A Lattice Green Function Introduction. Abstract

Transcription

1 August 5, 25 A Lattice Gree Fuctio Itroductio Stefa Hollos Exstrom Laboratories LLC, 662 Nelso Park Dr, Logmot, Colorado 853, USA Abstract We preset a itroductio to lattice Gree fuctios. Electroic address: stefa@exstrom.com; URL: 1

2 I. INTRODUCTION This is a short cocise itroductio to the cocept of a lattice Gree fuctio (LGF). The LGF is the discrete space couterpart to the more familiar cotiuous space Gree fuctio that has become such a versatile tool i may areas of theoretical physics. Some familiarity with the more commo uses of Gree fuctios, such as i the solutio of partial differetial equatios, is helpful i what follows but ot strictly ecessary. Excellet itroductios to Gree fuctios ca be foud i Barto [1], Duffy [2], ad Ecoomou [3]. The LGF is ofte used i codesed matter, ad statistical physics (radom walk theory). A good discussio of some uses of the LGF ca be foud i Cserti [4]. It also appears, although most ofte ot by ame, whe the fiite differece approximatio is used to solve partial differetial equatios. It is through this applicatio that we will itroduce the cocept of a LGF. I particular we will use the LGF to show how the discretized Poisso equatio ca be solved i a ifiite cubic (3 dimesioal) ad square (2 dimesioal) lattice. It is perhaps appropriate to itroduce the LGF i this way sice solvig the Poisso equatio was George Gree s origial motivatio for developig his epoymous fuctios [5]. A great deal of research has bee doe o lattice Gree fuctios over the last fifty years or so ad other itroductios do exist, see for example Katsura et al [6] ad the two recet papers by Cserti [4, 7]. The hope is that the simple examples give i this itroductio will be accessible to the widest possible audiece. The oly kowledge assumed o the part of the reader is some familiarity with Dirac vector space otatio ad a uderstadig of eigevalues ad eigevectors. II. THREE DIMENSIONAL DISCRETE POISSON EQUATION For the cubic lattice let ˆx 1, ˆx 2, ad ˆx 3 be a set of orthogoal uit vectors, so that ˆx i ˆx j = δ(i, j). If the lattice spacig is a the the primitive lattice vectors are a i = a ˆx i ad all poits i the lattice are give by the lattice vectors r = 1 a a a 3 i = iteger (1) Usig this otatio, the Poisso equatio o a cubic lattice takes the followig form. 3 i=1 [φ( r + a i ) 2φ( r ) + φ( r a i )] = f ( r ) (2) We will refer to this as the discrete Poisso equatio or DPE from here o. To fully defie the equatio, the size of the lattice ad the boudary coditios eed to be specified. 2 These are

3 however ot importat for the curret discussio ad will be specified later. I the limit as the lattice spacig goes to zero this becomes the cotiuous Poisso equatio. 3 2 φ( r) i=1 xi 2 = g( r) (3) Eq. 2 ca be regarded as a fiite differece approximatio of eq. 3 with f ( r ) = a 2 g( r ). Much of the followig developmet will be i terms of Dirac vector space otatio. I this otatio the DPE is Here L deotes the lattice Laplacia operator. L φ = f (4) If we let deote the lattice basis vector associated with the lattice poit r the φ = φ( r ) ad f = f ( r ). I the lattice basis, the vectors φ ad f are ad eq. 4 is φ = I terms of matrix ad vector elemets this becomes φ = φ( r ) (5) f = f = f ( r ) (6) l L φ = l f (7) L l φ( r ) = f ( r l ) (8) The matrix elemets L l ca be idetified by comparig eq. 8 with eq. 2. 6δ(l, ) L l = 1 if r r l = a otherwise (9) This ca also be expressed as follows. L l = 6δ( r l, r ) + Now we wat to solve eq. 4 for φ. At least formally, the solutio is 3 i=1 [δ( r l + a i, r ) + δ( r l a i, r )] (1) φ = L 1 f (11) 3

