techniques for translationally invariant matrix product states...2016/12/12 · techniques for...

Techniques for translationally invariant matrixproduct states

Ian McCulloch

University of QueenslandCentre for Engineered Quantum Systems (EQuS)

12/12/2016

Ian McCulloch (UQ) iMPS 12/12/2016 1 / 51

Outline

1 Scaling relations in the thermodynamic limit

2 Expectation values of iMPO’sHigher momentsBinder cumulant

3 Entanglement spectrum

4 Projective representations

5 Triangular Heisenberg model example

6 Binder ratio for symmetric states


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)〉 =∑

i

ai λ|y−x|i


50 100 150 200Number of states kept

0

0.2

0.4

0.6

0.8

1λ

(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

1

10

100

Cor

rela

tion

leng

th

(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

1

10

100

50 100 200

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

ξ

mIan McCulloch (UQ) iMPS 12/12/2016 6 / 51

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(Vid Stojevic, IPM et al, Phys. Rev. B 91, 035120 (2015))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)=

(ξ2

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 fitIan McCulloch (UQ) iMPS 12/12/2016 8 / 51

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

0.03125

0.0625

pref

acto

r of t

he sp

in o

pera

tor a

t thi

s mod

e

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

Heisenberg model fit for the scaling dimension


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

= λW

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

ji W

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 :

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 (e1) and perpendicular (E′1) to the identity:

E1(L + 1) = 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


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


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

〈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

σ2

-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


Example 2 - momentum distribution

Momentum-dependent operators have a simple form,

b†k =∑

x

eikxb†x

Wb†k=

(I

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

1

10

100

1000N

(k)

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 =∑

i

Mi

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

0.25

0.5

0.75

1

0.18 0.2 0.22 0.24

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

|〈∆NL/2〉|

Ω

0

0.2

0.4

0.6

0.8

0.18 0.2 0.22 0.24

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

UL

Ω

0.4

0.425

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

∑i

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

0

0.2

0.4

0.6

0.8

κ1

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

0

10

20

30

40

50

κ2

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

0

κ4

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

3

if κ1 = 0, then m4(L) = 3k22L2 + k4L

Hence

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

0.3

0.4

0.5

0.6

0.7

Um

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 m=4m=8m=12

Order parameter


Scaling functions

Critical exponentsExponent relationν ξ ∝ |t|−νβ 〈M〉 ∝ (−t)β

α C ∝ |t|αγ χ ∝ |t|γ

Finite-size scaling functions

UL = U0(t L1/ν)〈M〉 = L−β/νM0(t L1/ν)

Finite-entanglement scaling functions

Um = U0(t mκ/ν)〈M〉 = m−βκ/νM0(t mκ/ν)


-3 -2 -1 0 1 2

(λ−λc) m

κ/ν

0.2

0.3

0.4

0.5

0.6

0.7U

m

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

Binder Cumulant scaling function collapse

Fixes κ/ν=2


-3 -2.5 -2 -1.5 -1 -0.5 0

(λ−λc)κ/ν

0.3

0.4

0.5

0.6

0.7

<m

>κ

β/ν

Magnetization scaling function below λc

fixes β=1/8


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 =

∑i

pi

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, New J. Phys. 17 (2015) 092001)

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

P =

(II (−1)n

)Hence we can calculate higher moments of P.


1 1.1 1.2 1.3Jperp

0

0.2

0.4

0.6

0.8

Op

2

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

Bose-Hubbard LadderString order parameter

1 1.1 1.2 1.30.001

0.01

0.1

1


100 1000m

1

10

100

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

0.1

0.15

0.2

0.25

0.3

0.35

Uξ

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

Bose-Hubbard LadderString parameter Binder cumulant


Entanglement spectrum spectroscopy

Bipartite cut – reduced density matrix ρEntanglement Hamiltonian H = − ln ρReduced dimension – the entanglement Hamiltonian lives on theboundaryLabel states by global symmetries spin, transverse momentumDegeneracy structure provides information about symmetries

For SU(2) symmetry broken states:If the symmetry is allowed to break: bulk gapunbroken SU(2): Goldstone mode, separated band of excitations belowthe bulk gap


Symmetries

Infinite cylinder of circumference Ly

Map onto one-dimensional chain:

Global symmetries

Preserved in the MPS structure:internal symmetries, such asparticle number, spin, . . .

Broken by the real-space structure:spatial symmetries, such astranslation, reflection of thecylinder


density matrix as the eigenvector of the transfer operatorLeft eigenvectoridentity matrix

Right eigenvectordensity matrix

=ρ ρ

=

Symmetry-resolved density matrix

ρ ρ=’ ’

Eigenvalues of ρ′ are λieiθi

λi are the density matrix eigenvaluesθi is the corresponding symmetry angle up to a global phase


Allowed symmetries of the ES

Let U be a global symmetry of the wavefunction

U|ψ〉 = eiθ|ψ〉

What is the action of U on the reduced density matrix ρ?Bipartition:

|ψ〉 =∑

ab ψab |a〉 ⊗ |b〉U = UA ⊗ UB

Action on ρ is UAρU†A

UA is distinguishable only up to a global phaseUA is a projective representation of the symmetry


