iq1 x1 3 3 hδǫin (x)δǫjm (x )ie d xd x1 = h. . .iei(q1 −q)X+iy(q1 +q)/2 d3 Xd3 y = k ′ k1′

k ′ k1′

=

X k ′ k1′

(2π)

3

Z

iy(q1 +q)/2

h. . .ie

3

δ(q1 − q)d y =

Z X k

eiqy h. . .id3 y (37)

Therefore hEi′ (Ej′ )∗ i

∝

Z X k

eiqy hδǫii′ (y)δǫjj ′ (0)iEi′ Ej ′ d3 y = V hδǫii′ (q)δǫjj ′ (−q)iEi′ Ej ′

(38)

b) It is interesting to apply this result to scattering in a nematic liquid crystal2 caused by fluctuations of the nematic order parameter n. In this case, the fluctuating part of dielectric permeability δǫij is proportional to the traceless tensor ni (x)nj (x) − 31 δij . Consider a liquid crystal in the presence of a uniform static electric field. The polarizability of anisotropic rod-like molecules is much higher along the molecule axis than in the transverse direction, αk ≫ α⊥ . Hence the field is trying to align the molecules, which is described by adding a term − α2 (E · n)2 to the liquid crystal free energy (Eq.(13), lecture 1). (Here α = αk − α⊥ is a positive constant.) Show that with the field applied, the lowest energy state is indeed n k E. c) Consider the fluctuations n = n(0) + δn, with δn ⊥ n(0) , by expanding the free energy to the lowest nonvanishing order in δn. Find the correlator hδnk δn−k i and show that applied electric field suppresses scattering. The possibility to change the medium from opaque to transparent by an external field has various practical applications (such as the liquid crystal displays). Expand n = n0 (1 − δn2 )1/2 + δn, δn · n = 0. To order O(δn2 ), obtain  Z  1 1 2 2 2 2 2 K1 (div δn) + K2 (n0 · curl δn) + K3 (n0 × curl δn) + αE δn d3 x = H(δn) = 2 2   X 1 1 2 2 2 2 2 = K1 |q · δnq | + K2 |n0 · (q × δnq )| + K3 |n0 × (q × δnq )| + αE |δnq | (39) 2 q 2 since n0 × (q × δn) = q(n0 · δn) − δn(q · n0 ) = −δn(q · n0 ), H can be simplified as H(δn) = 2

1X 1 Kij (q)δnq,i δn−q,j + αE 2 δnq · δn−q 2 q 2

(40)

Scattering in nematic liquid crystals is discussed in: Landau & Lifshits, Electrodynamics of Continuous Media, §122

where Kij (q) = K1 qi qj + K2 (n0 × q)i (n0 × q)j + K3 (n0 · q)2 δij It follows from the equipartition theorem that  −1 1 2 hδnq,i δn−q,j i = Kij (q) + αE δij 2

(41)

(42)

At zero field, E = 0, the fluctuations h|δnq |2 i ∝ 1/q 2 diverge at q → 0. Finite field suppresses h|δnq |2 i at q = 0 to 2/αE 2

8.334: Statistical Mechanics II

1

Last modified: February 19, 2003

Lecture 4. Universality classes. Susceptibility and fluctuations.

1.1

Isomorphic phase transitions

One useful consequence1 of the symmetry approach is that it allows to identify isomorphic phase transitions in which the symmetry change is the same. Isomorphic transitions may take place in very different systems, and despite that, they have the same macroscopic characteristics. Our reason to emphasize symmetry is that the change of the state that occurs at phase transition is macroscopic. The characteristic length scale for thermodynamic fluctuations near the phase transition, as well as below it in the ordered state, is large. (This will be discussed below.) The microscopic origin of the degrees of freedom participating in the transition plays a relatively minor role, since on relevant length scales, much larger than interparticle or lattice spacing, a coarse grained picture emerges. For concreteness, let us talk now about a magnetic system. The order parameter variable used to describe the ordering is magnetization m(x), representing microscopic magnetic moments coarse grained from lattice length scale up to a large length scale at which thermodynamic fluctuations take place. In that, we treat x as a continuous variable with the understanding that the function m(x) exhibit variations on length scales larger than the lattice spacing a. The free energy expansion in powers of m near the transition, known as Landau functional, has a universal form dictated solely by symmetry: H=

1 1 − aτ m2 + bm4 + K(∇m)2 + O(m6) d3 x , 2 2

Z 



τ = (Tc − T )/Tc

(1)

(a, b > 0). Compared to the discussion in Lecture 3, we added a gradient term 12 K(∇m)2, sometimes called Lifshits term. On phenomenological grounds, this term gives the energy associated with spatially nonuniform fluctuations. The rigidity constant in this term is positive, K > 0. Such a term can also be derived from the microscopic mean field approach, as we saw in Lecture 1. There are many phase transition varieties described by the free energy (1). In general, m is an n component vector. One can relate phase transitions in different systems which are described by the same symmetry change, and hence by an order parameter with the same number of components: • n = 1 describes liquid-gas critical point, binary mixtures (e.g., alloys), as well as uniaxial magnets (e.g. Ising model); 1

Moved here from Lec. 3 and expanded.

1

• n = 2 describes superfluidity, superconductivity, planar magnets (i.e. Heisenberg magnets with the the so-called easy plane anisotropy); • n = 3 corresponds to Heisenberg ferromagnets, etc. There are many other transitions not included in this brief list. For instance, the problem (1) in the limit of zero number of components, n = 0, is used to describe disordered systems. Another important characteristic is space dimension. It turns out that the behavior near transition, and even the mere existence of the transition, is very sensitive to the dimensionality of the problem. Most of the systems of interest have d = 3, but it is very useful to consider different space dimensions, d = 1, 2 and even d = 4 and higher. The transitions in the same space dimension, described by the same order parameter symmetry are said to belong to the same universality class. Many universality classes are possible, not limited to the above three varieties. To illustrate how the correspondence between the systems of the same universality class can be established, let us discuss the first item in the above list, n = 1. The Ising phase transition, the liquid-gas critical point, and the binary alloy, have already been described in Lecture 2. In all three cases, the ordering is describes by a scalar order parameter. The symmetry is exact in the case of the Ising problem (m → −m), and in the case of a binary alloy (a ↔ b), but only approximate near the liquid-gas critical point (however, it becomes exact asymptotically right at this point). For convenience, we list analogous quantities in a table: System de- Curie point of a LiquidCritical point of a biscribed by Ising ferromagnet gas critical point of a nary mixture model one component liquid Density-like Magnetization M = Density Concentration x = variable −(∂Φ/∂H)T ρ = −(∂P/∂µ)T −(∂µ1 /∂µ∗ )T Field-like Magnetic field H = Chemical po- The difference of the variable −(∂F/∂µ1 )T tential µ = F + P V = chemical potentials of [∂(ρF )/∂ρ]T the components, µ∗ = µ2 − µ1 = (∂F/∂x)T Thermodynamic F (T, M) = Φ + Mh, ρF (T, ρ) = −P + ρµ , F (T, x) = µ1 + µ∗ x , potential for the dF = −SdT + hdM d(ρF ) = −ρSdT +µdρ dF = −SdT + µ∗ dx density variable Thermodynamic Φ(T, h) = F − Mh , −P (T, ρ) , −dP = µ1 (T, µ∗ ) = F − µ∗ x , potential for the dΦ = −SdT − Mdh ρSdT − ρdµ dµ1 = −SdT − xdµ∗ field variable Order parameter M (ρ − ρc )/ρc x − xc φ Ordering field h h (µ − µ(ρc , T ))/Tc (µ∗ − µ∗ (ρc , T ))/Tc Susceptibility (∂M/∂h)T (Tc /ρc )(∂ρ/∂µ)T Tc (∂x/∂µ∗ )T (∂φ/∂h)T Specific heat Cφ CM = −T (∂ 2 F/∂T 2 )M ρCV = −T (∂ 2 ρF/∂T 2 )V Cx = −T (∂ 2 ρF/∂T 2 )x 2