4 The problem therefore ivolves fidig L 1 = G, where the operator G is what we will call the lattice Gree fuctio. We ca get a expressio for the matrix elemets of G by usig a eigevector expasio. It is easy to show that G ad L have the same eigevectors, ad if λ is a eigevalue of L the 1/λ is a eigevalue of G. The first step the is to fid the eigevalues ad eigevectors of L. Util ow, o assumptios have bee made about the size of the lattice or the boudary coditios. We begi by assumig a fiite lattice with N i poits i the directio a i ad periodic boudary coditios. Periodic boudary coditios mea that for ay lattice fuctio v( r ), the followig will be true v( r + N i a i ) = v( r ) i = 1,2,3 (12) With these assumptios the eigevalue problem for L ca ow be solved. Write the eigevalue equatio for L as follows. I the lattice basis the eigevalue equatio is or i terms of matrix elemets Usig eq. 9 for the matrix elemets of L, eq. 15 becomes. L v m = λ m v m (13) l L v m = λ m l v m (14) L l v m ( r ) = λ m v m ( r l ) (15) v m ( r l + a 1 )+v m ( r l a 1 )+v m ( r l + a 2 )+v m ( r l a 2 )+v m ( r l + a 3 )+v m ( r l a 3 ) 6v m ( r l ) = λ m v m ( r l ) (16) We will ow show that periodic boudary coditios, v m ( r l +N i a i ) = v m ( r l ), require that v m ( r l ) have the followig form v m ( r l ) = Ae i k m r l (17) We set the vector k m equal to k m = m 1 N 1 b 1 + m 2 N 2 b 2 + m 3 N 3 b 3 (18) where m i =,1,2,...,N i 1 ad the vectors b i are reciprocal lattice vectors equal to b i = 2π a ˆx i (19) 4

5 so that we have b i a j = 2πδ(i, j) (2) With this defiitio of k m it is easy to show that eq. 17 obeys the periodic boudary coditios v m ( r l + N i a i ) = Ae i k m r l e i k m N i a i = Ae i k m r l e i2πm i (21) = v m ( r l ) The costat A is chose so that the eigevector is ormalized. v m v m = v m l l v m (22) l = Ae i k m r l Ae i k m r l l = A 2 N 1 N 2 N 3 therefore let A = 1/ N 1 N 2 N 3. We ca ow fid the eigevalues by substitutig eq. 17 ito eq. 16. This gives ( ) λ m = 2 cos k m a 1 + cos k m a 2 + cos k m a 3 3 ( = 2 cos 2πm 1 + cos 2πm 2 + cos 2πm ) 3 3 N 1 N 2 N 3 Now that the eigevalue problem has bee solved we ca express L ad G = L 1 i terms of the eigebasis. For L we have The matrix elemets of L are the (23) L = λ m v m v m (24) m L l = λ m l v m v m (25) m 1 = N 1 N 2 N 3 λ m e i k m ( r l r ) m It is ot too difficult to show that this equatio gives the same results as i eq. 9. For G = L 1 we have ad the matrix elemets are G l = m G = m v m v m λ m (26) l v m v m = 1 e λ m N 1 N 2 N 3 i k m ( r l r ) (27) m λ m 5

6 Note that G l depeds oly o the differece r l r so that G has a circulat matrix represetatio. Let r p = r l r the ad Usig this otatio, eq. 27 becomes 1 G l = G( r p ) = N 1 N 2 N 3 r p = (l 1 1 ) a 1 + (l 2 2 ) a 2 + (l 3 3 ) a 3 (28) = p 1 a 1 + p 2 a 2 + p 3 a 3 k m r p = 2π m 1p 1 N 1 + 2π m 2p 2 N 2 + 2π m 3p 3 N 3 (29) N 1 1 m 1 = N 2 1 m 2 = N 3 1 m 3 = e i 2πm 1 p 1 N 1 e i 2πm 2 p 2 N 2 e i 2πm 3 p 3 N 3 ( ) (3) 2 3 cos 2πm 1 N 1 cos 2πm 2 N 2 cos 2πm 3 N 3 Eq. 3 gives the matrix elemets of the lattice Gree fuctio of the DPE for a fiite lattice with periodic boudary coditios. Note that this is essetially a Fourier series expasio of the matrix elemets which is possible because of the periodic boudary coditios. For other boudary coditios such as G(N i a i ) =, the expasio would have to be i terms of a sie series. We will ow go from a fiite lattice to a ifiite lattice by lettig N i, i = 1,2,3. This meas goig from the Fourier series represetatio of eq. 3 to a Fourier trasform represetatio of the matrix elemets. I eq. 3 let Whe m i is icremeted by 1 the chage i x i is x i = 2πm i N i (31) x i = 2π N i The summatios i eq. 3 ca the be writte as or 1 N i = x i 2π (32) 1 N i N i 1 m i = 2π(1 1 N i ) x i = x i 2π (33) ad i the limit N i the summatio becomes a itegral. For a ifiite lattice eq. 3 the becomes G( r p ) = 1 (2π) 3 Z 2π Z 2π Z 2π 1 2π Z 2π dx i (34) e ix 1 p 1 e ix 2 p 2 e ix 3 p 3 2(3 cosx 1 cosx 2 cosx 3 ) dx 1dx 2 dx 3 (35) 6

