Propagators for TDDFT

Transcription

1 Propagators for TDDFT A. Castro 1 M. A. L. Marques 2 A. Rubio 3 1 Institut für Theoretische Physik, Fachbereich Physik der Freie Universität Berlin 2 Departamento de Física, Universidade de Coimbra, Portugal. 3 Departamento de Física de Materiales, Centro Mixto CSIC-UPV/EHU and Donostia International Physics Center, San Sebastián, Spain. Benasque 2008 TDDFT School, September 3 rd, 2008

2 Outline Introduction 1 Introduction The Problem Time discretization Self consistency Notation 2 3 4

3 Outline Introduction The Problem Time discretization Self consistency Notation 1 Introduction The Problem Time discretization Self consistency Notation 2 3 4

4 Definition of the problem The Problem Time discretization Self consistency Notation i t φ j( r, t) = { v ext ( r, t) + d 3 r n( r, t) r r + δe xc δn( r, t) }φ j( r, t) n( r, t) = Φ(t) j δ( r ˆ r j ) Φ(t) = j φ j ( r, t) 2. φ j (t = 0) = φ 0 j. i d φ(t) = Ĥ(t) φ(t). dt φ(t = 0) = φ 0.

5 Definition of the problem The Problem Time discretization Self consistency Notation i t φ j( r, t) = { v ext ( r, t) + d 3 r n( r, t) r r + δe xc δn( r, t) }φ j( r, t) n( r, t) = Φ(t) j δ( r ˆ r j ) Φ(t) = j φ j ( r, t) 2. φ j (t = 0) = φ 0 j. i d φ(t) = Ĥ(t) φ(t). dt φ(t = 0) = φ 0.

6 Definition of the problem The Problem Time discretization Self consistency Notation i d φ(t) = Ĥ(t) φ(t). dt φ(t = 0) = φ 0. Ĥ is very large, sparse, typically Hermitian, and unbounded. The Hamiltonian Ĥ(t) that appears in TDDFT is necessarily time-dependent. We do not know a priori this time-dependence: the Hamiltonian is both the problem and the solution: The problem is to obtain φ(t + t), given the knowledge of φ(τ) and Ĥ(τ), for τ t.

7 The solution Introduction The Problem Time discretization Self consistency Notation φ(t) = Û(t, t 0 ) φ(t 0 ). Û is a non-linear operator. It is unitary: Û = Û 1. There is (usually) time-reversal symmetry: Û 1 (t, t 0 ) = Û(t 0, t). The evolution equation may be equivalently rewritten for Û: i d dt Û(t, t 0) = Ĥ(t)Û(t, t 0 ) Û(t 0, t 0 ) = ˆ1. Or in integral form: t Û(t, t 0 ) = ˆ1 i dτĥ(τ)û(τ, t 0 ). t 0

8 The Solution Introduction The Problem Time discretization Self consistency Notation By making use of the previous integral equation, one may derive an explicit form for the propagator: Û(t, t 0 ) = ( i) n n=0 n! t t 0 dt 1 t t 0 dt 2... t t 0 dt n T [ Ĥ(t 1 )Ĥ(t 2 )...Ĥ(t n ) ]. To beautify the expression, the time-ordering exponential is defined: Û(t, t 0 ) = Texp{ i t t 0 dτĥ(τ)}. (But this is only an aesthetic trick, that does not help to design numerical propagators) If [Ĥ(t), Ĥ(t )] = 0, Û(t, t 0 ) = exp{ i t t 0 dτĥ(τ)}. If the Hamiltonian is time-independent, Û(t, t 0 ) = exp{ iĥ(t t 0 )}.

9 The Solution Introduction The Problem Time discretization Self consistency Notation By making use of the previous integral equation, one may derive an explicit form for the propagator: Û(t, t 0 ) = ( i) n n=0 n! t t 0 dt 1 t t 0 dt 2... t t 0 dt n T [ Ĥ(t 1 )Ĥ(t 2 )...Ĥ(t n ) ]. To beautify the expression, the time-ordering exponential is defined: Û(t, t 0 ) = Texp{ i t t 0 dτĥ(τ)}. (But this is only an aesthetic trick, that does not help to design numerical propagators) If [Ĥ(t), Ĥ(t )] = 0, Û(t, t 0 ) = exp{ i t t 0 dτĥ(τ)}. If the Hamiltonian is time-independent, Û(t, t 0 ) = exp{ iĥ(t t 0 )}.

