Techniques for translationally invariant matrix product states

Transcription

1 Techniques for translationally invariant matrix product states Ian McCulloch University of Queensland Centre for Engineered Quantum Systems (EQuS) 7 Dec 2017 Ian McCulloch (UQ) imps 7 Dec / 33

2 Outline 1 Scaling relations in the thermodynamic limit Correlation funtions Transfer matrix Scaling dimensions 2 Expectation values of impo s Higher moments Energy variance Example = momentum distribution Binder cumulant Ian McCulloch (UQ) imps 7 Dec / 33

3 Correlation Functions The form of correlation functions are determined by the eigenvalues of the transfer operator Expansion in terms of eigenspectrum λ i : O(x)O(y) = i All eigenvalues 1 One eigenvalue equal to 1, corresponding to the identity operator a i λ y x i Ian McCulloch (UQ) imps 7 Dec / 33

4 1 Hubbard Model transfer matrix spectrum Half-filling, U/t = (0,0) Singlet (1,0) Spin triplet (0,1) Holon Triplet (1/2,1/2) Single-particle λ Number of states kept Ian McCulloch (UQ) imps 7 Dec / 33

5 Hubbard model transfer matrix spectrum Half-filling, U/t=4 100 (0,0) Singlet (1,0) Spin triplet (0,1) Holon Triplet (1/2,1/2) Single-particle Correlation length Number of states kept Ian McCulloch (UQ) imps 7 Dec / 33

6 Critical scaling example Two-species bose gas with linear tunneling Ω, from F. Zhan et al, Phys. Rev. A 90, ξ m Ω = Ω = Ω = Ω = Ω = Ω = Ian McCulloch (UQ) imps 7 Dec / 33

7 CFT Parameters For a critial mode, the correlation length increases with number of states m as a power law, ξ m κ [T. Nishino, K. Okunishi, M. Kikuchi, Phys. Lett. A 213, 69 (1996) M. Andersson, M. Boman, S. Östlund, Phys. Rev. B 59, (1999) L. Tagliacozzo, Thiago. R. de Oliveira, S. Iblisdir, J. I. Latorre, Phys. Rev. B 78, (2008)] This exponent is a function only of the central charge, κ = 6 12c + c [Pollmann et al, PRL 2009] Note: in practice this usually isn t a good way to determine c better to use entropy scaling Ian McCulloch (UQ) imps 7 Dec / 33

8 Scaling dimensions (Vid Stojevic, IPM et al, Phys. Rev. B 91, (2015)) Suppose we have a two-point correlator that has a power-law at large distances O(x)O(y) = y x 2 As we increase the number of states kept m the correlation length increases, so the region of validity of the power law increases. Take two different calculations with m 1 and m 2 Correlation lengths ξ 1 and ξ 2 ( ) ξ2 We expect: O(ξ 2) O(ξ 1 ) = ξ 1 for x large, we have: O(x) a λ x (with ξ = 1/ ln λ) Prefactor a is overlap of operator O with next-leading eigenvector of transfer operator a ξ This gives directly the operator scaling dimensions by direct fit Ian McCulloch (UQ) imps 7 Dec / 33

9 Scaling dimensions (Vid Stojevic, IPM et al, Phys. Rev. B 91, (2015)) Suppose we have a two-point correlator that has a power-law at large distances O(x)O(y) = y x 2 As we increase the number of states kept m the correlation length increases, so the region of validity of the power law increases. Take two different calculations with m 1 and m 2 Correlation lengths ξ 1 and ξ 2 ( ) ξ2 We expect: O(ξ 2) O(ξ 1 ) = ξ 1 for x large, we have: O(x) a λ x (with ξ = 1/ ln λ) Prefactor a is overlap of operator O with next-leading eigenvector of transfer operator a ξ This gives directly the operator scaling dimensions by direct fit Ian McCulloch (UQ) imps 7 Dec / 33

