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

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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)]  eikx 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

We evaluate the factor

 I  eikn . n

It is evident that I is not equal to zero only when k = 2m (m; integer). Therefore I can be expressed by

 I  A  (k  2m) , m where A is the normalization factor. Then

 1    f (x  n)   F(k)dk[A  (k  2m)] n 2  m A    F(k) (k  2m)dk 2   m  A    F(k  2m) 2 m A    F(k  2n) 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  2n)  en 2 ______((Mathematica))

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  2n) or

  I  eikn  2  (k  2m) . (2) n m

Using this formula, we have

   f (x  n)  2  F(k  2n) . (Poisson sum formula) n n

11S.2 Summary When we put k  2x in I of Eq.(2)

  e2ixn  2  (2x  2m) n m 2    (x  m) 2 m    (x  m) m or

  e2ixn   (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)]   eikx  (x  na)dx  eikna . 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) eikna . 2 n

We use the Poisson summation formula;

   ka eikna  eikna    (  m) n n m 2  a 2m   [ (k  )] m 2 a 2  2m   (k  ) a m a where

  e2ixn   (x  m) . n m

ka with x  . Then we get 2

1  1  F[ f * g]  F(k) eikna  F(k) eikna 2 n 2 n 1 2  2m    (k  )F(k) 2 a m a 1 2  2m 2m    (k  )F(k  ) 2 a m a a

The inverse Fourier transform of F[ f * g] is obtained as

1  1  1 2  2m 2m f * g  F[ f * g]eikxdk   eikxdk   (k  )F(k  ) 2  2  2 a m a a 1  2  2m   F(k  m)  eikxdk (k  ) a m a  a 2m 1  2 i x   F(k  m)e a a m a where

 2 2 1 i mx F(k  m)   e a f (x)dx . a 2 

Finally we get

2m 1  1  2m i x f * g   f (x  na)   F(k  )e a . (5) 2 n a m a or

2m  2  2m i x  f (x  na)   F(k  )e a . n a m a

When a = 1, we get

   f (x  n)  2  F(k  2m)ei2mx . (6) n m

When x = 0,

   f (x  n)  2  F(k  2m) . (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

2m  2  2m i x N(x)   f (x  na)   F(k  )e a n a m a iGx   NGe G

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  eiGx f (x)dx  eiGx f (x)dx a  a a / 2 where f(x) is just like a Gaussian distribution function around x = 0.

N x

 

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,

1 x2 f (x)  exp( ) . 2 2 2

Then we get

1 a / 2 1 1 1 2iG 2 1 2iG 2 N  eiGx 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)  et 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

Suppose that the function N(x) is a periodic function of x with the periodicity a. Then we have

2m  2  2m i x N(x)   f (x  na)   F(k  )e a n a m a iGx   NGe G

2 where G  m (m: integer). a

((Example-1))

f(x) = x for |x|

1 1 2i(1)n F(k  G  n)   xexp(inx)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)   eimx   eimx  n n  n n n0

We make a plot of N(x) as a function of x for the summation for n = -nmax and nmax with nmax = 50.

Fig. nmax = 50. The Gibbs phenomenon is clearly seen.

((Example-2))

f(x) = 0 for -1

1 1 i(1)n[1 (1)n ] F(k  G  n)  exp(inx)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)    eimx 2  n 2n n0

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

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

1  F(k)   f (x)eikxdx 2  1    f (m) (x  m)eikxdx 2    m 1    f (m)  (x  m)eikxdx   2 m 1   eikm f (m) 2 m

We note that

1  1  F(k  2 )  ei(k 2 )m f (m)  eikm 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 (mm')dk  2 , (10)  m,m'  since

1  1 1    F(k)eikmdk   f (m')  eik (mm')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

1  1  1  f (x)   F(k)eikxdk   dkeikx eikm f (m) 2  2  2 m 1    f (m) dkeik (xm) 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  (2n1) f (x)   F(k)eikx dk    F(k)eikx dk 2  2 n (2n1)

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    ei2nx  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

  ei2xn   (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)]. the particle starts from the origin x = 0. The particle can move from x = m to x = m+1, or from x = m to x = m-1 for each jump. We assume that the probability W(m, N) such that the particle reaches at the position x = m after jumps with N times. The probability W(m, N+1) can be described by

m-1 m m+1 N N+1 N Fig. Random walk model.