10 The Solution Introduction The Problem Time discretization Self consistency Notation By making use of the previous integral equation, one may derive an explicit form for the propagator: Û(t, t 0 ) = ( i) n n=0 n! t t 0 dt 1 t t 0 dt 2... t t 0 dt n T [ Ĥ(t 1 )Ĥ(t 2 )...Ĥ(t n ) ]. To beautify the expression, the time-ordering exponential is defined: Û(t, t 0 ) = Texp{ i t t 0 dτĥ(τ)}. (But this is only an aesthetic trick, that does not help to design numerical propagators) If [Ĥ(t), Ĥ(t )] = 0, Û(t, t 0 ) = exp{ i t t 0 dτĥ(τ)}. If the Hamiltonian is time-independent, Û(t, t 0 ) = exp{ iĥ(t t 0 )}.

11 Outline Introduction The Problem Time discretization Self consistency Notation 1 Introduction The Problem Time discretization Self consistency Notation 2 3 4

12 The Problem Time discretization Self consistency Notation Time discretization: short-time propagation The exact propagator verifies the well-known property: Û(t 1, t 2 ) = Û(t 1, t 3 )Û(t 3, t 2 ) Û(t, t 0 ) = N 1 j=0 This permits us to work for short time propagations: φ(t + t) = Û(t + t, t) φ(t) = Texp{ i t+ t t Û(t j, t j + t j ). dτĥ(τ)} φ(t). The time-dependence of the Hamiltonian is small. The norm of the time-ordered exponential argument is proportional to t. One normally needs to monitor the evolution: If we want to discern frequencies up to ω max, the the time step t has to be no larger than 1 ω max.

13 The Problem Time discretization Self consistency Notation Time discretization: short-time propagation The exact propagator verifies the well-known property: Û(t 1, t 2 ) = Û(t 1, t 3 )Û(t 3, t 2 ) Û(t, t 0 ) = N 1 j=0 This permits us to work for short time propagations: φ(t + t) = Û(t + t, t) φ(t) = Texp{ i t+ t t Û(t j, t j + t j ). dτĥ(τ)} φ(t). The time-dependence of the Hamiltonian is small. The norm of the time-ordered exponential argument is proportional to t. One normally needs to monitor the evolution: If we want to discern frequencies up to ω max, the the time step t has to be no larger than 1 ω max.

14 The Problem Time discretization Self consistency Notation Time discretization: short-time propagation The exact propagator verifies the well-known property: Û(t 1, t 2 ) = Û(t 1, t 3 )Û(t 3, t 2 ) Û(t, t 0 ) = N 1 j=0 This permits us to work for short time propagations: φ(t + t) = Û(t + t, t) φ(t) = Texp{ i t+ t t Û(t j, t j + t j ). dτĥ(τ)} φ(t). The time-dependence of the Hamiltonian is small. The norm of the time-ordered exponential argument is proportional to t. One normally needs to monitor the evolution: If we want to discern frequencies up to ω max, the the time step t has to be no larger than 1 ω max.

15 Outline Introduction The Problem Time discretization Self consistency Notation 1 Introduction The Problem Time discretization Self consistency Notation 2 3 4

16 Self-consistency Introduction The Problem Time discretization Self consistency Notation For most methods, one needs to know Û(τ) for t τ t + t. Two possibilities: Perform some kind of predictor-corrector, ideally in a self consistent manner or not. Extrapolate Û(τ) from the knowledge or a given number of previous steps Û(t i ).

