(3)球方势阱:考虑位势为 0 r<a 0 r> a u(r,e,中)=R(r)Ylm(e,中) d (rR)+ 2uEI(1+1) h2 (rR)=0r< dr 小2(rR+人2(V E)I(+1 I(rR=0 r>a
（3）球方势阱：考虑位势为 令 ⎩⎨⎧ >< = V r a 0 r a V(r) 0 u(r, , ) R(r)Y ( , ) θ φ = θφ lm 2 2 22 d 2 E l(l 1) (rR) [ ](rR) 0 r a dr r μ + + − =< h 2 0 2 22 d l(l 1) 2 (V E) (rR) [ ](rR) 0 r a dr r −μ − + + − => h
V(r) 0
A, E<V 2 2mE 2_2m(V0-E) K A 则有 kJ,(kr A r<a R(r)= Ici(iKr)+c2,(iKr)r>a 1.由波函数在无穷远处为0,即当r→>00
A. 令 则有 1. 由波函数在无穷远处为 0，即当 E < V0 2 2 2mE k O = 2 2 0 2m(V E) O − κ = kl l 1l 2 l A j (kr) r a R(r) c j (i r) c n (i r) r a ⎧ < = ⎨⎩ κ +κ> r → ∞
T T Sln(Kr一 coS(IKr 2 IKr IKI lπ lπ Kr+I e 2 -e lπ c, (e K-12+c Kr十1 2 air 2 T T C1-C2 Kr-1- Kr+1 )e (以!+c2)e2 air →)0
l l ri ri l l 2 2 ri ri 1 2 2 2 1 c (e e ) [ c (e e )] 2i r i π π −κ − κ + π π − −κ − κ + = −+ κ c )e ] ic c )e ( ic [( 2i r 1 2 l r i 2 2 1 l r i 2 1 π κ π κ κ − − + = − − + i r ) 2l cos(i r c i r ) 2l sin(i r c1 2 κ π κ κ π κ − − − → 0
c=0 2 =B, C2 B 于是有 R(r=Blj(iKr)+in ikr) =Bh((iKr)
则 于是有 1 B l c = κ 2 l c = iBκ R(r) B [j (i r) in (i r)] = κll l κ+ κ (1) = B h (i r) κl l κ c 0 i c 2 1 + =