Splittings, commutators and the linear Schrödinger equation

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Splittings, commutators and the linear Schrödinger equation

1 UNIVERSITY OF CAMBRIDGE CENTRE FOR MATHEMATICAL SCIENCES DEPARTMENT OF APPLIED MATHEMATICS & THEORETICAL PHYSICS Splittings, commutators and the linear Schrödinger equation Philipp Bader, Arieh Iserles, Karolina Kropielnicka & Pranav Singh Minneapolis May 202

2 THE PROBLEM We consider the linear Schrödinger equation i h u t = h2 2 u V (x)u, t 0, x, 2m x2 where 0 < h, and the interaction potential V is a smooth periodic function, given with smooth periodic initial and boundary conditions. We can contemplate this equation on a multivariate torus, i h u t = h2 u V (x)u. 2m However, our purpose here is to explore ideas, rather than presenting them in the greatest possible generality. It is convenient to set ω = h, whereby the equation becomes u t = u icω 2 + iωv (x)u, c = x2 2m. 2

3 EXPONENTIALS AND SPLITTINGS... The conventional approach: replace 2 u x2 by a linear combination of function values (or Fourier modes). This results in the linear system whose exact solution is y = (ω K + ωd)y, t 0, y(0) = y 0, y(t) = e t(ω K+ωD) y 0. It is usual to split the exponential, e.g. with the Strang splitting e 2 tω K e tωd e 2 tω K = e t(ω K+ωD) + O ( t 3) or higher-order splittings. Typically, such high-order splittings are obtained either via the Yošida device or similar techniques, and result in palindromic expressions of the form e α tω K e β tωd e α 2tω K e β rtωdb e α 2tω K e β tωd e α tω K. {}}{{}}{{}}{ (Palindromy implies even order and has other welcome features.) 3

4 In one sense the above expression is good: it separates scales. Thus, exponentials with ω and ω are kept apart. In another sense it is bad: we need plenty of exponentials to attain reasonable order. Typically, the order of semidiscretization is very high: it is wasteful to accompany it with very low order of the time-stepping method! And in yet another sense it is ugly: ideally, we would have liked to combine numerics with asymptotics, namely for the arguments of exponentials to become progressively smaller while in the present case they are all of roughly the same order of magnitude. The challenge is to develop a methodology which attains all the advantages without any disadvantages: A high-order splitting that separates scales and produces an asymptotic expansion in increasing powers of h. 4

5 ALGEBRA OF OPERATORS The vector field is a linear combination of two linear operators: 2 x and (multiplication by) V. Let us consider the free Lie algebra F generated by 2 x and V. Note that [V, x] 2 = ( xv 2 ) 2( x V ) x, [[V, x], 2 x] 2 = ( xv 4 ) + 4( xv 3 ) x + 4( xv 2 ) x, 2 [[V, x], 2 V ] = 2( x V ) 2, [[[V, x], 2 x], 2 x] 2 = ( xv 6 ) 6( xv 5 ) x 2( xv 4 ) x 2 8( xv 3 ) x 3 and so on. In general, F g, where g is the Lie algebra g = n k=0 y k (x) k x : n Z +, y 0, y,..., y n smooth & periodic ht n k=0 is the height of an element in g. y k (x) x k = n. 5

6 PROPOSITION ht([x, Y ]) = (ht(x) + ht(y ) ) + for all X, Y g. COROLLARY Any nested commutator in F which has at least two more Vs than 2 xs is necessarily zero. For example, out of the 4 terms in the Hall basis of all the elements of F with at most 7 letters, are all zero. [[[V, x 2 ], V ], V ], [[[[V, 2 x ], V ], V ], V ], [[[[[V, 2 x ], 2 x ], V ], V ], V ], [[[[[V, x 2 ], V ], V ], V ], V ], [[[[V, 2 x ], V ], V ], [V, 2 x ]], [[[[[[V, x 2 ], 2 x ], V ], V ], V ], V ], [[[[[[V, 2 x ], V ], V ], V ], V ], V ], [[[[[V, x 2 ], V ], V ], V ], [V, 2 x ]], [[[[V, 2 x ], V ], V ], [[V, 2 x ], 2 x ]], [[[[V, x 2 ], V ], V ], [[V, 2 x ], V ]] But this is not the real reason why all this is important! The real reason is that V never walks on its own, it is always multiplied by a large parameter ω. The elimination of ω-rich terms allows us to derive asymptotic splittings! 6