17 Outline Introduction The Problem Time discretization Self consistency Notation 1 Introduction The Problem Time discretization Self consistency Notation 2 3 4

18 Some notation Introduction The Problem Time discretization Self consistency Notation The key parameters that deterimine any algorithm s accuracy are the time step t, and the norm of the Hamiltonian. This is determined by the highest eigenvalue. And for a Kohn-Sham Hamiltonian discretized on a real space or plane wave mesh, this is determined quadratically by the grid spacing h. The mesh ratio is defined to be t/h 2. For a given approximate propagator U(t, t + t), the local truncation error is defined as: E = 1 t [ψ exact(t + t) U(t, t + t)ψ(t)] (1)

19 Some notation Introduction The Problem Time discretization Self consistency Notation A propagator U is consistent if the truncation error is O( t p ) for some integer p. All reasonable methods should be consistent. A propagator is stable if there exists a t max such that, for any t < t max, U(t, t + t) n is uniformly bound for any n > 0. A propagator is contractive if U <= 1. It is unitary if U = 1. Being unitary implies being contractive; being contractive involves being stable. A propagator is unconditionally (consistent, stable, contractive, unitary) if it is such independently of the mesh ratio. A good propagator should also be time-reversible.

20 Classical propagators Crank-Nicholson, or Implicit-Midpoint Rule: Û CN (t + t, t) = 1 i t 2 Ĥ(t + t/2) 1 + i 2 tĥ(t + t/2) The method is implicit since it requires the solution of a linear system: It is unitary, and b = ˆLφ(t + t) = b, ˆL = ˆ1 + i t Ĥ(t + t/2), [ 2 ˆ1 i t ] Ĥ(t t/2) φ(t). 2 If Ĥ(t + t/2) is exact, it preserves time-reversal symmetry. Other classical propagators: implicit or explicit Runge-Kutta, Euler s method, etc.

21 Classical propagators Crank-Nicholson, or Implicit-Midpoint Rule: Û CN (t + t, t) = 1 i t 2 Ĥ(t + t/2) 1 + i 2 tĥ(t + t/2) The method is implicit since it requires the solution of a linear system: It is unitary, and b = ˆLφ(t + t) = b, ˆL = ˆ1 + i t Ĥ(t + t/2), [ 2 ˆ1 i t ] Ĥ(t t/2) φ(t). 2 If Ĥ(t + t/2) is exact, it preserves time-reversal symmetry. Other classical propagators: implicit or explicit Runge-Kutta, Euler s method, etc.

22 Classical propagators Crank-Nicholson, or Implicit-Midpoint Rule: Û CN (t + t, t) = 1 i t 2 Ĥ(t + t/2) 1 + i 2 tĥ(t + t/2) The method is implicit since it requires the solution of a linear system: It is unitary, and b = ˆLφ(t + t) = b, ˆL = ˆ1 + i t Ĥ(t + t/2), [ 2 ˆ1 i t ] Ĥ(t t/2) φ(t). 2 If Ĥ(t + t/2) is exact, it preserves time-reversal symmetry. Other classical propagators: implicit or explicit Runge-Kutta, Euler s method, etc.

23 Exponential Midpoint Rule Û EM (t + t, t) = exp{ i tĥ(t + t/2)}. If the exponential is calculated exactly: It is unitary, and If Ĥ(t + t/2) is exact, it preserves time-reversal symmetry. In our experience, it performs better than Û CN although this depends on the quality of the methods to calculate the exponential or to solve linear systems. Note that the most simple-minded approximation to the propagator, exp{ i tĥ(t)} would not preserve time-reversal symmetry.

24 Exponential propagator based on TRS Assuming the simplest approximation to the propagator, time-reversal symmetry can be enforced by stating: exp{+i t 2 which defines the propagator: Ĥ(t + t)}φ(t + t) = exp{ i t Ĥ(t + t)}φ(t), 2 Û ETRS (t + t, t) = exp{ i t Ĥ(t + t)} exp{ i t 2 2 Ĥ(t)} It is also unitary, but contains once again one unknown in the definition, Ĥ(t + t). This Hamiltonian must be either extrapolated or approximated with a previous evolution. The procedure, ideally, should be self-consistent.

25 Exponential propagator based on TRS Assuming the simplest approximation to the propagator, time-reversal symmetry can be enforced by stating: exp{+i t 2 which defines the propagator: Ĥ(t + t)}φ(t + t) = exp{ i t Ĥ(t + t)}φ(t), 2 Û ETRS (t + t, t) = exp{ i t Ĥ(t + t)} exp{ i t 2 2 Ĥ(t)} It is also unitary, but contains once again one unknown in the definition, Ĥ(t + t). This Hamiltonian must be either extrapolated or approximated with a previous evolution. The procedure, ideally, should be self-consistent.