1.2

Susceptibility

An efficient probe of the physics near transition is provided by susceptibility of the order parameter to various external fields. It is particularly illuminating to introduce the so called ordering field that couples to the order parameter linearly: H(m) = H0 (m) −

Z

h · m d3 x

(2)

where H0 is the field independent functional (1). In magnetic systems, the field h is just the ordinary magnetic field. In other problems it may have a different meaning (see Table above). Susceptibility is defined as χ = dM/dh ,

M = hmi

(3)

Here h...i, as usual, stands for thermodynamic averaging Z1 tr (...e−βH ). For illustration, let us consider the Landau theory for an Ising ferromagnet, ignoring the effects of fluctuations: 1 H(m) = − aτ m2 + bm4 − hm 2

(4)

(here τ = (Tc − T )/Tc ). The average hmi in this case is simply given by the extremal m value. At h = 0,  0, at τ < 0 m0 = hmi = (5) (aτ /4b)1/2 at τ > 0 In a small field h, the extremal value changes: m = m0 + δm ,

h = δm = ′′ H (m0 )



h/(−aτ ), at τ < 0 h/(2aτ ) at τ > 0

(6)

where H′′ = ∂ 2 H/∂m2 . Hence the susceptibility χ diverges as 1/τ near the transition. Similar result comes from the mean field theory. Indeed, above T c one can linearize the selfconsistent equation for magnetization, m = tanh(βJm + βh), derived in Lecture 2. This gives a linear relationship, m = h/(T −T c ), which yields the Curie law χ = 1/(T −T c ). Comparison with experiment. A power law divergence is seen, but the form of singularity is often somewhat different: 1/τ γ with γ > 1. To understand this, we need to look into the role of fluctuations.

1.3

Fluctuations. Structure factor.

The appearance of a long range order is associated with the change in the behavior of the correlation function C2 (x − x′ ) ≡ hm(x)m(x′ )i = 3

 1  tr m(x)m(x′ )e−βH Z

(7)

Here, for simplicity, we discuss the case of one component order parameter, n = 1. Generalization to other forms of order parameter will be straightforward. Above the transition, fluctuations of m at different points are practically uncorrelated, and thus the correlation function (7) falls of rapidly as a function of the distance |x − x ′ |. As the temperature approaches the transition, the range in which the fluctuations of m are correlated increases, typically as a power law in 1/τ . At the transition, the order proliferates making the range of correlations infinite. One can use, as an illustration of this behavior, the Landau theory (1), and show that C2 (x − x′ ) ∝ exp (|x − x′ |/ξ) ,

ξ(τ ) ∝ 1/|τ |1/2

(8)

This behavior is analyzed in more detail in the homework problem 4, PS#3. Fluctuations can be easily measured experimentally in a scattering experiment. Remind of the liquid-gas system becoming opaque near the critical point. (This phenomenon is called critical opalescence.) Magnetic ordering is usually probed by neutron scattering: unlike light, neutrons have spin which couples to electron spins (via dipole interaction) and to to nuclear spins (via exchange interaction). There are two characteristics that one can look for in a scattering experiment, the intensity and the angular distribution of scattered light (or other waves used in the experiment). To illustrate the relation of these characteristics with the correlation function, let us consider neutron scattering in an Ising magnetic system, with a contact interaction R Hint = γ σn m(x) d3 x. Taking the incident neutron state to be a plane wave with momentum p, assuming the scattering to be elastic, and using Born scattering approximation, the angular distribution of scattered neutrons is 2

dN(θ) ∝ γ h|

Z

e−ik·x m(x) d3 x|2 i ,

|k| = 2 sin(θ/2)|p|

(9)

where k = p′ − p is the momentum change at scattering, |p ′ | = |p|. This expression can R −ik·x be rewritten in terms of Fourier harmonics m k = e m(x) d3 x as dN(θ) ∝ γ 2 h|mk |2 i

(10)

Also, Eq.(9) can be brought to the form where the relation with the correlation function (7) is more explicit. For that, we change the integration variables x and x ′ in (9) to X = 21 (x + x′ ) and y = x − x′ : ZZ

=

′

e−ik·x+ik·x hm(x)m(x′ )i d3xd3 x′

ZZ

e−ik·y hm(x)m(x′ )i d3 yd3X = V

(11) Z

e−ik·y C2 (y) d3y

(12)

where V is the volume of the system. We see that in the scattering measurement one can probe the so-called structure factor, defined as a Fourier transform of the correlation function (7).

4

As an illustration, consider again the n = 1 problem (1). Above the transition, at τ > 0, the mean value m0 = hmi = 0. Expanding (1) to the second order in the fluctuations m = m0 + δm, we have H(δm) =

Z 

 X1 1 −aτ (δm) + K(∇δm)2 d3 x = K k2 + a|τ | |δmk |2 2 k 2 

2

(13)

Here we use the standard notation k = ...d3 k/(2π)3 From that, using equipartition theorem, obtain P

R

h|δmk |2 i =

K

k2

T + a|τ |

(14)

Thus, the scattering intensity in this system will have angular distribution given by a Lorentz function of the width ∝ τ 1/2 . In a more general situation, q the width of the scattering distribution is given by the inverse correlation radius 1/ξ = a|τ |/K.

1.4

An intuitive meaning of C2 (x − x′ ).