Projective representationsPhysical states are rays in Hilbert space

a|ψ〉, for a ∈ C represents the same physical state

Symmetry transformations that are ambiguous up to some scalar areprojective representationsAssuming a unitary representation, only the phase is unknown: a = eiθ.A projective representation is defined by:

UxUy = eiθ(x,y)Uxy

The function θ(x, y) defines the representationExample:

Z2 has no non-trivial (unitary) projective representationsZ2 × Z2 has has two projective representations

U2x = I

U2y = I

UxUy = ±UyUx


Triangular lattice – frustration

Frustrated lattice – suppress ordering

But ordering may be possibleGroundstate widely established as SU(2)-broken 120 order

Beyond nearest neighbor: H = J1∑〈i,j〉

~Si~Sj + J2∑〈〈i,j〉〉

~Si~Sj


Z2 spin liquid

The simplest spin liquid that doesn’t break time reversal symmetryElementary excitations are spinons with non-trivial braiding statistics(anyons)Four anyon types:

1 : vacuumb : bosonic spinon (electric charge e)v : vison (magnetic charge m)f : fermionic spinon (fermion ψ)

Fusion rules for combining excitations:b × b = 1v × v = 1f × f = 1b × v = fb × f = vv × f = b


Anyonic statistics

R matrix: counter-clockwise exchangeof particles

S matrix: intertwining two vacuumexcitations

Sab =1D

∑c

dc Tr(Rab

c Rbac

)Rbb

1 = Rvv1 = 1

Rff1 = (Rvb

f )2 = −1

In basis (1, b, v, f ):

S =12

1 1 1 11 1 −1 −11 −1 1 −11 −1 −1 1


Identifying the topologyGroundstate degeneracy on a torus

construct a vison or spinon that threads down the torusspinon flips between time-reversal symmetric andantisymmetric statesvison inserts −1 flux – antiperiodic boundary conditionsSymmetry fractionalization – these correspond toprojective symmetries

How to construct these fluxes?translationally invariant infinitesystem and manipulate theboundary conditions


Antiperiodic boundary conditions

Dihedral group symmetries around the cut:Ty – translation by one site around the cylinderRy – reflection about a site or bond

This forms the dihedral group D2×Ly

TLyy = R2

y = 1

RyTLy/2y R−1

y = TLy/2y

(a π rotation commutes with a reflection)

The dihedral group has a non-trivial projective representation:

TLyy = R2

y = 1

RyTyR−1y = −T−1

y

(a π rotation anti-commutes with a reflection)This is the subgroup of D2×2×Ly taking only the odd momenta (antiperiodic BC)


Antiperiodic boundary conditionsNariman Saadatmand, IPM, Phys. Rev. B 94, 121111 (2016)

We see two (nearly) degenerate statessmaller Ly: the groundstate has a 2-fold degenerate entanglementspectrumfor larger Ly, we obtained a non-degenerate entanglement spectrum

In the two-fold degenerate sector, we measure

〈RyTLy/2y R†yTL/2†

y 〉 = −0.9999970 · · ·

-5π/6 -3π/6 -π/6 π/6 3π/6 5π/6

k, momenta

0

2.5

5

7.5

10S=0S=1S=2

Ly = 6 entanglement spectrumIan McCulloch (UQ) iMPS 12/12/2016 46 / 51

Detecting ordered phases

J20.105(5)

only for YC6: algebraic

0.140(5) 0.200(5)0.170(5)

topological spin liquids

columnar order columnar

order columnar

order 120° order

Using SU(2) symmetry the magnetic order parameter is identically zero.can we see signatures of symmetry breaking in higher moments of theorder parameter?Binder cumulant U4 = 1− 〈M4〉

3〈M2〉Cumulant expansion of the 4th moment:

〈M4〉L = κ41L4 + 6κ2κ

21L3 + (4κ1κ3 + 3κ2

2)L2 + κ4L

Symmetry implies all odd cumulants vanish

〈M4〉L = 3κ22L2 + κ4L

Need L length scale – use correlation length ξ from transfer matrixBinder ratio

BR =κ4

κ22ξ


Binder ratio – 120 order

0.06 0.08 0.1 0.12J

2

-5

-4

-3

-2

-1

0

1B

inder

Rati

os o

f M

tri

m=800

m=900

m=1000

m=1200

SL v-sector

120° order

YC6


Binder ratio – staggered order

0.1 0.12 0.14 0.16 0.18 0.2J

2

-40

-30

-20

-10

0B

ind

er R

atio

s o

f M

stag

m=1200m=1400m=1600m=1800m→∞

columnar order

SL b-sector

YC10


Binder ratio – staggered order YC6

0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24J

2

-20

-10

0

10

20

30

40B

ind

er R

atio

s o

f M

stag

m=800m=900m=1000m=1200m→∞

500 1000 1500 2000m

0

20

40

SL v-sectorstripedorder

algebraicSL

stripedorder

YC6


SummaryDownload the code: https://people.smp.uq.edu.au/IanMcCulloch/mptoolkit

finite entanglement scalinghigher momentsBinder cumulantscaling functionssymmetries of the entanglement spectrumBinder ratio for symmetric states


techniques for translationally invariant matrix product states...2016/12/12 · techniques for...

Documents