Session 1 - XY model in two dimensions

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1 Ecole de Physique des Houches August, 007 Predoctoral school in statistical physics Quantum Statistical Mechanics Session 1 - XY model in two dimensions The XY model is defined as H = J r ˆµ=ˆ ŷ S r+ˆµ S r = J rˆµ cos(θ r+ˆµ θ r ) where the spins S are vectors of length 1 oriented in the y plane and can then be defined by the angle θ between the spin and the ais (see figure). S L ŷ θ ˆ a Figure 1: Sketch of the XY model The spins are located on the nodes of a two dimensional square lattice, the distance between two sites being a. ˆ and ŷ are vectors of length a along the two aes. 1 - Fluctuations around an order At T = 0, to minimize the energy, all the spins will align themselves along a given direction, thus breaking the rotation symmetry. For simplicity, we will say that the -ais is along that direction. At low T, the spins should fluctuate a bit around this direction and we can establish the change in thermodynamic quantities due to these fluctuations. a. We rewrite the spin as S r = ( 1 σ r, σ r) where σ r is the fluctuation perpendicular to the ordering of the spins. Develop the Hamiltonian to second order in σ and show that it becomes: H = E 0 + J L being the side of the system. rˆµ (σ r+ˆµ σ r ) where E 0 = J L a 1

2 b. Show that, it we σ s are not varying a lot from site to site, one can write in a continuous limit H = E 0 + J ( ) d r σ(r) c. The system has periodic boundary conditions and the field σ(r) can be written as a Fourier serie σ(r) = A(k)e ik r k where k = (k, k y ) and where k (y) are multiples of π/l. Why should we have the condition k < 1/a? Show that the hamiltonian can be rewritten as H = H E 0 = J L k A(k) k d. The k modes being independent, calculate the partition function for one mode and show that A(k) = k BT JL k - Magnetisation We will now try to calculate how the magnetisation of the system is reduced by thermal fluctuations. a. Show the following relation for the fluctuation correlation function C(R) = 1 d r σ(r + R)σ(r) = A(k) e ik R L k 0 b. By taking (once again) a continuous limit for the sum over k, show the following epression C(R) = k BT J 1/a 1/L d k e ik R (π) k c. The magnetisation along the ordering direction is given by m = 1 σ 1 σ / where σ = C(0). Show that m 1 k BT 4πJ ln(l/a) d. For fied L, how is the magnetisation evolving with temperature? For a fied low T, how is the magnetisation evolving with the size of the system? What happens in the thermodynamic limit? What would happen in three dimensions? 3 - Correlation function We now want to find directly the spin-spin correlation function G(R) = 1 L d r S(r + R) S(r) a. Show that G(R) 1 C(0) + C(R)

3 b. The Bessel function J 0 being defined as J 0 () = 1 π dφ e icos(φ) π 0 show that G(R) = 1 k BT J R/a R/L d π [1 J 0 ()] c. We now consider the case where L R a. The integral in the preceding equation converges in zero and, for large, we have the approimation J 0 () cos( π/4) π Show that the leading term in this case leads to G(R) = 1 k BT πj ln(r/a) d. Finally, show that, at low T, the correlation function is approimately where η = k B T/πJ 4 - Stiffness G(R) R η We now come back to the description of the XY model in terms of the angles θ. We apply a twist on the boundary conditions θ(l, y) = θ(0, y) + ϕ This is equivalent to impose locally a phase difference aϕ/l on each of the links oriented along. At zero temperature, the energy of the system will be increased because of the twist. For symmetry reasons, E(ϕ) = E( ϕ). Then the energy as a function of ϕ should be written as E(ϕ) = E(ϕ = 0) + αϕ. If L becomes large, the phase difference imposed on each link becomes negligable and the energy difference between the periodic and the twisted system should be microscopic. If the imposed gradient aϕ/l remains non zero, we should on the contrary observe a macroscopic change in the energy E(ϕ) = E(ϕ = 0) + L ( aϕ L ) ρs / where the remaining parameter ρ s is called the stiffness. At finite temperature, this parameter is defined in the same way, but with the free energy F replacing the energy. a. Show that ρ s = 1 a F ϕ b. F is defined as F = k B T ln(z). Show that [ ( F 1 ϕ = k BT Z ϕ=0 ) Z 1 ϕ Z ] Z ϕ 3

