dmrg in the thermodynamic limit · ian mcculloch (uq) idmrg 23/6/2015 7 / 54. matrix product...

DMRG in the Thermodynamic limitWorkshop and Symposium on DMRG Technique for Strongly Correlated

Systems in Physics and Chemistry

Ian McCulloch

University of QueenslandCentre for Engineered Quantum Systems (EQuS)

23/6/2015

Ian McCulloch (UQ) iDMRG 23/6/2015 1 / 54

Outline

1 Constructing Matrix Product States - alternative viewsFrom classical to quantum statesSequential generation

2 Matrix Product Operators

3 Infinite size DMRG

4 Broken symmetries

5 Scaling relations in the thermodynamic limit

6 Expectation values of iMPO’sHigher momentsBinder cumulant

7 Infinite Boundary Conditions

Method 1: quantize a classical state

Start from a classical (product) state

|ψ〉 = |s1〉 |s2〉 |s3〉 |s4〉 · · ·

Each |si〉 is a classical vector, with real (or c-number) coefficients in somebasis

|si〉 = axi |x〉+ ay

i |y〉+ azi |z〉

Turn our (commuting) numeric coefficients into a matrix

|si〉jk = Axjk|x〉+ Ay

jk|y〉+ Azjk|z〉

We can recover an amplitude at the end by taking the trace, or arranging thatthe boundary matrices are 1× D and D× 1.

|ψ〉 = Tr∑

As1 As2 As3 As4 · · ·|s1〉 |s2〉 |s3〉 |s4〉 · · ·

Method 1: quantize a classical state

Start from a classical (product) state

|ψ〉 = |s1〉 |s2〉 |s3〉 |s4〉 · · ·

Each |si〉 is a classical vector, with real (or c-number) coefficients in somebasis

|si〉 = axi |x〉+ ay

i |y〉+ azi |z〉

Turn our (commuting) numeric coefficients into a matrix

|si〉jk = Axjk|x〉+ Ay

jk|y〉+ Azjk|z〉

We can recover an amplitude at the end by taking the trace, or arranging thatthe boundary matrices are 1× D and D× 1.

|ψ〉 = Tr∑

As1 As2 As3 As4 · · ·|s1〉 |s2〉 |s3〉 |s4〉 · · ·

Method 2: quantum finite-state machines

What is a Matrix Product State?

Another way to visualizing them (from Greg Crosswhite)

A finite-state machineis a model of a systemthat can transitionbetween a finitenumber of states.

A classical finite-state machine is always in one discrete state.

In a quantum finite-state machine, we choose every possible transition withsome probability amplitude

(from Crosswhite and Bacon, Phys. Rev. A 78, 012356 (2008))

|ψ〉 =

| ↑〉| ↓〉

|ψ〉 =

| ↑↑〉| ↓↑〉+ | ↑↓〉

|ψ〉 =

| ↑↑↑〉| ↓↑↑〉+ | ↑↓↑〉+ | ↑↑↓〉

|ψ〉 = | ↓↑↑↑〉+ | ↑↓↑↑〉+ | ↑↑↓↑〉+ | ↑↑↑↓〉

Matrix Product StatesThis quantum finite-state machine has a transition matrix associated with it

W-state

|ψ〉 =1√N

(| ↓↑↑↑ . . .〉+ | ↑↓↑↑ . . .〉+ | ↑↑↓↑ . . .〉+ . . .)

(| ↑〉 0| ↓〉 | ↑〉

)Practically all prototype wavefunctions studied in quantum information have alow-dimensional MPS representation

GHZ state – long-range entangled, S = ln 2

|ψ〉 =1√2

(| ↑↑↑ . . .〉+ | ↓↓↓ . . .〉)

(| ↑〉 00 | ↓〉

)AKLT state

( √1/3|0〉 −sqrt2/3|+〉√2/3|−〉 −

√1/3|0〉

)Ian McCulloch (UQ) iDMRG 23/6/2015 6 / 54

W-state

|ψ〉 =1√N

(| ↓↑↑↑ . . .〉+ | ↑↓↑↑ . . .〉+ | ↑↑↓↑ . . .〉+ . . .)

