Many-body physics 2: Homework 8

Last update: 215.1.31 Many-body physics 2: Homework 8 1. (1 pts) Ideal quantum gases (a)foranidealquantumgas,showthatthegrandpartitionfunctionz G = Tre β(ĥ µ ˆN) is given by { [ ] 1 Z G = i=1 for bosons, 1 e β(ε i [ µ) 1+e β(ε i µ) ] for fermions. i=1 (b) From the result of (a), we can express the average number of particles as N = i=1 n i where n i is the average occupation number in the ith state. Find the expression of n i for bosons and fermions, respectively. Further reading: Fetter and Walecha, 5. 2. (2 pts) Bose-Einstein condensation Let s consider an ideal Bose gas with spin zero. (a) Express the average occupation number for the zero-energy state, N, in terms of the fugacity z = e βµ. From the expression of N, discuss the condition for chemical potential µ that Bose-Einstein condensation (BEC) occurs. (b) Show that the average occupation number is given by N = N + ρ(ε) dε z 1 e βε 1 where ρ(ε) = V d d k (2π) d δ(ε ε k ) is the density of states (DOS) and V is the volume of the system. (c) Assume that the DOS is given by ρ(ε) = C α ε α 1. Then show that N = N +C α (k B T) α Γ(α)g α (z) where Γ(α) = dtt α 1 e t is the Gamma function and g α (z) = n=1 zn n α. (d) For a three-dimensional gas with ε k = h2 k 2 2m, show that where λ T = 2π h 2 mk B T N = z 1 z + V λ 3 g 3/2 (z) T ( ) 2 h2 is the thermal wavelength. Note that 2π 2m λ T = πkb T, thus λ T is the de Brogile wavelength of a quantum particle with kinetic energy πk B T and it is a measure of the spread of a wavefunction of the particles. (e) In the thermodynamic limit V, show that C α N V = V (k B T) α Γ(α)g α (z) for T > T c, C α V (k B T c ) α Γ(α)g α (1) for T = T c, N V + Cα V (k BT) α Γ(α)g α (1) for T < T c. For T T c, find the fraction of particles in the condensed phase η = N N. For any questions or comments, send email to Hongki Min, hmin@snu.ac.kr

(f) Similarly, we can find the average energy E and the thermodynamic potential Ω = k B T lnz G. Show that ε i E = e β(εi µ) 1 = ρ(ε)ε dε i z 1 e βε 1 = C α(k B T) α+1 Γ(α+1)g α+1 (z), ( ) Ω = k B T ln 1 e β(ε i µ) = k B T [ln(1 z) C α (k B T) α Γ(α)g α+1 (z)]. i=1 (g) From (f), we can calculate thermodynamic properties such as the specific heat C = E T. First, for the condensed phase below T c, find C analytically. Show that in three dimensions, C T 3/2 for the parabolic energy dispersion, whereas C T 3 for the linear energy dispersion. (h) Draw C/(Nk B ) as a function of < T/T c < 2 numerically by taking derivative with respect to T directly for the parabolic energy dispersion and for the liner dispersion, respectively in three dimensions. In DOS, obtain α for a d-dimensional gas with the energy dispersion ε k k J. Further reading: Pethick and Smith, Ch.2.2-4. 3. (1 pts) Canonical transformation: Interacting bosons Consider the simplified form of Hamiltonian in a Bose system ) h = ε (a a+b b +g (ab+b a ) with [a,a ] = [b,b ] = 1 and other commutators zero. The basic idea for the canonical transformation is to introduce a new set of operators α and β such that the Hamiltonian has only terms proportional to α α and β β. (a) Let s define new operators α and β as a = uα vβ, b = uβ vα where coefficients u and v are assumed to be real and will be determined later. The transformation is canonical if the new operators also obey the same canonical commutation relations. Show that u and v should satisfy u 2 v 2 = 1, thus we can parametrize u and v as u = cosh(θ/2) and v = sinh(θ/2), respectively. (b) Next, substitute the above equation in (a) to the Hamiltonian. Show that the term proportional to αβ +β α can be made to vanish by choosing u and v as Find coshθ and sinhθ, and show that u 2 = 1 2 ( ε ε +1 ) tanhθ = 2uv u 2 +v 2 = g ε., v 2 = 1 2 ( ε ε 1 ) (c) Show that the Hamiltonian can be rewritten as h = ε(α α+β β) (ε ε). Next, let s consider the ground state of h., uv = g 2ε, ε = ε 2 g2.