7 THE SYMMETRIC ZASSENHAUS SPLITTING We commence from the symmetric Baker Campbell Hausdorff formula where e tx e ty e tx = e sbch(t;x,y ), sbch(t; X, Y ) = t(2x + Y ) 6 t3 [Y, [X, Y ]] + t 5 ( [X, [X, [X, [X, Y ]]]] [Y, [Y, [Y, [X, Y ]]]] 90 [X, [Y, [Y, [X, Y ]]]] + 45 [Y, [X, [X, [X, Y ]]]] + 60 [X, [X, [Y, [X, Y ]]]] + 30 [Y, [Y, [X, [X, Y ]]]]) + O ( t 7). It is possible to expand as far as we wish, noting that all the expansion terms live in the free Lie algebra generated by the letters X and Y. Such algebras possess convenient bases that can be constructed algorithmically, e.g. the Hall basis, the Lyndon basis and the Dynkin basis. 7

8 The classical Zassenhaus splitting is where e t(x+y ) = e tx e ty e t2 U 2 (X,Y ) e t3 U 3 (X,Y ) e t4 U 4 (X,Y ), U 2 (X, Y ) = 2 [X, Y ], U 3 (X, Y ) = 3 [Y, [X, Y ]] + 6 [X, [X, Y ]], U 4 (X, Y ) = 24 [[[X,Y ],X],X] 8 [[[X,Y ],X],Y ] 8 [[[X,Y ],Y ],Y ]. We are, however, interested in symmetric (i.e., palindromic) splittings. Moreover, powers of t are misleading: we need to reckon with three parameters: ω (large!), the number N of degrees of freedom once we semi-discretize (large!) and the time step t (small!). It makes sense to reduce this to a single currency by requiring N O(ω ρ ), t O ( ω σ) for some ρ, σ > 0. Note that x scales like O(N) = O(ω ρ ). In the sequel we assume that ρ = σ = 2. 8

9 SYMMETRIC ZASSENHAUS Setting for simplicity c =, we seek a splitting of the form e i( t)(ω 2 x +ωv ) e R 0e R e Rs e T s+e Rs e R e R 0, {}}{{}}{{}}{ where (recalling, in line with our currency conversion, that x O ( ω /2) ) R k = R k ( t, ω, V ) = O ( ω k+/2), k = 0,,..., s, T s+ = T s+ ( t, ω, V ) = O ( ω s /2). There are two options: we can start with the derivative or the potential. In this talk we ll start with the potential. To this end we let X = 2 τωv, Y = τω 2 x + τωv, where τ = i t = O ( ω /2) and rewrite the sbch formula as e Y = e X e sbch(x,y ) e X. 9

10 In what follows we restrict ourselves to an O ( ω 7/2) expansion, i.e. s = 2. After long calculation: Expanding sbch( 2 τωv, τω 2 x + τω) with the sbch formula and replacing elements from F by terms from g; 2 Throwing away all zero terms, but also all terms which are O ( ω α) for α 7 2 ; 3 Expressing everything in the Hall basis we obtain the asymptotic expansion sbch(x, Y ) = τω C 2 τ 3 ω C 4 24 τ 3 ωc τ 5 ω C τ 5 ωc 80 τ 5 ω C τ 5 ωc τ 7 ωc τ 7 ωc τ 7 ωc τ 7 ωc 40 + O ) (ω 7/2, 0

