Lanczos Algebra

m

Preliminaries

Krylov subspaces

Iteration scheme

Application: S=1 BAHC

Application: S=1/2 XYchain

Application: S=1/2 Ising intransversal field (ITF)

Lanczos Algorithm: Theory and Aplications

J. Almeida1,2

1Department of Theoretical Physics, University of Ulm (Ulm, Germany)2 these notes are largely based on the book "Matrix Computations", by G. H. Gollub and C. F. Van

Loan

York (United Kingdom), April 2012

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Preliminaries

Krylov subspaces

Iteration scheme




Outline

1 Preliminaries

2 Krylov subspaces

3 Iteration scheme

4 Application: S=1 Bond-alternating Heisenberg chain

5 Application: S=1/2 XY-Heisenberg chain

6 Application: S=1/2 ITF

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Preliminaries

Krylov subspaces

Iteration scheme




Eigensystem solution in many-body systems

Problem: Huge Hilbert spaces in many-body systems

Sztotal S = 3

2 S = 1 S = 12

0 1.703.636 73.789 9241 1.650.792 69.576 7922 1.501.566 58.278 4953 1.281.280 43.252 2204 1.024.464 28.314 665 766.272 16.236 126 534.964 8.074 17 347.568 3.4328 209.352 1.2219 116.336 352

10 59.268 7811 27.456 1212 11.440 113 4.22414 1.35315 36416 7817 1218 1

Table: Dimension of sectors with well defined total z-axis spin projection in aL = 12 spin chain. Note: an 8-byte representation of a 11585× 11585 matrixoccupies 1GB of memory.

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Preliminaries

Krylov subspaces

Iteration scheme




Lanczos algorithm: Preliminaries

Theorem (Courant-Fischer Minimax Theorem):

If A ∈ Rn×n is symmetric then, for k = 1 . . . n:

λk(A) = maxdim(S)=k

min0 6=y∈S

yT AyyT y

(1)

with λn(A) ≤ λn−1(A) ≤ · · · ≤ λ2(A) ≤ λ1(A) the egenvalues of A.

Extremal Eigenvalues:

i/ k = 1:

λ1(A) = maxyT AyyT y

(2)

ii/ k = n:

λn(A) = minyT AyyT y

(3)

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Preliminaries

Krylov subspaces

Iteration scheme




Lanczos algorithm: Preliminaries

Rayleigh quotient

r(x;A) =xT AxxT x

x 6= 0 (4)

Rewritting expressions above

λ1(A) = max‖y‖=1

r(y;A) (5a)

λn(A) = min‖y‖=1

r(y;A) (5b)

where y ∈ Rn and n is the dimension of the matrix A.

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Preliminaries

Krylov subspaces

Iteration scheme




Lanczos algorithm: preliminaries

Eigenvalues in a truncated space:

Let {qi} ⊆ Rn be a sequence of orthonormal vectors and the matrixQj ≡ [q1, . . . , qj].

Mj ≡ λ1(QTj AQj) = max

‖y‖=1

yT(QTj AQj)y

yT y= (6a)

= max‖y‖=1

(Qjy)T A(Qjy)(Qy)T(Qy)

= max‖y‖=1

r(Qjy;A)

mj ≡ λj(QTj AQj) = min

‖y‖=1r(Qjy;A) (6b)

Extremal approximation of the eigenvalues:

Notice that y ∈ Rj and Qjy ⊂ Rn. Taking into account the definitionsabove,

mj ≥ λn(A), Mj ≤ λ1(A).

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Preliminaries

Krylov subspaces

Iteration scheme




Lanczos algorithm: preliminaries

Eigenvalues in a truncated space:

Let {qi} ⊆ Rn be a sequence of orthonormal vectors and the matrixQj ≡ [q1, . . . , qj].

Mj ≡ λ1(QTj AQj) = max

‖y‖=1

yT(QTj AQj)y

yT y= (7a)

= max‖y‖=1

(Qjy)T A(Qjy)(Qy)T(Qy)

= max‖y‖=1

r(Qjy;A)

mj ≡ λj(QTj AQj) = min

‖y‖=1r(Qjy;A) (7b)

Extremal approximation of the eigenvalues:

The lanczos algorithm can be derived by considering how togenerate the qj vectors so that mj and Mj are increasingly betterestimates of λn(A) and λ1(A).

