Quantum Cluster Methods

Transcription

1 Quantum Cluster Methods An introduction David Sénéchal Université de Sherbrooke Département de physique CIFAR - PITP International Summer School on Numerical Methods for Correlated Systems in Condensed Matter May 2008

2 Outline Exact Diagonalizations Clusters and Cluster Perturbation Theory (CPT) The Self-Energy Functional Approach The Variational Cluster Approximation (VCA) Cluster Dynamical Mean Field Theory (CDMFT)

3 Part I Exact Diagonalizations

4 An old Persian Legend, revisited

5 The Hubbard Model on finite cluster Simple Hubbard model (conserves N and N separately): H = a,b,σ t ab c aσc bσ + U a n a n a µ a n a Typical cluster (L sites): e 2 e 1

6 Hamiltonian matrix : 2 sites Half-filled Hubbard model L = 2 U 2µ t t 0 t 2µ 0 t t 0 2µ t 0 t t U 2µ

7 Hamiltonian matrix : 6 sites Sparse matrix structure

8 Hamiltonian matrix : example Dimension of the Hilbert space (half-filled Hubbard model): ( d = L! [(L/2)!] 2 ) 2 2 4L πl One double-precision vector means 1.23 GB of memory L dimension

9 Steps 1. Building a basis 2. Constructing the Hamiltonian matrix 3. Finding the ground state (e.g. by the Lanczos method) 4. Calculating the one-body Green function

10 Coding of the states Basis of occupation number eigenstates: (c 1 )n 1 (c L )n L (c 1 )n 1 (c L )n L 0 Binary representation of basis states: b where b = b + 2 L b Example : b = ( ) = = 349, 866 Need a direct table:... and a reverse table: consecutive label b = B (i ) b = B (i ) mod i = I (b ) + d N I (b ) i = i% d N i = i/ int. division d N

11 Coding of the states (2) Tensor product structure of the Hilbert space: V = V N V N dimension: Example (6 sites): d = d(n )d(n ) d(n σ ) = L! N σ!(l N σ )!

12 Constructing the Hamiltonian matrix Form of Hamiltonian: H = K K + V int. K = a,b t ab c ac b K is stored in sparse form. V int. is diagonal and is stored. Matrix elements of V int. : bit count(b & b ) Two basis states b σ and b σ are connected with the matrix K if their binary representations differ at two positions a and b. b K b = ( 1) M ab t ab M ab = b 1 c=a+1 n c

13 The Lanczos algorithm Finds the lowest eigenpair by an iterative application of H Start with random vector φ 0 An iterative procedure builds the Krylov subspace: K = span { φ 0, H φ 0, H 2 φ 0,, H M φ 0 } Lanczos three-way recursion: φ n+1 = H φ n a n φ n b 2 n φ n 1 a n = φ n H φ n φ n φ n b 2 n = φ n φ n φ n 1 φ n 1 b 0 = 0

14 The Lanczos algorithm (2) In the basis of normalized states n = φ n / φ n φ n, the projected Hamiltonian has the tridiagonal form a 0 b b 1 a 1 b H = 0 b 2 a 2 b a N At each step n, find the lowest eigenvalue of that matrix Stop when the lowest eigenvalue E 0 has converged ( E 0 /E 0 < ) Then re-run to find eigenvector ψ = n ψ n n as the φ n s are not kept in memory.

15 Lanczos method: characteristics Typical required number of iterations: from 20 to 200 Extreme eigenvalues converge first Rate of convergence increases with separation between ground state and first excited state Cannot resolve degenerate ground states : only one state per ground state manifold is picked up If one is interested in low lying states, periodic re-orthogonalization may be required, as orthogonality leaks will occur For degenerate ground states and low lying states (e.g. in DMRG), the Davidson method is generally preferable

16 Lanczos method: illustration of the convergence 100 iterations on a matrix of dimension 600: eigenvalues of the tridiagonal projection as a function of iteration step