Let us discuss here the properties of the correlation function C 2 (x − x′ ). Although it was defined in a somewhat formal way, we now show that one can think of it in terms of linear response to an external field. To build the relation with linear response, let us consider C2 (x − x′ ) for the Ising model: 

C2 (x − x′ ) = hσx σx′ i = tr σx σx′ e−βH(σ)



(15)

Let us note that, in the absence of external magnetic field, the Hamiltonian is invariant under changing signs of all spins, σ x → −σx . Since the function C2 (x−x′ ) is also invariant under a global sign change, we can rewrite (15) as 

C2 (x − x′ ) = tr σx e−βH(σ)



σx′ =+1

(16)

(we changed the sign of all spins so that σ r′ = +1). The result (16) can be interpreted as the magnetization at the point x due to a spin at the point x ′ polarized by an external field. Thus the correlation function C 2 (x − x′ ) has a meaning of a nonlocal response function. This interpretation of the correlation function in terms of nonlocal response yields a useful relation between fluctuations and the susceptibility χ in a uniform field. One can prove a theorem: χ=β

Z

C2 (y) d3y

(which is just C2 (k = 0))

(17)

This result is completely general. It is related to the fluctuation-dissipation theorem 2 . 2

see: §124, Chap. XII, Landau & Lifshits, Statistical Physics, part I

5

Proof: Z 2   1 ∂ 2 ln(Z(h)) β ZZ ′ −βH 3 3 ′ 3 χ = dM/dh = = tr m(x)m(x )e d xd x − β m(x)d x V ∂2h V ZZ Z h i β = tr (m(x) − hmi)(m(x′ ) − hmi)e−βH d3 xd3 x′ = β C2 (y) d3y (18) V

Here we used the partition function in magnetic field Z(h) = tr exp (−βH + β m(x)hd3 x) Q.E.D. This theorem has numerous applications in the theory of critical phenomena and beyond. R

1.5

Summarize

• Isomorphic phase transitions correspond to the same symmetry change. They are described by order parameter with the same symmetry and the same coarse grained free energy (Landau functional). • Susceptibility characterizes the response of the order parameter to an external field. It diverges near type II transition due to fluctuations of soft modes with long wavelengths. • Correlation function of the order parameter has a finite range (correlation radius) in the disordered state. The correlation radius increases as we approach the transition and ordering takes place. The correlation function is measurable in a scattering experiment.

6

8.334: Statistical Mechanics II

Problem Set # 4

Due: 3/5/03

Real space renormalization. Scaling theory. 1. Renormalization for 1D Ising model Consider a 1D Ising model with the nearest neighbor interaction in external magnetic field H=−

X

(Jsi si+1 + hsi )

(1)

i

To apply the renormalization group to this problem, generalize the decimation procedure discussed in class to include the coupling to magnetic field. Obtain a renormalization mapping J ′ = f1 (J, h) ,

h′ = f2 (J, h)

(2)

Linearize the mapping in h in order to study the renormalization flow at small h. Draw schematically the flow in the (J, h) plane. What are the fixed points of this flow? Which of them are stable or unstable? Which states correspond to the fixed points? 2. Migdal-Kadanoff renormalization for Ising model a) Apply real space renormalization to Ising ferromagnet in magnetic field. Use the bond-moving algorithm, as discussed in class. Sketch the RG flow on the (J, h) plane, identify fixed points, stable and unstable, and discuss their physical meaning. b) Calcualte the temperature and magnetic field eigenvalues y t , yh of the RG flow at the fixed point found in part a). Assuming scaling of the singular part of the free energy, f (τ, h) = b−d f (byt τ, byh h)

(3)

with small h, τ , and b = 2 the rescaling factor, predict the form of singularity in the specific heat and susceptibility. Compare to the exact results (Eq.2, Lecture 5). 3. Exact renormalization on a hierarchical lattice Renormalization group theory of a phase transition aren’t necessarily aproximate. Consider an Ising model on a hierarchical lattice shown in Fig. and apply to it the decimation RG, as in problem 1. Find the RG transformation, idenify its fixed points, and sketch the flow on the (J, h) plane.

Figure 1: Hierarchical Berker lattice for which decimation can be carried out exactly.

4. Ising model on a fractal Consider the Ising problem on a fractal lattice shown in the figure. By using spin blocking, find the renormalization flow for the coupling constant. Consider the fixed points and discuss the ordering types and phase transitions possible in this system.

Figure 2: A fractal lattice: Sierpinsky gasket 5. Relationship between critical indices Consider the principal theromodynamical indices α, β, γ, δ, η, ǫ, µ defined in Lecture 5. Based on the scaling of the Ising model free energy near RG fixed point and on the correlation length scaling, find the relation in terms of the RG flow eigenvalues y t and yh for as many of the indices α, ..., µ, as you can. Show that all of these indices can be expressed in terms of two “independent” indices, say α and β.

8.334: Statistical Mechanics II Problem Set # 4 Real space renormalization. Scaling theory. 1. Renormalization for 1D Ising model Consider a 1D Ising model with the nearest neighbor interaction in external magnetic field X H=− (Jsi si+1 + hsi )

(1)

i

To apply the renormalization group to this problem, generalize the decimation procedure discussed in class to include the coupling to magnetic field. Obtain a renormalization mapping J ′ = f1 (J, h) ,

h′ = f2 (J, h)

(2)

Linearize the mapping in h in order to study the renormalization flow at small h. Draw schematically the flow in the (J, h) plane. What are the fixed points of this flow? Which of them are stable or unstable? Which states correspond to the fixed points? Summation over si with i = 2n + 1 yields  2J+h + e−2J−h si + si+2 = 2  e X exp(Jsi si+1 + Jsi+1 si+2 + hsi+1 ) = eh + e−h (3) si + si+2 = 0   si+1 =±1 e−2J+h + e2J−h si + si+2 = −2

where notations βJ → J and βh → h are used. Compare this to  J ′ +2δh si + si+2 = 2   Z0 e ′ −J ′ Z0 exp(J si si+2 + δh(si + si+2 )) = Z0 e si + si+2 = 0   J ′ −2δh Z0 e si + si+2 = −2

(4)

Therefore

1 + tanh 2J tanh h cosh(2J + h) = cosh(2J − h) 1 − tanh 2J tanh h cosh(2J + h) cosh(2J − h) = = cosh2 (2J) − sinh2 (2J) tanh2 h cosh2 h

e4δh =

(5)

′

(6)

e4J

To the order O(h), obtain 2δh = tanh(2J)h 4J ′ = 2 ln cosh(2J)

(7) (8)

Hence   1 h = 1 + tanh(2J) h 2 1 J ′ = ln cosh(2J) 2 ′

(9) (10)

Under RG, J decreases, while h grows. Fixed points are h = 0, J = 0, ∞. There is no phase transition. 2. Migdal-Kadanoff renormalization for Ising model a) Apply real space renormalization to Ising ferromagnet in magnetic field. Use the bond-moving algorithm, as discussed in class. Sketch the RG flow on the (J, h) plane, identify fixed points, stable and unstable, and discuss their physical meaning. b) Calculate the temperature and magnetic field eigenvalues yt , yh of the RG flow at the fixed point found in part a). Assuming scaling of the singular part of the free energy, f (τ, h) = b−d f (byt τ, byh h)

(11)

with small h, τ , and b = 2 the rescaling factor, predict the form of singularity in the specific heat and susceptibility. Compare to the exact results (Eq.2, Lecture 5). After the first stage, J˜x = 2Jx 1 J˜y = ln cosh 2Jy 2  1 ˜ h = 1 + tanh(2Jy ) h 2

(12) (13) (14)

and after the second, 1 ln cosh 2J˜x 2 Jy = 2J˜y   1 ˜ ˜ h = 1 + tanh(2Jx ) h 2

Jx′ =

(15) (16) (17)

