文档介绍：第四章
0
{Ψ},正交。归一条件为< Ψ| Ψ 0 0 0 0 >= δ(i, i ).lm
nlmms nlmms n l m ms
是轨道角动量及投影的量子数,ms为自旋投影。n为径向波函数的量子数,试用相应于(nlmms)的
−→2
产生消灭算符写出轨道角动量的 L 及Lz
P P +
解:L = 0 hΨ| l | Ψ 0 0 0 0 i = ma a
z i,i nlmms 3 n l m ms nlmms nlmms nlmms
X X X X
−→2 −→−→−→2 −→−→
L = ( li ) · ( li ) = li + li · lj (1)
i i i i6=j
−→ P P −→−→
故 L 2 = l(l+1)a+ a + hΨ(1)Ψ(2) | l ·l | Ψ(1)Ψ(2)ia+ a+ a a
nlmms nlmms nlmms i1i2i3i4 i1 i2 1 2 i3 i4 i1 i2 i4 i3
2.
X Z Z X Z
−→−→−→0 −→ˆ+ −→−→+
P = d p d p < Ψ−→| p | Ψ−→0 0 > a−→ a−→0 0 = d p ( p a−→ a−→) (2)
p ms p ms p ms p ms p ms p ms
msms0 ms
2
X Z Z −→ˆ X Z −→2
−→−→0 p + −→ p +
T = d p d p < Ψ−→p m | | Ψ−→p 0m0 > a−→ a−→p 0m0 = d p ( a−→ a−→p m ) (3)
s 2m s p ms s 2m p ms s
msms0 ms
3.
r
2! X
Ψ(ξξξ) = P (Ψ(ξ)Ψ(ξ)Ψ(ξ))
N1N2 1 2 3 3! E 1 1 1 2 2 3
PE
1
= √{Ψ1(ξ1)Ψ1(ξ2)Ψ2(ξ3) + Ψ1(ξ3)Ψ1(ξ2)Ψ2(ξ1) + Ψ1(ξ1)Ψ1(ξ3)Ψ2(ξ2)}(4)
3
Z
∗
ΨN1N2 (ξ1ξ2ξ3){f(1) + f(2) + f(3)}ΨN1N2 (ξ1ξ2ξ3)dξ1dξ2dξ3
1
Z X
1 ∗
= √ dξ1dξ2dξ3( PEΨ1(ξ1)Ψ1(ξ2)Ψ2(ξ3)) {f(1) + f(2) + f(3)}ΨN N (ξ1ξ2ξ3)
3 1 2
PE
√ Z
∗∗∗
= 3 dξ1dξ2dξ3Ψ1(ξ1)Ψ1(ξ2)Ψ2(ξ3){f(1) + f(2) + f(3)}ΨN1N2 (ξ1ξ2ξ3)
Z
∗∗∗
= dξ1dξ2dξ3Ψ1(ξ1)Ψ1(ξ2)Ψ2(ξ3){f(1) + f(2) + f(3)}Ψ1(ξ1)Ψ1(ξ2)Ψ2(ξ3)
Z Z Z
∗∗∗
= dξ1Ψ1(ξ1)f(1)Ψ1(ξ1) + dξ2Ψ1(ξ2)f(2)Ψ1(ξ2) + dξ2Ψ2(ξ2)f(3)Ψ2(ξ3) (5)
4.
X
+
F = < i | f | k > ai ak (6)
ik
X
1 2 +