1 1 W (m, N 1)  W (m 1, N)  W (m 1, N) , 2 2 where one is a jump from x = m-1 to x = m for the (N+1)-th jump, and the another is a jump from x = m+1 to x = m for the (N+1)-th jump. The probability for these jumps is 1/2. In order to find the expression of W (m, N) we define the Fourier transform

1  (k, N)  eikmW (m, N) . 2 m

Note that (k, N) is a periodic function of k with a periodicity 2, and

1  1  eikmW (m 1, N)  eik eik (m1)W (m 1, N) 2 m 2 m ,  eik (k, N)

1  1  eikmW (m 1, N)  eik eik (m1)W (m 1, N) 2 m 2 m .  eik(k, N)

Then we have

1  (k, N 1)  eikmW (m, N 1) 2 m 1  (eik  eik )(k, N) 2  (cosk)(k, N) or

(k, N)  (cosk)N (k, N  0) .

We use the initial condition that the particle is located at x = 0 at the probability of 1;

W (m, N  0)   m,0 . or

  1 ikm 1 ikm 1 (k, N  0)  e W (m, N  0)  e  m,0  . 2 m 2 m 2

So we get

1 (k, N)  (cosk)N . 2

F k,N 0.4

 

0.3 N=200

0.2 N=400

0.1

N=1000 k -0.3 -0.2 -0.1 0.1 0.2 0.3 Fig. Plot of (k, N) vs k, where N = 200, 400, 600, 800, and 1000. In the large limit of N, (k, N ) approaches a Gaussian distribution.

The inverse Fourier transform of (k, N) is

1  1  1 W (m, N)  (k, N)eikmdk  dkeikm (cosk) N 2  2  2   1    dkeikm (cosk) N 2 

Here we use the binomial theorem to get

N  1  (cosk)N    (eik  eik )N  2 

N  1  N N!     eik (2l  N )  2  l 0 (N  l)!l!

Then we have

N 1   1  N N! W (m, N)   dkeikm    eik (2l N ) 2   2  l0 (N  l)!l! N  1  N N! 1       dkeik (m2lN )  2  l0 (N  l)!l! 2  N  1  N N! 1     (2 ) m2lN ,0  2  l0 (N  l)!l! 2 N  1  N N!      m2lN ,0  2  l0 (N  l)!l! N  1  N N!      m2lN ,0  2  l0 (N  l)!l! 1 N!  2 N N  m  N  m  ( )! ! 2  2  where

N  m m  N  2l , or l  . 2

This implies that for N = even, m should be even. In other words, W(N = even, m = odd) = 0. For N = odd, m should be odd. In other words, W(N = odd, m = even) = 0.

11S.13 Numerical calculation of W(m, N) Using the Mathematica, we calculate numerically the value of W (m, N) given by

1  1  W (m, N)   dkeikm (cosk) N   dk cos(km)(cosk) N , 2   0 for the cases with N = 12 (even) and N = 13 (odd).

(i) N = 12.

W m,N

0.20  N=12

0.15

0.10

0.05

m -10 -5 5 10

This figure clearly shows that W(m, N = 12) = 0 for m = -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, and 11 (m = odd).

(ii) N = 13. W m,N

0.20   N=13

0.15

0.10

0.05

m -10 -5 5 10

This figure clearly shows that W(m, N = 13) = 0 for m = -12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, and 12 (m = even).

((Mathematica))

Clear "Gobal`" ;

1  J1 m_, N1_ : Cos k m Cos k N1 k   0

d12  Table m,J1 m,12 , m, 12, 12, 1 ;        f1  ListPlot d12, PlotStyle  Red, PointSize 0.02 , AxesLabel  "m", "W m,N " ; f2  Graphics Text Style "N12", Black, 15 , 6, 0.18 ;      Show f1, f2         W m,N         0.20  N=12

0.15

0.10

0.05

m -10 -5 5 10

d13  Table m,J1 m,13 , m, 13, 13, 1 ; g1  ListPlot d13, PlotStyle  Red, PointSize 0.02 , AxesLabel  "m", "W m,N " ; g2  Graphics Text Style "N13", Black, 15 , 6, 0.18 ;      Show g1, g2         W m,N         0.20   N=13

