2019 HKU-UCAS Computational Study Group

Exact Diagonalization

Han-Qing Wu (邬汉青)

School of Physics, Sun Yat-Sen University, Guangzhou

Outline

• Eigen problem

• Full Exact Diagonalization

• Lanczos Exact Diagonalization

• Summary

Eigen problem• Mathematic

• Quantum Mechanic

( ) 0λx A λI xAx ⇔ −= =

H EΨ = Ψ 1 2: , , , Nbasis Φ Φ Φ1

ˆi i

N

iI

=

= Φ Φ∑ˆ ˆ ˆ ˆIHI EIΨ = Ψ

1 1 1

ˆN N N

i j ii i j j i iH E

= = =

Φ Φ Φ Φ Ψ = Φ Φ Ψ∑ ∑ ∑

1 1 1j i i i

N N N

iji j i

H E= = =

Φ Φ = Φ Φ

∑ ∑ ∑

1j i

N

ijj

H E=

⇒ Φ = Φ∑

lengthen, shorten and reverse directions

complete (and orthogonal) basis

Matrix mechanics

Eigen problemDensity matrix and partition function,

Many interesting physical variables,

ˆ1 1ˆ iβEβHi i

iρ e e

Z Z−− Ψ== Ψ∑

ˆ( ) iβEβH

iZ tr e e−−= = ∑

( )

( )

0

0

ˆ ˆˆ ˆˆ( )

ii

i i

β EβEi i i i

i iβE β E

E

E

i i

e O e OO tr ρO

e e

−−

− −

−

−

Ψ Ψ Ψ Ψ= = =

∑ ∑∑ ∑

0 01

ˆ1 m mm

iβ O

m =

→ ∞ Ψ Ψ∑ m is the number of degenerate ground states.

Finding roots of high-degree polynomials is a numerically tricky task

A better idea is to make use of the orthogonal or unitary transform

Unitary transform does not change the eigenvalue of a matrix!

† † †, ,H HU U UU HU IU E′ ′= = ′Φ == ′Φ† ( ) ( )HU HE U UU E′ ′ ′ ′⇒ Φ = Φ ⇒ Φ = Φ

† † † † †1 1 1 1 1 12 2 2 2HU HU U HH U U U U U U U→ → →→ Λ

1 32† , U UHU U U UΛ = = Jacobi iterative method, power method, …

Theorem: Suppose matrix A is real and symmetric, then

(1) All of the eigenvalues are real;

(2) Any two eigenvectors with different eigenvalues are orthogonal to

each other;

(3) There exists a orthogonal matrix U that transforms A into a diagonal

matrix.

For eigen problem of a hermitian matrix C=A+iB,

it can be mapped into the eigen problem of a real symmetric matrix

A B u uB A v v

λ−

=

† ,T TC C A A B B= ⇒ = = −

Full Exact Diagonaliaztion1. Jacobi method

10 1, ,T T

k k k k k kH H H J H J J J−+= = =

1 1 0k kJU J JJ−=

U is the product of all rotation matrices.

0

1 0 20 2 12 1 1

A =

0

0.707 0 0.7070 1 0

0.707 0 0.7071 0 0 0

3 0.7070.707 2 0.707

0 0 1

0

0 .7 7

TJ

A J A J

− =

→ = = −

1

0.888 0.460 00.460 0.888 0

0 0 12 1 1 1

3.366 0.00

3251.634 0.628

0.325 0.628 1

JTA J A J

− =

→ = = −

2

1 0 00 0.975 0.2260 0.226 0.975

3 2 2 2

3.366 0.0735 0.3170.0735 1.7800.325 0 1.

0145

TJ

A J A J

= −

→ = = −

Example:

2. QR or QU factorization——widely used method

Decomposition of a matrix H into a product of an orthogonal matrix Q

and an upper triangular matrix R,

It can be proved that Hk+1 is similar to Hk,

0 1 1 1 1k k k k k k k k kH H H Q R H R Q H Q R+ + + += → = → = → = →

1T T

k k k k k k k k k kH R Q Q Q R Q Q H Q+ = = =

1 1 1 1 0 0 0 1T T T T Tk k k k k k k k kQ Q H Q Q Q Q H QQ Q Q− − − − −= = =

1 2,TH U U Q QU Q∞⇒ =Λ=

Two strategies are used to improve covergence.