Therefore after the two steps 1 ln cosh 4Jx 2 Jy = ln cosh 2Jy    1 1 h = 1 + tanh(4Jx ) 1 + tanh(2Jy ) h 2 2 Jx′ =

(18) (19) (20)

Trivial fixed points are J = 0, ∞. Nontrivial (and unstable) fixed point is Jx∗ =

1 ln cosh 4Jx∗ 2

Jy∗ = ln cosh 2Jy∗

(21)

Hence Jx∗ = Jy∗ /2. Eigenvalues are given by dJy′ 2 = = 2 tanh 2Jy∗ dJy yt

2

yh

=



1 1 + tanh 2Jy∗ 2

2

(22)

Fixed point J∗

Renormalized coupling J’

1

0.8

0.6

0.4

0.2

0 0

0.2

0.4

0.6 Coupling J

0.8

1

1.2

Figure 1: Renormalized coupling J ′ versus J for 1D Ising model. Numerical solution (see Fig. 1) gives Jy∗ = 0.6094 and yt = 0.7472, yh = 1.0111 . Using results obtained in Lecture 7 with d = 2, obtain the critical exponents α = 2 − d/yt = 2 − 2/yt = −0.6767

γ = (2yh − d)/yt = (2yh − 2)/yt = 0.0297

(23)

3. Exact renormalization on a hierarchical lattice Renormalization group theory of a phase transition aren’t necessarily approximate. Consider an Ising model on a hierarchical lattice shown in Fig. 2 and apply to it the decimation RG, as in problem 1. Find the RG transformation, identify its fixed points, and sketch the flow on the (J, h) plane. 2 3

4 1

Figure 2: Hierarchical Berker lattice for which decimation can be carried out exactly. First consider the problem at h = 0. Decimation is achieved by summing over the spins with two neighbors: !2 X X X X ′ Js3 (s1 +s2 ) = (2 cosh J(s1 + s2 ))2 = 4 cosh 2J eJ s1 s2 (24) Z♦ = e s1 ,s2

s3

s1 ,s2

s1 ,s2

where exp(J ′ ) = cosh 2J. The partition function for n-level lattice is !4  4
4 Z = . . . (4 cosh 2J)4 4 cosh 2J ′ 4 cosh 2J ′′ . . . 4 cosh 2J (n)

(25)

The recursion relations for J and F = −T ln Z are Jn = ln cosh 2Jn−1

Fn = 4Fn−1 − T ln(4 cosh 2Jn )

(26)

The fixed point is J∗ = 0.6094 (see Problem 2). Tc = J/0.6094. The temperature dependence of the specific heat C = −T ∂ 2 F/∂ 2 T is shown on Fig. 3 16

x 10

Ising model on a Berker lattice

4

14

Specific heat

12 10 8 6 4 2 0 0

Tc 0.5

1

1.5 Temperature

2

2.5

3

Figure 3: Specific heat for Ising model on Berker lattice. At h 6= 0, Z♦ =

X

ehs1 +hs2

s1 ,s2

X

eJ(s1 +s2 )s3 +hs3

s3

!2

= Z0

X

′

eJ s1 s2 +(h+2δh)s1 +(h+2δh)s2

(27)

s1 ,s2

where J ′ and δh are given by expressions derived in Problem 1. At small h, h′ = (1 + tanh(2J)) h J ′ = ln cosh(2J) At the fixed point J = J∗ , the magnetic eigenvalue is  ′ yh = ln(h /h) = ln (1 + tanh(2J∗ )) = ln

e2J∗ cosh 2J∗

(28) (29)



= ln(eJ∗ ) = J∗ = 0.6094

(30)

The thermal eigenvalue is yt = ln(dJ ′ /dJ) = ln 4. Ising model on a fractal


2 sinh 2J∗ = ln eJ∗ − e−3J∗ = 0.5180 cosh 2J∗

(31)

Consider the Ising problem on a fractal lattice shown in the figure. By using spin blocking, find the renormalization flow for the coupling constant. Consider the fixed points and discuss the ordering types and phase transitions possible in this system. Z˜123 =

X

s4,5,6

exp (Ks4 (s1 + s2 ) + Ks5 (s2 + s3 ) + Ks6 (s1 + s3 ) + Ks4 s6 + Ks6 s5 + Ks4 s5 )

(32)

2 4

5

1

3 6

Figure 4: A fractal lattice: Sierpinsky gasket The new partition function obtained by summing over s4 , s5 , s6 is ′

Z = Z0 eK (s1 s2 +s2 s3 +s3 s1 )

(33)

To find Z0 and K ′ , consider configurations s1 s2 s3 = + + + and s1 s2 s3 = + + −. For s1 s2 s3 = + + + X
Z123 = exp (2K(s4 + s5 + s6 ) + K(s4 s5 + s6 s5 + s4 s6 )) = e9K + e−3K + 3 eK + e−3K (34) s4 ,s5 ,s6

and for s1 s2 s3 = + + − X Z123 = exp (2Ks4 + K(s4 s5 + s6 s5 + s4 s6 )) = e5K + 4eK + 3e−3K

(35)

s4 ,s5 ,s6

′

′

Hence Z0 e3K = Z+++ and Z0 e−K = Z++− . Therefore
9K −3K K −3K 5K e + e + 3 e + e + 3e−3K + 4e−7K ′ 4K e = e e4K = e5K + 4eK + 3e−3K e5K + 4eK + 3e−3K

(36)

3 and Z04 = Z+++ Z++− . Note that K ′ is always less than K which means that there is no phase transition in this system. The partition function for the n-level structure !3  3
3 Z = . . . Z03 (K)Z0 (K ′ ) Z0 (K ′′ ) Z0 (K ′′′ ) . . . Z0 (K (n) ) (37)

The specific heat C = T ∂ 2 (T ln Z)/∂T 2 is shown on Fig. 5.

Ising model on a Sierpinsky gasket 8000 7000

Specific heat

6000 5000 4000 3000 2000 1000 0 0

0.5

1

1.5

2 2.5 Temperature

3

3.5

4

Figure 5: Specific heat for Ising model on a Sierpinsky gasket 5. Relationship between critical indices Consider the principal theromodynamical indices α, β, γ, δ, η, ǫ, µ defined in Lecture 5. Based on the scaling of the Ising model free energy near RG fixed point and on the correlation length scaling, find the relation in terms of the RG flow eigenvalues yt and yh for as many of the indices α, ..., µ, as you can. Show that all of these indices can be expressed in terms of two “independent” indices, say α and β. The relations between critical exponents α. . . η and RG flow eigenvalues are found in Lecture 7. To find ǫ and µ, we can use scaling argument as follows. Specific heat C ∝ ∂ 2 F/∂τ 2 is a function of τ and h/τ yh /yt . Since C ∝ τ −α at h = 0 and C ∝ h−ǫ at τ = 0, the following relation must hold: (τ yh /yt )−ǫ = τ −α , i.e. ǫ = yt /yh α = (2yt − d)/yh . Similar argument for correlation length gives µyh /yt = ν, i.e. µ = yt /yh ν = 1/yh . The thermal and magnetic eigenvalues can be expressed in terms of two “independent” indices: yt =

d 2−α

hence all the indices can be expressed in terms of α and β.

yh =

2−α−β d 2−α

(38)

8.334: Statistical Mechanics II

1

Last modified: February 23, 2003

Lecture 5. Fluctuations near phase transition. Renormalization group.

