sin(kr) cos(kr) = 0, a σ (k) a σ (k) = 1 R a Semi-momentum states cos(kr), σ = +1 sin(kr), σ = 1. C σ k (r)t r a N a k = m 2π N

Transcription

1 Semi-momentum states Mix momenta +k and k for k 0,π. Introduce function C σ k (r) = Semi-momentum state { cos(kr), σ = +1 sin(kr), σ = 1. Useful trigonometric relationships C ± k ( r) = ±C± k (r), C ± k (r + d) = C± k (r)c+ k (d) C k (r)c k (d). a σ (k) = 1 N 1 Na k = m 2π N r=0 C σ k (r)t r a, m =1,..., N/2 1, σ = ±1 States with same k, different σ are orthogonal a σ (k) a σ (k) = 1 R a N a r=1 sin(kr) cos(kr) = 0, 17

2 Normalization of semi-momentum states N a = ( N R a ) 2 R a [Ck σ (r)] 2 = N 2 r=1 Hamiltonian: ac with H H a ± (k) = N j=0 h j a The matrix elements are 2R a b τ (k) H j a σ (k) = τ (σ τ)/2 h j a R ( a C + k R (l j) b ± j (k) C k (l j) b j ), (k) bj N bj N a C στ k (l j ) σ is mot a conserved quantum number H and T mix σ=+1 and σ= 1 states the H matrix is twice as large as for momentum states 18

3 Why are the semi-momentum states useful then? Because we can construct a real-valued basis: Semi-momentum states with parity This state has definite parity with p=+1 or p= 1 a σ (k, p) = 1 N 1 N σ a r=0 C σ k (r)t r (1 + pp ) a (k, 1) and (k,+1) blocks roughly of the same size as original k blocks but these states are real, not complex! For k 0,π, the p=-1 and p=+1 states are degenerate P,T transformations example: N=8;note that T 5 P a>= a> such P,T relationships will affect normalization and H-elements 19

4 Normalization: We have to check whether or not T m P a = a for some m {1,..., N 1} Simple algebra gives Na σ = N 2 { 1, T m P a a R a 1+σpcos(km), T m P a = a In the latter case the σ=-1 and σ=+1 states are not orthogonal Then only one of them should be included in the basis convention: use σ=+1 if 1+σpcos(km) 0, else σ=-1 If both σ=+1 and σ= 1 are present: we store 2 copies of the same representative we will store the σ value along with the periodicity of the representative 20

5 Pseudocode: semi-momentum, parity basis construction do s =0, 2 N 1 call checkstate(s, R, m) do σ = ±1 (do only σ = +1 if k = 0 or k = N/2) if (m 1) then if (1 + σp cos(ikm2π/n ) = 0) R = 1 if (σ = 1 and 1 σp cos(ikm2π/n ) 0); R = 1 endif if R>0 then a = a + 1; s a = s; R a = σr; m a = m endif In the subroutine checkstate(), we now find whether T m P a = a for some m {1,..., N 1} m= 1 if there is no such transformation if m -1, then the σ=+1 and σ=-1 states are not orthogonal use only the σ=+1 state if it has non-zero normalization use the σ=-1 state if σ=+1 has normalization=0 R= 1 for not including in the basis 21

6 the subroutine checkstate() is modified to gives us: periodicity R (R= 1 if incompatible) m>0 if T m P s>= s> m= 1 if no such relationship check all translations of s> { construct reflected state P s> { check all translations of P s> subroutine checkstate(s, R, m) R = 1 if ( P i s[i] n ) return t = s do i =1,N t = cyclebits(t, N) if (t <s) then return elseif (t = s) then if (mod(k, N/i) 0) return R = i; exit endif t = reflectbits(s, N); m = 1 do i =0,R 1 if (t <s) then R = 1; return elseif (t = s) then m = i; return endif t = cyclebits(t, N) 22

