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11S Poisson Summation Formula Masatsugu Sei Suzuki Department of Physics, SUNY at Bimghamton (Date: January 08, 2011)

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11S Poisson Summation Formula Masatsugu Sei Suzuki Department of Physics, SUNY at Bimghamton (Date: January 08, 2011) Chapter 11S Poisson summation formula Masatsugu Sei Suzuki Department of Physics, SUNY at Bimghamton (Date: January 08, 2011) Poisson summation formula Fourier series Convolution Random walk Diffusion Magnetization Gaussian distribution function 11S.1 Poisson summataion formula For appropriate functions f(x), the Poisson summation formula may be stated as f (x n) 2 F(k m) , (1) n m where m and n are integers, and F(k) is the Fourier transform of f(x) and is defined by 1 F(k) F[ f (x)] eikx f (x)dx . 2 Note that the inverse Fourier transform is given by 1 f (x) eikx F(k)dk . 2 The factor of the right hand side of Eq.(1) arises from the definition of the Fourier transform. ((Proof)) The proof of Eq.(1) is given as follows. 1 f (x n) F(k)dk eikn n 2 n 1 We evaluate the factor I eikn . n It is evident that I is not equal to zero only when k = 2m (m; integer). Therefore I can be expressed by I A (k 2m) , m where A is the normalization factor. Then 1 f (x n) F(k)dk[A (k 2m)] n 2 m A F(k) (k 2m)dk 2 m A F(k 2m) 2 m A F(k 2n) 2 n The normalization factor, A, is readily shown to be 2 by considering the symmetrical case n2 f (x n) e , 1 2 F(k 2n) en 2 ________________________________________________________________________ ((Mathematica)) 2 f x_ Exp x2 ; FourierTransform f x , x, k, FourierParameters 0, 1 2 k 4 2 ______________________________________________________________________ Since 2 f (x n) A 2 . F(k 2n) or I eikn 2 (k 2m) . (2) n m Using this formula, we have f (x n) 2 F(k 2n) . (Poisson sum formula) n n 11S.2 Summary When we put k 2x in I of Eq.(2) e2ixn 2 (2x 2m) n m 2 (x m) 2 m (x m) m or 3 e2ixn (x m) . (3) n m 11S.3. Convolution of Dirac comb: another method in the derivation of Poisson sum formula The convolution of functions f(x) and g(x) is defined by 1 1 f * g f (x )g( )d f ( )g(x )d 2 2 The Fourier transform of the convolution is given by F[ f * g] F[ f ]F[g]. Here we assume that g(x) (x na) . (Dirac comb) n The Fourier transform of g(x) is 1 1 G(k) F[g(x)] eikx (x na)dx eikna . 2 n 2 n The convolution f*g is obtained as 1 1 f * g f (x ) ( na)d f (x na) . (4) 2 n 2 n The Fourier transform of the convolution is 1 F[ f * g] F(k)G(k) F(k) eikna . 2 n We use the Poisson summation formula; 4 ka eikna eikna ( m) n n m 2 a 2m [ (k )] m 2 a 2 2m (k ) a m a where e2ixn (x m) . n m ka with x . Then we get 2 1 1 F[ f * g] F(k) eikna F(k) eikna 2 n 2 n 1 2 2m (k )F(k) 2 a m a 1 2 2m 2m (k )F(k ) 2 a m a a The inverse Fourier transform of F[ f * g] is obtained as 1 1 1 2 2m 2m f * g F[ f * g]eikxdk eikxdk (k )F(k ) 2 2 2 a m a a 1 2 2m F(k m) eikxdk (k ) a m a a 2m 1 2 i x F(k m)e a a m a where 5 2 2 1 i mx F(k m) e a f (x)dx . a 2 Finally we get 2m 1 1 2m i x f * g f (x na) F(k )e a . (5) 2 n a m a or 2m 2 2m i x f (x na) F(k )e a . n a m a When a = 1, we get f (x n) 2 F(k 2m)ei2mx . (6) n m When x = 0, f (x n) 2 F(k 2m) . (7) n m This is the Poisson sum formula. 