1.1

Critical exponents

There are several quantities that have singular behavior, typically of a power law divergence form, near a type II phase transition. For reference we list the most often used scaling relations with a standard notation for the critical indices: • Specific heat CV ∝ |τ |−α ; • Susceptibility χ ∝ |τ | −γ ; • Order parameter hmiτ >0 ∝ |τ |β (hmiτ >0 = 0); • Correlation radius ξ ∝ |τ | −ν ; • Correlation function G(x) ∝ |x| −d+2−η for |x| ≪ ξ. (τ = (Tc − T )/Tc ). Several other scaling relations arise for thermodynamic quantities vs. ordering field h at the transition, τ = 0: • Specific heat CV ∝ |h|−ǫ ; • Susceptibility χ ∝ |h| −1/δ−1 ; • Order parameter hmi ∝ |h|−1/δ ; • Correlation radius ξ ∝ |h| −µ . The experimentally measured values of critical indices are typically different from the mean field values. They depend on system universality class (space dimension, the number of order parameter components, and the symmetry class). The same is true for the indices found numerically. For reference, we list the results found from simulation of the 3D Ising model on a computer 1 : α = 0.104 ± 0.003, β = 0.325, γ = 1.2385 ± 0.0015, ν = 0.632 ± 0.025, η = 0.039 ± 0.004. (1) 1 L. P. Kadanoff, Chap. 11, §8, p.244, Statistical Physics. Statics, Dynamics, and Renormalization. (World Scientific, 1999)

1

and the exponents obtained from Onsager’s exact solution of the 2D Ising model α = +0, β = 1/8, γ = 7/4, ν = 1, η = 1/4.

(2)

are also totally different from the mean field theory prediction. (We write α = +0 to indicate the logarithmic singularity of the specific heat, C ∝ ln(1/|τ |).) However, in some experimental situations, such as ferroelectrics and superconductors, the agreement with the mean field theory is nearly perfect. To understand this, we need to analyze the role of fluctuations.

1.2

When mean field theory fails. Upper critical dimension

Whether the mean field theory2 is consistent can be verified by estimating the magnitude of fluctuations. There are many ways of doing this. We first present a simple argument due to Ginzburg. The correlation function or order parameter fluctuations is ′

C2 (r − r ) = h(σr − m)(σr′ − m)i =

Z

′

eq·(r−r ) C2 (q)

dd q , (2π)d

C2 (q) =

K(q2

T (3) + ξ −2)

with the correlation length ξ ∝ |τ | −1/2 . For the sake of generality we consider a system of arbitrary dimension d. The dependence of the results on d will be quite interesting. The Fourier transform C2 (q) was obtained in Lecture 4 from equipartition theorem applied to fluctuations at T > Tc . One can derive a very similar expression at T < T c (see homework problem 4, PS#3). The integral in (3) can be estimated from a dimensional argument by setting |q| ≃ |r − r′ |−1 : |r − r′ | C2 (r − r′ ) ≃ exp − ξ

!

ad−2 |r − r′ |d−2

(4)

with a = T /K. This expression, correct up to a numerical factor, is all we need now. (In fact, one can evaluate the integral exactly — see problem 4, PS#3.) Let us suppose that the crucial fluctuations occur at the characteristic length scale ξ. According to the above, the characteristic size of the fluctuation is hδσr δσr′ i ∼ |τ |d/2−1

(5)

This fluctuation should be compared with the average value hσr i ∼ |τ |1/2

(6)

2 All the same is true for Landau theory. In fact, in the discussion of the effects of fluctuations, the distinction between these two theories – one being microscopic, the other purely phenomenological, based on symmetry, – is totally unimportant. You can often see in the texts on critical phenomena both of these theories discussed under the same name.

2

To see which is bigger, we divide fluctuations by the mean magnetization: hδσr δσr′ i ∼ |τ |d/2−2 hσr ihσr′ i

(7)

Near the transition, at small τ , the relative size of the two terms will depend on whether the exponent d/2 − 2 is positive or negative. If it is negative, the fluctuations diverge and make the mean field picture invalid near the transition. This is the case in the space dimension d is less than 4. The space dimension above which the fluctuations can be ignored is called the upper critical dimension. A more explicit relation between the effects of fluctuations and thermodynamic quantities can be obtained. Let us revisit the mean field estimate of the free energy near the transition (Lecture 2) and study fluctuation corrections. For that, we estimate the contribution to the free energy of the term δH = −

X

J(r − r ′ )δσr δσr′ ,

δσr = σr − hσr i,

(8)

r6=r ′

which was neglected in the mean field calculation. Taking the average of δH, express the result through spin correlation function defined hδHi = −

X

J(r − r ′ )hδσr δσr′ i = −

X

J(r − r ′ )C2 (r − r ′ )

(9)

r6=r ′

r6=r ′

The correlation function C 2 (r − r ′ ) was defined in Lecture 4. Using its Fourier transform G(q) = a/(q2 + ξ −2 ) and the correlation length ξ ∝ |τ | −1/2 , we obtain hδHi = −

Z

BZ

J(q)

a dd q q2 + ξ −2 (2π)d

(10)

where the integral is taken over the Brillouin zone of the lattice, i.e. over the period of the reciprocal lattice. Here again we consider a lattice of arbitrary dimension d. We have to extract the contribution of the long wavelength modes with q ≃ ξ −1 . A convenient way to do this is to subtract and add an expression for hδHi with ξ −1 = 0, which gives Z Ja dd q hδHi = const + (11) (2π)d ξ 2 q2 (q2 + ξ −2 ) Here we note that the integral is determined by small q ≃ ξ −1 , much smaller than lattice spacing, allows to replace J(q) by a constant J = J(0) and to extend integration limits to infinity. For d < 4 the integral can be estimated by setting q ∼ ξ −1, which gives aξ 2−d ≈ (τ /K)d/2−1 /K. Using this result, let us find the correction to the specific heat: δC = dhδHi/dT ≈ aτ d/2−2 /K d/2

(12)

This expression should be compared with the mean field specific heat, which has a finite jump at the transition (Lecture 3). The fluctuation correction is significant when aτ d/2−2 /K d/2 ≫ 1 or aξ 4−d > K 2 3

(13)

(we recall that ξ ∝ (K/τ )1/2 ). We conclude that at d < 4 the effect of fluctuations dominates near the critical point, while at d > 4 the role of fluctuations at the transition is small. Thus in d = 3, 2, 1 the mean field treatment is inadequate near T c . However, even for d < 4, the region where fluctuations dominate can be small when the rigidity K is large. This is the case for superconductors and ferroelectrics, where the discrepancy with the mean field theory can be detected only extremely close to the transition, at τ = (Tc − T )/Tc ∼ 10−4 − 10−8 .

