(1) Consider a lattice of noninteracting spin dimers, where the dimer Hamiltonian is

Size: px

Start display at page:

Download "(1) Consider a lattice of noninteracting spin dimers, where the dimer Hamiltonian is"

Ethel Griffin
5 years ago
Views:

1 PHYSICS 21 : SISICL PHYSICS FINL EXMINION 1 Consider a lattice of noninteracting spin dimers where the dimer Hamiltonian is Ĥ = H H τ τ Kτ where H and H τ are magnetic fields acting on the and τ spins respectively Each spin is a two-state Ising variable ie = ±1 and τ = ±1 a Compute the partition function ξ for a single dimer [ points] he single dimer partition function is H + H τ H ξ = 2cosh e K/ H τ + 2cosh e K/ b Find m [ points] = F H = k B ξ ξ sinh H = cosh H +H τ H +H τ e K/ + 2sinh e K/ + 2cosh H H τ H H τ e K/ e K/ Now consider an interacting model Ĥ = 1 2 Jij i j 1 2 Jij τ τ i τ j K ij ij i i τ i reat the first two terms by mean field theory writing i = m + δ i and τ i = n + δτ i with m = i and n = τ i are presumed to be independent of i You may assume that all interactions are ferromagnetic reat the third term exactly and do not make the mean field approximation for this term c What is the mean field H eff for the spins? [ points] he mean fields are H eff = Ĵ m H τ eff = Ĵτ n where Ĵ = j J ij and Ĵτ = j Jτ ij d What is the mean field free energy per site F/N? Hint : Once you have the mean fields H and H τ you can make use of the results of part a [ points] F = 1 2 NĴ m NĴτ n 2 { Ĵ Ĵ } m + N 2cosh Ĵτ n e K/ m + 2cosh Ĵτ n e K/kB 1

2 e Find the mean field equations [ points] Setting the variation of F with respect to m and n to zero we find sinh m +n τ e K / + sinh m n τ m = cosh m +n τ e K / + cosh m n τ e K / e K / and n = sinh m +n τ e K / sinh m n τ cosh m +n τ e K / + cosh m n τ e K / e K / where Ĵ τ Ĵτ and K K f he mean field free energy has a Landau expansion of the form fmn = f a m a ττ n2 + a τ mn + O m 4 m nm 2 n 2 mn n 4 Focusing only on the quadratic terms find an equation for the temperature where the curvature of the free energy first changes sign as is decreased from infinity If there is no preempting first order transition this gives you the critical temperature c You will need to first find the coefficients a a ττ and a τ [1 points] Expanding f = F/N in powers of m and n we must go to work on the log expanding the hyperbolic cosine terms each to fourth order in their arguments and then using the power series for the logarithm itself his is tedious but you were only asked to go to the lowest nontrivial order which isn t so difficult Here I will show how to obtain the Landau coefficients up to fourth order We have f = ln [ 4cosh K / ] m τ n 2 + f where f = ln cosh m +n τ e K / + cosh m n τ e K / 2cosh K / Working on the log term [1 + m2 2 + n2 τ 2 + 2mn τ tanh K / f = ln m4 4 + m2 n 2 2 τ 2 + n4 τ m n2 τ 2 m n τ tanh K / ] 24 + o carry out the Landau expansion to quadratic order as you were asked one needs only the first nontrivial term inside the big brackets o go to fourth order however we must 2