7 Note that the itegrad has a period of 2π i each of the variables so that the limits of itegratio ca be chaged to the more symmetric G( r p ) = 1 (2π) 3 Z π π Z π Z π π π e ix 1 p 1 e ix 2 p 2 e ix 3 p 3 2(3 cosx 1 cosx 2 cosx 3 ) dx 1dx 2 dx 3 (36) The itegral ca be further simplified by lookig at the parity properties of the itegrad. Multiplyig the e ix i p i = cosx i p i + isix i p i factors ad leavig out the resultig odd terms reduces the itegral to G(p 1, p 2, p 3 ) = 1 2π 3 Z π Z π Z π cosx 1 p 1 cosx 2 p 2 cosx 3 p 3 3 cosx 1 cosx 2 cosx 3 dx 1 dx 2 dx 3 (37) Clearly G is a fuctio oly of the parameters p 1, p 2, ad p 3 ad it is a eve fuctio of these parameters. G is also symmetric uder ay permutatio of the parameters. All the uique values of G are therefore cotaied i the wedge p 1 p 2 p 3. We will ow derive a recurrece equatio that the matrix elemets of G obey. By defiitio we have LG = I which i the lattice basis is Substitutig i eq. 1 for L l gives 6G( r l r m ) + l L G m = l m (38) L l G m = δ(l,m) L( r l r )G( r r m ) = δ(l,m) 3 i=1 [G( r l + a i r m ) + G( r l a i r m )] = δ(l,m) (39) Now usig the otatio, r l r m = (l 1 m 1 ) a 1 + (l 2 m 2 ) a 2 + (l 3 m 3 ) a 3 = p 1 a 1 + p 2 a 2 + p 3 a 3, eq. 39 becomes 6G(p 1, p 2, p 3 )+G(p 1 +1, p 2, p 3 )+G(p 1 1, p 2, p 3 )+G(p 1, p 2 +1, p 3 )+G(p 1, p 2 1, p 3 )+G(p 1, p 2, p 3 +1)+ Eq. 4 simplifies cosiderably for some specific values of p 1, p 2, ad p 3. I particular for p 1 = p 2 = p 3 = we get G(1,,) = G(,,) 1 6 Where the symmetry properties of G have bee used, i.e. G(1,,) = G( 1,,) = G(,1,) = G(, 1,) = G(,,1) = G(,, 1). Lettig p 1 = p 2 = p 3 = p i eq. 4, we have (4) (41) 2G(p, p, p) = G(p + 1, p, p) + G(p, p, p 1) (42) 7