1.3

The renormalization group idea.

The renormalization group is a framework for analyzing problems where strong interactions happen on a range of length scales. Although the RG will be discussed here mainly in the framework of the theory of phase transitions, we point out that nowdays these ideas are being applied much more broadly. Comment on history and (unfortunate) terminology. The word ‘group’ is misleading, the underlying mathematical structure is certainly not a group, rather a nonlinear dynamical flow in a parameter space. The word ‘renormalization’ indicates that in the high energy physics, where prehistoric forms of renormalization group ideas were developed, it was believed that numerical values of physical constants are determined by fundamental symmetries, rather than by dynamical effects. Thus, when dynamical effects change a value of a physical constant (e.g. corrections to electron mass in QED vacuum fluctuations), it should be rescaled back to the physical value, i.e. the value observed in a low energy experiment. The renormalization group ideas were initially brought in statistical mechanics by K. Wilson, L. Kadanoff and M. Fisher in early 70’s, and then rapidly evolved and transformed. Even the article ‘the’ is misleading, because the renormalization group is not a rigid framework or definite prescription, but rather a loose system of concepts, a flexible framework (a philosophy, in contrast with ideology). Whether or not a RG approach is quantitatively successful depends on the nature of the problem. However, even when the RG approach is not too accurate quantitatively, it can provide a useful qualitative picture. All RG studies have in common the idea of re-expressing the parameters which define the problem in terms of some other, perhaps simpler, set, while keeping unchanged those physical aspects of the problem which are of interest. In the theory of critical phenomena this is accomplished by coarse-graining of the short-distance degrees of freedom. [comment on other situations: many-body quantum problems, turbulence in fluid dynamics, multiscale dynamics, etc.] We shall start with the example of real space renormalization for spin systems on a lattice. This situation is particularly illuminating conceptually, although not always the best from a quantitative point of view. The approximations made in a real space RG sometimes are difficult to control. Later we shall discuss other varieties of RG, more field theory-based, where renormalization is carried out in momentum space. In such approaches, one can develop methods, such as the 4 − ǫ, 2 + ǫ and 1/n expansion, using small parameters to yield systematically improvable quantitative results. 4

1.4

Block spin transformation

Consider coarse-graining in the Ising model on a 2D square lattice. Divide lattice in blocks, say 3 × 3, each containing 9 spins (Fig. 1). To each block assign a new variable σ ′ = ±1, depending on whether the spins in the block are predominantly up or down. The simplest way to do it is by the ‘majority rule’: σ ′ = ±1 if there are more spins down than up, and vice versa. Figure 1:

After defining block variables, we rescale the whole picture by a factor of 3 so that one block is of the same size as original squares. If we start from a spin configuration at the Ising model critical point T = T c , we observe that the new picture is statistically equivalent to the old one, as long as the behavior at large distances (in this case, larger than 3 lattice periods) is concerned. We can continue this blocking procedure, and at all pictures will look pretty much the same. This observation illustrates the scale invariance at the critical point. On the other hand, let us start from the spin configuration with temperature a little bit above or below Tc . In this case, even though the initial configuration is very similar to the first one, the result of the blocking repeated several times, looks very different. At T 6= Tc the system is not scale invariant. These qualitative observations are the essential basis of the renormalization group approach. More quantitatively, the renormalization procedure is built by introducing a block Hamiltonian. The initial configurations are drawn prom the probability distribution p ∝ e−βH defined by the Ising model Hamiltonian H(σ). We can try to interpret the statistics of the block spin configurations in terms of some new Hamiltonian H ′ (σ ′ ). Of course, this can always be achieved by choosing a sufficiently complicated form of H ′ (σ ′ ), perhaps including not only nearest neighbor, but also long-range interactions, as well as various 3-spin, 4-spin, and higher order interactions, characterized by a set of couplings Ki . Except some special cases, the block Hamiltonian looks very different from the initial Hamiltonian. The underlying idea, however, will be to focus on dominant interactions and replace the exact Hamiltonian by an approximate form, perhaps close algebraically 5

to the initial Hamiltonian, and leave out all unimportant interactions. This is the scheme we will use in practice. However, before considering examples, it is worthwhile to discuss the renormalization method in a general form, without making any approximations. Let us consider the initial partition function Z = tr σ e−H(σ) (14) where the factor β is absorbed in H. (Redefining H in this way is convenient, because it allows to deal with dimensionless coupling constants, such as spin exchange J ∗ = βJ and magnetic field h∗ = βh.) To go from initial spin configurations to block spins, we insert a projection operator under the trace in (14) as follows. Suppose the block spins are given by a majority rule, as discussed above. We define T (σ ′ ; σ1 , ...σ9 ) =



1, if σ ′ ( i σi ) > 0; 0, otherwise. P

(15)

T (σ ′ ; σi )e−H(σ)

(16)

The block Hamiltonian is defined by ′

′

e−H (σ ) ≡ tr σ

Y

blocks

We note that, because of the identity σ′ T (σ ′ ; σi ) = 1 (unity decomposition), the partition functions for H and H′ are the same: P

′

′

tr σ′ e−H (σ ) = tr σ e−H(σ)

(17)

Moreover, all correlation functions that depend on the block spin variables σ ′ , σ ′′ , σ ′′′ , ... at higher levels of blocking will be left invariant. Thus all macroscopic properties remain unchanged under the blocking procedure. The only difference is that the quantities have to be expressed in terms of blocked, or renormalized, spins, rather than the original, or bare, spins. It is useful to think of all couplings in the reduced Hamiltonian H ′ as a multicomponent vector {K} ≡ (K1 , K2 , ...). In the original Hamiltonian one may have only one nearest neighbor coupling, say K 1 , with all other K i = 0. But the RG procedure will in principle generate all other possibilities. We may therefore think of the RG transformation as acting in the space of all couplings: {K ′ } = R{K}

(18)

The problem is thereby reduced to the problem of understanding the mapping (18). In practice, since the sums involved in the trace in (16) are usually intractable, one has to make approximations by identifying the relevant couplings and leaving out irrelevant ones.

6

1.5

Summarize

• Fluctuations invalidate the mean field theory in the space dimension d < 4. Understanding phase transitions for d < 4 requires tackling a nontrivial problem of interaction of fluctuations. • The renormalization group is a mathematical framework for expressing scale invariance at the critical point. It allows to coarse-grain physical variables (spins) and their interactions.