(d) First, prove [ a,(a ) n] = n(a ) n 1. Then show that [ a,e v u b a ] ( v = b u) e v u b a. (e) Using the result of (d), show that ( ua+vb ) ue v u b a = α Ω = where are the ground state of a and b, and Ω = ue v u b a. Similarly, show that β Ω =. This means that Ω is the ground state of h. Further reading: Fetter and Walecha, 35; Coleman, Exercise 3.4. 4. (1 pts) Canonical transformation: Interacting fermions Consider the simplified form of Hamiltonian in a Fermi system: ( ) ( h = ξ a a+b b + ab+b a ) = ξ +Ψ ( ) ξσ 3 σ 1 Ψ where Ψ = (a b) a spinor with {a,a } = {b,b } = 1 and other commutators zero. (a) Let s define new operators α and β such that the Hamiltonian has only terms proportional to α α and β β: a = uα+vβ, b = uβ vα. Show that u and v should satisfy u 2 +v 2 = 1, thus we can parametrize u and v as u = cos(θ/2) and v = sin(θ/2), respectively. (b) Substitute the above equation in (a) to the Hamiltonian. Show that the term proportional to αβ +β α can be made to vanish by choosing u and v as tanθ = 2uv u 2 v 2 = ξ. Find cosθ and sinθ, and show that u 2 = 1 ( 1+ ξ ), v 2 = 1 ( 1 ξ ) 2 E 2 E, uv = 2E, E = ξ 2 + 2. (c) Show that the Hamiltonian can be rewritten as h = E(α α+β β) (E ξ). (d) Express h in a matrix form using Ψ, and obtain E, u and v directly by diagonalizing the matrix. In the bosonic case in Prob. 3, can we directly diagonalize the Hamiltonian as in the fermionic case? If not, what is the reason? (e) Next, let s consider the ground state of h. Similarly as the bosonic case, we can expect that the ground state of h is given by ( Ω = ue v u b a = u vb a ). Show that Ω satisfies α Ω = β Ω =. We will use the result of Prob. 4 when we learn superconductivity. Further reading: Fetter and Walecha, 37; Coleman, Ch.14.4.

5. (2 pts) Weakly interacting Bose gas Consider the grand-canonical Hamiltonian for a weakly interacting homogeneous Bose gas with repulsive short-range interaction: [ ˆK = d d x ˆψ (x)(h (x) µ) ˆψ(x)+ 1 (ˆψ ) ] 2 2 g (x)ˆψ(x) where H (x) = h2 2m 2 is the kinetic energy. Assume that temperature is near zero and interaction is weak so that the system is close to its ground state with the small depletion of the condensate. (a) Split ˆψ(x) = ψ (x)+δˆψ(x) into a condensed part ψ (x) = ˆψ(x) and a noncondensed part δˆψ(x), and expand ˆK in powers of δˆψ(x) and δˆψ (x). Find the (time-independent) Gross-Pitaevskii equation that describes the ground state. (b) Show that the second-order expansion of ˆK becomes [ ˆK 2 = d d x δˆψ (x)(h (x) µ)δˆψ(x) + g ( )] ψ 2 2 (x)δˆψ 2 (x)+4 ψ (x) 2 δˆψ (x)δˆψ(x)+ψ 2 (x)δˆψ 2 (x). (c) Assume that the Bose gas is uniform and ψ (x) = n. Show that for µ >, condensation occurs. Then show that in momentum space ˆK 2 becomes ˆK 2 = 1 2 k [ ] (ε () k +n g)(â kâk +â kâ k)+n g(â kâ k +â kâ k ). (d) Referring to Prob. 3, find the energy spectrum. Show that for small k the energy spectrum becomes linear. Explain that the interaction could lead to superfluidity. (e) The number operator is given by ˆN = N + 1 2 k ( ) â kâk +â kâ k where N is the average occupation number for the zero-energy state. Then show that the depletion density n d at zero temperature is given by n d = n n = 1 V vk 2 = k d 3 k (2π) 3 v2 k = k3 24π 2 = (mn g) 3/2 3π 2 h 3 where N = Ω ˆN Ω evaluated in the ground state Ω, n = N/V, n = N /V and h 2 k 2 2m = 2n g. (For the definitions of v k, refer to Prob. 3.) Note that the quantity v k can be interpreted as the probability amplitude that the state k is occupied. Comparing with BEC for an ideal Bose gas, we can see that the interaction removes particles from the zero-momentum condensate. You can use the following integral: dxx ( x 2 + 1 2 x 2 +1 x ) = 1 6. Further reading: Annet Ch.5.6; Fetter and Walecha, 55; Nagaosa I, Ch.4.2.