(1) shifted QR

(2) two steps

symmetricmatrix

tridiagonalmatrix

diagonalmatrix

Householdertransformation

QRfactorization

Lanczos Exact Diagonalization1. Power method

Eigenvalues: |λ1| > |λ2| > |λ3| >...,

Eigenvector: u1, u2, u3,….

10 1 0,i i

n

iv αuα

=

= ≠∑2

1 2 0k

k k kHv H vv H v− − == = =

( )1 12

11

1/n

kk ki i i i i

ii

n

iu uλ α λ uα λ λ α

= =

= = +

∑ ∑ 1

1

1

lim

lim

k

kk

kk

vv

u v

λ→∞

−

→∞

∞

∞

=⇒

=

2. Krylov subspace

For H∈Cnxn and 0≠v0∈Cnxn,

Ill-conditioned: Hm-1v0 point more and more in the direction of the dominant

eigenvector for increasing m, and hence the basis vectors become dependent

in finite precision arithmetic. It is better to work with an orthonormal basis.

(1) symmetric matrices——Lanczos algorithm;

(2) unsymmetric matrices——Arnoldi algorithm.

{ }2 10 0 0 0, , , : Krylov sequence, jv Hv H v vH −

{ }2 10 0 0 0 0( ; ) , , , : Krylov subspac, espanm mH v v Hv H v vH −=

3. (Modified) Lanczos algorithm

Lanczos method: construct a special orthogonal basis by numerically efficient

recursion scheme where the Hamiltonian has a tridiagonal representation.

( )( )

( )( )( )

( )

11 1

1 1 1

01 1

1 0

0

,,

,, 0

,

0,choose ran omly

,,

d

n n n n n n

n nn

n

ni

n

n nn

n

nn n i

i

n

i i

v Hv a v b v

v Hva

v v

v vb b

v

v vv v v

v

v v

v v

+ − −

− −

++ +

=

−

′ = − − = = =

′′=

−

=

∑ Gram–Schmidt process cost time and memory

Since we use floatingpoint arithemetic, whenthe basis size becomeslarge, round-off errorsaccumulation makes theorthogonality lost.

Every iterative step generates one Krylov vector,

These vectors form Krylov matrix V with orthonormal columns. This

matrix can project the Hamiltonian matrix into a tridiagonal matrix in

Krylov subspace.

1 2 1

1 1 2 2 1

0

0 0 1

, , ,( , ) ( , ) ( , ) ( ,

,)

M

M M

v v v vv v v v v v v v

−

− −

0 1

1 1 2

2 2†

2 1

1 1

0 0 00 0

0 0 0

0 0 00 0 0

M M

M M

a bb a b

b aT V HV

a bb a

− −

− −

= =

†

† †

TUU HVU

V U=

Λ =

Eigenvalues

Eigenvectors

How to get the orthogonal matrix U ?

QR(QL), MRRR, …~ 100M

4. Sparse matrix storage format: Compressed Row Storage (CRS)

MPI .vs. OPENMP

5. Degenerate states and excited states

vk is a superposition of u1 and u2。

(1) Gram–Schmidt process: each new Lanczos vector constructed isexplicitly orthogonalized with respect to all previous basis vectors.

(2) Instead of converging several excited states in the same run, one canalso target excited states one-by-one, starting each time from avector which is orthogonal to all previous ones.

13

1 1 2 2 1( / )k kk i i

n

iiλ α α λ λ αv u u u

=

= + +

∑ u1,u2 are degenate states.

Anders W. Sandvik, AIP Conference Proceedings 1297, 135 (2010)

target excited states one-by-one

6. Download and run the source code

Download the source code: https://github.com/hqwu/Lanczos-exact-diagonalization

Finite-size goundstate energies of 1D Heisenberg model.

Summary1.Full exact diagonalization

(1)similarity transform to get the tridiagonal or diagonal matrix;

(2)QR or QL factorization to diagonal form.

2. Lanczos exact diagonalization

(1)Krylov subspace;

(2)degenerate states need to be careful ;

(3) excited states can be got by targeting one by one or using reorthogonalization.

