数理解析研究所講究録
第 1824 巻 2013 年 51-67

51

The Solutions to The Radial Schrodinger Equationof The Hydrogen Atom by Means of
N-Fractional Calculus Operator
Katsuyuki Nishimoto
Institute for Applied Mathematics, Descartes Press Co.
2-13-10 Kaguike, Koriyama 963-8833, Japan
Abstract
In this article, the solutions to the radial Schrodinger equation of theHydrogen atom (in the Coulomb field)$\varphi_{2}\cdot x^{2}+\varphi_{1}\cdot 2x+\varphi\cdot\{-(1/4)x^{2}+\nu x-l(l+1)\}=0$

are discussed by means of-fractional calculus operator.
A particular solution to the equation above is shown as follows forexample.
$N$$\varphi=\varphi_{[1]}=x^{t}e^{X/2}(e^{-x}\cdot x^{v-(t+1)})_{l+v}$(fractional differintegrated form)$=(e^{i\pi})^{t+\nu}e^{-x/2}x^{v-1}{}_{2}F_{0}(-l-\nu,l+1-\nu;-1/x)$$(|-1/x|<1)$

where${}_{p}F_{q}(\cdots\cdots)$; Generalized Gauss hypergeometric functions.

52$辷_{}\overline{1}$Notice that$(\prime\downarrow)$$\overline{r_{\overline{1}}}s2\vee\cdot\cdot$

奪．
$\wedge--,$is reduced to $Goursat\cdot s$ integral for

uced to the famous $\subset audsinte_{o}$$o_{\sim}^{\mathcal{L}}($

Cau$\subset hy^{\sigma}$

inte ral$an\tilde{d}$offor $v=0$$Go\iota xrsa_{L^{1}}^{A}sone_{r}$That is.

conversel$y^{}$and is redis an exten$v=n(\in\dot{Z}^{A})$$(\prime\downarrow)$$Cauchy^{-\prime}s$$\overline{a}or.$andones are speciai cases of (I).
Moreonotice thatffie representation whi 何 h unifi es ffie deriva ト$Goursat^{r}s$$ver_{r}$$\overline{J^{-}.v}esa_{\lambda}\urcorner di\sigma.$$(\prime\downarrow)i_{5}$53On the fractional calculus operator $N^{v}|3|$
Theorem A.
cuIus operator Nishimoro(II)$Le \tau_{J}\dagger ac\tau ionalca\int$.

$($$(v\not\in Z^{\sim})$$N^{\iota^{r}}=( \frac{\Gamma(\nu+1)}{2\pi\dot{r}}\int_{c}\frac{dc^{arrow}\sim}{(\tilde{\underline{c}}-z)^{VA}1}))$$N^{-r}=\varliminf_{v-r}.N^{v}$

be

(3)

$]^{-}$(4)

operation $oa5$$bir za\bigcap_{d}j$$N^{\beta}oN^{\alpha}f=N^{\beta}N^{\sigma}f=N^{\beta}(N^{\alpha}f)$$fhe\prime i$

Refer to (1)$1V^{1’}$$(m\in Z^{A})_{7}$$wi_{\hat{L}}l\tau$and define the$[$rator)$sop\dot{e}$$(\alpha, \beta\in R)$(5),が詑 set(6)$\{N^{1’}\}=\{N^{t’}|v\in R\}$is an Abelian Productransform operator$\iota iorrf$$\sigma roup$$c$}(havirlg $corlti_{tl}uous$ index v) which $ha5$ the inverse$(N^{v})^{-i}=N^{-v}torhe_{J}\not\simeq actio/\iota al$$\hat{\iota}$such that$f\in F=\{f|0\neq|f_{\iota},$$|<\infty_{i}v\in R\}_{i}$where(For our $convenience_{f}$ we call

Theorem B. The
nuous

”

$i_{ti}dexv\prime$$cor_{L}’f\dot{\tau}$group) [3]Theorem C-$\overline{l.}ef$

’

F.O.

for

$G.$

$\hat{}he$

ser

$F$

is

. (

as product of$fhese_{1}^{\angle}S$$N^{\beta}$$ar\iota$(i)$(ii)$$F-O.(_{\mathcal{F}-r}^{\neg}$is a$comr_{\Gamma}\tau utativerirLgfo7^{\cdot}\mathcal{E}hef\tau.\ell$

nctionf$(N^{\sigma},N^{\beta}-N^{r}\prime\in S)$$(\log(z-c))_{\sigma}=-e^{-iir\alpha}\Gamma(\alpha)(z-c)^{-\alpha}$$( iii)((z-c)^{-\alpha})_{-\alpha}=-e^{j_{\overline{lu}G}}\frac{\prime\perp}{\Gamma(\alpha)}\log(z-\cdot c)$$(i\cdot Y7)$$(i)_{r}$$(|\Gamma(\alpha)|<\infty)$$([I^{-}(\alpha)|<\infty)$

and $z-c\neq 0,1$ in ( ii).

$(\mathcal{U}^{-v)_{\sigma};=\sum_{k\cdot 0}^{\infty}\frac{\Gamma(\alpha^{\perp_{l}}1)}{k!\Gamma(\alpha+1-k)}u_{\alpha-k}\iota_{k}}$(7).

