Задачи по квантовой механике. Часть 2. Корнев А.С. - 15 стр.

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0
x
V x( )
E
V
max
V
min
j
ïàä
j
îòð
j
ïðîø
R =
j
j
¯
¯
¯
¯
x→−∞
,
j = |j |; j = |j |
D =
j |
x+
j |
x→−∞
, j = |j |.
x +
x +
Ψ(x)|
x+
= e
ik
0
x
,
k
0
=
p
2m(E V )
}
.
j |
x+
=
}k
0
m
.
                                                                        0       9
                                                       V(x)
                       jîòð                                         Vmax                                              jïðîø
                              jïàä
                                                              E
                                                        Vmin
                                                              0                                                           x
                                                              ‚&ƒU„Y=†Y W           




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                                                      +                                                                     9 0 7 [ ?
                         ¯
                 j ¯¯
              R=            ,                                    j              = |j   |;     j   = |j  |
                 j ¯x→−∞
       g G d @"@(B c €B > E F G \                                                                    c5
                                         R€G[ZDG N K > E BQP^ ;# !#6- 2 T + !#NQ  T ,6
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    !#X
                            D=
                                    j      |
                                         
                                                x→+∞
                                                        ,
                                                                  +  j             = |j         |.
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     M -%!8 ;"OH;#;-+c  ! 2 ) 2 E  `!#;*!#" * 6- 
                                                                                                          
                                     j | 
                                                x→−∞

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-.E  ;*-.%!#% ? 7 M(] % !8;!#"Q  -Q &("O
   Xa; x → +∞ ;# !#6- 2  L!#NQ ;-%!#"!c -SA
                                                 Ψ(x)|x→+∞ = eik x ,
                                                                                             0                                9 0 7 Z5 ?
+ 
                                                                                                            9 0 7 Z0/?
                                                             p
                                                                 2m(E − V )
 o "X 2 9 0 7 Z0/? %6 !#;*  ;-%!8T;"O ;,6\#)6TQ8
                                                k0 =                            .
                                                                     }
!#X4!("O-! !#!#"QH!#"%%!#.,c 9 !  7;  1 7 Z !# 0/? A

                                                 j       |x→+∞ =
                                                                                          }k 0
                                                                                               .
                                                                                                                              9 0 7 Z ?
                                                                                          m