Bibliography(1)Numerical Simulations of Frustrated Systems, Andreas M. Läuchli, Springer

Berlin Heidelberg, 2011.

(2)Anders W. Sandvik, AIP Conference Proceedings 1297, 135 (2010).

(3)Numerical Recipes: The Art of Scientific Computing, Third Edition (2007), 1256

pp. Cambridge University Press ISBN-10: 0521880688.

(4) Templates for the Solution of Algebraic Eigenvalue Problems: A Practical Guide,

Zhaojun Bai et al., Society for Industrial and Applied Mathematics, 2000

Appendix AFour-site plaquette

Basis of Mz=0 sector:

( )( ) ( )1 3 2 4 2 1 3 2 41ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ S S S S S S S SH J J= + + + +

1 1 2 3 4 1 2 3 4 1 2 4 1 2 4

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

1 2 3 4 1 2

2 3 3

3

3 4 1 2 3 4 2 3

4

5 1 46

= 11 0 0 =3, 1 0 1 0 5,

0 1 1 0 6, 1 0 0 1 9,

0 1 0 1 10, 0 0 1 1 12

ψ ψ

ψ ψ

ψ ψ

= ↑ ↑ ↓ ↓ = ↑ ↓ ↑ ↓ = =

= ↓ ↑ ↑ ↓ = = = ↑ ↓ ↓ ↑ = =

= ↓ ↑ ↓ ↑ = = = ↓ ↓ ↑ ↑ = =

2 1 2 2 12 3 4 5

1 2 1 1 12 1 1 2 3 4 6

2 1 2 1 23 1 2 3 5 6

2 1 2 1 24 1 2 4 5 6

1 1 1 2 15 1 3

1

4 1 5 6

6

12 2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2 2 2

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

2

J J J J J

J J J J JJ

J J J J J

J J J J J

J J J

H

H

H

H

H J

H

JJ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ ψ

ψ

= − + + + +

= + + − + + + +

= + + − + +

= + + − + +

= + + + + − + +

= 1 2 2 1 22 3 4 5 62 2 2 2 2

J J J J Jψ ψ ψ ψ ψ+ + + + −

2 1 2 2 1

1 2 1 1 11

2 1 2 1 2

2 1 2 1 2

1 1 1 2 11

1 2 2 1 2

02 2 2 2 2

02 2 2 2 2

02 2 2 2 2

02 2 2 2 2

02 2 2 2 2

02 2 2 2 2

J J J J J

J J J J JJ

J J J J J

HJ J J J J

J J J J JJ

J J J J J

−

− + −

= − − + −

J2

nE

( )( ) ( )

( )

( )

ˆ22

2ˆ

ˆ

ˆˆ

ˆ2 2

ˆ2 2ˆ

ˆˆ ˆ

ˆˆ ˆ

ˆˆ

1

ˆ

1

1

n

n

n

n

n

H

V HB

EHn

nHEH

E

nn

n n

nn

Hn

HEH

TrE HeH H

eC

T N T Nk TNTr

n n ETr

Z n n

n n

He eH He

ee

H e eH H e

ee

ETr

Z n n

β

β

ββ

βββ

ββ

βββ

−

−

−−

−−−

−−

−−−

∂ ∂ = = = − ∂ ∂

= = =

= = =

∑ ∑∑∑

∑ ∑∑∑

Appendix B

( )

( )

ˆ

22ˆ

2 2 2ˆ ˆ

2ˆˆ

ˆ

ˆˆ ˆ

ˆ ˆˆ

ˆ

1

ˆ

n

n

i

i n i

H zi

zu z zH

H z EH z zi i i

z EHH

H z

n i

nn

ii

z

Tr eM

H N N H NTr e

Tr e n e

SM M

S S SM

e

n e

en nTr e

Tr e SM

β

β

β ββ

βββ

β

βχ

−

−

− −−

−−−

−

∂ ∂ = = = − ∂ ∂

= = =

=

∑

∑ ∑ ∑ ∑ ∑∑∑

∑

( )

ˆ

ˆˆ

ˆ n

n

E

n i

H z

n i

nn

zi i

EHH

n e n e

en

S S

e nTr e

ββ

βββ

−−

−−−

= =

∑ ∑ ∑ ∑∑∑