(| ↑〉 0| ↓〉 | ↑〉

|ψ〉 =1√2

(| ↑↑↑ . . .〉+ | ↓↓↓ . . .〉)

(| ↑〉 00 | ↓〉

)AKLT state

( √1/3|0〉 −sqrt2/3|+〉√2/3|−〉 −

√1/3|0〉

W-state

|ψ〉 =1√N

(| ↓↑↑↑ . . .〉+ | ↑↓↑↑ . . .〉+ | ↑↑↓↑ . . .〉+ . . .)

(| ↑〉 0| ↓〉 | ↑〉

|ψ〉 =1√2

(| ↑↑↑ . . .〉+ | ↓↓↓ . . .〉)

(| ↑〉 00 | ↓〉

)AKLT state

( √1/3|0〉 −sqrt2/3|+〉√2/3|−〉 −

√1/3|0〉

The Matrix Product Ansatz: beyond groundstatesThe key advantage of MPS formulation: arithmetic manipulationsThe sum (superposition) of two matrix product states is also a matrix productstate

|C〉 = |A〉+ |B〉

C =∑si

Tr Cs1 Cs2 . . .CsL |s1s2 . . . sL〉

whereCsi = Asi ⊕ Bsi

The dimension of the matrices increases: dim(C) = dim(A) + dim(B)Action of an operator on a state: if the operator is a product of local terms:

O = O1 ⊗ O2 ⊗ . . .

|C〉 = O|A〉 is a Matrix Product State, with

Csi =∑

Osi,s′ii As′i

The Matrix Product Ansatz: beyond groundstatesThe key advantage of MPS formulation: arithmetic manipulationsThe sum (superposition) of two matrix product states is also a matrix productstate

|C〉 = |A〉+ |B〉

C =∑si

Tr Cs1 Cs2 . . .CsL |s1s2 . . . sL〉

whereCsi = Asi ⊕ Bsi

The dimension of the matrices increases: dim(C) = dim(A) + dim(B)Action of an operator on a state: if the operator is a product of local terms:

O = O1 ⊗ O2 ⊗ . . .

|C〉 = O|A〉 is a Matrix Product State, with

Csi =∑

Osi,s′ii As′i

Matrix Product OperatorsIPM J. Stat. Mech. P10014 (2007), arXiv:0804.2509

At each iteration we have a set of block operators, acting on them-dimensional auxiliary spaceIt is natural to use a Matrix Product approach to constructing the blockoperators used in DMRG

Ising model H =∑<i,j>

zj + λ

Sxi , adding a site to the block:

(identity operator) I → I ⊗ Ilocal(z-spin acting on right-most site) Sz → I ⊗ Sz

local(block Hamiltonian) H → λI ⊗ Sx

local + Sz ⊗ Szlocal + H ⊗ Ilocal

In matrix form:

(H Sz I)′︸︷︷︸new block operators

= (H Sz I)︸︷︷︸old block operators

λSx Sz I

︸︷︷︸

Matrix Product OperatorsThis form can represent many operators

fermionic c†k=0: Wc†k=0=

(Ic† P

), P = (−1)N , J-W string

finite momentum b†k : Wb†k=

b† eikI

)Advantages of the MPO representation: arithmetic operations!

H1 + H2 direct sum of the MPO representationsH1 × H2 direct product of the MPO representations

also derivatives, etcThis preserves the lower triangular form.Can we evaluate an expectation value of an MPO in the thermodynamic limit?

〈A〉L = polynomial function of L

Examples:Energy: 〈H〉L = L εHamiltonian block operator matrix elements to restart a calculationSingle-mode approximation: 〈S−k HS+

k 〉L/〈S−k S+

k 〉LIan McCulloch (UQ) iDMRG 23/6/2015 9 / 54

Matrix Product OperatorsThis form can represent many operators

fermionic c†k=0: Wc†k=0=

(Ic† P

), P = (−1)N , J-W string

finite momentum b†k : Wb†k=

b† eikI

)Advantages of the MPO representation: arithmetic operations!

