Задачи по квантовой механике. Часть 2. Корнев А.С. - 6 стр.

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a 0 a + 0
}
2
2m
0
(a + 0) Ψ
0
(a 0)] = 0.
Ψ(a 0) = Ψ(a + 0); Ψ
0
(a 0) = Ψ
0
(a + 0).
Ψ(a ±0)
x = a
Ψ
0
(x)
Ψ(x)
¯
¯
¯
¯
a0
=
Ψ
0
(x)
Ψ(x)
¯
¯
¯
¯
a+0
,
d
dx
Ψ(x)
¯
¯
¯
¯
a0
=
d
dx
Ψ(x)
¯
¯
¯
¯
a+0
.
¤
x = a
x = a
Ψ(a 0) = Ψ(a) = 0; Ψ(x > a) 0.
x = a
                                                                  ?
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+  H9 0 7 0/? ;#- 8  a − 0  a + 0 A
                                       }2
                                     −    [Ψ0 (a + 0) − Ψ0 (a − 0)] = 0.
         b "                      2m
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                        Ψ(a − 0) = Ψ(a + 0);                           Ψ0 (a − 0) = Ψ0 (a + 0).
                                                                                                                               9 0 7 ?
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    (
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                                                       O-                                                                     9 0 7Z ?
                   ¯           ¯                                              ¯              ¯
           Ψ0 (x) ¯¯   Ψ0 (x) ¯¯                                       d      ¯       d      ¯
                     =           ,                                       Ψ(x)¯¯    =    Ψ(x)¯¯    .
           Ψ(x) ¯a−0   Ψ(x) ¯a+0                                      dx       a−0   dx       a+0

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         R!
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 %!#H; x = a -  2 &(%"X 2 -L3N.%! 2  *-.A
                                Ψ(a − 0) = Ψ(a) = 0;                   Ψ(x > a) ≡ 0.
                                                                                                             9 0 7 1?
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