m Preliminaries Krylov subspaces Iteration scheme Application: S=1 BAHC Application: S=1/2 XY chain Application: S=1/2 Ising in transversal field (ITF) Lanczos Algorithm: Theory and Aplications J. Almeida 1,2 1 Department 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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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
m
Preliminaries
Krylov subspaces
Iteration scheme
Application: S=1 BAHC
Application: S=1/2 XYchain
Application: S=1/2 Ising intransversal field (ITF)
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.
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: 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)
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: 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.
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: 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).
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: 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).
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: 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
Application: S=1/2 Ising intransversal field (ITF)
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
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: 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.
m
Preliminaries
Krylov subspaces
Iteration scheme
Application: S=1 BAHC
Application: S=1/2 XYchain
Application: S=1/2 Ising intransversal field (ITF)
Application: S=1/2 Ising intransversal field (ITF)
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· · ·
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: 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
Application: S=1/2 Ising intransversal field (ITF) Application 1
S=1 Bond-alternated Heisenberg chain:measuring the hidden AKLT order
m
Preliminaries
Krylov subspaces
Iteration scheme
Application: S=1 BAHC
Application: S=1/2 XYchain
Application: S=1/2 Ising intransversal field (ITF)
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).
Phases of the model:γ < γc (1,1)-VBS (Haldane)γ > γc (2,0)-VBS (Dimer)
m
Preliminaries
Krylov subspaces
Iteration scheme
Application: S=1 BAHC
Application: S=1/2 XYchain
Application: S=1/2 Ising intransversal field (ITF)
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.
m
Preliminaries
Krylov subspaces
Iteration scheme
Application: S=1 BAHC
Application: S=1/2 XYchain
Application: S=1/2 Ising intransversal field (ITF)
AKLT Physics
Hidden order of the ground state
In the usual z-axis projection basis, the AKLT ground state reveals ahidden antiferromagnetic order: