Задачи по квантовой механике. Часть 2. Алмалиев А.Н - 11 стр.

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t =
2Z
a
0
r; dt =
2Z
a
0
dr
s l = 0 m
Ψ
1s
(r) = Ψ
1s
(r, θ, ϕ) =
s
Z
3
πa
3
0
exp
µ
−
Zr
a
0
,
Y
00
(θ, ϕ) =
1
√
4π
¤
1s
Ze
1s
Ψ
1s
(r) = f
10
(r)Y
00
(θ, ϕ),
f
10
(r) Y
00
(θ, ϕ) =
1
√
4π
1s
Z
|Ψ
1s
(r)|
2
d
3
r =
∞
Z
0
r
2
dr
Z
|Ψ
1s
(r)|
2
dΩ =
∞
Z
0
f
2
10
(r)r
2
| {z }
w
10
(r)
dr = 1.
r
2
d
3
r = r
2
dr sin θ dθ dϕ
r dr
w
10
(r) = f
2
10
(r)r
2
= R
2
10
(r)
R(r) = r
−1
f(r)
w
10
(r) ∼ r
2
exp
µ
−
2Zr
a
0
.
÷òî íàõîäèòñÿ â ñîãëàñèè ñ (1.14) (ïðîâîäèëàñü çàìåíà ïåðåìåííûõ t =
2Z         2Z
   r; dt =    dr).
a0         a0
   Ïîñêîëüêó â s-ñîñòîÿíèè l = 0, òî m òàêæå ðàâíî íóëþ. Ïîëíàÿ âîë-
íîâàÿ ôóíêöèÿ íàõîäèòñÿ â ñîîòâåòñòâèè ñ (1.10):
                                          s         µ     ¶
                                            Z3         Zr
                Ψ1s (r) = Ψ1s (r, θ, ϕ) =        exp −      ,  (1.25)
                                            πa30       a0

                              1
ãäå ó÷òåíî, ÷òî Y00 (θ, ϕ) = √ .                                         ¤
                              4π
Ïðèìåð 1.4. Íàéòè íàèâåðîÿòíåéøåå óäàëåíèå ýëåêòðîíà â 1s-
ñîñòîÿíèè îò ÿäðà ñ çàðÿäîì Ze.
Ðåøåíèå.
  1 ñïîñîá.
  Ñîãëàñíî (1.10), ïîëíàÿ âîëíîâàÿ ôóíêöèÿ 1s-ñîñòîÿíèÿ

                             Ψ1s (r) = f10 (r)Y00 (θ, ϕ),

ãäå ðàäèàëüíàÿ ôóíêöèÿ f10 (r) îïðåäåëåíà â (1.17). Ïîñêîëüêó Y00 (θ, ϕ) =
  1
√ , 1s-ñîñòîÿíèå èçîòðîïíî, ò.å. ñôåðè÷åñêè ñèììåòðè÷íî, è îòâåò íå
  4π
èçìåíèòñÿ, åñëè ïî òåëåñíîìó óãëó ïðîâåñòè èíòåãðèðîâàíèå. Èñïîëüçóÿ
óñëîâèÿ íîðìèðîâêè (1.7), (1.14), ïîëó÷àåì:
     Z                    Z∞    Z               Z∞
         |Ψ1s (r)|2 d3 r = r2 dr |Ψ1s (r)|2 dΩ = f10
                                                   2
                                                     (r)r2 dr = 1.   (1.26)
                                                  | {z }
                        0                            0   w10 (r)


Íàëè÷èå ìíîæèòåëÿ r2 â (1.26) îáóñëîâëåíî òåì, ÷òî â ñôåðè÷åñêîé ñèñòå-
ìå êîîðäèíàò ýëåìåíò îáúåìà d3 r = r2 dr sin θ dθ dϕ. Ïîñêîëüêó êâàäðàò
ìîäóëÿ âîëíîâîé ôóíêöèè èìååò âåðîÿòíîñòíûé ñìûñë, îáâåäåííîå ôè-
ãóðíîé ñêîáêîé â (1.26) ñëåäóåò ðàññìàòðèâàòü êàê ïëîòíîñòü âåðîÿòíîñòè
íàõîæäåíèÿ ýëåêòðîíà â ñôåðè÷åñêîì ñëîå ðàäèóñà r òîëùèíîé dr:
                                       2
                            w10 (r) = f10 (r)r2 = R10
                                                   2
                                                      (r)            (1.27)

(íàïîìíèì, ÷òî R(r) = r−1 f (r)). Èç (1.17) ñëåäóåò, ÷òî
                                        µ       ¶
                                   2        2Zr
                      w10 (r) ∼ r exp −           .
                                             a0


                                         11

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