6. (15 pts) Path integral method Consider a weakly interacting Bose gas subject to a repulsive constant interaction of strength g >. Then Z = D ψdψe 1 h S[ ψ,ψ] where β h [ S[ ψ,ψ] = dτ d d x ψ ( h τ h2 2m 2 µ )ψ + g2 ] ( ψψ) 2. (a) For ψ(x,τ) = ρ(x,τ)e iθ(x,τ), show that β h S[ρ,θ] = dτ d d x [i hρ τ θ + h2 ( ( ρ) 2 +ρ( θ) 2) µρ+ g ] 2m 2 ρ2. (b) From δs[ρ,θ] =, show that the equations of motion for ρ and θ are given by δs δρ = τ it h θ t µ gρ, δs δθ = τ it ρ t = J, where J(x,τ) = hρ(x,τ) m θ(x,τ) is the particle current density. Note that the first equation (obtained ignoring gradient terms) tells us that the system adjusts the spatial fluctuations of the density by a dynamic fluctuation, and the second equation represents a continuity equation. Next, expand the action to second order in the density field ρ(x,τ) around the reference mean-field ρ. (c) Assuming a homogeneous system, find ρ. (d) Then split the density field as ρ(x,τ) = ρ +δρ(x,τ). Ignoring gradients acting on ρ(x,τ), show that β h [ S[δρ,θ] dτ d d x i hδρ τ θ + h2 ρ 2m ( θ)2 + g ] 2 δρ2. (e) After Gaussian integration over the field δρ(x, τ), show that the effective action for the field θ(x,τ) becomes S[θ] 1 β h dτ d d 2 ] [ h x 2 g ( τθ) 2 + h2 ρ m ( θ)2. Find the energy spectrum ω k. Note that this mode represents the Goldstone mode. Further reading: Altland and Simons, Ch.6.3; Nagaosa I, Ch.4.2. 7. (15 pts) Berezinskii-Kosterlitz-Thouless transition Consider a two-dimensional square lattice with a phase θ i which parameterizes the direction of a classical spin at a site i: H = J i,j cos(θ i θ j ) with J >. (This Hamiltonian is called the XY model.) Then the partition function is given by dθ i Z = 2π ek i,j cos(θ i θ j ) i

where K = βj. (a) At low temperatures, K 1, fluctuations of neighboring phases are small due to large energy cost. Then show that the lattice model can be approximated by the continuum model with Z = Dθe S[θ] where S[θ] = K d 2 x( θ) 2. 2 (b) Show that the order parameter correlation becomes S() S(r) = cos(θ() θ(r)) = Re e i θ(r) = e 1 2 ( θ(r)) 2 where θ(r) = θ() θ(r). (c) Show that 1 ( θ(r)) 2 = 1 2 K q <Λ d 2 q 1 e iq r (2π) 2 q 2 = 1 K Λ qdq 2π 1 J (qr) q 2 1 ( r ) 2πK ln a where Λ a 1 is the momentum cutoff given by the inverse of the lattice constant a and J (x) is the Bessel function of the first kind for order. (d) From (b) and (c), show that the correlation is given by the following form: ( r ) η. S() S(r) a Find η. This means that at low temperatures, there is no long-range order and the correlation decays algebraically with a non-universal exponent. At high temperatures, K 1, we can expect that there is no long-range order and the correlation decays exponentially. Thus, there must be a phase transition at some intermediate T c below which correlations have power-law decay while above which they have exponential decay. Note that from the Mermin-Wagner theorem there is no long-range order in both phases. The transition is associated with the unbinding of vortices, which are topological defects with singular spin (or phase) configurations. (e) The energy cost of a single vortex of topological charge n has contributions from the core region En core (a ) as well as from the relatively uniform distortions away from the center. First, show that θ(x) = nˆθ. r Then show that E n = En core (a )+ J L ( ) L d 2 r( θ) 2 En core (a )+πjn 2 ln 2 a a where a is a radius of the core and L is the size of the system. (f) Explain that the entropy of a single vortex in an area L 2 is S k B ln Then show that the free energy change to add a vortex is ( ) L F n En core (a )+(πjn 2 2k B T)ln. (g) At low temperatures, the free energy cost of creating a vortex diverges as L, while at high temperatures, it is favorable to create isolated vortices. In-between, there occurs a phase transition, which is associated with unbinding of vortices. Then show that the vortex unbinding occurs at a transition temperature k B T n=1 c = π 2 J. Further reading: Goldenfelt Ch.11.2; Altland and Simons, Ch.8.6. a ( L a ) 2.