17 Lanczos method for the Green function Zero temperature Green function: G µν (ω) = G µν,e (ω) + G µν,h (ω) 1 G µν,e (ω) = Ω c µ c ω H + E ν Ω 0 G µν,h (ω) = Ω c 1 ν c µ Ω ω + H E 0 Consider the diagonal element Use the expansion φ µ = c 1 µ Ω = G µµ,e = φ µ φ µ ω H + E 0 1 z H = 1 z + 1 z 2 H + 1 z 3 H2 +

18 Lanczos method for the Green function (2) Truncated expansion evaluated exactly in Krylov subspace generated by φ µ if we perform a Lanczos procedure on φ µ. Then G µµ,e is given by a Jacobi continued fraction: G µµ,e (ω) = ω a 0 φ µ φ µ ω a 1 b 2 1 b 2 2 ω a 2 The coefficients a n and b n are stored in memory What about non diagonal elements G µν,e? See, e.g., E. Dagotto, Rev. Mod. Phys. 66:763 (1994)

19 Lanczos method for the Green function (3) Trick: Define the combination G + 1 µν,e(ω) = Ω (c µ + c ν ) (c µ + c ν ) Ω ω H + E 0 G + µν,e(ω) can be calculated like G µµ,e (ω) Since G µν,e (ω) = G νµ,e (ω), then G µν,e (ω) = 1 [ G + 2 µν,e (ω) G µµ,e (ω) G νν,e (ω) ] Likewise for G µν,h (ω)

20 Lehman representation Lehmann representation of the Green function G µν (ω) = m + n 1 Ω c µ m m c ω E m + E ν Ω 0 Ω c 1 ν n n c µ Ω ω + E n E 0 Define the matices Then Q (e) µm = Ω c µ m Q (h) µn = Ω c µ n G µν (ω) = m = r Q (e) µmq (e) νm ω ω m (e) Q µr Q νr ω ω r + n Q (h) µn Q (h) νn ω ω n (h)

21 Alternate way : The Band Lanczos method Define φ µ = c µ Ω, µ = 1,..., L. Extended Krylov space : { φ 1,..., φ L,H φ 1,..., H φ L,..., } (H) M φ 1,..., (H) M φ L States are built iteratively and orthogonalized Possible linearly dependent states are eliminated ( deflation ) A band representation of the Hamiltonian (2L + 1 diagonals) is formed in the Krylov subspace. It is diagonalized and the eigenpairs are used to build an approximate Lehmann representation dongarra/etemplates/node131.html

22 Lanczos vs Band Lanczos The usual Lanczos method for the Green function needs 3 vectors in memory, and L(L + 1) Lanczos procedures. The Band Lanczos method requires 3L + 1 vectors in memory, but requires only 2 iterative procedures ((e) et (h)). If Memory allows it, the band Lanczos is much faster.

23 Cluster symmetries Clusters with C 2v symmetry Clusters with C 2 symmetry

24 Cluster symmetries (2) Symmetry operations form a group G The most common occurences are : C1 : The trivial group (no symmetry) C2 : The 2-element group (e.g. left-right symmetry) C2v : 2 reflections, 1 π-rotation C 4v : 4 reflections, 1 π-rotation, 2 π/2-rotations C 3v : 3 reflections, 3 2π/3-rotations C 6v : 6 reflections, 1 π, 2 π/3, 2 π/6 rotations States in the Hilbert space fall into a finite number of irreducible representations (irreps) of G The Hamiltonian H is block diagonal w.r.t. to irreps.

25 Group characters C 2 E C 2 A 1 1 B 1 1 C 2v e c 2 σ 1 σ 2 A A B B C 4v e c 2 2c 4 2σ 1 2σ 2 A A B B E

26 Taking advantage of cluster symmetries... order of the group Reduces the dimension of the Hilbert space by G Accelerates the convergence of the Lanczos algorithm Reduces the number of Band Lanczos starting vectors by G But: complicates coding of the basis states Make use of the projection operator: dimension of irrep. See, e.g. Poilblanc & Laflorencie cond-mat/ P (α) = d α χ G g (α) g g group character