4 c. We now consider the change of variable θ r = θ r ϕ. Show that the Hamiltonian becomes L H = J rˆµ cos ( aφ/l δˆµ ˆ + θ r+ˆµ θ r ) where the difference between the θ should be small. d. Deriving the partition function, show that F = Ja cos(θ ϕ L r+ˆ θ r) ϕ=0 + Ja L βj Ja L βj r sin(θ r+ˆ θ r ) r ( ) sin(θ r+ˆ θ r ) e. Arguing that the sin are small, show that F ϕ Ja ϕ=0 L L G(0) and finally that ρ s = J Ak B T where A is a constant that you do not need to calculate. Discuss the difference between J and ρ s and the physical meaning of this quantity. r Bibliography The XY model and the non-linear σ model are discussed in Lectures on phase transitions and the renormalization groupe by N. Goldenfeld, Addison-Wesley (199) and in The theory of critical phenomena by J.J. Binney, N.J. Dowrick, A.J. Fisher, and M.E.J. Newman, Oford (199). Many eercizes concerning the XY model and the related clock model can be found in Equilibrium and Non-Equilibrium Statistical Thermodynamics by M. Le Bellac, F. Mortessagne, and G.G. Batrouni 4

5 Ecole de Physique des Houches August, 007 Predoctoral school in statistical physics Quantum Statistical Mechanics Session - Quantum Heisenberg Models The anisotropic Heisenberg model (also known as XXZ model) is defined as ( H = J y σ r+ˆµ σr + σ y r+ˆµ r) σy + Jz rµ rµ σ z r+ˆµσ z r where J y and J z can be negative (ferromagnetic) or positive (antiferromagnetic). where the σ s are Pauli matrices ( ) ( ) ( ) i 1 0 σr =, σr y 1 0 =, σr z i 0 =, [σr 0 1, σy r ] = δ rr iσr z 1 - Symmetries a. We define the total magnetisation along z as σ z tot = r σz r. Show that [H, σz tot ] = 0. b. In the same way, show that [H, σ tot] = 0 only if J y = J z. c. What is then the commutator [H, σ tot ]? d. If we have N spins (N even), what is the number of states corresponding to σtot z = 0? What are the maimum number of sites for which we can diagonalize the Hamiltonian if we are able to diagonalize matrices of size M M? e. If we have a lattice where the sites can be separated into two classes A and B where the sites A are always connected to sites B (and never to sites other sites A), the lattice is called a bipartite lattice (see figure for eamples). A B A B A B A B A B A B A B A A B A Figure 1: Eamples of bipartite lattices On such a lattice, an additionnal symmetry eist. We define a new set of operators σ σ r = ( 1)+y σ r σ y r = ( 1)+y σ y r σ z r = σz r where the factor ( 1) +y is +1 on sites A and -1 on sites B. Show that the σ still obey the commutation relations of Pauli matrices. f. Show that, under the transformation the Hamiltonian is modified with a resulting change of the sign of J y. 1

6 - Mean field theory at zero temperature We will now use a mean field theory to study the low temperature properties of this system. We will suppose that the spin on site r is submitted to the mean values of the surrounding spins. a. Show that the mean field Hamiltonian is H MF 4N constant = J y ( σ MF σ r + σy MF σy r ) + J z σ z MF σz r where we defined a vector σ MF = ( σ MF, σy MF, σz MF ) of norm 3. b. Rewrite the mean field Hamiltonian as H = h MF σ r and calculate h MF c. What is the state ψ of lowest energy for this Hamiltonian? What will be the mean value σ = ψ σ ψ? d. For the approach to be consistent, we must have σ = σ MF. What are the different possible consistent solutions? Which is the one with the lowest energy, depending on the values of J y and J z? e. Discuss the approimate phase diagram of this system as a function of J y /J z and the temperature T. Bibliography The quantum Heisenberg model is discussed in Statistical Mechanics by S-K. Ma, World Scientific (1985). Introduction to the theory of quantum phase transitions can be found in Continuous quantum phase transitions by S.L. Sondhi, S.M. Girvin, J.P. Carini, and D. Shahar, Rev. Mod. Phys. 69, 315 (1997) or in Quantum Phase Transitions by S. Sachdev, Cambridge (1999).