3 invoke ln1+ε = ε 1 2 ε2 + Oε and then combine the square of the second term and the third term inside the big brackets he result to fourth order is f = ln [ 4cosh K / ] m τ 1 τ n 2 k B τ + k 4 B 12 m4 + k 4 B τ 12 n4 + k 2 B τ 2 4 tanh 2 K / m 4 + m n 2 2 τ t any rate the coefficients of the quadratic form are a = 1 a ττ = τ 1 τ a τ = τ tanh K / m n τ m n tanh K / + tanh K / If K = the model decomposes into two independent Landau theories with critical temperatures and τ respectively When K we need to find the eigenvalues of the matrix a a M = τ a τ a ττ which are λ ± = 1 2 a + a ττ ± 1 4 a a ττ 2 + a 2 τ Setting the lower of the two eigenvalues to zero we obtain the equation a a ττ = a 2 τ his gives rise to the transcendental equation 2 + τ + τ sech 2 K / = the solution of which is the critical temperature c his is as far as you were asked to go Proceeding further let us adimensionalize by defining θ / τ θ K = K / τ and ε τ / We may without loss of generality assume ε 1 We then have θ ε + ε 1 θ = sech 2 θ K /θ In the top panel of Fig 1 we plot the left hand side LHS of this equation in black and the right hand side in different colors corresponding to three different values of θ K all for ε = 7 Defining Dθ θ 2 ε + ε 1 θ + sech 2 θ K /θ the determinant detm is proportional to Dθ/θ 2 For high temperatures Dθ > and both eigenvalues λ ± of M are positive What happens as we lower θ then depends on the value of θ K If θ K > θ ε then there is a single second order phase transition at the unique root of the equation Dθ = s we pass through this temperature the lower eigenvalue λ passes through zero and becomes negative he low temperature phase exhibits a spontaneous moment where both m and n are nonzero s θ is further decreased toward θ = we still have λ < < λ + and no additional transitions are encountered If θ K < θ ε then Dθ = has three roots θ > θ B > θ C For θ > θ we have < λ < λ + For θ B < θ < θ we have λ < < λ + t θ = θ B the upper eigenvalue λ + changes sign and for θ C < θ < θ B we have λ < λ + < ie both eigenvalues are negative

4 Figure 1: op panel: Graphical solution to the transcendental equation from problem 1f for ε = 7 Blue curve: θ K = 1; green curve: θ K = θ ǫ = 214; red curve: θ K = Bottom panel: Critical value of θ K below which there are three solutions to detm = Finally at θ = θ C the upper eigenvalue changes sign once again and for θ < θ C we have λ < < λ + once more hroughout the entire region θ < θ the system is ordered and there is a spontaneous moment in which both m and n are nonzero Despite the additional roots of Dθ = when θ K < θ ε there are no additional phase transitions It is worth emphasizing two things that are not happening here he first thing is a spontaneous breaking of an additional symmetry For K there is only one global symmetry in this problem which is the Z 2 Ising symmetry under which i i and τ i τ i for every spin Many systems in Nature exhibit a cascade of symmetry breaking transitions Not this one s soon as the spins develop a spontaneous moment this acts through the K term as an external field which immediately polarizes the τ spins and there is no phase where m and n = and vice versa he second thing which isn t happening is a reentrant transition If the lower eigenvalue λ were to change sign at θ B and θ C as well as at θ then throughout the range θ C < θ < θ B we would have both eigenvalues again positive and the minimum of fmn would again lie at m = n = as it does in the high temperature phase s θ finally passed through θ C the moment would spontaneously reappear he sign change of λ + at θ = θ B however is inconsequential since λ is negative and the minimum of fmn already lies away from the origin for θ < θ 4

5 2 three-dimensional gas of particles obeys the dispersion εk = k /2 here are no internal degrees of freedom ie the degeneracy factor is g = 1 he number density is n We will solve the problem for the more general dispersion εk = k and then indicate the result for = 2 a Compute the single particle density of states gε [ points] WIth ε = k we have kε = ε/ 1/ and gε = 1 k 2 2π 2 ε k = ε 1 ε1/ k=kε 2π 2 = / π 2 / b For bosons compute the condensation temperature BEC n [ points] he number density nzn for bosons in the grand canonical ensemble is nzn = gε z 1 e ε/ 1 + n where n is the condensate density For < BEC we have z = 1 and n > For > BEC we have z < 1 and n = Precisely at = BEC both conditions apply: z = 1 and n = hus hus n = gε e ε/ = Γ c 1 2π 2 BEC n = 2π 2 n Γ kb / c = Γ π 2 / = π 2 n Γ / kb / c c For fermions compute the ground state energy density e n [ points] he ground state energy density for spinless ie g = 1 fermions is he number density is ε = E d V = k 2π k Θk F k = k + F 2π 2 + n = N V = d k 2π Θk F k = k F π 2 = k F = π 2 n 1/ hus ε n = π 2 / 1 + n 1+ = 11 π 2 / n 11/

