Задачи по квантовой механике. Часть 2. Корнев А.С. - 50 стр.

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cos θ cos
2
θ s
0
1
3
cos
2
θ p m = 0, ±1
3
5
1
5
Ψ
200
Ψ
210
Ψ
100
Ψ
210
f
20
(r) f
21
(r) f
10
(r) f
21
(r)
r
n
1s
n > 2
(n + 2)!
2
n+1
³
a
0
Z
´
n
r
n
2s
n > 2
1
8
³
a
0
Z
´
n
(n
2
+ 3n + 4)(n + 2)!
r
n
2p
n > 4
1
24
³
a
0
Z
´
n
(n + 4)!
p
2
r
1
n
p
2
= µ
Z
2
n
2
e
2
a
0
r
1
=
Z
n
2
a
0
U(r) =
Ze
2
r
+
α
r
2
E
nl
=
1
2
Z
2
(n µ
l
)
2
e
2
a
0
µ
l
=
αµ
}
2
(l +
1
2
)
1s
Z
Z = 1 Z > 1
ϕ(r) =
Ze
r
e
r
½
1
·
2Z
2
a
2
0
r
2
+
2Z
a
0
r + 1
¸
exp
µ
2Zr
a
0
¶¾
Ze
a
0
µ
2Ze
a
0
+ 1
exp
µ
2Zr
a
0
.
                                               95
 "%`"/%#_'VŒŒ"#v&Ž2ŒaŒ*2"!DŽ #
 #UXG ,Q cos θ cos θ  s !#%!# 2 >; !#!# +
                                 2
                                                                                            67B9, F^C
4I>fF A 0 ^    17?

 #Qg G ,6Q cos θ  p!8%!# 2 4;%!8!# + L6W!
               3
                                                                                                                        7
  9 F24?>fF A 3 ^ 1 7 ?
                          2                                                                         m = 0, ±1


# G M("O)*56. 5 +/-.%!8. E 8 &("XQ Ψ                        ^
                7 N#.%! 2   + 6-. !# f (r)             ^      Ψ210 Ψ100
                                                                                                        7
 4
                                                                                          200

  NΨ210
     2 !#.          ;7                                 20       f21 (r) f10 (r) f21 (r)

   B G ,6Q !## T)  rn  1s !#%!8 2 `*;N+  6% 
        9 n > −2? 7:9  F24I>fF A (n + 2)! ³ a0 ´n 7 ?
      B#)G ,6Q !## T)  rZn  2s !#%!8 2 `*;N+  6% 
                                       2n+1

         9 n > −2? 7:9  F24I>fF A 1 ³ a0 ´n (n2 + 3n + 4)(n + 2)! ?
    B)BG ,Q !## T) 8*Z  rn  2p !#%!8 2 `;N+  6% 
       9 n > −4? 7:9  F24I>fF A 1 ³ a0 ´n (n + 4)!?
     BV G ,QT!8# ) 24* Z 2 p2 r−1  ;)-. a!#6X%a!#%!#
2  ;N+ \%  !F+ -E  "%E  !8-% n 7S9  F24I>fF A
           2
                Z 2 e2 −1               Z 7?
p =µ                         r =
              n2 a 0            n2 a 0
 B\ G ,Q !#" EFQ !8; ">- 2 ]- "3;- U (r) = − Ze + α 7
    ∗
                                                                                                          2


   9 F24?>fF A E = − 1 Z e ^ +  µ = αµ 7 ?
                                                                                                     r         r  2
                                      2           2



  B)3 G ,Q'!##Q ; X6- ]- "%!#6*!#"%+ 4;- 2 ^ !#)#  6*+
                     nl                       2             l      2        1
                            2 (n − µ ) a  l       0            } (l + )     2

;NE H%4 1s!8%!# 2  ^ "O"&(%"XUc ! !8 2  2 (
    ∗

 +  2  ! )* 2         76I * !#%!#;X;6-. - !8-% 2
             Z > 1                Z
 9  F24?>fF A
Z=1

                              ½   · 2               ¸    µ      ¶¾
          Ze e                     2Z 2 2Z                  2Zr
   ϕ(r) =   −                  1−       r +    r + 1 exp −          −
          r   r                     a20     a0               a0
                                                         µ        ¶   µ      ¶
                                                      Ze 2Ze             2Zr
                                                   −           + 1 exp −       .
                                                      a0   a0             a0