11 where all that survives of the Hall basis is C = 2 x, C 4 = ( 4 x V ) + 4( 3 x V ) x + 4( 2 x V ) 2 x, C 5 = 2( x V ) 2, C 0 = {6( 5 x V )( xv ) + 2( 4 x V )( 2 x V ) + 8( 3 x V )2 } 24{( 4 x )( xv ) + ( 3 x V )( 2 x V )} x 24( 3 x )( xv ) 2 x, C = 8( 2 x V )( xv ) 2, C 3 = {2( 5 x V )( xv ) 4( 4 x )( 2 x V ) 4( 3 x V )2 } + 8{( 4 x V )( xv ) ( 3 x V )( 2 x V )} x + 4( 3 x V )( xv ) 2 x, C 4 = 4( 2 x V )( xv ) 2, C 27 = 48( 3 x V )( xv ) 3, C 32 = 6( 3 x V )( xv ) ( 2 x V )2 ( x V ) 2, C 35 = 8( 3 x V )( xv ) 3 6( 2 x V )2 ( x V ) 2, C 40 = 32( 2 x V )2 ( x V ) 2. 4 We next express the remaining commutators as terms in the larger Lie algebra g. But here we have a problem!!! Ultimately, we will replace derivatives by (finite-dimensional) differentiation matrices. Even derivatives yield symmetric matrices and multiplication by i results in skew-hermitian matrices good! But, by the same token, odd derivatives skew-symmetric matrices (after multiplication by i) Hermitian matrices bad! We must somehow get rid of odd derivatives!!!

12 5 Replacement of odd derivatives: The main idea is to replace odd derivatives by linear combinations of even derivatives: x [ x ] y(x) x = 2 y(x) 3 x = y (x) 2 x x and so on. The outcome is 0 y(ξ) dξ 2 x 2 y (x) x [ x 0 x 0 y(ξ) dξ 0 y(ξ) dξ 4 x + 4 y (x) 2 2 x[y (x) ] y(ξ) dξ ] sbch(x, Y ) = τω 2 x 2 τ 3 ω { ( 4 x V ) x [( 2 x V ) ] + 2( 2 x V ) 2 x ] + 2 τ 3 ω( x V ) τ 5 ω {6( 5 x V )( xv ) + 2( 4 x V )( 2 x V ) + 4( 3 x V )2 2 2 x [( 3 x V )( xv ) ] 2( 3 x V )( xv ) 2 x } 90 τ 5 ω( 2 x V )( xv ) 2 90 τ 5 ω {3( 5 x V )( xv ) 2( 4 x V )( 2 x V ) 4( 3 x V )2 + 2 x 2 [( 3 x V )( xv ) ] 4 x 2 [( 2 x V )2 ] + 4( x 3 )( xv ) x 2 8( 2 x )2 x 2 } + 72 τ 5 ω( x 2 V )( xv ) τ 7 ω( x 3 V )( xv ) τ 7 ω{2( x 3 V )( xv ) 3 + 4( x 2 V )2 ( x V ) 2 } τ 7 ω{( x 3 V )( xv ) 3 + 2( x 2 V )2 ( x V ) 2 } ) τ 7 ω( x 2 V )2 ( x V ) 2 + O (ω 7/2., 2

13 6 Arranging in decreasing powers of ω: = sbch(x, Y ) ω /2 {}}{ τω 2 x + 2 τ 2 ω( x V ) 2 } ω 3/2 {}}{ 360 τ 5 ω( xv 2 )( x V ) 2 6 τ 3 ω { x[( 2 xv 2 ) ] + ( xv 2 ) x} 2 ω { 5/2 }}{ 2 τ 3 ω ( xv 4 ) ω 5/2 {}}{ 80 τ 5 ω { 5 x[( 2 xv 3 )( x V ) ] + 4 x[( 2 xv 2 ) 2 ] 5( x)( 3 x V ) x 2 + 4( xv 2 ) 2 x} 2 ω { 5/2 }}{ 520 τ 7 ω{09( xv 3 )( x V ) ( xv 2 ) 2 ( x V ) 2 } +O ( ω 7/2). 3