6 d For photon statistics compute the temperature n [ points] he photon density is n = gε e ε/ 1 which is the same expression as in part b above! hus / 2π 2 n n = Γ = k B π 2 n Γ / e For photon statistics compute the entropy density sn = S/V [ points] he grand potential is ΩV = V gε ln 1 e ε/ = V where gε = H ε Integrating gε to obtain Hε we have ΩV = V π 2 / he entropy density is then s = 1 V ε / e ε/ 1 = Γ + 1 Ω = Γ π 2 he number density as we have seen is n = Γ 2π 2 kb Hε e ε/ 1 π 2 / V 1+ kb / / hence sn = n = nk B On dimensionful grounds we knew a priori that sn n f For bosons and fermions compute the second virial coefficient B 2 [ points] We have n = p = gε z 1 e ε/ 1 = ± Γ 2π 2 Hε z 1 e ε/ 1 = ± Γ 2π 2 kb kb / ±z / +1±z

7 where the top sign is for bosons and the bottom for fermions It helps to define the thermal wavelength 2π 2 1/ λ Γ kb 1/ π 2 1/ 2/ = Γ k B so From the first of these we have nλ = ± ±z = z ± 2 / z 2 + O z pλ = ± +1±z = z ± 2 1 / z 2 + O z z = nλ 2 / n 2 λ + O n λ 9 Substituting this into the second equation we obtain the lowest nontrivial term in the virial expansion of the equation of state: he second virial coefficient is then p = n 2 1 / n 2 λ + O n λ B 2 = 2 1 / λ = π2 2 11/ Γ / Provide clear accurate and brief answers for each of the following: a For the free energy density f = 1 2 am2 1 ym bm4 what does it mean to say that a first order transition preempts the second order transition? [ points] In the absence of a cubic term ie when y = there is a second order transition at a = assuming b > for stability he ordered phase for a < has a spontaneous moment m When the cubic term is present a first order ie discontinuous transition takes place at a = 2y2 9b which is positive hus as a is decreased from large positive values the first order transition takes place before a reaches a = hence we say that the second order transition that would have occurred at a = is preempted ypically we write a c where c is what the second order transition temperature would be in the case y = b system of noninteracting bosons has a power law dispersion εk = k What is the condition on the power and the dimension d of space such that Bose condensation will occur at some finite temperature? [ points] t = BEC we have the relation n = d d k 1 2π d e εk/ BEC 1 7

8 If the integral fails to converge then there is no finite temperature solution and no Bose condensation For small k we may expand the exponential in the denominator and we find the occupancy function behaves as BEC /εk k From the integration metric in d- dimensional polar coordinates we have d d k = Ω d k d 1 dk where Ω d is the surface area of the d-dimensional unit sphere hus the integrand is proportional to k d 1 For convergence then we require d > his is the condition for finite temperature Bose condensation c Which has a longer homas-fermi screening length: a metal with a high density of states at the Fermi level or a metal with a low density of states? Explain why [ points] small electrical potential φr will induce a local density change δnr = eφrgε F in a metal where gε F is the density of states at the Fermi energy We assume ε F which is very much the case for most metals at room temperature hen from Poisson s equation we derive 2 φ = 4πeδn = λ 2 F φ where λ F = 1/ 4πe 2 gε F is the homas-fermi screening length hus high density of states is associated with a shorter screening length there are more electrons present per unit volume to participate in screening he metal with the lower density of states has the longer F screening length d Sketch what the pair distribution function gr should look like for a fluid composed of infinitely hard spheres of diameter a How does gr change with temperature? [ points] he PDF for a hard sphere gas is shown in Fig 2 below he main features are gr = for r < a and a decaying oscillation for r > a Since the potential is either U = no two spheres overlapping or U = overlap of at least two spheres temperature has no effect because U/ is also either or he hard sphere gas is a reasonable model for the physics of liquid rgon see figure Figure 2: d Pair distribution functions PDF for hard spheres of diameter a at filling fraction η = π a n = 49 left and for liquid rgon at = 8K right Molecular dynamics data for hard spheres points is compared with the result of the Percus-Yevick approximation Experimental data on liquid argon are from the neutron scattering work of Yarnell et al 197 he data points are compared with molecular dynamics calculations by Verlet 197 for a Lennard-Jones fluid See fig 8 of the lecture notes 8

