4.E.4. Binomial Random Walk Consider a particle constrainted to move along the x-axis by steps of equal lengths Δ but random directions taken at fixed time intervals τ. Let the probability of its moving to the right and left be p and q 1 p, respectively. The movement at step i can therefore be decribed by a stochastic variable X i with realization x , probability density PXi x p x q x and characteristic function f Xi k dxe p x q x ikx peik qeik cosk if pq 1 2 N The net displacement after N step is given by YN X i . The characteristic i 1 function, according to (4.40-1), is N fYN k f X i k cos k N i 1 1 1 1 k 22 k 4 4 4! 2 N N N 1 N 4 4 1 1 Nk 2 2 k 2 8 4! (4.61) Hence, y y 2 k 0 2 1 fYN 2 i k 2 Y N 1 fYN i k N N 1 N 3 4 1 Nk 2 4 k i 8 4! k 0 y2 y 2 0 k 0 N N 1 N 2 4 1 N 2 12 k 2 i 8 4! 2 N k 0 N For a given N, there are 2 N possible realizations. Three of them for the case N 2000 are shown in Fig.4.4. Differential Equation Consider the limits , 0 with N . Setting fY k , N fYN k , we have fY k , N 1 fY k , N cos k cos k 1 cos k N 1 cos k N N 1 1 k 22 k 44 4! 2 Hence, with N t , we have fY k , N (4.62) fY k , t f k , t fY k , t lim Y 0 t lim fY k , N 1 fY k , N 0 1 2 2 1 4 4 lim k k 0 4! 2 Dk 2 fY k , t (4.63) fY k , t as 0 (4.64) where the diffusion coefficient D is defined as 2 D lim , 0 2 For a fixed k, the general solution to (4.64) is fY k , t fY k ,0 exp Dk 2t exp Dk 2t if fY k ,0 1 (4.66) Taking the inverse Fourier transform gives the probability distribution P y, t 2 exp iky Dk t dk 2 2 y 2 dk iy exp exp Dt k 2 Dt 4 Dt 2 y2 exp 4 Dt 4 Dt 1 (4.67) which is a Gaussian with 2Dt . Alternative Derivation Consider the stochastic variable Z N YN . The associated characteristic N function , according to (4.40-1), is N ik f Z N k dx j exp x j PX j x j N j 1 N j 1 k k fX j cos N N k2 k4 1 2 2 N 4! N N (4.68) N Hence k2 f Z k lim f Z N k exp N 2 PZ z dk 1 2 exp ikz 2 k 2 (4.69) z 2 dk 2 1 exp exp k iz 2 2 2 z2 1 exp 2 For very large N, we have PZ N z PYN y dz y N z PZ z 1 N dz z PZ z so that from (4.49), we have y N PZ z y2 exp 2 2 N 2 N 1 (4.70) If the partical takes n per unit time, we have N nt . Setting D P y, t lim PYN y N See Fig.4.5. y2 exp 4 Dt 4 Dt 1 (4.71) 1 2 n , we have 2