27 Taking advantage of cluster symmetries (2) Need new basis states, made of sets of binary states related by the group action: ψ = d α G g χ (α) Then matrix elements take the form ψ 2 H ψ 1 = d α G fermionic phase g g b g b = φg(b) gb g χ (α) h φ g (b) gb 2 H b 1 When computing the Green function, one needs to use combinations of creation operators that fall into group representations. Ex (4 1): c (A) 1 = c 1 + c 4 c (A) 2 = c 2 + c 3 c (B) 1 = c 1 c c (B) 2 = c 2 c 3

28 Taking advantage of cluster symmetries (3) Example : number of matrix elements of the kinetic energy operator (Nearest neighbor) on a 3 4 cluster with C 2v symmetry: A 1 A 2 B 1 B 2 dim. 213, , , , 248 value , 640 6, 208 7, 584 5, , 983, 264 2, 936, 144 2, 884, 832 2, 911, , , 168 1, 050, 432 1, 021, , 088 2, 304 3, 232 2,

29 Large dimensions : need for parallelization Memory needs exceed single cpu capacity beyond L 14 A half-filled 16-site system has dimension 165,636, GB for a state vector. Need to distribute the problem over many processors The main task is matrix-vector multiplication: a a a a a a a a a a a a a a a a a a a a a a a a

30 Part II Cluster Perturbation Theory

31 Clusters and superlattices e 2 e 1 K k k (π, π) (0, 0) ( π, π) 10-site cluster Reduced Brillouin zone

32 Basic Idea lattice Hamiltonian H hopping matrix cluster Hamiltonian = H + V inter-cluster hopping terms cluster hopping matrix t = t + V inter-cluster hopping Treat V at lowest order in Perturbation theory At this order, the Green function is G 1 (ω) = G 1 (ω) V cluster Green function C. Gros and R. Valenti, Phys. Rev. B 48, 418 (1993) D. Sénéchal, D. Perez, and M. Pioro-Ladrière. Phys. Rev. Lett. 84, 522 (2000)

33 Interlude : Fourier transforms i, j : lattice site index k : full wavevector m, n : lattice site index k : redcued wavevector a, b : cluster site index K : cluster wavevector f j = e ik r j f(k) f(k) = 1 e ik r j f j N k j f m = k e i k r m f( k) f( k) = L e i k r m f m N m f a = 1 e ik ra f K f K = 1 e ik ra f a L L K a

34 Basic Idea (cont.) More accurate formulation G 1 ( k, ω) = G 1 (ω) V( k). But G 1 = ω t Σ G 1 0 = ω t V, Thus : lattice self-energy is approximated as the cluster self-energy G 1 ( k, ω) = G 1 0 ( k, ω) Σ(ω), Example : 2-site cluster (1D): ( ) t 0 1 = t 1 0 V( k) = t ( ) 0 e 2i k e 2i k 0

35 Periodization CPT breaks translation invariance, which needs to be restored: G cpt (k, ω) = 1 e ik (ra rb) G ab ( k, ω). L Periodizing the Green function vs the self-energy (1D case): a,b (π) (π) (π/2) (π/2) (0) ω Green function periodization (0) ω Self-energy periodization

36 One-dimensional example Evolution of spectral function with increasing U/t:

37 Interlude : Relation with spectral function Lehmann representation: A(k, ω) = 2 lim Im G(k, ω + iη) η 0 + G αβ (ω) = 1 Ω c α m m c ω E m m + E β Ω 0 + Ω c β n 1 n c α Ω ω + E n n E 0 But : lim Im 1 η 0 + ω + iη = lim Therefore : A(k, ω) = m η 0 + η ω 2 + η 2 = πδ(ω) m c k Ω 2 2πδ(ω E m + E 0 ) + n n c k Ω 2 2πδ(ω + E n E 0 )

38 h-doped cuprates : Pseudogap from CPT Sénéchal and Tremblay., Phys. Rev. Lett. 92, (2004)

39 e-doped cuprates : Pseudogap from CPT Sénéchal and Tremblay., Phys. Rev. Lett. 92, (2004)