7 Ecole de Physique des Houches August, 007 Predoctoral school in statistical physics Quantum Statistical Mechanics Session 3 - Particles on lattices 1 - Second quantization We will talk mainly of fermions. A similar formalism eist for bosons and will be summarized at the end. We first introduce a state, called 0 that represents the empty system. Then we introduce a creation operator f ψ that creates a particle in a state ψ. f ψ 0 = ψ a. We can then create states with two particles by applying two creation operators ( ψ and φ are orthogonal states) f φ f ψ 0 = φψ We want these states to be physical, that is to be antisymmetric for fermions: φψ = ψφ. Show that it implies the anticommutation relation for these operators {f φ, f ψ } = 0 b. Show that this relation implies the Pauli principle c. Show that the adjoint operator f ψ is an operator that destroys a particle in state ψ. f ψ ψ = 0 d. By applying this operator on different states, show that {f ψ, f φ } = δ ψ φ e. Show that the operator n ψ = f ψ f ψ counts the number of particles in state ψ (0 or 1 for a fermion). f. A state ψ can always be developped on a position basis ψ = dψ() and the creation operator can then be developped as a sum of creation operators f creating particles at : f ψ = dψ()f Using this relation, show that, if we have two particles in states ψ and φ, the probability P to find one particle in and in one in is P = ψ()φ( ) ψ( )φ() 1

8 - Derivation of the Hubbard model Hubbard models are discrete versions of Schrödinger equations. We want to introduce a set of discrete positions (a lattice), separated by a small distance a, that will replace the continuous variable. ψ = dψ() ψ = ψ We want the variable ψ to be probabilities whereas ψ() is a probability density, we then use the following definition ψ() a = ψ ψ() = ψ a a. For one particle in a state ψ, with an eternal potential V () show that the mean value of the Hamiltonian is given in the discrete version by ψ H ψ = ) ] [ ( ψ ma ψ +a + ψ+a ψ ) + (V + ψ ma ψ (1) where V = V (). b. To describe the Hamiltonian in second quantized form, it is enough to replace wave functions by creation and destruction operators ψ f and ψ f. We obtain the following operator H = [ t (f f +1 + f +1f ) ] + V f f + tn f where t = /(ma ), where we have chosen a as the unit of length ( is then an integer), and where N f is the total number of fermions (we will neglect the constant tn f in the following). If we use the state ψ = ψ, show that the mean value of the Hamiltonian ψ H ψ is the previously obtained epression (1). c. Show that, if we introduce a two particles state (orthogonal states) ψφ, we find ψφ H ψφ = ψ H ψ + φ H φ. The second quantized Hamiltonian has always the same epression whatever is the number of particles. 3 - Kinetic term We will now study in detail the kinetic energy term, also denoted as the hopping term: this operator destroys a particle and recreates it on a neighboring site, thus producing a hop of the particle from one site to the other. We introduce a plane wave state and creation operator k = 1 e ik f k = 1 e ik f L L where k = πn/l, n = 0, 1, L 1 for a system with L sites and periodic boundary conditions. a. Show that the creation and destruction operators for plane waves are fermionic operators and form an orthonormal basis. b. Show that we can inverse the relation and that we have c. Show that n = k n k. f = 1 e ik f k L k

9 d. Show that the Hamiltonian can be rewritten as H = t k cos(k)f k f k + kq f k+q f kv q where V q = 1 V e iq L e. Considering the first term only show that H k = ǫ k k. Show that, when the plane wave has a long wavelength, ǫ k = t + k /m. Represent ǫ k for all k. 4 - For the bosons [b ψ, b φ ] = 0 [b ψ, b φ ] = δ ψφ A state nψ with n particles in the ψ state is given by nψ = (b ψ )n n! 0 where the factor is introduced for normalisation of the state nψ. The action of the operators is b ψ nψ = n + 1 (n + 1)ψ b ψ nψ = n (n 1)ψ Bibliography An introduction to the second quantization in the contet of condensed matter can be found in A quantum approach to condensed matter physics by P.L. Taylor and O. Heinonen, Cambridge (00) The Hubbard model is described as a quantum lattice gas in Statistical Physics by S- K. Ma, World Scientific (1985). 3

