Задачи по квантовой механике. Часть 2. Корнев А.С. - 71 стр.

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m ω
c
n
(p) =
1
p
2
n
n!p
0
π
H
n
(ξ) e
ξ
2
/2
ξ =
p
p
0
p
2
0
= m}ω n = 0, 1, . . .
Ψ
1
Ψ
2
ˆ
H
H
11
= a H
22
= b H
12
=
H
21
= c E
±
=
1
2
{a +
b ±
p
(a b)
2
+ 4c
2
}
V (r) =
α
r
µ
p
2
2m
E
Ψ
E
(p) +
1
2π
2
}
3
Z
α
|p
0
p|
2
Ψ
E
(p
0
) d
3
p
0
= 0
ˆ
U
|a
0
i =
ˆ
U |ai.
ˆ
U
=
ˆ
U
1
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                                                     n                 √ H (ξ) e          n        ξ =
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                                                                                                 > F  A
                                                                                                     22
                                                                                                             1
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                                                                                                        E± = {a +
b ± (a − b) + 4c }3
        p                           7                                                                        2

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                      2       2



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    r
    1 F24I>fF A
                  µ
                      p2
                            ¶
                                       1
                         − E ΨE (p) + 2 3
                                          Z
                                                α
                                                      Ψ   (p 0
                                                               ) d 3 0
                                                                    p  = 0
                                                                           73
V G ,Q
                                                        E
                      2m             2π }   |p0 − p|2
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!F;O-% ; N6)  6 2 9 Z 7 Z ? - 2 4I>fg;F GR <           ;#- * ;-%
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