40 CPT : characteristics Exact at U = 0 Exact at t ij = 0 Exact short-range correlations Allows all values of the wavevector But : No long-range order Controlled by the size of the cluster

41 Part III The self-energy functional approach

42 Motivation CPT cannot describe broken symmetry states, because of the finite cluster size Idea : add a Weiss field term to the cluster Hamiltonian H, e.g., for antiferromagnetism: H M = M a (π, π) e iq r a (n a n a ) This term favors AF order, but does not appear in H, and must be subtracted from V Need a principle to set the value of M : energy minimization? Better : Potthoff s self-energy functional approach

43 The Potthoff variational principle Variational principle for the Green function: Ω t [G] = Φ[G] Tr ((G 1 0t Where Φ[G] is the Luttinger-Ward functional: G 1 )G) + Tr ln( G). Tr (A) = P ω,α Aαα(ω) Φ = with the property M. Potthoff, Eur. Phys. J. B 32, 429?436 (2003) δφ[g] δg = Σ

44 The Potthoff variational principle (2) Here, Tr means a sum over frequencies, site indices (or wavevectors) and spin/band indices. The functional is stationary at the physical Green function (Euler eq.): δω t [G] δg = Σ G 1 0t + G 1 = 0. Approximation schemes: Type I : Simplify the Euler equation Type II : Approximate the functional (Hartree-Fock, FLEX) Type III : Restrict the variational space, but keep the functional exact

45 The Potthoff variational principle (3) Potthoff : Use the self-energy rather than the Green function Ω t [Σ] = F [Σ] Tr ln( G 1 0t + Σ) F [Σ] = Φ[G] Tr (ΣG) F is the Legendre transform of Φ: δf [Σ] δσ New Euler equation: = δφ[g] δg[σ] δg δσ ΣδG[Σ] δσ G = G δω t [Σ] δσ = G + (G 1 0t Σ) 1 = 0 At the physical self-energy, Ω t [Σ] is the thermodynamic grand potential

46 The Reference System To evaluate F, use its universal character : its functional form depends only on the interaction. Introduce a reference rystem H, which differs from H by one-body terms only (example : the cluster Hamiltonian) Suppose H can be solved exactly. Then, at the physical self-energy Σ of H, Ω = F [Σ] Tr ln( G ) by eliminating F : Ω t [Σ] = Ω + Tr ln( G ) Tr ln( G 1 0t + Σ) = Ω + Tr ln( G ) Tr ln( G) = Ω Tr ln(1 VG )

47 The Potthoff functional Making the trace explicit, one finds Ω t [Σ] = Ω T [ ] tr ln 1 V( k)g ( k, ω) ω k = Ω T [ ] ln det 1 V( k)g ( k, ω) ω The sum over frequencies is to be performed over Matsubara frequencies (or an integral along the imaginary axis at T = 0). The variation is done over one-body parameters of the cluster Hamiltonian H k In particular, the Weiss field M is to be varied until Ω is stationary

48 Calculating the functional I : exact form It can be shown that Tr ln( G) = T m poles of G ln(1 + e βω m) + T m ln(1 + e βζ m) zeros of G Use the Lehmann representation of the GF: G µν(ω) = r M. Potthoff, Eur. Phys. J. B, 36:335 (2003) Q µr Q νr G(ω) = Q 1 ω ω r ω Λ Q diagonal(ω r)

49 Calculating the functional I : exact form (2) A similar representation holds for the CPT Green function 1 G( k, ω) = G 1 V( k) = 1 [ ] 1 Q 1 ω Λ Q V( k) 1 = Q ω L( k) Q L( k) = Λ + Q V( k)q Let ω r ( k) be the eigenvalues of L( k). Then Ω(x) = Ω (x) ω r<0 variational parameters M. Aichhorn et al., Phys. Rev. B 74 : (2006) ω r + L N ω r ( k) k ω r( k)<0

50 Calculating the functional II : numerical integral Except for very small clusters (L 4), it is much faster to perform a numerical integration over frequencies: Ω(x) = Ω (x) 0 dx π L ln det(1 V( k)g (ix)) L(µ µ ) N k D. Sénéchal, proceedings of HPCS 2008, IEEE (2008)