H1 + H2 direct sum of the MPO representationsH1 × H2 direct product of the MPO representations

also derivatives, etcThis preserves the lower triangular form.Can we evaluate an expectation value of an MPO in the thermodynamic limit?

〈A〉L = polynomial function of L

Examples:Energy: 〈H〉L = L εHamiltonian block operator matrix elements to restart a calculationSingle-mode approximation: 〈S−k HS+

k 〉L/〈S−k S+

k 〉LIan McCulloch (UQ) iDMRG 23/6/2015 9 / 54

DMRG in the infinite size limit (arxiv:0804.2509)

Infinite-size translationally invariant MPS

The “infinite size” DMRG algorithm has existed since the start (1992)It doesn’t produce a translationally invariant MPS fixed pointNo prescription for constructing the initial wavefunction at next iterationiTEBD produces a translationally invariant MPS, but for groundstatesimaginary time evolution is not so fast

DMRG in the infinite size limit (arxiv:0804.2509)

Infinite-size translationally invariant MPS

The “infinite size” DMRG algorithm has existed since the start (1992)It doesn’t produce a translationally invariant MPS fixed pointNo prescription for constructing the initial wavefunction at next iterationiTEBD produces a translationally invariant MPS, but for groundstatesimaginary time evolution is not so fast

A recurrence relation for MPS

Suppose we have an initial state:0

Suppose we also have the MPS enlarged with an extra unit cell:R

Note: ΛL and ΛR are not necessarily diagonal

Now we can insert one more unit cell:Λ1

Λ1 = ΛR Λ−10 ΛL

Variant of the finite system algorithmDifferent treatment of the boundaries

Λ23 = Λ21 Λ−101 Λ03

Λ43 = Λ41 Λ−121 Λ23

Λ45 = Λ43 Λ−123 Λ25

Broken symmetries

Finite size MPS: No broken symmetries (to O(truncation error))Infinite size MPS: The Ansatz can break all symmetries

even continuous symmetries in one dimension

How to understand this?

Matrix elements connecting symmetry sectors vanish as∼ exp(−L) → 0Continuous symmetries cannot break in exact 1D because theassociated goldstone modes would destroy the order parametercompletely (percolation threshold!)But if the goldstone modes are gapped due to finite basis size, thesymmetry can breakAlternatively: in order to get a finite correlation length we must perturb theHamiltonian with a relevant perturbation. No reason why that perturbationshould not break any (or all) symmetry.

Broken symmetries

Finite size MPS: No broken symmetries (to O(truncation error))Infinite size MPS: The Ansatz can break all symmetries

even continuous symmetries in one dimension

How to understand this?

Matrix elements connecting symmetry sectors vanish as∼ exp(−L) → 0Continuous symmetries cannot break in exact 1D because theassociated goldstone modes would destroy the order parametercompletely (percolation threshold!)But if the goldstone modes are gapped due to finite basis size, thesymmetry can breakAlternatively: in order to get a finite correlation length we must perturb theHamiltonian with a relevant perturbation. No reason why that perturbationshould not break any (or all) symmetry.

Prototypical example: Bose-Hubbard model

Ni(Ni − 1)− J∑<i,j>

b†i bj + b†j bi − µN

Mean-field approximation: β = 〈b〉

HMF =U2

Ni(Ni − 1)− Jβ∑

(b†i + bi)− µN

Mean field Hamiltonian breaks U(1) particle number conservationGroundstate is an m = 1 infinite MPS (product state!)

|ψ〉 = (|0〉+ a1|1〉+ a2|2〉 . . .)⊗L

An iMPS with no symmetries reduces to mean-fieldImposing quantum number symmetries reduces the quality of thevariational state (for fixed m)But usually worth the cost in computational efficiency

Prototypical example: Bose-Hubbard model

Ni(Ni − 1)− J∑<i,j>

b†i bj + b†j bi − µN

Mean-field approximation: β = 〈b〉

HMF =U2

Ni(Ni − 1)− Jβ∑

(b†i + bi)− µN

Mean field Hamiltonian breaks U(1) particle number conservationGroundstate is an m = 1 infinite MPS (product state!)

