Задачи по квантовой механике. Часть 2. Корнев А.С. - 18 стр.

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Составители: 

x +
Ox
Ψ(x) = A
e
ikx
+ B
e
ikx
x −∞;
Ψ(x) = A
+
e
ik
0
x
+ B
+
e
ik
0
x
x +.
A
B
A
+
B
+
A
+
= αA
+βB
α β
V (x)
B
+
Ψ(x)
Ψ
(x)
Ψ
(x) = A
e
ikx
+ B
e
ikx
x −∞;
Ψ
(x) = A
+
e
ik
0
x
+ B
+
e
ik
0
x
x +
B
+
= αB
+ βA
B
+
= α
B
+ β
A
A
+
= αA
+ βB
; B
+
= β
A
+ α
B
.
B
+
= 0
B
/A
= β
A
= 0
A
+
/B
+
= β
R
1
=
¯
¯
¯
¯
B
A
¯
¯
¯
¯
2
=
¯
¯
¯
¯
β
α
¯
¯
¯
¯
2
, R
2
=
¯
¯
¯
¯
A
+
B
+
¯
¯
¯
¯
2
=
¯
¯
¯
¯
β
α
¯
¯
¯
¯
2
,
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18W;O-%!8"Q8 - ^ !8;%!#6 2 c 5\U8%! 2 N,!8E_%!# Ox A
                                                                                ; x → −∞;
                                     Ψ(x) = A− eikx + B− e−ikx                                                           9 0 7 1%5 ?
                                                    0
                                     Ψ(x) = A+ eik x + B+ e−ik x
                                                                        0          
                                                                                   ;  
                                                                                      
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                                                                                          x → +∞.

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        3 "%]*&,&(X 6 A ^ B A ^ B !#1 5T*!# H- Q 2 !# 2 ) .7
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B+
          !#  !#e. c !#6X+    A2 0 #+  49 0 7 0/? A*!8- Ψ(x) *!8.
  7 6T 3S+ a  *            2 0 #+   ^ 4 Ψ∗ (x) *!#.47 6T +
T,%  2 7K!#;*!#" E
                                     Ψ∗ (x) = A∗− e−ikx + B−            ∗ ikx
                                                                          e
                                                                                  ; x → −∞;
                                                        0
                                     Ψ∗ (x) = A∗+ e−ik x + B+
                                                                            0
                                                                         ∗ ik x      ; x → +∞
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                                                                           e

    ] %W * B ∗ = αB ∗ + βA∗ ^ O- B = α∗ B + β ∗ A 7 b "'N)   ^
"%]*&,&(X E +`9 0 7 1%5 −? !# 2 )*−E + !+ %+ %$−!#, 6T−  2 > 
                                                                           B + = β ∗ A − + α ∗ B− .
                                                                                                                        9 0 7 10/?
                 a - 2 !#X ^ /- 6 c 5TU8 'N .   ;-*L#-.% ;- 
                                     A+ = αA− + βB− ;

           9<!7 0 =7 9 ?8^  &("X 2 8D9 0 7 1%5? ! -#  ;-/. B = 0 ; ] %
                                         ∗ 7 I  ;;-*L% ! -%T;-+6*             ^ + 
                                      7 a - 2 "]*&,&(X >6T  2 4!# !# !H9 0 7 Z 9 ?
                                                                                                  +
                                  ∗
B− /A− = −β /α                                                                                             A− = 0
      * A
A+ /B+ = β/α∗
                                ¯    ¯  ¯ ¯                                   ¯    ¯  ¯ ¯
                                ¯ B− ¯2 ¯ β ∗ ¯2                              ¯ A+ ¯2 ¯ β ¯2
                          R1 = ¯¯    ¯ =¯ ¯ ,
                                        ¯ α∗ ¯                          R2 = ¯¯    ¯ =¯ ¯ ,
                                                                                      ¯ α∗ ¯
                                  A− ¯                                          B+ ¯