8 Lettig p 1 = p, p 2 = p 3 = i eq. 4 gives G(p + 1,,) = 6G(p,,) 4G(p,1,) G(p 1,,) (43) Lettig p 1 = p 2 = p, p 3 = i eq. 4 gives 3G(p, p,) = G(p + 1, p,) + G(p, p 1,) + G(p, p,1) (44) I geeral for p 1 = l, p 2 = m, p 3 = with l, m, ad ot all equal to zero, we have 6G(l,m,) = G(l +1,m,)+G(l 1,m,)+G(l,m+1,)+G(l,m 1,)+G(l,m,+1)+G(l,m, 1) Additioal recursio equatios were developed by Duffi ad Shelly. These recursio equatios, alog with some of those give above ad some relatios due to Horiguchi ad Morita, allowed Glasser ad Boersma to fid a expressio for the geeral matrix elemet G(l,m,) that ivolves kowig oly G(,,), which is give by the itegral G(,,) = 1 2π 3 Z π Z π Z π (45) dx 1 dx 2 dx 3 3 cosx 1 cosx 2 cosx 3 (46) This itegral was first evaluated by Watso i terms of complete elliptic itegrals. It was the show by Glasser ad Zucker to be expressible i terms of gamma fuctios as ( ) ( ) ( ) ( ) G(,,) = 96π 3 Γ Γ Γ Γ A idetity due to Borwei ad Zucker allows this to be simplified to ( ) ( ) G(,,) = 96π 3 Γ2 Γ Joyce [8] has also developed some recursio equatios that allow G(l,m,) to be calculated for arbitrary values of l,m,. He arrives at the same formula as Glasser ad Boersma via a differet method ad also derives a asymptotic formula for G(l,m,). I some very recet work, Joyce [9] gives some formulas that allow the diagoal elemets, G(,, ), to be calculated very accurately for arbitrary values of. He also gives asymptotic formulas for G(,,). (47) (48) III. TWO DIMENSIONAL DISCRETE POISSON EQUATION The same procedure give above ca be used to fid the lattice Gree fuctio for the two dimesioal Poisso equatio. I this case the lattice vectors are r = 1 a a 2 (49) 8

9 The matrix elemets of the lattice Laplacia are 4δ( r l, r ) L l = 1 if r r l = a otherwise (5) Which ca also be expressed as L l = 4δ( r l, r ) + δ( r l + a 1, r ) + δ( r l a 1, r ) + δ( r l + a 2, r ) + δ( r l a 2, r ) (51) The eigevector expasio of the lattice Laplacia matrix elemets are L l = 1 N 1 N 2 λ m e i k m ( r l r ) m (52) k m = m 1 N 1 b 1 + m 2 N 2 b 2 m i = iteger (53) ( ) λ m = 2 cos k m a 1 + cos k m a 2 2 ( = 2 cos 2πm 1 + cos 2πm ) 2 2 N 1 N 2 The matrix elemets of the lattice Gree fuctio are expaded i the eigebasis as which if we let r p = r l r, ca be expressed as G l = G( r p ) = 1 N 1 N 2 For the ifiite lattice this becomes (54) G l = 1 e N 1 N 2 i k m ( r l r ) (55) m λ m N 1 1 m 1 = N 2 1 m 2 = G( r p ) = G(p 1, p 2 ) = 1 2π 2 Z π e i 2πm 1 p 1 N 1 e i 2πm 2 p 2 N 2 ( ) (56) 2 2 cos 2πm 1 N 1 cos 2πm 2 N 2 Z π cosx 1 p 1 cosx 2 p 2 2 cosx 1 cosx 2 dx 1 dx 2 (57) G is a eve fuctio of the parameters p 1 ad p 2, ad it is symmetric uder ay permutatio of the parameters. All the uique values of G are therefore cotaied i the wedge p 1 p 2. There is oe problem with eq. 57. The itegral is diverget for all values of p 1 ad p 2. We ca fix this by usig the shifted Gree fuctio. g(p 1, p 2 ) = G(,) G(p 1, p 2 ) = 1 2π 2 Z π 9 Z π 1 cosx 1 p 1 cosx 2 p 2 2 cosx 1 cosx 2 dx 1 dx 2 (58)

