Задачи по квантовой механике. Часть 2. Алмалиев А.Н - 13 стр.

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∞
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 =
1
24
· 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.
3Z
2a
0
r = t
∞
Z
0
f
20
(r)f
10
(r)r
2
dr =
√
2
Z
3
a
3
0
∞
Z
0
µ
1 −
Zr
2a
0
exp
µ
−
3Zr
2a
0
r
2
dr =
=
8
27
√
2
∞
Z
0
t
2
µ
1 −
1
3
t
e
−t
dt =
8
27
√
2
µ
2! −
1
3
· 6!
= 0,
Ïîêàæåì òåïåðü, ÷òî è ðàäèàëüíûå ÷àñòè (1.28) òàêæå íîðìèðîâàíû íà
åäèíèöó. Èñïîëüçóÿ (1.28) ïîëó÷àåì:

  Z∞                     ∞µ       ¶2    µ     ¶
                      3 Z
     2      2     1 Z         Zr           Zr
    f20 (r)r dr =    3     1−        exp −      r2 dr =
                  2 a0        2a0          a0
  0                       0
                         Zµ
                         ∞                  ¶     µ     ¶
                 1 Z3               Z   Z2 2 2       Zr
               =                 1 − r + 2 r r exp −      dr =
                 2 a30              a0  4a0          a0
                         0
                          Z∞ µ           ¶           µ              ¶
                      1               1            1           1
                    =       t2 1 − t + t2 e−t dt =    2! − 3! + · 4! = 1;
                      2               4            2           4
                             0


  Z∞                    5 Z
                           ∞        µ     ¶         Z∞
     2              1 Z                Zr         1                1
    f21 (r)r2 dr =          r 4
                                exp  −      dr =      t 4 −t
                                                         e   dt =    · 4! = 1.
                   24 a50              a0        24               24
  0                              0                       0



                                                              Zr
 îáîèõ ñëó÷àÿõ ïðîâîäèëàñü çàìåíà ïåðåìåííûõ                    = t.                  ¤
                                                              a0
Ïðèìåð 1.6. Íåïîñðåäñòâåííûì ðàñ÷åòîì ïîêàçàòü îðòîãîíàëüíîñòü
âîëíîâûõ ôóíêöèé äëÿ 1s- è 2s-ñîñòîÿíèé âîäîðîäîïîäîáíîãî àòîìà.
Ðåøåíèå. Íåîáõîäèìî ïîêàçàòü, ÷òî èíòåãðàë îò ïðîèçâåäåíèÿ Ψ100 (r) è
Ψ200 (r) ðàâåí íóëþ:
       Z                                                   Z∞
  I=       Ψ∗200 (r, θ, ϕ)Ψ100 (r, θ, ϕ)r2 dr sin θ dθ dϕ = f20 (r)f10 (r)r2 dr = 0.
                                                          0



                                                                3Z
Èñïîëüçóÿ (1.25), (1.28) è äåëàÿ çàìåíó ïåðåìåííûõ                  r = t, ïîëó÷àåì:
                                                                2a0

  Z∞                            ∞µ        ¶    µ      ¶
                         √   3 Z
                           Z          Zr          3Zr
    f20 (r)f10 (r)r2 dr = 2 3      1−       exp −       r2 dr =
                           a0         2a0         2a0
  0                                      0
                                     Z∞ µ      ¶               µ         ¶
                        8√                   1            8 √       1
                     =     2           t2 1 − t e−t dt =      2 2! − · 6! = 0,
                       27                    3           27         3
                                     0


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