7

8.334: Statistical Mechanics II

Problem Set # 5

Due: 3/12/03

Fluctuations. Scaling. Gaussian statistics. 1. The fluctuation region width. Thermodynamic fluctuations control the behavior near the critical point. However, the size of the temperature interval, in which fluctuations are large, depends on microscopic details, and in some cases can be quite narrow. In Lecture 5 we discussed the mean field theory validity criterion, and used it to estimate the critical region. It is interesting to look a bit closer at several examples. a) For the Ising ferromagnet in space dimension d = 3, described by the mean field theory developed in Problem 1, PS#2 (see also Lecture 2), estimate the width of the temperature interval near critical point, τ = (T c − T )/Tc , where the contribution of fluctuations to the thermodynamic quantities is dominant. b) Consider a weakly interacting Bose gas (space dimension d = 3) with density n, particle mass m, and a short-range interparticle interaction g = 4π¯h2 a/m

U(r − r′ ) = gδ(r − r′ ) ,

(1)

The interaction is called weak, when atom scattering length a, related to the interaction constant by1 Eq.(1, is much less then the interparticle distance, a ≪ n −1/3 . In this weakly nonideal gas, the Bose-Einstein condensation transition temperature T c is accurately described by the ideal Bose gas theory. The mean field energy for the condensate amplitude ψ can be written in the form of Eq.(1), PS#1, with n being the temperature-dependent condensate density of an ideal gas. Estimate the fluctuation region τ ∗ for this system, using the parameters for magnetically trapped Rubidium from the experiment in JILA: n ≈ 10 14 cm−3 , a ≈ 100 aB , where aB = 0.5 × 10−8 cm is Bohr’s radius. 2. A relation between scaling dimensions of a field and an operator. Consider an RG transformation of a hamiltonian near critical point, H = H 0 + δH, where the hamiltonian H 0 is invariant under renormalization, R(H 0 ) = H0 , and δH is a perturbation. Suppose that δH is of the form Z δH = h(x)φ(x)d3 x (2)

where φ(x) is some density-type variable associated with the system (e.g. spin density, energy, charge or current density, etc.) and h is a field conjugate to φ. Suppose also that φ is a scaling variable. This means that the RG transforms the φ into itself, φ′ (x′ ) = byφ φ(x) ,

x′ = x/b,

(3)

i.e. the perturbation δH ∝ φ(x)dd x is an eigenvector of the RG transformation linearized at the fixed point H0 . a) Show that the scaling dimensions of φ and h are related as R

yh + yφ = d 1

(4)

This corresponds to the low energy limit of particle scattering. Although the real interaction between atoms is certainly not short-ranged, at low energy it can be effectively replaced by a δ–function pseudopotential expressed in terms of the scattering length by Eq.(1 — this is discussed in more detail, e.g., in “Statistical Mechanics,” by Parthia.

b) For the Ising model, by choosing φ and h to be spin density and magnetic field, and using Eq.(4), derive the RG equation for the pair spin correlation function (Eq.(17), Lecture 7) and the scaling form (Eq.(21), Lecture 7). c) How can one use the result (4), with properly chosen φ and h, to obtain the scaling equation for the free energy (Eqs.(5,6), Lecture 7)? 3. Gaussian statistics. Prove the basic facts about gaussian statistics of one variable, many variables, and fields (Eqs.(5,6,7,8), Lecture 8). 4. RG for quadratic hamiltonians. a) Apply the field-theoretic RG transformation described in Lecture 8 (coarse grain + rescale + renormalize hamiltonian) to a gaussian field problem with a quadratic hamiltonian H=

Z

1 K (∂µ φ)2 dd x 2

(5)

Show that, depending on the dimensionality d, the rigidity constant K grows or decreases upon RG transformation, with the critical dimension d = 2. b) Consider thermal fluctuations of the variable δφ 12 = φ(x1 ) − φ(x2 ). Show that the distribution is gaussian and relate its variance to the correlator hδφ212 i = h(φ(x1 ) − φ(x2 ))2 i c) How does the quantity (6) depend on the points 1 and 2 separation at |x1 − x2 | → ∞?

(6)

8.334: Statistical Mechanics II Problem Set # 5 Fluctuations. Scaling. Gaussian statistics. 1. The fluctuation region width. Thermodynamic fluctuations control the behavior near the critical point. However, the size of the temperature interval, in which fluctuations are large, depends on microscopic details, and in some cases can be quite narrow. In Lecture 5 we discussed the mean field theory validity criterion, and used it to estimate the critical region. It is interesting to look a bit closer at several examples. a) For the Ising ferromagnet in space dimension d = 3, described by the mean field theory developed in Problem 1, PS#2 (see also Lecture 2), estimate the width of the temperature interval near critical point, τ = (Tc − T )/Tc , where the contribution of fluctuations to the thermodynamic quantities is dominant. R Fluctuations in F = (−aτ m2 /2 + bm4 + K(∇m)2 /2)d3 x are strong when Tc b(aτ )3/2−2 ≫ K 3/2

(1)

for d = 3 (see Lecture 5). Therefore P the critical region is τ ≪ τ∗ = abTc /K 3 . In a ferromagnet (see Problem 1, PS2) a, b ≈ Tc , K = 1/6 r J(r)r 2 (i) For nearest neighbor interaction, K ≈ Tc . In this case τ∗ = 1, i.e. the fluctuation region width is of the order of Tc (ii) For interaction of large radius, K ≈ Tc N 2 where N is the number of spins within the interaction radius. Then τ∗ = 1/N 6 , i.e. the fluctuation region is narrow. b) Consider a weakly interacting Bose gas (space dimension d = 3) with density n, particle mass m, and a short-range interparticle interaction U(r − r′ ) = gδ(r − r′ ) ,

g = 4π¯ h2 a/m

(2)

The interaction is called weak, when atom scattering length a, related to the interaction constant by1 Eq.(2, is much less then the interparticle distance, a ≪ n−1/3 . In this weakly nonideal gas, the Bose-Einstein condensation transition temperature Tc is accurately described by the ideal Bose gas theory. The mean field energy for the condensate amplitude ψ can be written in the form of Eq.(1), PS#1, with n being the temperature-dependent condensate density of an ideal gas. Estimate the fluctuation region τ∗ for this system, using the parameters for magnetically trapped Rubidium from the experiment in JILA: n ≈ 1014 cm−3 , a ≈ 100 aB , where aB = 0.5×10−8 cm is Bohr’s radius. In a superfluid above TBEC , aτ = µ, the chemical potential, and b = g, the interaction. The constant K is h ¯ 2 /2m and a ≈ TBEC since at τ = 1 µ ≈ TBEC . Using τ∗ = abTc /K 3 obtain 2 2 τ∗ = TBEC g/(¯ h /2m)3 . Replacing TBEC by h ¯ 2 n2/3 /(2m) and g by 4π¯ h2 a/m. obtain τ∗ = an1/3 = 0.5 · 10−6 cm 1014/3 cm−1 ≈ 2 · 10−2 1

(3)

This corresponds to the low energy limit of particle scattering. Although the real interaction between atoms is certainly not short-ranged, at low energy it can be effectively replaced by a δ–function pseudopotential expressed in terms of the scattering length by Eq.(2 — this is discussed in more detail, e.g., in “Statistical Mechanics,” by Parthia.

