講者: 許永昌 老師 1 Contents Preface Generating function Legendre Polynomials Example: multipoles Recurrence relations Special values Upper and Lower bounds for Pn. 2 Preface Generating Function Special values Upper & Lower Bound Legendre Equation Recurrence relations Legendre Polynomial Multipoles Rodrigues’ Formula Orthogonality Series Expansion Associated Legendre functions Spherical Harmonics Addition Theorem Green’s function Atomic Orbitals Orthogonality Angular momentum Helmholtz Eq. Group theory 3 Generating Function (請預讀P741~P744) rl 1 From l 1 Pl rˆ rˆ r r ' l 0 r 1 r 2r r ' r ' 2 , 2 If tr</r>, we get g x, t 1 1 2tx t 2 Series form: 1 1 2tx t 2 Pn x t n . n 0 2n ! n 0 22 n n ! n ! 2n ! n 1 2n x , 1 x n 0 2 n ! n ! 2tx t , 2 n n ! 1 2 x 2n p ! n p ! n 0 2 n ! n ! p 0 2n ! m t 2m m 0 2 m 2 n m 2 p n p t n p , a b n n p 0 22 p 2m 2 p ! 1 2 x , p ! m 2 p ! m p ! p 0 p 1 2m 2 p ! We get Pm x m x m 2 p . p 0 2 p ! m p ! m 2 p ! m 2 p n! a p bn p , p ! n p ! m n p m, p 0 ~ , 2 p 4 Multipoles (請預讀P744~P747) If 2V 2 , we get G r , r ' r r ' & V r G d . 0 0 Green’s function: 1 1 G r , r ' 4 r r ' 4 V(r) rn P rˆ rˆ ' , n 1 n r n 0 If |r| > |r’|, we get Multipole expansion: r ' 1 1 n V r G r , r ' d ' r ' Pn rˆ rˆ ' r ' d '. n 1 0 4 0 n 0 r Remember, in physical systems we do not encounter pure multipoles. 5 Recurrence Relations (請預讀P749~P751) 1. From tg: g x, t 3 x t g x, t nPn x t n 1 , t n 0 x t Pn x t 1 2 xt t n 2 n 0 nP x t n 0 n n 1 , xPn Pn 1 n 1 Pn 1 2nxPn n 1 Pn 1 , For each t n 2n 1 xPn nPn 1 n 1 Pn 1 . 2. From xg : g x, t x tg 3 x, t P 'n x t n , n 0 t Pn x t 1 2 xt t n n 0 1 2 P ' xt n 0 Pn P 'n 1 2 xP 'n P 'n 1 , n n , 2 For each t n 1 6 Recurrence Relations (continue) From Eqs. (1) & (2) we can get 1 ' n 1 2 P 'n1 xP 'n nPn , 1 ' n 2 P 'n 1 xP 'n n 1 Pn , 3 4 Let’s try to get the Legendre Eq.: x 3 4 nn 1 1 x 2 P 'n nPn 1 nxPn 5 2 2 1 x 5 n 3 1 x n n 1 Pn 0. x x x 7 Special Values & Parity (請預讀P752~P753) g(1,t)=1/|1-t| Pn(1)=1. g(-1,t)=1/|1+t| Pn(-1)=(-1)n. n 1 2n ! 2 n 1 g(0,t)= 2n t Pm 0 t m , 2 2 n!n! 1 t n 0 m 0 P2n+1(0)=0, P2n(0)= (-1/4)n C2nn. g(x,t)=g(-x,-t) Pn(x)=(-1)nPn(x). Even or odd function. 8 Upper and Lower Bounds for Pn(cosq) (請預讀P753~P754) n If we can prove that Pn cos q am cos mq m 0 Pn cos q Proof: Pn cos q t n n 0 n a m 0 m 1 am 0, Pn 1 . 1 1 1 2t cos q t 2 1 teiq 1 te iq 1 i p q q 2 p 2 q C p2 p Cq2 q t p q e , if p q n, p 0 q 0 2 n 1 2 p 2 n 2 p n i 2 p n q C p Cn p t e 2n n 0 p 0 2 n 1 1 Pn cos q 2 n C p2 p Cn2n p2 p ei 2 p n q 2 n 2 p 0 2 n 2 p 2 n 2 p C p Cn p cos n 2 p q . p 0 9 Homework 12.1.6 (11.1.6e) 12.1.7 (11.1.7e) 12.2.1 (11.2.1e) 12.2.5 (11.2.5e) 12.2.7 (11.2.7e) 10
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