|ψ〉 = (|0〉+ a1|1〉+ a2|2〉 . . .)⊗L

An iMPS with no symmetries reduces to mean-fieldImposing quantum number symmetries reduces the quality of thevariational state (for fixed m)But usually worth the cost in computational efficiency

0 0.05 0.1 0.15 0.2 0.25J/U

m=10m=20m=30m=40m=50m=60m=70m=80

Bose-Hubbard Model Mott-Superfluid Transitionµ=0.25

0.219 0.2192 0.2194 0.2196 0.21981e-06

Correlation Functions

The form of correlation functions are determined by the eigenvalues of thetransfer operator

All eigenvalues ≤ 1One eigenvalue equal to 1,corresponding to the identityoperator

Expansion in terms of eigenspectrum λi:

〈O(x)O(y)〉 =∑

ai λ|y−x|i

50 100 150 200Number of states kept

(0,0) Singlet(1,0) Spin triplet(0,1) Holon Triplet(1/2,1/2) Single-particle

Hubbard Model transfer matrix spectrumHalf-filling, U/t = 4

64 128Number of states kept

(0,0) Singlet(1,0) Spin triplet(0,1) Holon Triplet(1/2,1/2) Single-particle

Hubbard model transfer matrix spectrumHalf-filling, U/t=4

Critical scaling exampleTwo-species bose gas with linear tunneling Ω, from F. Zhan et al, Phys. Rev. A 90, 023630 2014

50 100 200

Ω = 0.2148Ω = 0.215Ω = 0.2152Ω = 0.2154Ω = 0.2156Ω = 0.2158

mIan McCulloch (UQ) iDMRG 23/6/2015 19 / 54

CFT Parameters

For a critial mode, the correlation length increases with number of states m asa power law,

ξ ∼ mκ

[T. Nishino, K. Okunishi, M. Kikuchi, Phys. Lett. A 213, 69 (1996)M. Andersson, M. Boman, S. Östlund, Phys. Rev. B 59, 10493 (1999)L. Tagliacozzo, Thiago. R. de Oliveira, S. Iblisdir, J. I. Latorre, Phys. Rev. B 78, 024410 (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 useentropy scaling

Scaling dimensions

Suppose we have a two-point correlator that has a power-law at largedistances

〈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 m1 and m2

Correlation lengths ξ1 and ξ2

We expect:O(ξ2)

O(ξ1)=

for x large, we have: O(x) ' a λx (with ξ = −1/ lnλ)Prefactor a is overlap of operator O with next-leading eigenvector oftransfer operator

a ∝ ξ−∆

This gives directly the operator scaling dimensions by direct fit

Scaling dimensions

〈O(x)O(y)〉 = |y− x|−2∆

We expect:O(ξ2)

O(ξ1)=

a ∝ ξ−∆

Scaling dimensions

〈O(x)O(y)〉 = |y− x|−2∆

We expect:O(ξ2)

O(ξ1)=

a ∝ ξ−∆

0.0078125 0.015625transfer matrix eigenvalue 1 - λ = 1 / ξ

0.03125

0.0625

r of t

iDMRG data for m=15,20,25,30,35y = 0.45126 * x^0.480

Heisenberg model fit for the scaling dimension

Generalized Scaling

Alternative viewpoint (Vid Stojevic et al, Phys. Rev. B 91, 035120 (2015))

Scaling relation for large s:

O(sξ) ' aλsξ = aλ−s/ lnλ = ae−s

So we obtain ∆ by scaling O(sξ) versus sξ

However, this also works for s small, eg s 1,

O(sξ) ∝ (sξ)−∆

because for s 1, the correlation function is already (approximately)power-law.

the scaling relation works for any 0 < s <∞ !

But: O(sξ) ' aλsξ only for s 1

Generalized Scaling

O(sξ) ∝ (sξ)−∆

Generalized Scaling

O(sξ) ∝ (sξ)−∆

Expectation values of MPO’s - arxiv:0804.2509

We have seen that we can write many interesting operators in the form of amatrix 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

For a lower triangular MPO, this doesn’t work.

