Physics 562: Statistical Mechanics Spring 2002, James P. Sethna Final Exam, due Monday, May 13 Latest revision: May 9, 2002, 9:21

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1 Physics 562: Statistical Mechanics Spring 2002, James P. Sethna Final xam, due Monday, May 13 Latest revision: May 9, 2002, 9:21 Open ook xam Work on your own for this exam. You may consult your notes, homeworks and answer keys, books and published work, or Web pages as you find useful. The problems have been designed to be doable given only material already presented in the course. If you find something in the literature or on the Web that is particularly helpful (e.g., solvesthe problem), feel free to use it: but, just as in a publication, give a citation. Problems (F.1) nsembles and Statistics: Three Particles, Two Levels. (48 points) system has two single-particle quantum states, with energies (measured in degrees Kelvin) 0 /k = 10 and 2 /k = 10. xperiments are performed by adding three non-interacting particles to these two states, either identical spin 1/2 fermions, identical spinless bosons, distinguishable particles, or spinless identical particles obeying Maxwell- oltzmann statistics. Please make a table for this problem, giving your answers for the four cases (Fermi, ose, ist., and M) for each of the three parts. S/k onstant, N ntropies for Three Particles in Two States Log(8) Log(4) Log(3) Log(2) Log(1) Log(1/2) /k (degrees K) Log(1/6) Figure (F.1.) (a) The system is first held at constant energy. In figure (F.1.), which curve represents the entropy of the fermions as a function of the energy? osons? istinguishable particles? Maxwell-oltzmann particles? 1

2 20 onstant T, N nergies of Three Particles in Two States verage nergy /k (degrees K) T (degrees K) Figure (F.1.) (b) The system is now held at constant temperature. In figure (F.1.), which curve represents the mean energy of the fermions as a function of temperature? osons? istinguishable particles? Maxwell-oltzmann particles? onstant T, µ hemical Potentials for Three Particles in Two States hemical Potential µ/k F F Temperature T (degrees K) Figure (F.1.) (c) The system is now held at constant temperature, with chemical potential set to hold the average number of particles equal to three. In figure (F.1.), which curve represents the chemical potential of the fermions as a function of temperature? osons? istinguishable? Maxwell-oltzmann? alculations may be needed. 2

3 (F.2) Topological efects in Nematic Liquid rystals. (12 points) S=1/2 S= 1/2 efects in Nematic Liquid rystals. The molecules in nematic liquids have long-range order in the orientation of their long axes, but the direction of the heads of the molecules do not order. The dots on each molecule are not physical. Think of the dots as added after the snapshot was taken in order to trace the orientations. The winding number S of a defect is θ net /2π, whereθ net is the net angle of rotation that the order parameter makes as you circle the defect. The winding number is positive if the order parameter rotates in the same direction as the traversal (left above), and negative if you traverse in opposite directions (right). s you can deduce topologically (middle figure above), the winding number is not a topological invariant in general. It is for superfluids S 1 and crystals T, but not for Heisenberg magnets or nematic liquid crystals (shown). If we treat the plane of the paper as the equator of the hemisphere, you can see that the S =1/2 defect rotates around the sphere around the left half of the equator, and the S = 1/2 defect rotates around the right half of the equator. These two paths can be smoothly deformed into one another: the path shown on the order parameter space is about half-way between the two. Which of the following figures represents the defect configuration in real space halfway between S = 1/2 and S = 1/2, corresponding to the intermediate path shown in the middle above? () () () () () 3