26 Splitting techniques Typically, Ĥ = ˆT + ˆV, where ˆT is diagonal in Fourier space, and ˆV is diagonal in real space (or almost). This is indeed the case in TDDFT The split-operator technique makes use of this fact, and traditionally splits the exponential that approximates the propagator (i.e. via the EMR) into three exponentials: exp{ i t(ˆt + ˆV)} = exp{ i t/2ˆt} exp{ i t ˆV} exp{ i t/2ˆt}. This may be generalized and obtain numerous methods. The root to derive and analyze them is the well-known Baker-Campbell-Hausdorff relation, and generalizations. 1 e A e B = exp{a + B + 1 [A, B] +...}, 2 1 T. Y. Mikhailova and V. I. Pupyshev, Phys. Lett. A 257, 1 (1999)

27 Splitting techniques Typically, Ĥ = ˆT + ˆV, where ˆT is diagonal in Fourier space, and ˆV is diagonal in real space (or almost). This is indeed the case in TDDFT The split-operator technique makes use of this fact, and traditionally splits the exponential that approximates the propagator (i.e. via the EMR) into three exponentials: exp{ i t(ˆt + ˆV)} = exp{ i t/2ˆt} exp{ i t ˆV} exp{ i t/2ˆt}. This may be generalized and obtain numerous methods. The root to derive and analyze them is the well-known Baker-Campbell-Hausdorff relation, and generalizations. 1 e A e B = exp{a + B + 1 [A, B] +...}, 2 1 T. Y. Mikhailova and V. I. Pupyshev, Phys. Lett. A 257, 1 (1999)

28 Splitting techniques Typically, Ĥ = ˆT + ˆV, where ˆT is diagonal in Fourier space, and ˆV is diagonal in real space (or almost). This is indeed the case in TDDFT The split-operator technique makes use of this fact, and traditionally splits the exponential that approximates the propagator (i.e. via the EMR) into three exponentials: exp{ i t(ˆt + ˆV)} = exp{ i t/2ˆt} exp{ i t ˆV} exp{ i t/2ˆt}. This may be generalized and obtain numerous methods. The root to derive and analyze them is the well-known Baker-Campbell-Hausdorff relation, and generalizations. 1 e A e B = exp{a + B + 1 [A, B] +...}, 2 1 T. Y. Mikhailova and V. I. Pupyshev, Phys. Lett. A 257, 1 (1999)

29 Splitting techniques A split operator scheme method taylored for TDDFT: 2 Û SO (t + t, t) = exp{ i t/2ˆt} exp{ i tˆv } exp{ i t/2ˆt}. ˆV is the Kohn-Sham potential built from the orbitals that result of applying the first kinetic-energy term: φ i = exp{ i t/2ˆt} φ i (t) n ( r, t) = i φ i( r, t) 2 ˆV = V KS [n ]. This scheme permits to maintain an O( t 2 ) order in the error, without the need of performing a self-consistent obtention of ˆV(t + /2). 2 N. Watanabe and M. Tsukada, Phys. Rev. E 65, (2002).

30 Splitting techniques: higher orders Suzuki 3 generalized the split-operator splitting to higuer orders. For example, to fourth order. e xh = S 4 (x) = 5 S 2 (p j x), where p j is a set of real numbers, and S 2 (x) is the normal split-operator splitting. The number of FFTs is multiplied by five. It is thus unclear that this method involves overall increase in speed over normal Strang splitting. j=1 3 M. Suzuki, J. Phys. Soc. Jpn. 61, L3015 (1992); O. Sugino and Y. Miyamoto, Phys. Rev. B 59, 2579 (1999).

31 Magnus expansions Once again: Û(t + t, t) does not reduce to a simple exponential unless the Hamiltonian is time-independent. Question: Is there any operator ˆΩ(t + t, t) such that: Û(t + t, t) = exp{ˆω(t + t, t)}? Answer: Yes. There exists an infinite series, convergent for small enough t, such that the seeked operator may be written as: 4 ˆΩ(t + t, t) = Ω k (t + t, t). k=1 Moreover, there exists a procedure to generate recursively the terms of the infinite sum. 5 For example: Ω 2 (t + t, t) = Ω 1 (t + t, t) = t+ t t+ t t dτ 1 [ iĥ(τ1 ) ]. t+ t [ dτ 1 dτ 2 iĥ)(τ1 ), iĥ(τ 2 ) ]. t t 4