But we can make use of the triangular structure ISz

λSx Sz I

index by index, each component is a function only of the previouslycalculated terms

Expectation values of MPO’s - arxiv:0804.2509

We have seen that we can write many interesting operators in the form of amatrix 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

For a lower triangular MPO, this doesn’t work.

But we can make use of the triangular structure ISz

λSx Sz I

index by index, each component is a function only of the previouslycalculated terms

Choose bond indices i, j of Wij, and denote TWij

Example: ISz

λSx Sz I

Eigentensor is (E1 E2 E3)

Starting from E3:E3 = TI(E3) = I

is equivalent to the orthogonality condition - E3 is just the identityE2:

E2 = TSz(E3) = TSz(I) = Sz

E1: doesn’t reach a fixed point, E1 = E1(L) depends on the number ofiterations L

E1(L + 1) = TI(E1(L)) + TSz(E2) + TλSx (E3)= TI(E1(L)) + C

where C is a constant matrix, C = TSz(Sz) + λSx

Fixed point equations for E1:

E1(L + 1) = TI(E1(L)) + C

Eigenmatrix expansion of TI :

m2∑n=1

λn|λ〉〈λ|

giving

E(n)1 (L + 1) = λnE(n)

1 (L) + C(n)

Since λ1 = 1 by construction, this motivates decomposing intocomponents parallel and perpendicular to the identity:

E1(L) = E′1(L) + e1(L) I

where Tr E′1(L)ρ = 0

Component in the direction of the identity:

e1(L + 1) = e1(L) + Tr Cρ

Has the solutione1(L) = L Tr Cρ

is the energyComponent perpendicular to the identity:

E′1(L + 1) = TI(E′1(L)) + C′

where C′ = C − (Tr Cρ) I

E′1(L + 1)n = λnE′1(L)n + C′n

Since all |λn| < 1 here, this is a geometric series that converges to a fixedpoint (independent of L),

(1− TI)(E′1) = C′

Linear solver for the unknown matrix E′1

Component in the direction of the identity:

e1(L + 1) = e1(L) + Tr Cρ

Has the solutione1(L) = L Tr Cρ

is the energyComponent perpendicular to the identity:

E′1(L + 1) = TI(E′1(L)) + C′

where C′ = C − (Tr Cρ) I

E′1(L + 1)n = λnE′1(L)n + C′n

Since all |λn| < 1 here, this is a geometric series that converges to a fixedpoint (independent of L),

(1− TI)(E′1) = C′

Linear solver for the unknown matrix E′1

Summary:

Decompose eigentensor into components parallel and perpendicular tothe identityThe component parallel to the identity is the energy per siteThe perpendicular components reach a fixed point and give theHamiltonian matrix elements

E3 = I Identity operatorE2 = Sz Sz block operatorE1(L) = E′ + Le1 Hamiltonian operator + energy per site

Generalization to arbitrary triangular MPO’sarXiv:1008.4667

At the ith iteration, we have

Ei(L + 1) = TWii(Ei(L)) +∑j>i

TWji(Ej(L))︸︷︷︸= C(L)

Basic idea:if Wii = 0, then Ei = C

if Wii 6= 0, then solve (1− TWii)(Ei) = 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 ofmagnitude 1If any eigenvalues are complex, then expand also in fourier modes

Generalization to arbitrary triangular MPO’sarXiv:1008.4667

At the ith iteration, we have

Ei(L + 1) = TWii(Ei(L)) +∑j>i

TWji(Ej(L))︸︷︷︸= C(L)

Basic idea:if Wii = 0, then Ei = C

if Wii 6= 0, then solve (1− TWii)(Ei) = 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 ofmagnitude 1If any eigenvalues are complex, then expand also in fourier modes

Examples 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

〈H2〉L − 〈H〉2L = 〈(H − E)2〉L = Lσ2

A universal measure for the quality of a variational wavefunctionlower bound for the energy: there is always an eigenstate within σ of E