14 We now set R 0 = X old and X new = 2 τω 2 x 24 τ 3 ω( x V ) 2, Y new = sbch(x old, Y old ). We repeat steps 6 and this results in {}}{ sbch(x, Y ) = 360 τ 5 ω( x 2 V )( xv ) 2 6 τ 3 ω { x 2 [( 2 x V ) ] + ( 2 x V ) 2 x } ω 5/2 {}}{ 2 τ 3 ω ( 4 x V ) ω 3/2 ω 5/2 {}}{ 80 τ 5 ω { 5 2 x [( 3 x V )( xv ) ] x [( 2 x V )2 ] 5( 3 x )( xv ) 2 x + 4( 2 x V )2 2 x } ω 5/2 {}}{ ) 520 τ 7 ω{09( x 3 V )( xv ) ( x 2 V )2 ( x V ) 2 } +O (ω 7/2. 4

15 Thus, R = X old and we again set X new as minus half the leading term of Y new = sbch(x old, Y old ). This is the last such step, leading to R 2 and the remainder, T 3. To sum up, R 0 = 2 τωv = O( ω /2), R = 2 τω 2 x + 24 τ 3 ω( x V ) 2 = O ( ω /2), R 2 = 2 τ 3 ω { 2 x[( 2 xv ) ] + ( 2 xv ) 2 x} τ 5 ω( 2 xv )( x V ) 2 = O ( ω 3/2), T 3 = 2 τ 3 ω ( 4 xv ) + 80 τ 5 ω { 5 2 x[( 3 xv )( x V ) ] x[( 2 xv ) 2 ] 5( 3 x)( x V ) 2 x + 4( 2 xv ) 2 2 x} τ 7 ω{09( 3 xv )( x V ) ( 2 xv ) 2 ( x V ) 2 } = O ( ω 5/2). 5

16 Note that, once we replace derivatives by (even-degree!) differentiation matrices, e R 0 is a diagonal matrix, while matrix/vector products with R, R 2 and T 3 can be computed with FFTs as part of a Krylov-subspace-based computation of their exponentials. Stability Suppose that each (2r)th derivative is replaced by the Hermitian differentiation matrix K 2r. Since τ = i t always features with an odd power, K 2r, as well as the diagonal matrix that we obtain from discretizing multiplication operators, are multiplied by ±i and become skew-hermitian. For example R 0 2 ( t)ωid V, R 2 ( t)ω ik 2 24 ( t)3 ωid V 2 x. However... Because of replacement of odd derivatives, we have expressions of the form idh and ihd where D is diagonal and H Hermitian and they are not skew-hermitian! Is this a problem? 6

17 Not at all! Recalling that each term in F is multiplied by ±i, we note that FLA(i 2 x, iv ) su. All our operations, using sbch but also the replacement of odd derivatives, are Lie-algebra compliant (linear combinations and commutators), therefore everything we have obtained in the course of our algorithm stays within su. Thus, as a sanity check, K symmetric, D diagonal i(kd + DK) skew-hermitian, hence R 2 2 ( t)3 ω i(k 2 D V 2 x + D V 2 x K 2 ) ( t)5 ωid Vxx V 2 x T 3 2 τ 3 ω D 4 x V + 80 τ 5 ω {4(K 2 D ( 2 x V ) 2 + D ( 2 x V ) 2 K 2 ) 5(K 2 D ( 3 x V )( x V ) + D ( 3 x V )( xv ) K 2)} τ 7 ω{09d ( 3 x V )( x V ) D ( 2 x V ) 2 ( x V ) 2 } and it is easy to verify that bothr 2 and T 3 are skew-hermitian. THEOREM It is true that R k su, k = 0,,..., s, and T s+ su. 7

18 LEADING WITH THE DERIVATIVE We have commenced the splitting with X = τωv an alternative is to start from X = τω x 2. On the face of it, this makes little sense, because we want to start with the larger O ( ω /2) term. However, let s try... R 0 = 2 τω 2 x = O ( ω /2), R = 2 τωv = O( ω /2), R 2 = 2 τ 3 ω( x V ) 2 = O ( ω /2), R 3 = 24 τ 3 ω { 2 x[( 2 xv ) ] + ( 2 xv ) 2 x} 480 τ 5 ω( 2 xv )( x V ) 2 = O ( ω 3/2), T 4 = 24 τ 3 ω ( xv 4 ) τ 5 ω { 30 2 x[( xv 3 )( x V ) ] + 30 ( 3 xv )( x V ) x x[( xv 2 ) 2 ] 60 ( 2 xv ) 2 x} 2 τ 7 ω{ ( 3 xv )( x V ) ( 2 xv ) 2 ( x V ) 2 } = O ( ω 5/2). On the face of it, more terms. However... R 0 is a circulant and e R 0 can be computed with a single FFT, R and R 2 are diagonal only R 3 = O ( ω 3/2) and T 4 = O ( ω 5/2) require Krylov subspace methods to compute their exponential! 8