32 Magnus expansions An approximation of order 2n to the exact Û(t + t, t) is achieved by: Truncating the Magnus series to n-th order. Approximating the integrals in time through an n-th order quadrature formula. This gives us the EMR for the order-two Magnus expansion: ˆΩ M(2) (t + t, t) = iĥ(t + t/2) t. The first interesting result is thus the order-four Magnus expansion: ˆΩ M(4) (t + t, t) = i t 2 [Ĥ(t1 ) + Ĥ(t 2 ) ] 3 t 2 [Ĥ(t2 ), Ĥ(t 1 ) ], 12 t 1,2 = t + (1/2 3/6) t.

33 Magnus expansions We have applied this result for our TDDFT case. The propagator then takes the form: Û M(4) (t + t, t) = exp{ i tĥ M(4) (t, t)}, Ĥ M(4) (t, t) = H(t, t) + i [ˆT + ˆV nonlocal ext, V(t, t) ], H(t, t) = ˆT {ˆV KS (t 1 ) + ˆV KS (t 2 )}, V = 3 12 t{ˆv KS (t 2 ) ˆV KS (t 1 )}.

34 Magnus expansions <µ(t)> (a.u.) Exact M(4) EM=M(2) ETRS Time (a.u.) Evolution of the dipole moment of Na 8 subject to an intense laser field, calculated via the methods indicated in the legend.

35 The solutions Introduction Û CN (t + t, t) = 1 i t 2 Ĥ(t + t/2) 1 + i t 2 Ĥ(t + t/2) Û EM (t + t, t) = exp{ i tĥ(t + t/2)} Û ETRS (t + t, t) = exp{ i t Ĥ(t + t)} exp{ i t 2 2 Ĥ(t)} Û SO (t + t, t) = exp{ i t/2ˆt} exp{ i tˆv } exp{ i t/2ˆt} 5 Û ST (t+ t, t) = exp{ ip j t/2ˆt} exp{ ip j tˆv(t+p j t)} exp{ ip j t/2ˆt} j=1 n Û M(2n) (t + t, t) = exp{ Ω j (t + t, t)} j=1

36 The solutions Introduction Û CN (t + t, t) = 1 i t 2 Ĥ(t + t/2) 1 + i t 2 Ĥ(t + t/2) Û EM (t + t, t) = exp{ i tĥ(t + t/2)} Û ETRS (t + t, t) = exp{ i t Ĥ(t + t)} exp{ i t 2 2 Ĥ(t)} Û SO (t + t, t) = exp{ i t/2ˆt} exp{ i tˆv } exp{ i t/2ˆt} 5 Û ST (t+ t, t) = exp{ ip j t/2ˆt} exp{ ip j tˆv(t+p j t)} exp{ ip j t/2ˆt} j=1 n Û M(2n) (t + t, t) = exp{ Ω j (t + t, t)} j=1

37 The solutions Introduction Û CN (t + t, t) = 1 i t 2 Ĥ(t + t/2) 1 + i t 2 Ĥ(t + t/2) Û EM (t + t, t) = exp{ i tĥ(t + t/2)} Û ETRS (t + t, t) = exp{ i t Ĥ(t + t)} exp{ i t 2 2 Ĥ(t)} Û SO (t + t, t) = exp{ i t/2ˆt} exp{ i tˆv } exp{ i t/2ˆt} 5 Û ST (t+ t, t) = exp{ ip j t/2ˆt} exp{ ip j tˆv(t+p j t)} exp{ ip j t/2ˆt} j=1 n Û M(2n) (t + t, t) = exp{ Ω j (t + t, t)} j=1

38 The solutions Introduction Û CN (t + t, t) = 1 i t 2 Ĥ(t + t/2) 1 + i t 2 Ĥ(t + t/2) Û EM (t + t, t) = exp{ i tĥ(t + t/2)} Û ETRS (t + t, t) = exp{ i t Ĥ(t + t)} exp{ i t 2 2 Ĥ(t)} Û SO (t + t, t) = exp{ i t/2ˆt} exp{ i tˆv } exp{ i t/2ˆt} 5 Û ST (t+ t, t) = exp{ ip j t/2ˆt} exp{ ip j tˆv(t+p j t)} exp{ ip j t/2ˆt} j=1 n Û M(2n) (t + t, t) = exp{ Ω j (t + t, t)} j=1