0.15

0.10

0.05

m -10 -5 5 10

11S.14 Limit of long distance and long time Using the Stirling's formula,

1 1 ln(x!)  ln(2 )  (x  )ln(x)  x for large x, 2 2 and the Mathematica, we have the series expansion to the order of (m/N)8,

N  m N  m lnW (m, N)  N ln(2)  ln(N!)  ln( )!ln( )! 2 2 2 4 1 N 1  m  1  m    ln( )  (1 N)   (3  N)  2 2 2  N  12  N  6 8 1  m  1  m   (5  N)   (7  N)   ... 30  N  56  N  1 N 1 m2   ln( )  2 2 2 N or

1 N 1 m2 W (m, N)  exp[ ln( )  ] 2 2 2 N 1 2 N  1 m  exp[ln( ) 2  ] 2 2 N 1 2 N  1 m  ( ) 2 exp( ) 2 2 N 1 2  2  2 1 m    exp( )  N  2 N

((Mathematica))

Clear "Gobal`" ;

1 1 f1 x_ : Log 2   x  Log x  x;  2  2 N  m N  m g1N Log 2  f1 N  f1 f1 ; 2 2 rule1  m  N x ;         g11  g1 . rule1; Series g11, x,0,8 Simplify Normal  1 1 1  N x2  3  N x4     2 12 1 1 1 N  5  N x6  7  N x8  Log 30  56  2 2

11S.15  Diffusion     We assume that the distance between the nearest neighbor lattices is x and the time taken for each jump is t. Then we have

t  Nt , x  mx .

The probability of finding particle between x and x + dx is

1   2   x t dx 1 1 x 2 P(x,t)dx  W ( , )  dx  exp( ) x t 2x  (x)2  4t x 2  4 t   2t  2t 1 x 2  dx exp( ) 4Dt 4Dt where D is the diffusion constant and is defined by

(x)2 D  . 2t

x t dx Here we note that the factor 2x of the W ( , ) arises from the fact that (i) for x t 2x every jump (N = even), the probability of finding the particle at the sites with even m is zero, and that (ii) for every jump (N = odd), the probability of finding the particle at the sites with odd m is zero. The distance for the jump is 2x, but not x. The final form of P(x, t) is obtained as

1 x2 P(x,t)  exp( ) . 4Dt 4Dt

We note that P(x, t) satisfies the diffusion equation given by

P(x,t) 2P(x,t)  D , t x2 with the initial condition

P(x,t  0)   (x) .

11.S16 Gaussian distribution in magnetization; analogy of random walk We consider a system consisting of N independent spins. Each spin has a magnetic moment . In the absence of an external magnetic field, each spin has the magnetic moment (±) along the z axis. We assume that the number of spins having the z component magnetic moment (+) is N and the number of spins having the z-component magnetic moment (-):

1 1 N  (N  n) , N  (N  n)  2  2 where

N  N  N .

Here we discuss the probability distribution of total magnetic moment, M, which is given by

M  (N  N )  n .

The probability that the total magnetization has M = n is obtained as

1 N! 1 N! W (M )   2 N N !N ! 2 N 1 1   [ (N  n)]![( (N  n)]! 2 2

Using the Stirling's formula, we have

1 N n2 lnW (M )   ln( )  2 2 2N for n<

2 n2 W (M )  ( )1/ 2 exp( ). N 2N where M = n. The average of magnetization is equal to zero.

  M    n    nW (M )dn  0 . 

Since   M 2   2  n   2  n2W (M )dn  2 2 N ,  the standard deviation is obtained as

2 2 M   M    M   2N  .

Since

M 2N 1   2N 2N 2N

the relative width of the Gaussian distribution becomes sharp as N increases. We make a plot of W(M = n) as a function of N, where N = 100.

W M=nm 0.08

 

0.06

0.04

0.02

n -40 -20 20 40

Fig. Plot of W(M) vs n. M = 2n. N = 100. ______REFERENCES C. Kittel, Elementary Statistical Physics (Dover Publications, Inc, Mineola, NY, 2004). K. Kitahara, Nonequilibrium Statistical Physics, 2nd edition (Iwanami, Tokyo, 1998). [in Japanese]. R. Reif, Statistical Thermal Physics, McGraw-Hill, New York, 1965). C. Kittel and H. Kroemer, Thermal Physics, 2nd edition (W.H. Freeman and Company, New York, 1980).