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Preliminaries

Krylov subspaces

Iteration scheme




Lanczos algorithm: Krylov subspaces

Let

uj ∈ span{q1, . . . , qj} such that Mj = r(uj;A) (8a)

vj ∈ span{q1, . . . , qj} such that mj = r(vj;A) (8b)

We can make Mj+1 > Mj and mj+1 < mj if qj+1 holds that

∇r(uj;A) ∈ span{q1, . . . , qj+1} (9a)

−∇r(vj;A) ∈ span{q1, . . . , qj+1} (9b)

With the gradient of the Rayleigh quotient given by

∇r(x;A) =2

xT x(Ax− r(x;A)x) (10)

Krylov subspace: definition

Conditions (9) are then both satisfied provided that

span{q1, . . . , qj} = span{q1,Aq1, . . . ,Aj−1q1} ≡ K(A, q1, j) (11)

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Iteration scheme




Lanczos algorithm: Krylov subspaces

Krylov matrix: definition

We define the Krylov matrix of subspace K(A, q1, j) as

K(A, q1, n) ≡ [q1,Aq1,A2q1, . . . ,An−1q1]

QR decomposition of a Krylov subspace

Let Q be an orthornormal matrix such that Qe1 = q1 and T ≡ QT AQis tridiagonal. Substituting above gives

K(A, q1, n) = Q · [e1, Te1, T2e1, . . . , Tn−1e1]

= Q ·

1 • • • •0 • • • •0 0 · · · • •0 0 0 • •0 0 0 0 •

And hence,

range K(A, q1, n) = range Q

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Iteration scheme




Lanczos algorithm: recapitulation of main ideas

First idea

We have seen that the maximal eigevalues of a certainsymmetric matrix A restricted to the Krylov subspace K(A, q1, j) givebetter estimates upon increasing the dimension j (since it ’follows’the gradient of the Rayleigh quotient)

In general, the operations of truncation and diagonalisation of thematrix A in this new space are numerically costly.

Second idea

We have seen however that in principle it is possible to find anorthonormal basis of the Krylov space K(A, q1, j) such that thematrix A restricted to this space and written in this basis has atridiagonal form: this shots with the same stone the costs ofobtaining iteratively better estimates of the original eigensystem.

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Preliminaries

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Iteration scheme




Lanczos algorithm: iteration equations

By definition

T = QT AQ ≡

α1 β1 · · · 0

β1 α2. . .

.... . .

. . .. . .

.... . .

. . . βj−10 · · · βj−1 αn

with

{αi = qT

i Aqi

βi = qTi+1Aqi

By construction

QA = TQ⇐⇒ [Aq1, Aq2, . . . , Aqn] =

= [α1q1 + β1q2, β1q1 + α2q2 + β2q3, . . . , βn−1qn−1 + αnqn]

q2 = 1/β1(A− α1)q1

q3 = 1/β2{(A− α2)q2 − β1q1

}· · ·qk = 1/βk−1

{(A− αk−2)qk−1 − βk−2qk−2

}, 2 < k < n

0 = (A− αn)qn − βn−1qn−1

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Preliminaries

Krylov subspaces

Iteration scheme




Lanczos algorithm: iteration equations

Lanczos iteration

β0 = ‖r0‖; q0 = 0; j = 0

while(βj 6= 0) (12)

qj+1 = rj/βj;

j = j + 1;

αj = qTj Aqj;

rj = (A− αjI)qj − βj−1qj−1;

βj = ‖rj‖end

Begin with an arbitary vector r0 and compute β0

When j = 0 if β0 6= 0 we compute q1When j = 1 we compute α1, r1, β1 and if β1 6= 0 we also compute q2When j = 2 we compute α2, r2, β2 and if β2 6= 0 we also compute q3When j = 3 we compute α3, r3, β3 and if β3 6= 0 we also compute q4When j = 4 we compute α4, r4, β4 and if β4 6= 0 we also compute q5· · ·

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Iteration scheme




Lanczos algorithm: a simplest example

Matlab(R) code% Create a random symmetric matrixD=200for i=1:D,for j=1:i,A(i,j)=rand;A(j,i)=A(i,j);

endend

% Iteration with j=0r0=rand(D,1);b0=sqrt(r0’*r0);q1=r0/b0;a1=q1’*A*q1

% Iteration with j=1r1=A*q1-a1*q1;b1=sqrt(r1’*r1)q2=r1/b1;a2=q2’*A*q2

% Iteration with j=2r2=A*q2-a2*q2-b1*q1;b2=sqrt(r2’*r2)q3=r2/b2;a3=q3’*A*q3

% Create matrix QQ=[q1 q2 q3];% Check orthogonalityEYE=Q’*Q% Check the tridiagonal truncationT=Q’*A*Q