7 Hamiltonian : Act with an operator Hj on a representative state: H j a = h j ap q j T l j b j We can write H acting on a basis state as H a σ (k, p) = N j=0 h j a(σp) q j N σ a T m P b j = b j If, for some m, N σ b j 1 σp cos(km) Nb σ = j 1+σpcos(km) = N 1 r=0 then C σ k (r + l j )(1 + pp )T r b j Using the properties (trigonometry) of the C-functions: N N σ H a σ (k, p) = h j a(σp) q bj j j=0 N σ a ( cos(kl j ) b σ j (k, p) σ N σ b j N σ b j ) sin(kl j ) b σ (k, p) ) sin(km) 1+σp cos(km) b j (k, p) b± j (k, p) = p else the ratio is one and the + and states are orthogonal 23

8 The matrix elements are diagonal in σ N σ bj b σ j (k, p) H j a σ (k, p) = h j a(σp) q j off-diagonal in σ N σ a { cos(klj ), P b j T m b j cos(kl j )+σp cos(k[l j m]) 1+σp cos(km), P b j = T m b j b σ j (k, p) H j a σ (k, p) = h j a(σp) q j Na σ { σ sin(klj ), P b j T m b j, N σ b j σ sin(kl j )+p sin(k[l j m]) 1 σp cos(km), P b j = T m b j, 24

9 Pseudocode: semi-momentum, parity hamiltonian If 2 copies of the same representative, σ= 1 and σ=+1: do both in the same loop iteration examine the previous and next element carry out the loop iteration only if representative found for the first time do a =1,M if (a >1 and s a = s a 1 ) then cycle elseif (a <M and s a = s a+1 ) then n =2 else n =1 endif... n is the number of copies of the representative do i = a, a + n 1 H(a, a) =H(a, a) +E z diagonal matrix elements Ez = diagonal energy 25

10 s = flip(s a, i, j) call representative(s, r, l, q) call findstate(r, b) if (b 0) then if (b >1 and s b = s b 1 ) then m = 2; b = b 1 elseif (b < M and s b = s b+1 ) then m =2 else m =1 endif do j = b, b + m 1 do i = a, a + n 1 H(i, j) =H(i, j) +helement(i, j, l, q) endif subroutine representative(s, r, l, q)... t = reflectbits(s, N); q =0 do i =1,N 1 t = cyclebits(t, N) if (t <r) then r = t; l = i; q =1endif construct off-diagonal matrix elements helement() computes the values based on stored info and l,q find the representative r of s translation and reflection numbers l,q 26

11 Using spin-inversion symmetry Spin inversion operator: Z S z 1,S z 2,..., S z N = S z 1, S z 2,..., S z N In the magnetization block m z =0 we can use eigenstates of Z a σ (k, p, z) = 1 N 1 C N σ k σ (r)t r (1 + pp )(1 + zz) a a r=0 Z a σ (k, p, z) = z a σ (k, p, z), z = ±1 Normalization: must check how a representative transforms under Z,P,T 1) T m P a a, T m Z a a T m PZ a a 2) T m P a = a, T m Z a a T m PZ a a 3) T m P a a, T m Z a = a T m PZ a a 4) T m P a a, T m Z a a T m PZ a = a 5) T m P a = a, T n Z a = a T m+n PZ a = a For cases 2,4,5 only σ=+1 or σ= 1 included 1, 1) Na σ = 2N 2 1+σpcos(km), 2) 1+zcos(km), 3) R a 1+σpz cos(km), 4) [1 + σp cos(km)][1 + z cos(kn)], 5) 27

12 Hamiltonian: acting on a state gives a transformed representative H j a = h j ap q j Z g j T l j b j q j {0, 1}, g j = {0, 1}, l j = {0, 1,..., N 1} After some algebra... we can obtain the matrix elements diagonal in σ b σ j (k, p) H j a σ (k, p) = h j a(σp) q j z g j off-diagonal in σ b σ j (k, p) H j a σ (k, p) = h j a(σp) q j z g j N τ bj N σ a cos(kl j ), 1), 3) cos(kl j )+σp cos(k[l j m]) 1+σp cos(km), 2), 5) cos(kl j )+σpz cos(k[l j m]) 1+σpz cos(km), 4) N τ bj N σ a σ sin(kl j ), 1), 3) σ sin(kl j )+p sin(k[l j m]) 1 σp cos(km), 2), 5) σ sin(kl j )+pz sin(k[l j m]) 1 σpz cos(km), 4) 28