2. A relation between scaling dimensions of a field and an operator. Consider an RG transformation of a hamiltonian near critical point, H = H0 + δH, where the hamiltonian H0 is invariant under renormalization, R(H0 ) = H0 , and δH is a perturbation. Suppose that δH is of the form Z δH = h(x)φ(x)d3 x (4) where φ(x) is some density-type variable associated with the system (e.g. spin density, energy, charge or current density, etc.) and h is a field conjugate to φ. Suppose also that φ is a scaling variable. This means that the RG transforms the φ into itself, φ′ (x′ ) = byφ φ(x) ,

x′ = x/b,

(5)

R i.e. the perturbation δH ∝ φ(x)dd x is an eigenvector of the RG transformation linearized at the fixed point H0 . a) Show that the scaling dimensions of φ and h are related as yh + yφ = d

(6)

For partition function to be invariant under the transformations h → h′ = byh h, φ → φ′ = byφ φ and x → x′ = x/b, the term Z Z d −yh −yφ +d h(x)φ(x) d x = b h′ (x′ )φ′ (x′ ) dd x′ (7) must be invariant. Hence yh + yφ = d. b) For the Ising model, by choosing φ and h to be spin density and magnetic field, and using Eq.(6), derive the RG equation for the pair spin correlation function (Eq.(17), Lecture 7) and the scaling form (Eq.(21), Lecture 7). C2 (r, τ ) = hφ(0)φ(r)iτ = b−2yφ hφ(0)φ(r/b)i = b−2yφ C2 (r/b, τ ′ )

(8)

C2 (r) = b−2yφ C2 (r/b) = b−2nyφ C2 (r/bn )

(9)

with yφ = d − yh and τ ′ = byt τ — renormalized temperature perturbation. This agrees with Eq. (18), Lecture 7. At τ = 0 (exactly at the transition) for any n

Therefore C2 (r) ∝ r −2yφ = r −2(d−yh ) c) How can one use the result (6), with properly chosen φ and h, to obtain the scaling equation for the free energy (Eqs.(5,6), Lecture 7)? R The singular part of the free energy contributes2 δH = fs (τ, h)dd x to the hamiltonian. Hence the scaling dimension of fs is d. This gives the following RG scaling relations fs (h, τ ) = b−d fs (h/byh , τ /byt ) = b−nd fs (h/bnyh , τ /bnyt ) 2

Formally, the field conjugated to f is unity.

(10)

valid for any n. This means that fs (h, τ ) = τ d/yt Φ(h/τ yh /yt ).

(11)

3. Gaussian statistics. Prove the basic facts about gaussian statistics of one variable, many variables, and fields (Eqs.(5,6,7,8), Lecture 8). (i) One variable Z

∞

iJx−ax2 /2

e

dx = e

−∞

Therefore the average is iJx

he

i=

2

− J2a

R∞

Z

2 eiJx−ax /2 d x −∞ R∞ e−ax2 /2 d x −∞

∞

′ 2 /2

e−a(x )

d x′

(12)

  J2 = exp − 2a

(13)

−∞

The gaussian integral in eq. 12 can be evaluated as follows:  1/2  Z ∞ 1/2 Z Z  Z ∞ 1 2 2 −r 2 /2a −ax2 /2 = 2π e rd r = (2πa)1/2 (14) e dx = exp − a(x + y ) d x d y 2 0 −∞ (ii) in case of many variables, one needs to evaluate Z Z Y . . . eiJn xn −xi xj Aij /2 d xn .

(15)

where the summation over repeated indices is assumed. To integrate (15), diagonalize Anm using rotation xn = Unm x′m . Then the integral becomes Z Z Z Z Y Y ′ ′ ′ ′ iJn xn −xi xj Aij /2 d x′n = ... e d xn = . . . eiJn xn −xn xn an /2 det U   Y  Y Z (Jn′ )2 iJn′ x′ −x′ x′ an /2 1/2 ′ = e dx = (2πan ) exp − (16) 2a n n n where Jn′ = Jm Umn . Using U −1 = U T and X (J ′ )2 n

n

an

=

X i,j,n

Ji Uin

1 T U Jj = Ji (A−1 )ij Jj , an nj

(17)

find the average iJn xn

he


Q   1/2 exp − 12 J T A−1 J 1 T −1 n (2πan ) Q i= = exp − J A J 1/2 2 n (2πan )

(18)

(iii) Field average can be performed by discreetizing the space, replacing integrals by sums and using the above results ! P R X 1 ˜ x′ )J(x′ ) hei J(x)φ(x)d x i = hei xi J(xi )φ(xi ) i = exp − J(x)A(x, (19) 2 x,x′

Taking the limit of zero grid spacing gives i

he

R

J(x)φ(x)d x

  ZZ 1 ′ ′ ′ ˜ x )J(x ) dx dx J(x)A(x, i = exp − 2

(20)

4. RG for quadratic hamiltonians. a) Apply the field-theoretic RG transformation described in Lecture 8 (coarse grain + rescale + renormalize hamiltonian) to a gaussian field problem with a quadratic hamiltonian Z 1 H= K (∂µ φ)2 dd x (21) 2 Show that, depending on the dimensionality d, the rigidity constant K grows or decreases upon RG transformation, with the critical dimension d = 2. Starting with

1 H= 2

Z

K(∂µ φ)2 dd x =

1X Kq 2 φq φ−q 2 q

divide φ into fast and slow harmonics, φ = φsmooth + δφ, ! X 1 X Kq 2 φq φ−q . H= + 2 q 4, and discuss relevance of the cubic anisotropy at the critical point in each case.

8.334: Statistical Mechanics II

Problem Set # 9 Due: 4/16

Field-theoretic RG (continued). 1. RG for magnetic coupling. Consider Wilson’s RG scheme in d = 4 − ǫ near critical point in the presence of a weak magnetic field H=

Z 

 1 2 τ m + K(∇m)2 + u(m2 )2 − h · m dd x 2 

(1)

(here m is an n-component field). We would like to extend the RG analysis to find scaling transformation of the h · m term. a) Consider the gaussian problem with u = 0. Show that, since the term in question is linear, it does not change under coarse-graining. (Set m = m < + m> and average over m> in the partition function.) Find how this term is changed after rescaling q = b −1 q′ and field renormalization m = zm ′ with z chosen to maintain the rigidity K fixed. Put the recursion relation for h in a differential form (0)

dh/dl = yh h ,

l = ln(Λ/Λ′ )

(2)

b) At small u, perform cummulant expansion in ln(tr > e−Hint ) up to second order in u. Let us focus on the recursion relation for rigidity (Eq.26, Lecture 12), ˜ = K − u2 A′′ (q = 0) K

(3)

where A′′ is obtained from contributions of the form m 2< , as described in Lecture 12. (Fig.2, Lecture 12). Find the field renormalization factor z, taking into account this correction to K. Derive a recursion relation for h correct to the second order in u 2 . Assuming that A′′ (q = 0) = C ln(Λ/Λ′ ) (as will be found below), what is the correction (0) to the field scaling exponent y h ? c) Now we face the problem of calculating the correction to rigidity. NoteR that all diagrams in Fig.2 (Lecture 12), except the last one, give local contributions m2< dd x, and contribute only to τ , but not to K. The last diagram in Fig.2 (Lecture 12), and several other similar diagrams with different combinatorial factors (draw these diagrams and determine the combinatorial factors!) gives an expression 1 X δH = − u2 |m