Задачи по квантовой механике. Часть 2. Корнев А.С. - 47 стр.

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m = −1 Ψ
21−1
Ψ
200
(r) = f
20
(r)Y
00
(θ, ϕ) (2s)
Ψ
211
(r) = f
21
(r)Y
11
(θ, ϕ)
Ψ
210
(r) = f
21
(r)Y
10
(θ, ϕ)
Ψ
21−1
(r) = f
21
(r)Y
1−1
(θ, ϕ)
(2p)
f
20
(r) =
1
√
2
µ
Z
a
0
3
/
2
exp
µ
−
Zr
2a
0
µ
1 −
Zr
2a
0
;
f
21
(r) =
1
2
√
6
µ
Z
a
0
3
/
2
exp
µ
−
Zr
2a
0
µ
Zr
a
0
.
Z
∞
0
f
2
20
(r)r
2
dr =
1
2
Z
3
a
3
0
Z
∞
0
µ
1 −
Zr
2a
0
2
exp
µ
−
Zr
a
0
r
2
dr =
=
1
2
Z
3
a
3
0
Z
∞
0
µ
1 −
Z
a
0
r +
Z
2
4a
2
0
r
2
r
2
exp
µ
−
Zr
a
0
dr =
=
1
2
Z
∞
0
t
2
µ
1 − t +
1
4
t
2
e
−t
dt =
1
2
µ
2! − 3! +
1
4
· 4!
= 1;
Z
∞
0
f
2
21
(r)r
2
dr =
1
24
Z
5
a
5
0
Z
∞
0
r
4
exp
µ
−
Zr
a
0
dr =
1
24
Z
∞
0
t
4
e
−t
dt
|
{z }
4!
= 1.
Zr
a
0
= t
¤
1s 2s
Ψ
100
(r)
Ψ
200
(r)
I =
Z
Ψ
∗
200
(r, θ, ϕ)Ψ
100
(r, θ, ϕ)r
2
dr sin θ dθ dϕ =
Z
∞
0
f
20
(r)f
10
(r)r
2
dr = 0.
                                                        1   :
m = −1 Ψ21−1
               9         ? 7 ,3 !#%6 9 7 : ?   A

                                     Ψ200 (r) = f20 (r)Y00 (θ, ϕ)  (2s)
                                     Ψ211 (r) = f21 (r)Y11 (θ, ϕ) 
                                                                  
                                                                                                       9 7 :?
                                     Ψ210 (r) = f21 (r)Y10 (θ, ϕ)   (2p)
                                                                  
                                                                  
                                    Ψ21−1 (r) = f (r)Y (θ, ϕ)
                                                   21           1−1
+  ^ !8+ -!#H9 7 O0/? ^
                                       ¶ 3/2µ   µ      ¶µ         ¶
                                 1  Z              Zr         Zr
                       f20 (r) = √           exp −        1−        ;
                                  2 a0             2a0        2a0
                                   µ ¶ 3/2       µ      ¶µ ¶
                                 1   Z              Zr     Zr
                       f21 (r) = √            exp −             .
4    N   F( ^  ;%!8-#FW&("X-c! 2 -."%
                                2 6 a0              2a0    a0

 ! &( *!8"`&(%"X 2  ^ %E > #X ^ !#+ - !#H9YI 7@? ? 7
  M("O6T*  ; . ^ a /6-.EF\!# 9 7 : ? 6"T %E 
#X7G,!#;-.)  2 9 7 : ? ^ ;-%6* A
    Z     ∞                     Z µ            ¶2     µ      ¶
           2             1 Z3 ∞            Zr             Zr
          f20 (r)r2 dr=      3        1−          exp −         r2 dr =
        0                2 a0 0            2a0            a0
                         Z    µ                    ¶       µ        ¶
                    1 Z3 ∞            Z      Z2 2 2             Zr
                 =               1 − r + 2 r r exp −                  dr =
                    2 a30 0          a0      4a0                 a0
                          Z       µ            ¶             µ                  ¶
                        1 ∞2                1 2 −t        1               1
                     =         t 1 − t + t e dt =              2! − 3! + · 4! = 1;
                        2 0                 4             2               4
         Z ∞                      5 Z ∞
                                               µ      ¶           Z ∞
             2      2       1   Z        4         Zr          1
            f21 (r)r dr =         5     r exp −         dr =          t4 e−t dt = 1.
          0                 24 a0 0                a0         24 0
                                                                  |    {z     }

I NU8 !8- 2 W
                       8 ;*-!#.T)* 3; * E   8
                                                                                               4!
                                                                                     Zr
                                                                                        =t
                                                                                           7                 ¤
 f+(*,+ B G X G ) > G 5 R > K 5fF24?> EDE=T \ R fF G \ G g 
                         
                           /
                             \
                              - c T A                                                                  Ψ (r)
                                                                                                     100


          Z                                                                  Z       ∞
    I=        Ψ∗200 (r, θ, ϕ)Ψ100 (r, θ, ϕ)r 2 dr sin θ dθ dϕ          =             f20 (r)f10 (r)r2 dr = 0.
                                                                                 0

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