10 Scaling dimensions (Vid Stojevic, IPM et al, Phys. Rev. B 91, (2015)) Suppose we have a two-point correlator that has a power-law at large distances O(x)O(y) = y x 2 As we increase the number of states kept m the correlation length increases, so the region of validity of the power law increases. Take two different calculations with m 1 and m 2 Correlation lengths ξ 1 and ξ 2 ( ) ξ2 We expect: O(ξ 2) O(ξ 1 ) = ξ 1 for x large, we have: O(x) a λ x (with ξ = 1/ ln λ) Prefactor a is overlap of operator O with next-leading eigenvector of transfer operator a ξ This gives directly the operator scaling dimensions by direct fit Ian McCulloch (UQ) imps 7 Dec / 33

11 Heisenberg model fit for the scaling dimension prefactor of the spin operator at this mode idmrg data for m=15,20,25,30,35 y = * x^ transfer matrix eigenvalue 1 - λ = 1 / ξ Ian McCulloch (UQ) imps 7 Dec / 33

12 Expectation values of MPO s - arxiv: We have seen that we can write many interesting operators in the form of a matrix product operator Can we evaluate the expectation value of an arbitrary MPO? If the MPO has no Jordan structure, this is a simple eigenvalue problem W = λ For a triangular MPO, this doesn t work. But we can make use of the triangular structure I S z λs x index by index, each component is a function only of the previously calculated terms S z I Ian McCulloch (UQ) imps 7 Dec / 33

13 Expectation values of MPO s - arxiv: We have seen that we can write many interesting operators in the form of a matrix product operator Can we evaluate the expectation value of an arbitrary MPO? If the MPO has no Jordan structure, this is a simple eigenvalue problem W = λ For a triangular MPO, this doesn t work. But we can make use of the triangular structure I S z λs x index by index, each component is a function only of the previously calculated terms S z I Ian McCulloch (UQ) imps 7 Dec / 33

14 i W j Choose bond indices i, j of W ij, and denote T Wij Example: I Sz λs x S z I Eigentensor is (E 1 E 2 E 3 ) Starting from E 1 : E 1 = T I (E 1 ) = I is equivalent to the orthogonality condition - E 1 is just the identity E 2 : E 2 = T S z(e 1 ) = T S z(i) = S z E 3 : doesn t reach a fixed point, E 3 = E 3 (L) depends on the number of iterations L E 3 (L + 1) = T I (E 3 (L)) + T S z(e 2 ) + T λs x(e 1 ) = T I (E 3 (L)) + C where C is a constant matrix, C = T S z(s z ) + λs x Ian McCulloch (UQ) imps 7 Dec / 33

15 Fixed point equations for E 3 : Eigenmatrix expansion of T I : giving E (n) E 3 (L + 1) = T I (E 3 (L)) + C T I = λ n λ λ m 2 n=1 3 (L + 1) = λ ne (n) 3 (L) + C(n) Since λ 1 = 1 by construction, this motivates decomposing into components parallel (e 1 ) and perpendicular (E 1 ) to the identity: where Tr E 3 (L)ρ = 0 E 3 (L + 1) = E 3(L) + e 3 (L) I Ian McCulloch (UQ) imps 7 Dec / 33

16 Component in the direction of the identity: e 3 (L + 1) = e 3 (L) + Tr Cρ Has the solution e 3 (L) = L Tr Cρ is the energy Component perpendicular to the identity: E 3(L + 1) = T I (E 3(L)) + C where C = C (Tr Cρ) I E 3(L + 1) n = λ n E 3(L) n + C n Since all λ n < 1 here, this is a geometric series that converges to a fixed point (independent of L), (1 T I )(E 3) = C Linear solver for the unknown matrix E 3 Ian McCulloch (UQ) imps 7 Dec / 33

17 Component in the direction of the identity: e 3 (L + 1) = e 3 (L) + Tr Cρ Has the solution e 3 (L) = L Tr Cρ is the energy Component perpendicular to the identity: E 3(L + 1) = T I (E 3(L)) + C where C = C (Tr Cρ) I E 3(L + 1) n = λ n E 3(L) n + C n Since all λ n < 1 here, this is a geometric series that converges to a fixed point (independent of L), (1 T I )(E 3) = C Linear solver for the unknown matrix E 3 Ian McCulloch (UQ) imps 7 Dec / 33