We can easily construct an MPO representation of H2

0 1×10-6

2×10-6

3×10-6

4×10-6

5×10-6

6×10-6

-0.443148

-0.443147

-0.443146

-0.443145

-0.443144

-0.443143

-0.443142

ESpin 1/2 Heisenberg Model

Energy per site scaling with variance (exact energy = -ln 2 + 0.25 = -0.44314718056)

0 1×10-7

2×10-7

3×10-7

-0.44314720-0.44314715-0.44314710-0.44314705-0.44314700-0.44314695-0.44314690

Momentum distribution

Momentum-dependent operators have a simple form,

b†k =∑

eikxb†x

Wb†k=

b† eikI

Momentum occupation:

n(k) =1L

b†kbk

Broken U(1) symmetry: 〈b†〉 6= 0 hence n(k = 0) ∝ L (extensive)With U(1) symmetry: n(k = 0) is finite, but diverges with m in superfluidphase

-1 -0.5 0 0.5 1k/pi

U/J=3.2U/J=3.6U/J=4.0U/J=4.8U/J=6.4

Bose-Hubbard Model N(k)Infinite 1D, one particle per site

Higher moments

It is straight forward to evaluate a local order parameter, eg

M =∑

The first moment of this operator gives the order parameter,

〈M〉 = m1(L)

It is also useful to calculate higher moments, eg

〈M2〉 = m2(L)

or generally

〈Mk〉 = mk(L)

These are polynomial functions in the system size L.

For finite systems, the Binder cumulant of the order parameter cancels theleading-order finite size effects

UL = 1− 〈m4〉3〈m2〉2

0.18 0.2 0.22 0.24

L = 50L = 100L = 150L = 200iDMRGfinite size scaling

|〈∆NL/2〉|

0.18 0.2 0.22 0.24

L = 50L = 100L = 150L = 200

0.215 0.216

The 2-component Bose-Hubbard model, with a linear coupling betweencomponents, has an Ising-like transition from immiscible (small Ω) to miscible(large Ω).

H =∑

<i,j>,σ

b†i,σbj,σ +H.c.+ U∑i,σ

nσ(nσ−1)+ U12

n↑n↓+Ω∑<i,j>

b†i,↑bj,↓+H.c.

Cumulant expansions

Express the moments mi in terms of the cumulants per site κj,

m1(L) = κ1Lm2(L) = κ2

1L2 + κ2Lm3(L) = κ3

1L3 + 3κ1κ2L2 + κ3Lm4(L) = κ4

1L4 + 6κ21κ2L3 + (3κ2

2 + 4κ1κ3)L2 + κ4L

κ1 is the order parameter itselfκ2 is the variance (related to the susceptibility)κ3 is the skewnessκ4 is the kurtosis

The cumulants per site κk are well-defined for an iMPS

Note: the cumulants are normally written such that they are extensivequantities→ Lκk.

iMPS for two-component bose gas

The cumulant expansion already gives a lot of information

κ1 is the order parameter itself

0.1 0.12 0.14 0.16 0.18 0.2Ω

Ising transition in 2-component Bose gasOrder parameter |N_a - N_b|

The second cumulant gives the susceptibility

0.1 0.12 0.14 0.16 0.18 0.2Ω

2-component Bose gasSecond cumulant (susceptibility)

Different to a finite-size scaling, the susceptibility exactly diverges at thecritical point.Sufficiently close to the critical point, it looks mean-field-like (so will generallygive the wrong exponent!)

The fourth cumulant changes sign at the transition.

0.1 0.12 0.14 0.16 0.18 0.2Ω

-8e+05

-6e+05

-4e+05

-2e+05

2-component Bose gas

fourth cumulant

Binder Cumulant for iMPS

Naively taking the limit L→∞ for the Binder cumulant doesn’t produceanything useful:

if the order parameter κ1 6= 0,

UL = 1− 〈m4〉L

3〈m2〉2L→ 2

if κ1 = 0, then m4(L) = 3k22L2 + k4L

UL = 1− 3k22L2 + k4L3k2

2L2 → 0

Finally, a step function that detects whether the order parameter isnon-zero

Better approach, in the spirit of finite-entanglement scaling: Evaluate themoment polynomial using L ∝ correlation length