4 (F.3) Markov hains: ntropy Increases! (20 points) Let P αβ be the transition matrix for a Markov process, satisfying detailed balance with energy α at temperature T. The current probability of being in state α is ρ α. (a) Show that the free energy F = TS of the Markov chain can be written as F = F + k T ρ α log(ρ α /ρ α), (F.3.1) α where ρ α is the equilibrium distribution of probabilities, ρ α is the current non-equilibrium distribution of probabilities, and F is the equilibrium free energy. The Markov chain is implicitly exchanging energy with a heat bath at the temperature T. Thus to show that the entropy for the world as a whole increases, we must show that S /T increases, where S is the entropy of our system and /T is the entropy flow from the heat bath. Hence, showing that entropy increases for our Markov process is equivalent to showing that the free energy in (F.3.1) decreases. 0.4 ntropy is oncave (onvex downward) f( λa+(1-λ)b ) > λf(a) + (1-λ)f(b) -x log(x) f(a) λf(a) + (1-λ) f(b) f( λa+(1-λ)b ) f(b) x For x 0, f(x) = x log x is strictly convex downward (concave) as a function of the probabilities: for 0 <λ<1, the linear interpolation lies below the curve. The function f(x) = x log x is strictly concave for x 0 (defined in the subtitle of the figure below): this is easily shown by noting that its second derivative is negative in this region. (b) If α µ α = 1, and if for all α both µ α 0andx α 0, show by induction on the number of states M that if g(x)isconcaveforx 0, g( M α=1 µ αx α ) M α=1 µ αg(x α ). (c) Show that the free energy decreases for a Markov process. In particular, using part (b), show that the free energy for ρ α (n+1) = β P αβρ (n) β is less than or equal to the free energy for ρ (n). You may use the properties of the Markov transition matrix P, (0 P αβ 1and α P αβ = 1), and detailed balance (P αβ ρ β = P βαρ α ). 4

5 (F.4) orrelation Functions and Susceptibilities. (25 points) (dapted from Halperin s final exam, Harvard, 1977, when I took it.) spin has two states, s z = ±1/2. It can flip back and forth thermally between these two orientations. In an external magnetic field H, the spin has energy gs H. LetS(t) bethe z-component of the spin at time t. Given a time-dependent small external field H(t) along z, the expectation value of S satisfies d S(t) /dt = Γ S +Γχ 0 H(t) where Γ is the spin equilibration rate and χ 0 is the static magnetic susceptibility. (a) Use equilibrium statistical mechanics to derive the value of χ 0 as a function of temperature: it should not depend on H to linear order. (b) Use the Onsager regression hypothesis to compute (t) = S(t)S(0) at zero external field H = 0. What should it be for times t<0? What is (ω), the Fourier transform of (t)? (c) ssuming the classical fluctuation-dissipation theorem (classical if hω T, as in handler s chapter 8), derive the frequency dependent susceptibility χ(t) and χ(ω). (d) ompute the energy dissipated by the oscillator for an external magnetic field H(t) = H 0 cos(ωt). (F.5) ose ondensation. (15 points) The density of states g(ɛ) of a system of non-interacting bosons forms a band: the singleparticle eigenstates are confined to an energy range ɛ min <ɛ<ɛ max,sog(ɛ) is non-zero in this range and zero otherwise. The system is filled with a finite density of bosons. Which of the following is necessary for the system to undergo ose condensation at low temperatures? (a) g(ɛ)/(e β(ɛ µ) + 1) is finite as µ ɛ min. (b) g(ɛ)/(e β(ɛ µ) 1) is finite as µ ɛ min. (c) ɛ min 0. (d) g(ɛ)/(ɛ ɛ min ) dɛ is a convergent integral at the lower limit ɛ min. (e) ose condensation cannot occur in a system whose states are confined to an energy band. 5

6 (F.6) efect nergetics and Total ivergence Terms. (15 points) hypothetical liquid crystal is described by a unit-vector order parameter ˆn, representing the orientation of the long axis of the molecules. (Think of it as a nematic liquid crystal where the heads of the molecules all line up as well.) The free energy density is normally written F bulk [ˆn] = K 11 2 (div ˆn)2 + K 22 2 (ˆn curl ˆn)2 + K 33 2 (ˆn curl ˆn)2. (F.6.1) ssume a spherical droplet of radius R 0 contains a hedgehog defect (shown above) in its center, with order parameter field ˆn(r) =ˆr = r/ r =(x, y, z)/ x 2 + y 2 + z 2. The hedgehog is a topological defect, which wraps around the sphere once. (a) Show that curl ˆn = 0 for the hedgehog. alculate the free energy of the hedgehog, by integrating F[ˆn] over the sphere. ompare the free energy to the energy in the same volume with ˆn constant (say, in the ˆx direction). There are other terms allowed by symmetry that are usually not considered as part of the free energy density, because they are total divergence terms. ny term in the free energy which is a divergence of a vector field, by Gauss theorem, is equivalent to the flux of the vector field out of the boundary of the material. For periodic boundary conditions such terms vanish, but our system has a boundary. The order parameter is the same as the Heisenberg antiferromagnet, but the latter has a symmetry where the order parameter can rotate independently from the spatial rotations, which isn t true of liquid crystals. 6