18 Summary: Decompose eigentensor into components parallel and perpendicular to the identity The component parallel to the identity is the energy per site The perpendicular components reach a fixed point and give the Hamiltonian matrix elements E 1 = I Identity operator E 2 = Sz Sz block operator E 3 (L) = E + Le 3 Hamiltonian operator + energy per site Ian McCulloch (UQ) imps 7 Dec / 33

19 Generalization to arbitrary triangular MPO s arxiv: At the i th iteration, we have E i (L + 1) = T Wii (E i (L)) + T Wji (E j (L)) j<i }{{} = C(L) Basic idea: if W ii = 0, then E i = C if W ii 0, then solve (1 T Wii )(E i ) = C The result will be a polynomial function of L solve separately for the coefficient of the k-th power of n If the diagonal element is unitary, then obtain the eigenvalues of magnitude 1 If any eigenvalues are complex, then expand also in fourier modes Ian McCulloch (UQ) imps 7 Dec / 33

20 Generalization to arbitrary triangular MPO s arxiv: At the i th iteration, we have E i (L + 1) = T Wii (E i (L)) + T Wji (E j (L)) j<i }{{} = C(L) Basic idea: if W ii = 0, then E i = C if W ii 0, then solve (1 T Wii )(E i ) = C The result will be a polynomial function of L solve separately for the coefficient of the k-th power of n If the diagonal element is unitary, then obtain the eigenvalues of magnitude 1 If any eigenvalues are complex, then expand also in fourier modes Ian McCulloch (UQ) imps 7 Dec / 33

21 Example 1 - Variance How close is a variational state to an eigenstate of the Hamiltonian? Sometimes there is an algorithmic measure, often not. The square of the Hamiltonian operator determines the energy variance H 2 L H 2 L = (H E) 2 L = Lσ 2 A universal measure for the quality of a variational wavefunction lower bound for the energy: there is always an eigenstate within σ of E We can easily construct an MPO representation of H 2 Ian McCulloch (UQ) imps 7 Dec / 33

22 Spin 1/2 Heisenberg Model Energy per site scaling with variance (exact energy = -ln = ) E σ 2 Ian McCulloch (UQ) imps 7 Dec / 33

23 Example 2 - momentum distribution Momentum-dependent operators have a simple form, b k = x e ikx b x W b k ( e = ik I b I ) Momentum occupation: n(k) = 1 L b k b k Broken U(1) symmetry: b 0 hence n(k = 0) L (extensive) With U(1) symmetry: n(k = 0) is finite, but diverges with m in superfluid phase Ian McCulloch (UQ) imps 7 Dec / 33

24 1000 Bose-Hubbard Model N(k) Infinite 1D, one particle per site 100 U/J=3.2 U/J=3.6 U/J=4.0 U/J=4.8 U/J=6.4 N(k) k/pi Ian McCulloch (UQ) imps 7 Dec / 33

25 Higher moments It is straight forward to evaluate a local order parameter, eg M = i M i The first moment of this operator gives the order parameter, M = m 1 (L) It is also useful to calculate higher moments, eg M 2 = m 2 (L) or generally M k = m k (L) These are polynomial functions in the system size L. Ian McCulloch (UQ) imps 7 Dec / 33

26 For finite systems, the Binder cumulant of the order parameter cancels the leading-order finite size effects NL/ U L = 1 m4 3 m 2 2 L = 50 L = 100 L = 150 L = 200 idmrg finite size scaling Ω 0.8 L = 50 L = L = 150 L = 200 UL Ian McCulloch (UQ) imps 7 Dec / 33

