2e/2 2e"2e' (V1+V2) 2 12 鱼+庄 2 e r2(+r2_ r, cos 0) 设:Ho的基态为
设： 的基态为 2 2 22 2 2 1 2 1 2 12 2 ˆ e 2e e ( ) 2 rr H r − ′ ′ ′ ∇+ − − + μ = ∇ h 0 1 = H H ˆ ˆ + 2 2 2 2 12 12 1 2 12 1 e e r (r r 2r ˆ co ) H r s ′ ′ = = +− θ H 0 ˆ 0
s2Sz,n,20)=u100(巧1)u100(r2)00 1-(n1+r2) 4兀E。九 e a na m2e 2 (0) 2e -2 8 兀8na 4丌￡a
即 z 1 2 100 1 100 2 00 s,s ,r ,r 0 = u (r )u (r )χ 1(r r ) 1 2 a 3 00 1 e a − + χ π = 2 2 (0) 0 0 0 2e e E2 2 8a 4a =− =− πε πε 2 0 2 4 a m2e πε = ′ h
于是 E=(010 ∫dr」jea ([+r2)r] sin e,de,do, dr (ra3) r12 以1方向为Z方向,所以 0,=0 2
于是 以 方向为 Z 方向 ,所以 (1) 0 1 E 0H 0 = ˆ 12 2 2 2 2 2 2 2 1 3 2 2 1 2 r r sin d d dr dr e ( a ) e (r r ) a θ θ ϕ π ′ = ∫ ∫ − + 1r θ2 = θ
∑P1(co I1>r 1=0 2 ∑P(cos)() < I1=0 2 (1) 2πe dre a I (πa°) ∫n∑P(cos) d cos edr22 r1
⎪⎪⎩ ⎪⎪⎨⎧ ∑ < ∑ > = ∞ = ∞ = 1 2 l 0 l 2 1 l 2 1 2 l 0 l 1 2 l 1 12 ) r r r r P (cosθ)( r 1 ) r r r r P (cosθ)( r 1 r 1 1 2 1 2 2 2 r r r (1) 2 l a a 2 0 1 2l 2 3 2 0 0 1 1 l 0 2e 1 r E dr e [ e r P (cos )( ) d cos dr ( a) r r ∞ − − π = π ′ = ⋅− θ θ π ∫ ∫∫ ∑
enn21∑ os0)(1) sdr2 0 由于 ∫P(cos6)Pcos)dcos6 2 21+1 Po(cos 8) T (1)_2e 2 oo 2 ri dr. r ga
由于 2 2 1 2 0 0 2 2 2 1 1 2 ) dcos dr r r P (cos )( r e r l l l r r a − ∫ ∫ ∑ θ θ π ∞ = ∞ − l ll 0 l 2l 1 2 P (cos )P (cos )d cos ′ ′ + ∫ θ θ θ = δ π P (cos ) 1 0 θ = 12 2 1 1 2 22 2 rr r r (1) 2 2 aa a 0 1 22 22 6 0 r 1 2 2e 2 2 E dr e [ e r dr e r dr ] a rr ′ −− − ∞ =⋅ + π ∫∫ ∫