19 SPACE DISCRETIZATION In principle, we have three options to discretise 2s x to spectral accuracy: Spectral methods: We project in L 2 the solution on Nth-degree trigonometric polynomials: the unknowns are Fourier coefficients, K is diagonal and D a circulant. Spectral collocation: We interpolate at equally-spaced points by N th-degree trigonometric polynomials: the unknowns are nodal values, K is a circulant and D is diagonal. Pseudospectral method: We wrap around a finite-difference method an infinite number of times: again, the unknowns are nodal values, K is a circulant and D is diagonal. Except that, having committed already an O ( ω 7/2) error, we don t need spectral accuracy! 9

20 FINITE DIFFERENCES We discretise in space to the same O ( ω 7/2) = O ( ( x) 7) (actually, O ( ( x) 8) ) accuracy using finite differences: u m u(m x, ), m = N,..., N, x = N+, and 2 u m ( x) 2( 560 u m u m 3 5 u m u m u m Therefore: u m+ 5 u m u m u m+4).! K is a 9-diagonal circulant, and! D is a diagonal matrix. In that case all matrix-vector products are O(N) operations! 20

21 u(x) = 2 + sin πx 2

22 u(x) = e cos πx 22

23 So, does the error converge to a multiple of some mystery function? Yes: in general, for the rth finite difference of order 2r we have r m= r u (x) ( x) 2 r m= r a m u(x + m x) a m cos mθ = θ 2 + c r θ 2r+2 + O ( θ 2r+4), whereby the error is c r ( x) 2r u (2r+2) (x) + O ( ( x) 2r+2). In our case c 4 = 350, therefore the error is 350 ( x)8 u (0) (x) + O ( ( x) 0), x = N

24 The error (in blue) and the error estimate (in red) for u(x) = (2 + sin πx) and ω =

25 COMPUTING THE EXPONENTIALS Our method produces a dichotomy: Arguments of the exponentials are either trivial or small! They are trivial if the matrix is diagonal or is just a scaled differentiation matrix in both cases computing the exponential is straightforward. They are small if the argument is A = O ( ω α) for some α > 0. In that case the mth term in the Krylov basis {v, Av, A 2 v,, A 3 v...} is O ( ω mα). Therefore, using estimates of Hochbruck, Lubich & Selhofer, it is possible to prove that we need ridiculously small dimension of a Krylov subspace to compute e A v to O ( ω 7/2), say. Leading by derivative, R 3 = O ( ω 3/2) and T 4 = O ( ω 5/2) require dimensions 3 and 2 respectively! 25

26 CHALLENGES Other pairs (ρ, σ). While the generalisation is straightforward (as a concept: the computation is fiendish), the interesting bit is to explore what are good choices of ρ, σ (0, ). Time-dependent potentials. Here we need to marry Magnus expansions with symmetric Zassenhaus. Preliminary results indicate that, although the algebra is formidable, this can be done: the underlying Free Lie algebra is much more complicated but still embedded in g and height restrictions apply. Nonlinear Schrödinger. Can we extend our method to the nonlinear Schrödinger equation i hu t = h2 2m u V (x)u + λ u 2 u? Straightforward generalisation does not work but is there a clever way this set of ideas can be extended? Other equations. Can this work with the wave equation? With nonlinear wave equations, e.g. Klein Gordon? With Hamiltonian ODEs of the form H(p, q) = H (p, q) + H 2 (p, q), where H 2 H? 26

27 Happy birthday Peter!!! 27

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