39 The solutions Introduction Û CN (t + t, t) = 1 i t 2 Ĥ(t + t/2) 1 + i t 2 Ĥ(t + t/2) Û EM (t + t, t) = exp{ i tĥ(t + t/2)} Û ETRS (t + t, t) = exp{ i t Ĥ(t + t)}exp{ i t 2 2 Ĥ(t)} 5 Û ST (t+ t, t) = exp{ ip j t/2ˆt}exp{ ip j tˆv(t+p j t)}exp{ ip j t/2ˆt} j=1 n Û M(2n) (t + t, t) = exp{ Ω j (t + t, t)} j=1

40 The exponential exp exp{a} = a a a N If A is not diagonal, diagonalize: j=0 = A = VHV T, 1 j! Aj. e a e a e an and e A = Ve H V T.

41 The exponential exp exp{a} = a a a N If A is not diagonal, diagonalize: j=0 = A = VHV T, 1 j! Aj. e a e a e an and e A = Ve H V T.

42 The exponential exp exp{a} = a a a N If A is not diagonal, diagonalize: j=0 = A = VHV T, 1 j! Aj. e a e a e an and e A = Ve H V T.

43 The exponential: small matrices C. Moler and C. Van Loan, Nineteen Dubious Ways to Compute the Exponential of A Matrix, SIAM Review 20, 801 (1978). Taylor series. Padé approximations. Scaling and squaring. Chebyshev rational approximation. Ordinary differential equation methods. Polynomial methods. Matrix decomposition methods. Splitting methods. The focus is placed on the problem of calculating e A, which is only possible for small matrices. We have to be more modest, and look for methods to calculate e A v, for a given v.

44 The exponential: small matrices C. Moler and C. Van Loan, Nineteen Dubious Ways to Compute the Exponential of A Matrix, Twenty-Five Years Later, SIAM Review 45, 3 (2003). In principle, the exponential of a matrix could be computed in many ways. (... ) In practice, consideration of computational stability and efficiency indicates that some of the methods are preferable to others but that none are completely satisfactory. Two different problems: Given A, calculate e A, so that it can be applied to any vector: unfeasible for our TDDFT problem, wher A is huge and sparse. Given A and v, calculate e A v.

45 The exponential: small matrices C. Moler and C. Van Loan, Nineteen Dubious Ways to Compute the Exponential of A Matrix, Twenty-Five Years Later, SIAM Review 45, 3 (2003). In principle, the exponential of a matrix could be computed in many ways. (... ) In practice, consideration of computational stability and efficiency indicates that some of the methods are preferable to others but that none are completely satisfactory. Two different problems: Given A, calculate e A, so that it can be applied to any vector: unfeasible for our TDDFT problem, wher A is huge and sparse. Given A and v, calculate e A v.

46 N-th order expansion The most obvious way to approximate the exponential is to use its definition (standard expansion): exp ( i th) = ( i t) j j=0 taylor k { i th} = j! H j n (i t) j H j v. j! The error, for a given n, is O( t (n+1) ). This operator is not unitary. It may be proved that n = 4 is especially advantageous, since it is conditionally contractive, and thus stable for large values of t. 6 n = 2, for example, is unconditionally unstable; n = 6 is also conditionally stable, but only for smaller values of t. j=0 6 Jeff Giansiracusa, unpublished.

47 Chebyshev expansion Instead of using the standard polynomial basis, we may use other polynomial to expand the exponential. The Chebyshev basis is well known as a way to economize power series. Utilizing them is advantageous because: Since 1984, we know a closed form for the coefficients: 7 e ih t = k (2 δ k0 )J n ( t)( i) n T n (H). (2) n=0 7 H. Tal Ezer and R. Kosloff, J. Chem. Phys. 81, 3967 (1984).

48 Evaluation of the polynomials may be done at low cost (essentially the same than with standard expansion) thanks to Clenshaw s algorithm 8. Chebyshev expansion error vs standard expansion error 8 C. W. Clenshaw, MTAC 9, 118 (1955).

