Задачи по квантовой механике. Часть 2. Корнев А.С. - 11 стр.

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m
V
0
x = a
0
a
x
V x( )
E
II
I
V
0
0
a
x
w x( )
(à)
(á)
I (0 6
x 6 a) II (x > a)
I
Ψ
I
(x) = A sin kx,
A
k
E < V
0
,
II
}
2
2m
Ψ
00
II
(x) + V
0
Ψ
II
(x) = EΨ
II
(x).
x +
Ψ
II
(x) = B e
κx
,
κ =
p
2m(V
0
E)
}
=
q
K
2
0
k
2
,
K
0
=
p
2mV
0
/},
B
x = a Ψ(a 0) = Ψ
I
(a)
                                                                     00

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 ) < i F^B d E > R&H BDBe5[4 Pk?< E=EDT Z 5fF2< c B GME < R9EDT Z 5 G 5fF G P E B i := a)                                                                              I              II V0
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      ;] %a/W!8) EF;! .  7 6T             0                     a              x
        !  N*E ' !8- *-A
                                                                                        w(x)                    (á)
                                ΨI (x) = A sin kx,
                                                                     9 0 7 0*[?
       +  A  ;) -. 2  *- % 2 "%!#6O
     6 ^ 4- !8- k !# 2 )*6 ! ]  + Q
         !#, 6T *_9 0 7 05 ? 7
                 o\ T  )/L -."%               0                     a              x
    ;a%! -                                                                       ‚&ƒU„oD†oI
                                                                                                        9 0 70 m ?
   N;8   EF%.F;, 6T P  .
                                                                  E < V0 ,
                                                                           2 0 #+    FN-!#
II
      A
                                  −
                                     }2 00
                                          ΨII (x) + V0 ΨII (x) = EΨII (x).
                                                                                                         9 0 7 5 ?
  ( * N8. ! 2 ^  % \c 9 0 7 5 ? -  2  !8-#!c 5  2
                                    2m
   &(%"X 2 ^ N 53 c 5  2 ! 2 /-.; x → +∞ A
                                              ΨII (x) = B e−κx ,
                                                                                                          9 0 7 O0/?
 + 
                                                                                                               9 0 7 ?
                                                        p
                                                         2m(V0 − E)   
                                                                            q
                                                     κ=                =     K02 − k 2 ,
                                                            } p
                                                           K0 = 2mV0 /},
                                                                                                                9 0 7 Z ?
B    U7;6T ) 2 -9 0 .7 0*[ 2?
                           *-   2 "%!8667
                             9 0 7 O0*? NA8  ! 6T.    "%(" + T) E 
%H; X6- x = a  !8 !#$! 9 0 7 Z?8^ ; -  +# 2 Ψ(a − 0) = Ψ (a) ^
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