Задачи по квантовой механике. Часть 2. Корнев А.С. - 45 стр.

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1s
f
10
(r) = 2
µ
Z
a
0
3
/
2
exp
µ
−
Zr
a
0
Z
∞
0
f
2
10
(r)r
2
dr =
4Z
3
a
3
0
Z
∞
0
r
2
exp
µ
−
2Zr
a
0
dr =
1
2
Z
∞
0
t
2
e
−t
dt
|
{z }
2!
= 1,
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.
                                                                1   9
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c =B B         G 5 EDG 4 EDGWHJG 5 G 5 F G P E QB Ph4 GIKoGR€GIKoG GIKoGWXMEDGWHJG M< F G \w K B E B c 8Qn T DB95  R9G P9F E > i 2>[>h8 K <[6_> E B€>,d{6_>fg;F R€GYE < 4                         1s
                                                                                                                                      C
 5 G 5fF G P E B=B G F P VK R < 5k?< R P K[G \ Ze n
x(y{z|y{}~€yI
             0 #! ;%!#N 7
            @ + -!#H9 7 : ? ^ ;- 2 -  2 &("X 2       1s
                                                                                              !#%!# 2  2
                                                   Ψ1s (r) = f10 (r)Y00 (θ, ϕ),
 +  /6-. 2 &("X 2 f (r) ;#- F39 7 0/? 76M %!#"%-." Y (θ, ϕ) =
    1 ^         #
                 ! %
                    8
                     ! 
                         2 
                               )  
                                        
                                          
                                            
                                              
                                               ; 
                                                   ^ /
                                                        7 
                                                            7 ! (
                                                                 & 
                                                                    
                                                                        
                                                                             #
                                                                              ! 
                                                                                " '! #  
                                                                                            
                                                                                               
                                                                                                   
                                                                                                         ^ ' `
                                               10                                                             00

  √        1s
)*4π ! 2 ^ *!8-`;W-!# a+ ->;!#4 +  7 GP!#; -.) 2
 !8- 2 " 9 I @7 ? ?8^ 9 7 00/? ^ ;-* A
            Z
                           2 3
                |Ψ1s (r)| d r =
                                       Z       ∞
                                               r dr2
                                                       Z
                                                                        2
                                                           |Ψ1s (r)| dΩ =
                                                                                     Z       ∞
                                                                                              f 2 (r)r2 dr = 1.
                                                                                                                           9 7 9?
                                           0                                             0    | 10{z }
                                                                                                 w10 (r)

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