Задачи по квантовой механике. Часть 2. Корнев А.С. - 81 стр.

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L
(α)
n
(x) =
1
n!
x
−α
e
x
d
n
dx
n
[x
n+α
e
−x
];
L
(α)
n
(x) =
1
n!
Γ(n + α + 1)
Γ(α + 1)
1
F
1
(−n, α + 1, x).
ν
x
2
y
00
+ xy
0
+ (x
2
− ν
2
)y = 0.
J
ν
(x) =
³
x
2
´
ν
∞
X
k=0
(−x
2
/2)
k
k!Γ(ν + k + 1)
;
J
ν
(x) =
e
−iz
Γ(1 + ν)
³
z
2
´
ν
1
F
1
µ
ν +
1
2
, 2ν + 1, 2iz
.
j
l
(x) =
r
π
2z
J
l+
1
2
(x), l = 0, 1, . . .
j
0
(x) =
sin x
x
; j
1
(x) =
sin x
x
2
−
cos x
x
; j
2
(x) =
µ
3
x
3
−
1
x
sin x −
3
x
2
cos x.
(x)
y
00
− xy = 0
±
1
3
(x) =
1
3
√
x [I
−1/3
(ζ) − I
1/3
(ζ)]; (−x) =
1
3
√
x [J
−1/3
(ζ) + J
1/3
(ζ)],
I
ν
(ζ) = i
−ν
J
ν
(iζ) ζ =
2
3
x
3/2
                                                                       [O0

           0/? &(/-TU*+OA
                                                        1 −α x dn n+α −x
                                       Ln(α) (x)      =    x e     [x e ];
                                                        n!     dxn
           ? EF/   2 +; + % !#"O 2 &("X 2 A
                                                  1 Γ(n + α + 1)
                             Ln(α) (x) =                         1 F1 (−n, α + 1, x).
                                                  n! Γ(α + 1)
                 Ž        L $_#
                                   


           o\%"X 2 T> *! !##- 2 + '; 2 "O ) EF% c! 2  + *- 2 EF 76T  2
X -*!8"+ 3&,&(  X6-.+3  2 A
                                                  ν

                                                                                                        9 a 7 0*5 ?
      a /  "%EF 2 EFPEF6T  2 - 2 &("XQ>R*! !8- 2 A
                                   x2 y 00 + xy 0 + (x2 − ν 2 )y = 0.

       0/? 6)-*3P 2 :A
                                                                ∞
                                                         ³ x ´ν X              (−x2 /2)k
                                       Jν (x) =                                             ;
                                                            2                k!Γ(ν + k + 1)
                                                                      k=0
           ? EF/   2 +; + % !#"O 2 &("X 2 A
                                                      µ                 ¶
                                      e−iz ³ z ´ν         1
                           Jν (x) =               1 F1 ν + , 2ν + 1, 2iz .
                                    Γ(1 + ν) 2
           @ &( *!#"O 2 &("X 2          > *! !#- 2
                                                          2


                                                                                                                         9 a 7 00/?
                                                      r
                                                           π
                                       jl (x) =              J 1 (x),                l = 0, 1, . . .
                                                           2z l+ 2
EF6  ! 2  )P]-* 6EF ^ 6; ^
                                                                                            µ          ¶
         sin x                        sin x cos x                                               3    1           3
j0 (x) =       ;              j1 (x) = 2 −        ;                              j2 (x) =          −     sin x −    cos x.
           o\%"X 2 l Q
           x                           x
                                        
                                              x
                                                  2  - 2  ! 2  + *- 2 E '76\ *'%  2
                                                                                                x3   x           x2


                                                                                                                         9 a 7 0/?
                                            (x)


aEF66  ! 2   ) &("X>R*! !8- 2 ; 2 "
                                                             y 00 − xy = 0
                                                                                                ± 13
                                                                                                       A
                      1√                                                                 1√
            (x) =          x [I−1/3 (ζ) − I1/3 (ζ)];                             (−x) =       x [J−1/3 (ζ) + J1/3 (ζ)],
                        3                                                                   3
+            Iν (ζ) = i−ν Jν (iζ)                   ζ=    2
                                                                x3/2
                                                                        7
                                                            3

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