51 Evaluation of integrals For frequency integrals: Gaussian integration on three segments For wavevector integrals, adaptive mesh of points: Start with a coarse, regular grid On each plaquette, compare 4 and 9 point Gaussian integrals. Subdivide into 4 sub-plaquettes if necessary. Easy with recursive calls

52 Part IV The Variational Cluster Approximation

53 Basic Idea Set up a superlattice of clusters Choose a set of variational parameters, e.g. Weiss fields for broken symmetries Set up the calculation of the Potthoff functional: Ω t [Σ] = Ω T L N ω k [ ] ln det 1 V( k)g ( k, ω) Use an optimization method to find the stationary points Adopt the cluster self-energy associated with the stationary point with the lowest Ω and use it as in CPT

54 Example : Néel Antiferromagnetism Used the Weiss field H M = M a (π, π) e iq r a (n a n a ) Profile of Ω for the half-filled, square lattice Hubbard model: Ω U = 16 U = 8 U = 4 U = M Ω L = L = M

55 Example : Néel Antiferromagnetism (2) order parameter M (4 ) ordering energy U Best scaling factor : q = number of links 2 number of sites M Order Parameter x1 2x1 links 2L 2x1 2x2 2x1 2x2 1 1 L 2x3 2x4 B10 2x3 2x4 B10 2x2 3x4 4x4 2x2 3x4 4x4 2x3 2x4 B10 3x4 4 2x3 2x4 B10 3x

56 Example clusters 3 4 B10

57 Superconductivity Need to add a pairing field O sc = ij c i c j + H.c ij s-wave pairing: ij = δ ij d x 2 y 2 pairing: { 1 if ri r j = ±ˆx ij = 1 if r i r j = ±ŷ d xy pairing: { 1 if ri r j = ±(ˆx + ŷ) ij = 1 if r i r j = ±(ˆx ŷ) Ω d x 2 y 2 extended s-wave dxy cluster s-wave U = 8, µ =

58 Superconductivity (2) Pairing fields violate particle number conservation The Hilbert space is enlarged to encompass all particle numbers with a given spin In practice, on uses the Nambu formalism, i.e., particle-hole transformation on the spin-down sector : c a = c a and d a = c a Then the Hamiltonian looks like it conserves particle number, but not spin.

59 Superconductivity and Antiferromagnetism in the cuprates One-band Hubbard model for the cuprates: t = 0.3, t = 0.2, U = 8: order parameter pure d x 2 y 2 pure Néel coex. d x 2 y 2 coex. Néel n M. Guillot, MSc thesis, Univ. de Sherbrooke (2007)

60 Thermodynamic consistency The electron density n may be calculated either as n = Tr G or n = Ω µ The two methods give different results, except if the cluster chemical potential µ is treated like a variational parameter: cluster U = 8 normal state 0.8 n Ω µ cons. TrG TrG cons. Ω µ µ

61 Optimization procedure Need to find the saddle points of Ω(x) with the least possible evaluations of Ω(x) Use the Newton-Raphson algorithm: Evaluate Ω at a number of points at and around x0 that just fits a quadratic form Move to the stationary point x 1 of that quadratic form and repeat Stop when x i x i 1, or the numerical gradient Ω, converges The NR method is not robust : it converges fast when started close enough to the solution Proceed adiabatically through external parameter space (e.g. as function of U or µ)

62 Example: Homogeneous coexistence of dsc and AF orders

63 Example: dsc on a 4 4 cluster, spectral function spectral function : 4x4_C2v, U=8, mu=1.5, tx=1, Dx=0.035, (spin up) (27/5/2008) (0.5,0.5) (1,0) (1,0.5) (1,1) (0.5,0.5) (0,0) (0.5,0)