7 (b) onsider the effects of an additional term F div [ˆn] =K 0 (div ˆn), allowed by symmetry, in the free energy F[ˆn]. alculate its contribution to the energy, both by integrating it over the volume of the sphere and by using Gauss theorem to calculate it as a surface integral. ompare the total energy F bulk +F div d 3 r with that of the uniform state with ˆn = ˆx, and with the anti-hedgehog, ˆn(r) = ˆr. Which is lowest, for large R 0? How does the ground state for large R 0 depend on the sign of K 0? learly, the term K 0 from part (b) is not negligible! Liquid crystals in many cases appear to have strong pinning boundary conditions, where the relative angle of the order parameter and the surface is fixed by the chemical treatment of the surface. Some terms like K 0 are not included in the bulk energy because they become too large: they rigidly constrain the boundary conditions and become otherwise irrelevant. 7

8 (F.7) Scaling: ritical Points and oarsening. (20 points) Near critical points, the self-similarity under rescaling leads to characteristic power-law singularities. These dependences may be disguised, however, by less-singular corrections to scaling. You may find Yeoman s tables 2.3, 2.4, and 3.1 helpful in defining exponents. (a) n experiment measures the susceptibility χ(t ) in a magnet for temperatures T slightly above the ferromagnetic transition temperature T c. They find their data is fit well by the form χ(t )=(T T c ) (T T c )+(T T c ) (F.7.1) ssuming this is the correct dependence near T c, what is the critical exponent γ? When measuring functions of two variables near critical points, one finds universal scaling functions. The whole function is a prediction of the theory! (b) The pair correlation function G(r, T )= S(x)S(x + r) is measured in another, threedimensional system just above T c. It is found to be of the form G(r, T )=r f(r(t T c ) 0.59 ), (F.7.2) where the function f(x) is found to be roughly exp( x). What is the critical exponent ν? The exponent η? uring coarsening, we found that the system changed with time, with a length scale that grows as a power of time: L(t) t 1/2 for a non-conserved order parameter, and L(t) t 1/3 for a conserved order parameter. These exponents, unlike critical exponents, are easily derived from simple arguments. ssociated with these diverging length scales there are scaling functions. oarsening doesn t lead to a system which is self-similar to itself at equal times, but it does lead to a system which at two different times looks the same apart from a shift of length scales. (c) n Ising model with non-conserved magnetization is quenched to a temperature T well below T c. fter a long time t 0, the correlation function looks like G(r,T,t 0 )=g(r). ssume that the correlation function at short distances will be time independent, and that the correlation function at later times will have the same functional form apart from a rescaling of the length. Write G(r,T,2t 0 ) in terms of g(r). Write a scaling form G(r,T,t)=t ω G(r/t ρ,t). (F.7.3) What are the exponents ω and ρ? It was only recently made clear that the scaling function G does depend on temperature. Low-temperature coarsening isn t as universal as continuous phase transitions are: even in one model, different temperatures can have different scaling functions. 8

Physics 562: Statistical Mechanics Spring 2002, James P. Sethna Prelim, due Wednesday, March 13 Latest revision: March 22, 2002, 10:9

Physics 562: Statistical Mechanics Spring 2002, James P. Sethna Prelim, due Wednesday, March 13 Latest revision: March 22, 2002, 10:9 Open Book Exam Work on your own for this exam. You may consult your