Задачи по квантовой механике. Часть 2. Корнев А.С. - 80 стр.

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·
1
sin θ
θ
µ
sin θ
θ
+
1
sin
2
θ
2
ϕ
2
+ l(l + 1)
¸
Ψ(θ, ϕ) = 0;
·
i
ϕ
+ m
¸
Ψ(θ, ϕ) = 0
Z
2π
0
dϕ
Z
π
0
Y
l
0
m
0
(θ, ϕ) Y
lm
(θ, ϕ) sin θ dθ = δ
l
0
l
δ
m
0
m
;
X
l=0
l
X
m=l
Y
lm
(θ
0
, ϕ
0
) Y
lm
(θ, ϕ) = δ(cos θ
0
cos θ) δ(ϕ
0
ϕ).
cos θ Y
lm
(θ, ϕ) =
=
s
(l + 1)
2
m
2
(2l + 1)(2l + 3)
Y
l+1,m
(θ, ϕ) +
s
l
2
m
2
(2l 1)(2l + 1)
Y
l1,m
(θ, ϕ),
Y
00
(θ, ϕ) =
1
4π
;
Y
10
(θ, ϕ) =
r
3
4π
cos θ; Y
1±1
(θ, ϕ) =
r
3
8π
sin θ e
±iϕ
.
L
(α)
n
(x)
xy
00
+ (α + 1 x)y
0
+ ny = 0, n = 0, 1, . . .
α = 0
                                                                    [5

2 - 2 c 5TQ8%! 2  + *- 2 E H3#Q4! &((76\  2 H%6Q
                    ·            µ           ¶                         ¸
                          1 ∂            ∂          1 ∂2
                                   sin θ        +            + l(l + 1) Ψ(θ, ϕ) = 0;
                        sin θ ∂θ         ∂θ       sin2 θ ∂ϕ2
                                          ·           ¸
                                               ∂
                                            i     + m Ψ(θ, ϕ) = 0

aN) 1c T
               5 >T Qa;-!c  1c !#!#V
                                              ∂ϕ

                                                                                                               N ) !6A
                         Z    2π
                                   dϕ
                                        Z       π
                                                    Yl∗0 m0 (θ, ϕ) Ylm (θ, ϕ) sin θ dθ = δl0 l δm0 m ;
                                                                                                                       9YI 7@? ?

                                                                                                                       9YI 7 : ?
                          0                 0
               ∞
               X        l
                        X
                                ∗
                              Ylm (θ0 , ϕ0 ) Ylm (θ, ϕ) = δ(cos θ 0 − cos θ) δ(ϕ0 − ϕ).

           @ &( *!#" &(%"Xa-  2 c 3 " %a!#,6T Uc
                l=0 m=−l




          cos θ Ylm (θ, ϕ) =
            s                                    s
          =
                        2
                 (l + 1) − m  2
                                 Yl+1,m (θ, ϕ) +
                                                       l 2 − m2
                                                                    Yl−1,m (θ, ϕ),
                                                                                                                       9YI 7 [ ?
                (2l + 1)(2l + 3)                   (2l − 1)(2l + 1)
;# !#6- 2 c 5TH!8NQa)-*T , +- Qa!#4;LN%) !#>! &(
 !#"U8a&(%"XQ 7
      M(#* 2 EFQaO4 "%E 8a! &( *!#"Q8a&(%"XQ A
                              1
                                                                                                                       9YI 7@m ?
                Y00 (θ, ϕ) = √ ;
                              4π
                             r                                                      r
                                3                                                        3
                Y10 (θ, ϕ) =      cos θ;                      Y1±1 (θ, ϕ) = ∓              sin θ e±iϕ .
                               4π                                                       8π
               $Œ9%.2Ž}Ž2ŽR 2 Œ 2Ž}Œ('                          2   
           M(!## EF`;-%E < +                          2 - 2 c  ! 2 + *- 2 E 
76T  2  &,&(  X6-.+ T  2
                                                                             (α)
                                                                           Ln (x)


                               xy 00 + (α + 1 − x)y 0 + ny = 0,                     n = 0, 1, . . .
    M(                a;  8* 2  NEFEF(;-%E < +  7
            a /  "%EF 2  EFTEF6T  2 - 2 ;!#8E 48 ;-%
             α=0
<        + A