13 Example: block sizes k=0, mz=0 (largest block) (p = ±1,z = ±1) N (+1, +1) (+1, 1) ( 1, +1) ( 1, 1) Total spin S conservation more difficult to exploit complicated basis states calculate S using S 2 =S(S+1) N N S 2 = S i S j i=1 j=1 = 2 i<j S i S j N Full diagonalization; expectation values shorthand block label: j=(mz,k,p) or j=(mz=0,k,p,z) D 1 j H j D j = E j, n j A n j =[D 1 AD j ] nn T>0: sum over all blocks j and states in block n=0,mj 1 A = 1 Z j M j 1 n=0 e βe j,n [D 1 j A j U j ] nn, Z = j Ej = diagonal (energy) matrix, Ej,n = energies, n=0,...,mj 1 j M j 1 n=0 e βe j,n Full diagonalization limited to small N; N=

14 Example: Thermodynamics some quantities can be computed using only the magnetization mz=0 sector spin-inversion symmetry can be used, smallest blocks spin-s state is (2S+1)-fold degenerate (no magnetix field) weight factor possible spin dependence of expectation value average over mz= S,...,S C = d H = 1 dt χ z = d m z = 1 dh z T ( H 2 T 2 H 2) ( m 2 z m z 2) m z =0, m 2 z = m2 x + m 2 y + m 2 z 3 = S2 3 = S(S + 1) 3 Compared with leading high-t forms χ = (1/4)/T C = (3/13)/T 2 30

15 The Lanczos method If we need only the ground state and a small number of excitations can use Krylov space methods, which work for much larger matrices basis states with 10 7 states or more can be easily handled (30-40 spins) The Krylov space and projecting out the ground state Start with an arbitrary state ψ it has an expansion in eigenstates of H; act with a high power Λ of H H Λ Ψ = ( ( ) Λ c n En Λ n = E0 Λ E1 c c ) E n 0 For large Λ, if the state with largest En dominates the sum one may have to subtract a constant, H C, to ensure ground state even better to use linear combination of states generated for different Λ ψ a = Λ m=0 ψ a (m)h m Ψ, diagonalize H in this basis a =0,..., Λ In the Lanczos basis, H is tridiagonal, convenient to generate and use Normally M= basis states is enough; easy to diagonalize H 31

16 Constructing the Lanczos basis First: construct orthogonal but not normalized basis {fm}. Define N m = f m f m, H mm = f m H f m The first state f0> is arbitrary, e.g., random. The next one is f 1 = H f 0 a 0 f 0 Demand orthogonality f 1 f 0 = f 0 H f 0 a 0 f 0 f 0 = H 00 a 0 N 0 a 0 = H 00 /N 0 The next state and its overlaps with the previous states f 2 = H f 1 a 1 f 1 b 0 f 0 f 2 f 1 = H 11 a 1 N 1, f 2 f 0 = N 1 b 0 N 0 For orthogonal states a 1 = H 11 /N 1, b 0 = N 1 /N 0 All subsequent states are constructed according to f m+1 = H f m a m f m b m 1 f m 1 a m = H mm /N m, b m 1 = N m /N m 1 Easy to prove orthogonality of all these states (<fm+1 fm>=0 is enough) 32

17 The hamiltonian in the Lanczos basis Rewrite the state generation formula H f m = f m+1 + a m f m + b m 1 f m 1 Because of the orthogonality, the only non-0 matrix elements are f m 1 H f m = b m 1 N m 1 = N m f m H f m = a m N m f m+1 H f m = N m+1 But the f-states or not normalized. The normalized states are: φ m = 1 f m Nm In this basis the H-matrix is φ m 1 H φ m = b m 1 φ m H φ m = a m φ m+1 H φ m = b m 33