10 Ecole de Physique des Houches August, 007 Predoctoral school in statistical physics Quantum Statistical Mechanics Session 4 - Mappings We consider a bosonic Hubbard model H = t ( ) a r+ˆµ a r + a r+ˆµ a r + U rˆµ r n r (n r 1) + V rˆµ n r+ˆµ n r and we take the limit where U. This is equivalent to take a hardcore constraint : there can be only 0 or 1 particle per site. 1 - Bosons and spins With the hardcore constraint, the system has only two states per site, which eactly what one would have for a spin 1/ system. We can then define an empty site as a down spin and a filled site as an up spin 0 r = r 1 r = r a. What are the following operators in terms of spins (in term of Pauli matrices) : n r, a r, a r. b. Calculate how the Hamiltonian is transformed and show that it is equivalent to an anisotropic Heisenberg model. c. Discuss the relations between the different correlation functions in the bosons and spin systems and the correspondence between the phases. - Fermions and spins in one dimension Hardcore bosons have the same kind of constraint as fermions : two bosons cannot occupy the same state (same site). However, bosonic operators on different sites commute while fermionic operators should anticommute. A. Jordan-Wigner transformation The Jordan-Wigner transformation allows the transformation of hardcore bosons (or spins) into fermions in one dimension. The spin operators are related to the fermionic ones as follow: ( ) 1 σ z = n 1 σ + = f (iπ ep 1 n )σ = ep iπ n =0 a. Starting from [σ, σ+ ] = 1, first show that {f, f } = 1. b. Starting from [σ, σ+ +R ] = 0 show that {f, f +R } = 0 c. Calculate how a XXZ Hamiltonian is transformed into the fermion langage. d. We will now concentrate on the quantum XY (J z = 0) chain that we can solve eactly as it is equivalent to the free fermion system. We will concentrate on the zero magnetisation case, =0 f 1

11 that is, the half-filled fermion system. Show that, in this case, the fermionic Hamiltonian is diagonal in Fourier space and the epression of the energy of the fundamental state: E = 4J y k F k= k F cos(k) where k = πn/n B. Calculation of the ZZ correlation function We want to calculate the ZZ correlation function C ZZ (R) = 1 σ+r z N σz e. First show that it is equivalent (in the case of zero magnetisation along z) to C ZZ (R) = 4 n +R n 1 N By developping the number operators on the Fourier modes operators F k and F k you should find that 4 n +R n = 4 F N N k Q F k +Q F k F k e iqr + δ R,0 k,k,q f. The fundamental state in terms of fermions is ψ = where k F = π/. Using this state, show that k F k= k F F k 0 C ZZ (R) = 1 N sin (πr/n) + e iπr N sin (πr/n) This is an important result : although we have a XY Hamiltonian, we still have quasi long range correlations along the z direction, an effect of the quantum nature of the spins. It would be interesting to calculate the C XX (R) but this is difficult. C. Stiffness We now turn to the calculation of the stiffness of the 1D system. Applying a twist to the XY phase of the spin is equivalent to applying it to the fermionic operators: f N = f 0 eiφ f N = f 0 e iφ The Hamiltonian is also diagonalized in the Fourier space in that case BUT we have to change the values of the k to satisfy the new boundary conditions : k = πn N + φ N g. Show that, at half-filling, the energy as a function of φ equals E(φ) = 4J y cos(π/n + φ/n) sin(π/n)

12 h. Calculate the stiffness and show that, in the thermodynamic limit, it is equal to ρ s = 4J y π Once again, this is an interesting result: a one dimensional system has a non-zero stiffness at T = 0. In terms of bosons, this means that we have a non-zero superfluid density at T = 0. Bibliography The correspondence between spins and bosons is discussed in Statistical Physics by S.K. Ma, World Scientific (1985). The Jordan-Wigner transformation and discussions of 1D physics can be found in Quantum Physics in One Dimension by T. Giamarchi, Oford (004). The relation between the stiffness and superfluid density is presented in Path integral computation of superfluid densities by E.L. Pollock and D.M. Ceperley, Phys. Rev. B36, 8343 (1987). 3