Outputoctave:6> lanczosD = 200a1 = 3.6694b1 = 1.2412a2 = 0.18896b2 = 0.82261a3 = -0.38889EYE =

1.0000 0.0 0.00.0 1.0000 0.00.0 0.0 1.0000

T =

3.6694 1.2412 0.01.2412 0.1889 0.8226

0.0 0.8226 -0.3888

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Iteration scheme




Lanczos algorithm: othogonality and termination of theiteration

Theorem: othogonality and termination

Theorem: Let A ∈ Rn×n be symmetric and assume q1 ∈ Rn has unit2-norm. Then the Lanczos iteration (12) runs until j = m wherem = rank K(A, q1, n).

Moreover, for j = 1 : m we have that Qj = [q1, . . . , qj] hasorthonormal columns that span K(A, q1, j).

Corollary: βj = 0⇐⇒ span K(A, q1, j + 1) ⊆ span K(A, q1, j)

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Iteration scheme




Lanczos algorithm: headers of a complete Lanczos routine

Restart mode routine

void findEigenvectors_restartMode(int nev, int ncv, int& nit, double tol,const base_Obj& seed,vector<double>& evalue,vector<base_Obj*>& evector,double& min_beta, double& err,double zero_modes_shift=0.0 )

Low-level restart mode routine

void findEigenvectors(int nev, int ncv, int& nit, double tol,const base_Obj& seed,vector<double>& evalue,vector<base_Obj*>& evector,double& min_beta, double& err,int target_nev=-1,vector<double> rm_evalue=vector<double>(0),vector<base_Obj*> rm_evector=vector<base_Obj*>(0),double zero_modes_shift=0.0)

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Preliminaries

Krylov subspaces

Iteration scheme



Application: S=1/2 Ising intransversal field (ITF) Application 1

S=1 Bond-alternated Heisenberg chain:measuring the hidden AKLT order

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Preliminaries

Krylov subspaces

Iteration scheme




Example: S = 1 Bond-alternated Heisenbergantiferromagnetic chain

System Hamiltonian

H =

L−1∑i=1

(1− (−1)iγ

)SiSi+1

The ground states of this model are particular realizations of the valencebond solids (VBS) as described by Affleck, Kennedy, Lieb and Tasaki(see next slide).

Critical properties and phases

Location of critical point:NLσM estimate: γc = 1/2Numerical estimates: γc ' 0.259

Phases of the model:γ < γc (1,1)-VBS (Haldane)γ > γc (2,0)-VBS (Dimer)

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Iteration scheme




AKLT Physics

AKLT Hamiltonian

HAKLT =

L−1∑i=1

SiSi+1 + β(SiSi+1)2, β = − 1

3

The exact ground state of this model can be expressed using anextended S = 1/2 Hilbert space and is denoted as m = 1, n = 1 valencebond solid (i.e, a (1,1)-VBS).

Properties of the ground state

The AKLT antiferromagnet has continuous SU(2) symmetry,exponentially decaying correlations, a gap and a unique infinitevolume ground state.

with OBC there exist four orthogonal ground states (1 singlet, 1triplet).

with PBC the ground state is unique (1 singlet).

All these ground states converge to the same one in the infinitevolume limit.

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Iteration scheme




AKLT Physics

Hidden order of the ground state

In the usual z-axis projection basis, the AKLT ground state reveals ahidden antiferromagnetic order:

|ψ0〉 = C1| . . . 0+0−0+−+000−. . . 〉+C2| . . . 00+−+00−0+0−. . . 〉+. . .

We say it is topological order because:It is conditioned by the configuration of the degrees of freedom at

the endsIt can not be detected by means of local operators

String Order Parameter

The AKLT order can bedetected with a non local orderparameter mapped from theoryof preroughening surfacetransitions:

Ostr = lim|i−j|→∞

⟨S z

i

j−1∏k=i

eiπS zk S z

j

⟩

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Iteration scheme




Example: exactly solvable XY model

Computing the String Order Parameter

0.0

0.1

0.2

0.3

0.4

0.0 0.1 0.2 0.3 0.4

SO

P

γ

S=1 Bold-Alternating Heisenberg Chain (OBC)

DMRG, L=350

DMRG, L=90

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

γ

SO

P

S=1 Bond−alternating Heisenberg chain (PBC)