27 Cumulant expansions Express the moments m i in terms of the cumulants per site κ j, m 1 (L) = κ 1 L m 2 (L) = κ 2 1 L2 + κ 2 L m 3 (L) = κ 3 1 L3 + 3κ 1 κ 2 L 2 + κ 3 L m 4 (L) = κ 4 1 L4 + 6κ 2 1 κ 2L 3 + (3κ κ 1κ 3 )L 2 + κ 4 L κ 1 is the order parameter itself κ 2 is the variance (related to the susceptibility) κ 3 is the skewness κ 4 is the kurtosis The cumulants per site κ k are well-defined for an imps Note: the cumulants are normally written such that they are extensive quantities Lκ k. Ian McCulloch (UQ) imps 7 Dec / 33

28 imps for two-component bose gas The cumulant expansion already gives a lot of information κ 1 is the order parameter itself Ising transition in 2-component Bose gas Order parameter N_a - N_b κ Ω Ian McCulloch (UQ) imps 7 Dec / 33

29 The second cumulant gives the susceptibility 2-component Bose gas Second cumulant (susceptibility) κ Ω Different to a finite-size scaling, the susceptibility exactly diverges at the critical point. Sufficiently close to the critical point, it looks mean-field-like (so will generally give the wrong exponent!) Ian McCulloch (UQ) imps 7 Dec / 33

30 The fourth cumulant changes sign at the transition. 2-component Bose gas fourth cumulant 0-2e+05 κ 4-4e+05-6e+05-8e Ω Ian McCulloch (UQ) imps 7 Dec / 33

31 Binder Cumulant for imps Naively taking the limit L for the Binder cumulant doesn t produce anything useful: if the order parameter κ 1 0, U L = 1 m4 L 3 m 2 2 L 2 3 if κ 1 = 0, then m 4 (L) = 3k 2 2 L2 + k 4 L Hence U L = 1 3k2 2 L2 + k 4 L 3k 2 2 L2 0 Finally, a step function that detects whether the order parameter is non-zero Better approach, in the spirit of finite-entanglement scaling: Evaluate the moment polynomial using L correlation length Ian McCulloch (UQ) imps 7 Dec / 33

32 Transverse field Ising model Binder cumulant, scale factor s=5 M Order parameter m=4 m=8 m= λ U m m=4 m=5 m=6 m=7 m=8 m=9 m=10 m=11 m= λ Ian McCulloch (UQ) imps 7 Dec / 33

33 String parameters Order parameters do not have to be local Mott insulator string order parameter O 2 P = lim j i Πj k=i ( 1)nk We can write this as a correlation function of kink operators, p i = Π k<i ( 1) nk This turns the string order into a 2-point correlation function: O 2 P = lim p i p j j i Or as an order parameter: P = i p i Then O 2 p = 1 L 2 P 2 Ian McCulloch (UQ) imps 7 Dec / 33

34 Real example: 3-leg Bose-Hubbard model with flux phase (F. Kolley, M. Piraud, IPM, U. Schollwoeck, F. Heidrich-Meisner, New J. Phys. 17 (2015) ) For density n = 1/3 (one particle per rung), near flux φ π, there is a transition from a Mott to critical as a function of J P has a simple MPO representation ( ) e iπn I P = I Hence we can calculate higher moments of P. Ian McCulloch (UQ) imps 7 Dec / 33

35 Bose-Hubbard Ladder String order parameter O p m=100 m=150 m=200 m=300 m=400 m=500 m= Jperp Ian McCulloch (UQ) imps 7 Dec / 33

36 ξ Jperp = 0.99 Jperp = 1.00 Jperp = 1.01 Jperp = 1.02 Jperp = 1.03 Jperp = 1.04 Jperp = 1.05 Jperp = 1.06 Jperp = 1.10 Jperp = 1.14 Jperp = 1.25 Bose=Hubbard Ladder Scaling of correlation length m Ian McCulloch (UQ) imps 7 Dec / 33

37 Bose-Hubbard Ladder String parameter Binder cumulant U ξ m=200 m=250 m=300 m=350 m=400 m=450 m=500 m=550 m= Jperp Ian McCulloch (UQ) imps 7 Dec / 33

38 Summary Download the code: finite entanglement scaling higher moments Binder cumulant Ian McCulloch (UQ) imps 7 Dec / 33