49 Error Standard Expansion Chebyshev Expansion δt (a.u.) Std: e -ihδt =Σ (-iδth) n /n! Cheb: e -ihδt =Σ (2-δ n0 )J n (δt)(-i) n T n (H) (being H shifted to [-1,1]) 10 0 O(δt 2 ) Chebyshev vs Standard Standard Chebyshev Expansion Order O(δt 3 ) O(δt 4 ) O(δt 5 ) O(δt 6 ) O(δt 7 ) For an excited Na atom, error in the evaluation of the exponential of the Hamiltonian.

50 Split-Operator Approaches The Kohn-Sham hamiltonian H has the form H = T + V, where T is diagonal if reciprocal space, and V is diagonal in real space. This suggests the use of the Strang splitting (split-operator, split step,...): 9 exp ( i th) exp ( i t/2t) exp ( i tv) exp ( i t/2t) (3) The split-operator may be kinetic or potential referenced. The error is third order in t. The method is unitary and unconditionally stable. 10 A wealth of other splitting schemes are possible W. C. Strang, J. Numer. Anal. 6, 506 (1968). 10 R. Kosloff, J. Phys. Chem. 92, 2087 (1988)- 11 N. N. Yanenko, The Method of Fractional Steps, Springer, 1971.

51 Suzuki-Trotter Introduction Suzuki 12 generalized Strang splitting to higuer orders. For example, to fourth order. e xh = S 4 (x) = 5 S 2 (p j x), (4) where p j is a set of real numbers, and S 2 (x) is the normal Strang splitting. The number of FFTs is multiplied by five. It is thus unclear that this method involves overall increase in speed over normal Strang splitting. j=1 12 M. Suzuki, J. Phys. Soc. Jpn. 61, L3015 (1992); O. Sugino and Y. Miyamoto, Phys. Rev. B 59, 2579 (1999).

52 The method may be generalized to time-dependent Hamiltonians. So it is not only a way to approximate the Hamiltonian, but also a full algorithm to approximate the propagator (the same holds for the basic Strang splitting) Error t (a.u.) Comparison of the second-order split operator (SO, solid) and the fourth-order Suzuki-Trotter (ST, dashed) schemes.

53 Krylov subspace projection A given matricial function f(ta) may be Taylor expanded: m 1 f(ta)v = v + i=1 (ta) i + O(tA) m. i! This provides us with a polynomial approximation of degree m 1: m 1 f(ta)v = c i (ta) i. It is not the only possible polynomial approximation. All possibilities are elements of the Krylov subspace: i=0 K m (ta, v) = L{v,(tA)v,...,(tA) m 1 v}. What is the element of K m (ta, v) that optimally approximates f(ta)v?

54 Krylov subspace projection A given matricial function f(ta) may be Taylor expanded: m 1 f(ta)v = v + i=1 (ta) i + O(tA) m. i! This provides us with a polynomial approximation of degree m 1: m 1 f(ta)v = c i (ta) i. It is not the only possible polynomial approximation. All possibilities are elements of the Krylov subspace: i=0 K m (ta, v) = L{v,(tA)v,...,(tA) m 1 v}. What is the element of K m (ta, v) that optimally approximates f(ta)v?

55 Krylov subspace projection A given matricial function f(ta) may be Taylor expanded: m 1 f(ta)v = v + i=1 (ta) i + O(tA) m. i! This provides us with a polynomial approximation of degree m 1: m 1 f(ta)v = c i (ta) i. It is not the only possible polynomial approximation. All possibilities are elements of the Krylov subspace: i=0 K m (ta, v) = L{v,(tA)v,...,(tA) m 1 v}. What is the element of K m (ta, v) that optimally approximates f(ta)v?

56 Krylov subspace projection A given matricial function f(ta) may be Taylor expanded: m 1 f(ta)v = v + i=1 (ta) i + O(tA) m. i! This provides us with a polynomial approximation of degree m 1: m 1 f(ta)v = c i (ta) i. It is not the only possible polynomial approximation. All possibilities are elements of the Krylov subspace: i=0 K m (ta, v) = L{v,(tA)v,...,(tA) m 1 v}. What is the element of K m (ta, v) that optimally approximates f(ta)v?