11S.3 Fourier transform of periodic function We consider a periodic function N(x); N(x a) N(x) , where a is the periodicity. The function N(x) can be described by 2m 2 2m i x N(x) f (x na) F(k )e a a a n m iGx NGe G 6 Note that f(x) is defined only in the limited region (for example, -a/2≤x≤a/2). G is the reciprocal lattice defined by 2 G m . a The Fourier coefficient NG is given by 2 N F(k G) G a . 1 1 a / 2 eiGx f (x)dx eiGx f (x)dx a a a / 2 where f(x) is just like a Gaussian distribution function around x = 0. N x x a Fig. Plot of N(x) as a function of x. a is the lattice constant of the one-dimensional chain. ((Example)) Suppose that f(x) is given by a Gaussian distribution, 7 1 x2 f (x) exp( ) . 2 2 2 Then we get 1 a / 2 1 1 1 2iG 2 1 2iG 2 N eiGx f (x)dx exp( G2 2 )[erf ( ) erf ( )] , G a a / 2 2 2 2 2 2 2 where erf(x) is the error function and is defined by z 2 2 erf (z) et dt 0 2 2 Figure shows the intensity N vs n, where G n . G a 2 NG 1.0 0.8 0.6 0.4 0.2 n 2 4 6 8 10 2 2 Fig. a = 1. = 0.1. G n . The intensity N vs n (= integer). a G 11S.10 Fourier series 8 Suppose that the function N(x) is a periodic function of x with the periodicity a. Then we have 2m 2 2m i x N(x) f (x na) F(k )e a a a n m iGx NGe G 2 where G m (m: integer). a ((Example-1)) f(x) = x for |x|<a/2, with a = 2 1 1 2i(1)n F(k G n) xexp(inx)dx for n ≠ 0. 2 1 n 2 1 1 F(k G 0) xdx 0 for n = 0. 2 1 1 i(1)n 1 i(1)n N(x) eimx eimx n n n n n0 We make a plot of N(x) as a function of x for the summation for n = -nmax and nmax with nmax = 50. 9 N x 1.0 nmax = 50 0.5 x -3 -2 -1 1 2 3 -0.5 -1.0 Fig. nmax = 50. The Gibbs phenomenon is clearly seen. ((Example-2)) f(x) = 0 for -1<x<0 and 1 for 0<x<1. a = 2 1 1 i(1)n[1 (1)n ] F(k G n) exp(inx)dx for n ≠ 0. 2 0 n 2 1 1 1 F(k G 0) dx , for n = 0. 2 0 2 1 1 i(1)n[1 (1)n ] N(x) eimx 2 n 2n n0 10 N x 1.0 0.8 nmax = 100 0.6 0.4 0.2 x -2 -1 1 2 Fig. nmax = 100. The Gibbs phenomenon is clearly seen. _________________________________________________________________________ 11S.11 Fourier transform of function having values at inetger x-value We consider a function defined by f (x) f (m) (x m) . (8) m y x Fig. Plot of f(x) which is described by a combination of the Dirac delta function with f(m) at x = m. The Fourier transform of f(x) is given by 11 1 F(k) f (x)eikxdx 2 1 f (m) (x m)eikxdx 2 m 1 f (m) (x m)eikxdx 2 m 1 eikm f (m) 2 m We note that 1 1 F(k 2 ) ei(k 2 )m f (m) eikm f (m) F(k) , 2 m 2 m In other words, F(k) is a periodic function of k with a periodicity 2. We can also show that 1 f (m) F(k)eikm dk , (9) 2 where eik (mm')dk 2 , (10) m,m' since 1 1 1 F(k)eikmdk f (m') eik (mm')dk 2 2 2 m' 1 f (m')2 m,m' 2 m' f (m) Note that the inverse Fourier transform of F(k) is obtained as 12 1 1 1 f (x) F(k)eikxdk dkeikx eikm f (m) 2 2 2 m 1 f (m) dkeik (xm) 2 m 1 f (m)2 (x m) 2 m f (m) (x m) m ((Note)) The proof where the Poisson summation formula is used. We have another method for the derivation of Eq.(9): 1 1 (2n1) f (x) F(k)eikx dk F(k)eikx dk 2 2 n (2n1) We put k' k 2n Then we have 1 1 f (x) F(k)eikx dk F(k'2n )ei(k '2n )xdk' 2 2 n 1 ei2nx F(k')eik 'x dk' n 2 1 (x m) F(k)eikx dk m 2 1 (x m) F(k)eikmdk 2 m (x m) f (m) m Here we use the Poisson summation formula 13 ei2xn (x m) . n m 11S.11 Random walk in the one-dimensional chain We consider the case when a particle moves on the one dimensional chain. The particle can be located only on a discrete position [x = m (m; integer)].
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