0.98 1 1.02λ

m=4m=5m=6m=7m=8m=9m=10m=11m=12

Transverse field Ising modelBinder cumulant, scale factor s=5

0.98 0.99 1 1.01 1.02λ

00.10.20.30.40.50.60.7

m=4m=8m=12

Order parameter

String parametersOrder parameters do not have to be local

Mott insulator string order parameter

O2P = lim

|j−i|→∞〈Πj

k=i(−1)nk〉

We can write this as a correlation function of ‘kink operators’,

pi = Πk<i (−1)nk

This turns the string order into a 2-point correlation function:

O2P = lim

|j−i|→∞〈 pi pj 〉

Or as an order parameter:P =

Then O2p = 1

L2 〈P2〉

Real example: 3-leg Bose-Hubbard model with flux phase (F. Kolley, M. Piraud,IPM, U. Schollwoeck, F. Heidrich-Meisner, in preparation)

For density n = 1/3 (one particle per rung), near flux φ ∼ π, there is atransition from a Mott to critical as a function of J⊥

P has a simple MPO representation

(II (−1)n

)Hence we can calculate higher moments of P.

1 1.1 1.2 1.3Jperp

m=100m=150m=200m=300m=400m=500m=800

Bose-Hubbard LadderString order parameter

1 1.1 1.2 1.30.001

100 1000m

1000ξ

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 LadderScaling of correlation length

0.98 1 1.02 1.04 1.06 1.08Jperp

m=200m=250m=300m=350m=400m=450m=500m=550m=600

Bose-Hubbard LadderString parameter Binder cumulant

Infinite boundary conditionsH.N. Phien, G. Vidal, IPM, Phys. Rev. B 86, 245107 (2012), Phys. Rev. B 88, 035103 (2013)(see also Zauner et al 1207.0862, Milsted et al Phys. Rev. B 155116 (2013))

Local perturbation to a translationally invariant state

Window (N sites)Left Right

Map infinite system onto a finite MPS, with an effective boundary

Key point: Even if the perturbation is correlated at long range, only thetensors at the perturbation are modifiedDecompose the Hamiltonian

H = HL + HLW + HW + HWR + HR

We can calculate HL and HR by summing the infinite series of terms fromthe left and rightAway from the perturbation the wavefunction is approximately aneigenstate, so

exp itHL ∼ I

and we don’t leave the Hilbert space of the semi-infinite strip

Spin-1 Heisenberg chain, S+ initial perturbation

60 80 100 120 140

window size = 60

Infinite boundaries

Resize the window

We can do better - why keep the size of the window fixed?Window expansion - incorporate sites from the translationally-invariant sectioninto the window

Criteria for expanding: is the wavefront near the boundary?(Calculate from the fidelity of the wavefunction at the boundary)

−80 −60 −40 −20 0 20 40 60 80

t = 0.7

t = 2.25

t = 5.8

t = 7.55

t = 9.35

t = 11.15

t = 12.95

t = 14.75

t = 16.6

t = 18.45

t = 20.3

t = 22.15

t = 24

〈Sz(x,t)〉

Expanding windowExpanding windowFixed window

Window contraction

Window contraction - incorporate tensors from the window into the boundaryContract the MPS and Hamiltonian MPO

WWWW =

Follow the wavefront

60 40 20 0

t = 1.75

t = 3.95

t = 6.15

t = 8.35

t = 10.5

t = 12.7

t = 15

t = 17.35

t = 19.8

t = 22.2

t = 24.45

〈Sz(x,t)〉

Moving window

Fixed window

Summary

iDMRG – efficient algorithm for obtaining translationally invariant iMPSMany quantities are natural for iMPS but difficult to calculate for finitesystems (eg correlation length)Finite-entanglement scaling – often easier than finite-size scalingBinder cumulant for detecting phase transitionslocal perturbations – Infinite Boundary Conditions

Future:Equations for the moment expansion have the same structure as theequations for perturbations and excitations (see Frank’s talk!)

Thanks:Fei Zhan, Greg Crosswhite, Phien Ho, Guifre Vidal, lots more...

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