57 Krylov subspace projection To manipulate the elements of K m (ta, v), it is better to have an orthonormal base. This is the task of the Arnoldi (Lanczos) procedure: β = v v 1 = v/β for j = 1 to m do p = Av j for i = 1 to j do h ij = v T i p p = p h ij v i end h j+1,j = p v j+1 = p/h j+1,j end

58 Krylov subspace projection The result are two matrices: V m+1 = [v 1, v 2,...,v m+1 ] R n (m+1) H = (h ij ) R (m+1) m and H R m m, which is the square matrix formed by the first m rows of H. These matrices satisfy: V T mav m = H m AV m = V m+1 H V T mv m = 1

59 Krylov subspace projection The result are two matrices: V m+1 = [v 1, v 2,...,v m+1 ] R n (m+1) H = (h ij ) R (m+1) m and H R m m, which is the square matrix formed by the first m rows of H. These matrices satisfy: V T mav m = H m AV m = V m+1 H V T mv m = 1

60 Krylov subspace projection Using this recursion, each V j = [v 1,...,v j ] is an orthonormal base yes of K j (A, v). It may be proved that the optimal approximation to e ta v, in the least squares sense, within K m (A, v), is: w opt = βv m (V T me ta V m )ê 1 But we still have e ta in the way. The idea now is to do the following approximation: V T me ta V m = exp{tv T mav m } = e thm The final approximation is then: e ta v lanczos m (ta, v) = βv m e thm ê 1.

61 Krylov subspace projection Using this recursion, each V j = [v 1,...,v j ] is an orthonormal base yes of K j (A, v). It may be proved that the optimal approximation to e ta v, in the least squares sense, within K m (A, v), is: w opt = βv m (V T me ta V m )ê 1 But we still have e ta in the way. The idea now is to do the following approximation: V T me ta V m = exp{tv T mav m } = e thm The final approximation is then: e ta v lanczos m (ta, v) = βv m e thm ê 1.

62 Krylov subspace projection Using this recursion, each V j = [v 1,...,v j ] is an orthonormal base yes of K j (A, v). It may be proved that the optimal approximation to e ta v, in the least squares sense, within K m (A, v), is: w opt = βv m (V T me ta V m )ê 1 But we still have e ta in the way. The idea now is to do the following approximation: V T me ta V m = exp{tv T mav m } = e thm The final approximation is then: e ta v lanczos m (ta, v) = βv m e thm ê 1.

63 Krylov subspace projection Using this recursion, each V j = [v 1,...,v j ] is an orthonormal base yes of K j (A, v). It may be proved that the optimal approximation to e ta v, in the least squares sense, within K m (A, v), is: w opt = βv m (V T me ta V m )ê 1 But we still have e ta in the way. The idea now is to do the following approximation: V T me ta V m = exp{tv T mav m } = e thm The final approximation is then: e ta v lanczos m (ta, v) = βv m e thm ê 1.

64 For a given order m, the method is O(m + 1). For any m, the method is unitary. The computational cost grows linearly with m. The dimension m is increased recursively until some convergence criterium (β [exp( i th m )] m,m ). The decay of the error decays superlinearly with m. So the method is of arbitrary accuracy.

65 Normalized residue th order Chebyshev Lanczos t (a.u.) For an excited Na atom, error in the evaluation of the exponential of the Hamiltonian for both the Lanczos method (circles) for a fixed tolerance, and for the Chebyshev expansion of 8 th order (crosses). The numbers close to the circles show the Krylov basis dimension needed to achieve the desired accuracy.

66 Standard Chebyshev Lanczos Na [1s] p(δt)/δt (a.u. -1 ) C [2p] Au [5d] δt (a.u.) Number of Hamiltonian-wavefunction operations per unit time, as a function of t for the Taylor (solid) and Chebyschev (dashes) expansions, and for the Lanczos projection method (dotted)

67 Introduction There is not an always optimal algorithm for the propagation of the TDKS equations. For long time propagations, assuring time-reversal symmetry is very important. Some methods require the calculation of the action of the exponential of Hamiltonian matrices; there are efficient methods to perform this task. The Lanczos-Krylov subspace projection seems to be the best algorithm to calculate the action of exponentials. For problems involving very high frequencies, Magnus expansion are advantageous. Otherwise, a combination of the EM rule with Lanczos subspace projection is sufficient.