13 Ecole de Physique des Houches August, 007 Predoctoral school in statistical physics Quantum Statistical Mechanics Session 5 - Fermions in D Interacting electrons in cristals are often studied through the mean of Hubbard like fermionic models. In a cristal, the low energy physics is dominated by the movements of the electrons present in the eternal orbitals of atoms. If we consider this orbital to be an s-orbital, it can only contain two fermions (one with up spin, one with down spin). The simplest model then takes into account the possibility for these fermions to jump from one site to another one and the repulsion between fermions when they are present on the atom (site). U electrons t s orbital nucleus Figure 1: Sketch of the electrons in a square lattice cristal. The physics is dominated by the eternal s-orbital. Two electrons in this orbital repels each other with strength U. An electron can hop from one atom to another, which is associated to an energy t. This leads to a Hubbard like Hamiltonian (it is in fact the other way to find these kind of Hamiltonians) L H = t r=1 ˆµ=ˆ,ŷ σ=, ( ) c r+ˆµσ c rσ + c rσ c r+ˆµ σ + U r n r n r This model is a good candidate to describe high temperature superconductivity, especially because it develops an antiferromagnetic order which is present in the eperimental systems. 1 - Non interacting limit We introduce the plane wave creation (destruction) operators c kσ = 1 e ik r c rσ L r c kσ = 1 e ik r c rσ L where k = (k, k y ) and k,y = πn,y /L where n,y are integers belonging to [0, L 1]. r a. Find how c rσ can be written as a function of the c kσ kinetic term can be written in the diagonal form and, using this relation, show that the H K = kσ ǫ k c kσ c kσ 1

14 where ǫ k = t[cos(k ) + cos(k y )]. b. Find the Fermi surface in the two cases where the Fermi energy ǫ F is slightly above 4t and when ǫ F = 0. This latter case correspond to a half-filled system (half of the total number of states are occupied). - Strongly interacting limit We now consider the strongly interacting system at half-filling where the number of fermions is equal to the number of sites : N = N = N f / = L /. We first neglect the kinetic term a. What are the fundamental states of the system? b. What is the energy of the system with L + 1 particles? What is then the value of the compressibility ρ/ µ in this case? c. Now considering the kinetic term as a perturbation show that, for two neighbouring sites, the kinetic term generates an effective antiferromagnetic coupling of order t /U. How does this change the possible fundamental states of the system? 3 - Hartree Approimation We now want to estimate for which values of the interactions an antiferromagnetic state eists at half-filling. For this, we use a mean-field approimation. One can define the following fluctuations operators δn n rσ = δn rσ + n rσ and epand the interaction term to the first order in δn. a. Show that when we epand the interaction operator in this way, the Hamiltonian can be rewritten as H = H + H + E shift where H = t rˆµ ( ) c r+ˆµ c r + c r c r+ˆµ + U r (similar epression for H ) and E shift = U r n r n r n r n r b. The Hamiltonians for up and down particles can now be diagonalised independently. To do this analytically, we choose an antiferromagnetic ansatz for the n r, n r = 1 + ei π r m n r = 1 ei π r m where π = (π, π) Show that H can be written in Fourier space as: H = k ǫ k n k + UL /4 Um k c k+ π c k The only modes that are coupled are k and k + π. Rewrite the Hamiltonians in the convenient form H, = [ ( )] ǫ k (n k, n k+ π, ) Um c k+ π, c k, + c k, c k+ π, k HBZ with the following shift E shift = +U(m +1/4)L and where HBZ means that we sum only on of half the k vectors (half the Brillouin s zone).

15 c. Diagonalize this Hamiltonian and find that the eigenenergies are E ± (k) = ± ǫ k + U m d. We are interested in the eigenstates corresponding to the lowest energy E (k) (as we will only fill half of the states). The creation operator corresponding to E (k) is a k = cos(φ k )c k + sin(φ k )c k+ π Show that and tan(φ k ) = tan(φ k ) = ǫ k + ǫ k + U m Um (ǫ k + ǫ k + U m ) sin (φ k ) = sin (φ k ) = ( ǫ k + ǫ k + U m ) + U m e. Calculate m = n r n r. To do so calculate the wave function ψ k (r) of the eigenvector associated to the energy E (k). Each eigenvector then gives a contribution ψ k (r) to n r. f. You should finally arrive at the following self-consistent equation m = Um (1) L ǫ k + U m k HBZ This equation can easily be solved numerically and shows that there is (in the Hartree approimation at least) always an antiferromagnetic ground state at half-filling in the Hubbard model m 0. L = 1000, β = Bibliography U Figure : Numerical solution of equation (1) The mean field phase diagram of the Hubbard model was calculated in Two-dimensional Hubbard model: Numerical simulation study by J. Hirsch, Phys. Rev. B (1985). and in Stability Theory of the Magnetic Phases for a Simple Model of the Transition Metals by D.R. Penn, Phys. Rev. 14 (1966). A more pedagogical account is found in The Mott Metal-Insulator Transition by F. Gebhard, Springer (1997). 3