10 The itegral ow exists for all values of p 1 ad p 2. Usig g(p 1, p 2 ) istead of G(p 1, p 2 ) will provide a solutio to the DPE as log as the sum of the source terms, f ( r ), over all the lattice sites is equal to zero. To demostrate this, first ote that the solutio to the DPE i terms of G is give by Now if we have the eq. 59 ca also be writte as φ( r l ) = G l f ( r ) (59) f ( r ) = (6) φ( r l ) = (G ll G l ) f ( r ) (61) where G ll = G( r l r l ) = G(,), G l = G( r l r ) = G(p 1, p 2 ) ad G ll G l = g l. The solutio to the DPE i terms of the shifted Gree fuctio is the where g l = g( r l r ) = g(p 1, p 2 ) = G(,) G(p 1, p 2 ). φ( r l ) = g l f ( r ) (62) From the above discussio, you ca see that i a ubouded two dimesioal space or lattice the DPE is oly solvable if the sources add up to zero. A physical example of this is i two dimesioal electrostatics. The charge uits i two dimesioal electrostatics are actually parallel, ifiite lie charges embedded i a three dimesioal space. For a sigle lie charge, the potetial at ay fiite distace from the lie will be ifiite. For two lies of opposite charge the potetial is fiite i the space surroudig the lies. Note that we are assumig a ubouded space with the zero poit potetial at ifiity. Aother example comes from the theory of radom walks. I oe ad two dimesios a radom walker is guarrateed to evetually retur to its startig positio, while i three dimesios it may ever do so. To see how this is related to the DPE, see the excellet book by Doyle ad Sell [1] o radom walks i electrical etworks. For aother example see Cserti s paper [4] o usig the lattice Gree fuctio to calculate the resistace betwee two poits i a ifiite etwork of resistors. We will ow preset some recurrece equatios that the matrix elemets of the Gree fuctio obey. As i the three dimesioal case these ca easily be foud from the defiig relatio LG = I. This gives the geeral recurrece 4G(p 1, p 2 )+G(p 1 +1, p 2 )+G(p 1 1, p 2 )+G(p 1, p 2 +1)+G(p 1, p 2 1) = δ(p 1,)δ(p 2,) 1 (63)

11 For p 1 = p 2 = we have For p 1 = p, p 2 = we have G(,) G(1,) = 1 4 (64) 4G(p,) = G(p + 1,) + G(p 1,) + 2G(p,1) (65) For p 1 = p 2 = p we have 2G(p, p) = G(p + 1, p) + G(p, p 1) (66) Ad i geeral for p 1 = l ad p 2 = m we have 4G(l,m) = G(l + 1,m) + G(l 1,m) + G(l,m + 1) + G(l,m 1) (67) A additioal recurrece equatio for the diagoal elemets is [11] (2 + 1)G( + 1, + 1) 4G(,) + (2 1)G( 1, 1) = (68) Sice the coefficiets i each of these equatios adds to zero you ca see that the shifted Gree fuctio, g(p 1, p 2 ) = G(,) G(p 1, p 2 ) must obey the same recurrece equatios. These equatios for g are listed below g(1,) = 1 4 (69) 4g(p,) = g(p + 1,) + g(p 1,) + 2g(p,1) (7) 2g(p, p) = g(p + 1, p) + g(p, p 1) (71) 4g(l,m) = g(l + 1,m) + g(l 1,m) + g(l,m + 1) + g(l,m 1) (72) (2 + 1)g( + 1, + 1) 4g(,) + (2 1)g( 1, 1) = (73) Ackowledgmets Thaks. [1] G. Barto, Elemets of Gree s Fuctios ad Propagatio : Potetials, Diffusio, ad Waves (Oxford Uiversity Press, 1989). 11

12 [2] D. G. Duffy, Gree s Fuctios with Applicatios (Chapma ad Hall/CRC, 21). [3] E. Ecoomou, Gree s Fuctios i Quatum Physics (Spriger-Verlag, 1983). [4] J. Cserti, Amer. J. Phys. 68, 896 (2), cod-mat/ [5] G. Gree, A Essay o the Applicatio of Mathematical Aalysis to the Theories of Electricity ad Magetism (Nottigham, 1828). [6] S. Katsura, T. Morita, S. Iawashiro, T. Horiguchi, ad Y. Abe, J. Math. Phys. 12, 892 (1971). [7] J. Cserti, G. David, ad A. Piroth, Amer. J. Phys. 7, 153 (22), cod-mat/ [8] G. S. Joyce, J. Phys. A 35, 9811 (22). [9] G. S. Joyce ad R. T. Delves, J. Phys. A 37, 3645 (24). [1] P. G. Doyle ad J. L. Sell, Radom walks ad electric etworks (2), math.pr/157. [11] T. Morita, J. Math. Phys. 12, 1744 (1971). 12