64 Example: dsc on a 4 4 cluster, Fermi surface plot

65 VCA vs Mean-Field Theory Differs from Mean-Field Theory: Interaction is left intact, it is not factorized Retains exact short-range correlations Weiss field order parameter More stringent that MFT Controlled by the cluster size Similarities with MFT: No long-range fluctuations (no disorder from Goldstone modes) Yet : no LRO for Néel AF in one dimension Need to compare different orders yet : they may be placed in competition / coexistence

66 Part V Cluster Dynamical Mean Field Theory

67 Basic Idea To add variational degrees of freedom in the form of a bath of uncorrelated sites H = µ,ν t µν c µc ν + U a n a n a + µ,α θ µα (c µa α + H.c.) + hybridization matrix α ε α a αa a bath energies 7 8 ε 2 t 2 t 2 ε 2 ε 2 t 2 t 2 ε ε 1 t 1 t 1 ε 1 ε 1 t 1 t 1 ε

68 The hybridization function If we trace over the bath degrees of freedom, the cluster Green function takes the form Γ(ω) is the hybridization function: G 1 = ω t Γ(ω) Σ(ω) Γ µν (ω) = α θ µα θ να ω ε α

69 The hybridization function (2) Proof: (U = 0) G 1 full (ω) = 1 ω T T = ( ω t θ θ ω ε ) Given A = G 1 full, need to find B 1 11 : ( ) ( A 11 A 12 B 11 B 12 = A 21 A 22 B 21 B 22 Simple manipulations lead to ) 1 ( A11 A 12 A 1 22 A ) 21 B11 = 1 G 1 = ω t θ 1 ω ε θ U 0 : simply add the free energy (by definition)

70 The hybridization function (3) Γ(ω) embodies the effect of the rest of the lattice on the cluster, in some effective dynamics. The action would take the form β β S = dτ dτ c µ(τ)gµν 1 (τ τ )c ν (τ ) 0 0 µν + U d β 0 τ n a (τ)n a (τ) a where G (iω n ) = β 0 e iωnτ G (τ) G 1 µν (iω n ) = iω n δ µν t µν Γ µν (iω n )

71 Baths and the SFA The Potthoff functional approach carries over unchanged in the presence of a bath The bath makes a contribution to the Potthoff functional: Ω bath = ε α<0 On can in principle use the same methods as in VCA The presence of the bath increases the resolution of the approach in the time domain, at the cost of spatial resolution, for a fixed total number of sites (cluster + bath). ε α

72 The CDMFT Procedure 1. Start with a guess value of (θ µα, ε α ). 2. Calculate the cluster Green function G(ω) (ED). 3. Calculate the superlattice-averaged Green function Ḡ(ω) = k 1 G 1 0 ( k) Σ(ω) and G 1 0 (ω) = Ḡ 1 + Σ(ω) 4. Minimize the following distance function: d = ω,ν,ν over the set of bath parameters. 5. Go back to step (2) until convergence. ( ω + µ t Γ(ω) G0 1 (ω) ) 2 νν

73 The CDMFT Procedure (2) Initial guess for Γ Cluster Solver: Compute G Ḡ = k[g 1 0 ( k) Σ(ω)] 1 G 1 0 = Ḡ 1 + Σ update Γ by minimizing d No Γ converged? Yes Exit

74 Example : the 1D Hubbard model 1 n exact VCA β = 20, ω c = 1.5 β = 20, ω c = 10, 1/ω weight β = 50, ω c = 1.5 β = 50, ω c = µ

75 Example : the 1D Hubbard model (2) ε t µ µ

76 Example : dsc and AF in the 2D Hubbard model Nine bath parameters Homogeneous coexistence of d x 2 y2 SC and Néel AF order parameter dsc ( 5) AF n

77 Example : The Mott transition The CDMFT is well suited to detect the Mott transition This transition manifests itself as a jump in the double occupancy n n In an exact SFA solution : discontinuity in the bath parameters (first order transition). in CDMFT : hysteresis is possible, because of the method s own dynamics for finding solutions D (a) 0.05 solid: t /t=0.7 dashed: t /t= U/t B. Kyung and A.-M. S. Tremblay. Physical U c /t Review Letters, 97 : (2006)

78 QUESTIONS?