9 e For the cluster γ shown in Fig identify the symmetry factor s γ the lowest order virial coefficient B j to which this contributes and write an expression for the cluster integral b γ in terms of the Mayer function [ points] Figure : Left: the connected cluster γ for problem e Right: a labeled version of this cluster used in expressing the cluster integral b γ he symmetry factor is 2!! = 12 because consulting the right panel of Fig vertices 2 and can be exchanged and vertices 4 and can be permuted in any way here are six vertices hence the lowest order virial coefficient to which this cluster contributes is B he cluster integral is b γ = 1 12V d d x 1 d d x 2 d d x d d x 4 d d x d d x f 12 f 1 f 2 f 24 f 2 f 2 f 4 f f f 4 f 4 f where f ij = e ur ij / 1 See Fig for the labels Whee! f Explain the following terms in the context of dynamical systems: recurrent ergodic and mixing How are these classifications arranged hierarchically? [ points] recurrent dynamical system exhibits the property that within any finite region of phase space one can find a point which will return to that region in a finite time Poincaré recurrence is guaranteed whenever the dynamics are invertible and volume-preserving on a finite phase space n ergodic system is one where time averages are equal to phase space averages For the dynamical system ϕ = V ϕ ergodicity means fϕ = lim 1 dt f ϕt = rfϕδ E Hϕ rδ E Hϕ = fϕ S where fϕ is any smooth function on phase space mixing system is one where any smooth normalized distribution ϕt satisfies lim r ϕtfϕ = fϕ t S hus the distribution spreads out evenly over the entire energy surface he hierarchy is mixing ergodic recurrent 9

10 g What is the significance of the Ginzburg criterion? [ points] he Ginzburg criterion tells how close we can come to a second order phase transition before fluctuation effects start to overwhelm the mean field contribution to the heat capacity he criterion for the sufficiency of mean field theory is t t G where t = c / c is the reduced temperature and t G = 2d 4 d a/r where a is the lattice spacing and R is a length scale set by the interactions For most magnetic transitions R is on the order of the lattice spacing t G 1 and mean field theory breaks down rapidly For most superconductors R is on the order of the coherence length which is large on the scale of the lattice spacing and t G and mean field exponents are quantitatively accurate essentially all the way up to the transition Some useful formulas: α z = k=1 z k k α ε α 1 z 1 e βε 1 = Γα α z α ln1 + x = x 1 2 x2 + 1 x 1 4 x4 + coshx = ! x ! x4 + 1

PHYSICS 210A : EQUILIBRIUM STATISTICAL PHYSICS HW ASSIGNMENT #4 SOLUTIONS

PHYSICS 210A : EQUILIBRIUM STATISTICAL PHYSICS HW ASSIGNMENT #4 SOLUTIONS PHYSICS 0A : EQUILIBRIUM STATISTICAL PHYSICS HW ASSIGNMENT #4 SOLUTIONS () For a noninteracting quantum system with single particle density of states g(ε) = A ε r (with ε 0), find the first three virial