Lanczos, L=14

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Krylov subspaces

Iteration scheme




S=1/2 Heisenberg XY chain: an integrablesystem

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Iteration scheme





Hamiltonian (with Periodic Boundary Conditions)

HXY = −12

L−1∑j=1

(σxi σ

xi+1 + σy

i σyi+1)− (σx

1σxL + σy

1σyL)

= −σ+1 σ−L − σ

−1 σ

+L +

L−1∑j=1

{− σ+

i σ−i+1 − σ

−i σ

+i+1

}

with σ±j = 12 (σ

xj ± iσy

j )

Jordan-Wigner transformation

HXY = −L−1∑j=1

(C†i Ci+1 + CiC†i+1)− (−1)NF+1(C†1 CL + C1C†L)

with C±m =

m−1∏j=1

(−σzj )σ±m , and NF the total number of fermions

.

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Preliminaries

Krylov subspaces

Iteration scheme





Fourier-Transforming Hamiltonian

aq =1√

L

L∑m=1

eiqmCm, Cm =1√

L

∑q

e−iqmaq

L−1∑j=1

C†j Cj+1 =1L

L−1∑j=1

∑qq′

a†q aq′e−i(q′−q)je−iq′

L−1∑j=1

C†j+1Cj =1L

L−1∑j=1

∑qq′

a†q aq′ei(q−q′)jeiq

±(C†1 CL + C1C†L) =1L

∑qq′

(a†q aq′eiqLe−iq′ e−iq′L︸︷︷︸±1

+a†q aq′eiqe−iq′L eiqL︸︷︷︸±1

)

with the choice of momentum boundary conditions:{if NF odd eiqL = +1⇒ q = 2π

L kif NF even eiqL = −1⇒ q = 2π

L k + πL

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Iteration scheme





Diagonal form of the Hamiltonian

HXY = −1L

∑qq′

a†q aq′

e−iq′( L−1∑

j=1

e−i(q′−q)j + e−i(q′−q)L)

+

+ eiq( L−1∑

j=1

ei(q−q′)j + ei(q−q′)L) = −2

∑q

cos(q)a†q aq

Ground state energy of a XY-spin chain with PBC

The lowest energy configuration corresponds exactly to L/2fermions with momenta in the semicircle with positive cosine.

Since boundary conditions are periodic (in contrast with anti-periodic) we have to consider L even and hence momenta of theform q = 2πk/L + π/L.

The momenta that fulfill at the same time the conditions above canbe parametrized as q̃k = −π/2 + (2k − 1)π/L, with k = 1 . . . L/2

Hence, E0XY = −2

L/2∑k=1

cos(q̃k)a†q aq =−2

sin(π/L)

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S=1/2 XY model ground state energy

L E0XY Time (s)

10 -6.4721359 0.0212 -7.7274066 0.0614 -8.9879184 0.1116 -10.2516618 0.3718 -11.5175410 1.7120 -12.7849064 8.5722 -14.0533483 35.3324 -15.3225952 158.02

Table: Computations carried out in an Intel(R) Xeon(R)X5570 @ 2.93 GHz processor

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Iteration scheme




S=1/2 Ising in transversal field: a mostparadigmatic model

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Krylov subspaces

Iteration scheme




Example: S = 1 Bond-alternated Heisenbergantiferromagnetic chain

System Hamiltonian

H =

L−1∑i=1

(1− (−1)iγ

)SiSi+1

Location of critical point:NLσM estimate: γc = 1/2Numerical estimates: γc ' 0.259

Phases of the model:γ < γc (1,1)-VBS (Haldane)γ > γc (2,0)-VBS (Dimer)

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Magnetic field (x−axis)

Ga

p (

E1

−E

0)

Ferromagnetic ITF Model

L=10

L=20

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

Magnetic field (x−axis)z−

axis

to

tal m

ag

ne

tiza

tio

n


L=10

L=12

L=14

L=16

L=18

L=20

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Preliminaries

Krylov subspaces

Iteration scheme





Ground state energy computed with Lanczos

0 0.5 1 1.5 20

50

100

150

200

Magnetic field (x−axis)

S(q

=0

) (a

. u

.)


Lanczos, L=20

0 0.2 0.4 0.6 0.8 10

50

100

150

200Ferromagetic ITF model

q/π

S(q

)

h=0.0

h=0.8

h=1.0

h=1.2

h=1.4

h=1.6

Lanczos, L=20

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Lanczos Algebra

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