于是 ekz= 2i(21+1)j(kr)P,(cos 8) 1=0 C.当k在任意方向,则 ● elk=2i(21+1)ji ( kr)P(COSY) 1=0
于是 C. 当 在任意方向，则 e i (2l 1)j (kr)P (cos ) l l l 0 ikz l ∑ θ ∞ = = + k e i (2l 1)j (kr)P (cos ) l l l 0 ik r l ∑ γ ∞ = ⋅ = +
y为k和之间的夹角 P(coy)=∑ Ym(Ok,okYm(8,o) 2I+1 el=∑∑im(kr)Ym(,中)Ym(,) l=0m=-1 IK.r e Im(ek,k)ukIm(r, 0, o) (2兀 3/2 m
为 和 之间的夹角 r l * l lm k k lm m l 4 P (cos ) Y ( , )Y ( , ) =− 2l 1 π γ = θ φ θφ + ∑ l ik r l * l lm lm k k l 0m l e i 4 j (kr)Y ( , )Y ( , ) ∞ ⋅ = =− = π θφ θ φ ∑ ∑ γ k = ∑ ⋅ l , m k k klm ( r , , ) * lm l i k r 3 2 Y ( , ) u k i e ( 2 ) 1 θ ϕ θ ϕ π
现可求ukm的归一化因子 ik·r-ik dr=(2π)8(k-k) =(27)328(k-k∑ Im(Ok, k)Im(ek,OK) 而根据展开有
现可求 的归一化因子： 而根据展开有 e e d r ( 2 ) ( k k ) i k r i k r 3 = − ′ ∫ ⋅ − ′⋅ π δ k k kk 3 * 2 lm( , ) lm( , ) lm 1 (2 ) (k k ) Y Y k θ ′ ′ φ θφ =π δ− ′ ∑ uklm
∫dr∑4mikr)Ym(,9)Ym( ∑4π(-i)i(kr)Yma(,4)Ym(, .m ∫rd∑(4n3jkr)i(k)Ym( Imk k (27) <e-8(k-k,) ∑Yn Im(ek,)"lm(e4,φ lr
22 * l l lm k k lm k k l,m r dr (4 ) j (kr)j (k r)Y ( , )Y ( , ) ′ ′ = π θφ θ φ ∑ ′ ∫ kk k k 3 * 2 lm( , ) lm( , ) lm 1 (2 ) (k k ) Y Y k θφ θ φ′ ′ =π δ− ′ ∑ l * l lm lm k k l ,m l * l lm lm k k l,m dr[ 4 i j (kr)Y ( , )Y ( , ) 4 ( i) j (k r)Y ( , )Y ( , )] ′ ′ ′′ ′′ ′ ′ ′ ′ π θφ θ φ ⋅ π− θφ θ φ ′ ∫ ∑ ∑
从而有 r(4r32jk)i(kr)=(2)36(k-k) 即 r drj (kr)jj(k'r) 2 6(k-k 2k 于是有 ukm(r,0,p)=k=j,(kr)Y,m(0, d)
从而有 即 于是有 ( k k ) k 1 r dr ( 4 ) j (kr ) j ( k r ) ( 2 ) 2 3 l l 2 2 ′ = − ′ ∫ π π δ ( k k ) 。 2 k r dr j (kr ) j ( k r ) 2 l l 2 ′ = − ′ ∫ δ π klm l lm 2 u (r, , ) k j (kr)Y ( , ) θ φ = θφ π