. We have [1]in)$p7^{r}$$((z-c)^{\beta})_{\sigma}=e^{-i_{\overline{d}}\alpha} \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}(z-c)^{1,-\ulcorner\wedge}$

where $\cdot\sim$$N^{\alpha}.$“Action oduct group
Fractional calcuIus operator

holds. [5]$Len_{\grave{t}}ma$

and

be

$N^{\alpha}+N^{\beta}=N^{\gamma}$

(IID

-

$\tau x’ hich\dagger ws$$S;=\{-arrow N^{\tau^{t}}\}\cup\{0\}=\{N^{\nu}\}\cup\{-N^{\nu}\}^{()}\sim\{()\}(v\in R)$

Thenident 吻for the fiurlc$N_{i}^{v}$$(v\dot{\overline{\iota}}z\cdot<\cdot$$f=f(\tilde{\mathcal{L}})art^{}d^{t}z\in C.$$N^{\beta}oN^{\sigma}$$\{N^{v}\}^{rr}$caldtus operator$(_{\nu=v(z)}^{u=u(z)\prime}$$\in F$, when the

(8)

54
\S 1. Preliminary(I) The theorem below is reported by the author already (cf. J.F.C.Vol. 27,
May $(2005),83-88.)[31]$Theorem D. Let be$P=P( \alpha,\beta,\gamma):=\frac{\sin\pi\alpha\cdot\sin\pi(\gamma-\alpha-\sqrt{})}{\sin\pi(\alpha+\sqrt{})\cdot\sin\pi(\gamma-\alpha)}$$(|P(\alpha,\beta,\gamma)|=M<\infty)$

and$Q=Q(\alpha,\beta,\gamma):=P(\beta,\alpha,\gamma),$$t\phi 7_{l}en$$\alpha,$(i)$\beta,$$\gamma\not\in Z_{0}^{+}$,(2)$(|P(\beta,\alpha,\gamma)|=M<\infty)$

we have(3)$((z-c)^{\alpha} \cdot(z-c)^{\beta})_{\gamma}=e^{-i\pi\gamma}P(\alpha,\beta,\gamma)\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\sqrt{})}(z-c)^{\alpha+\beta-\gamma}$$({\rm Re}(\alpha+\beta+1)>0, (1+\alpha-\gamma)\not\in Z_{0}^{-})$

(ii)

,

$((z-c)^{\beta} \cdot(z-c)^{\alpha})_{\gamma}=e^{-i\pi\gamma}Q(\alpha,\beta,\gamma)\frac{\Gamma(\gamma-\prime\alpha-\sqrt{})}{\Gamma(-\alpha-\beta)}(z-c)^{\alpha+\beta-\gamma}$$({\rm Re}(\alpha+\beta+1)>0, (1+\beta-\gamma)\not\in Z_{0})$

(iii)

(1)

,

(4)

,

$((z-c)^{a+\beta})_{\gamma}=e^{-i\pi\gamma} \frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\sqrt{})}(z-c)^{\sigma+\beta-\gamma}$

(5)

,

where
$z-c\neq 0,$

Then the inequahties below are established from this theorem.
Corollary 1. We have the inequalities
(i)

(6)

$((z-c)^{a}\cdot(z-c)^{\beta})_{\gamma}\neq((z-c)^{\beta}\cdot(z-c)^{a})_{\gamma’}$and$(ii)$$((z-c)^{a}\cdot(z-c)^{\beta})_{\gamma}\neq((z-c)^{a+\beta})_{\gamma}$

(7)

,

where
$\alpha,\beta,\gamma\not\in Z_{0}^{+}, \alpha\neq\beta, z-c\neq 0$

Corollar 2.
(i)When
we have$Y$$\alpha,\beta,\gamma\not\in Z_{0}^{+}$(8), and $P(\alpha,\beta,\gamma)=Q(\beta,\alpha,\gamma)=1$ ,$((z-c)^{\alpha}\cdot(z-c)^{\beta})_{\gamma}=((z-c)^{\beta}\cdot(z-c)^{a})_{\gamma}=((z-c)^{\alpha+\beta})_{\gamma}$,$({\rm Re}(\alpha+\beta+1)>0, (1+\alpha-\gamma)\not\in Z_{0}^{-}, (1+\beta-\gamma)\not\in Z_{0}^{-})$,(ii) When$\gamma=m\in Z_{0}^{+}$(9), we have ;$((z-c)^{a}\cdot(z-c)^{\beta})_{m}=((z-c)^{\beta}\cdot(z-c)^{\alpha})_{m}=((z-c)^{\alpha+\beta})_{m}$

(10)

55
\S 2. The Solutions to The Radial Schrodinger Equationof The Hydrogen Atom

Theorem 1. Let be, then the homogeneous linear
ordinary differential equation (Radial Schrodinger equation of theHydrogen atom in the Coulomb field)$\varphi=\varphi(x)\in F$$L[\varphi;x;\nu,t]=\varphi_{2}\cdot x^{2}+\varphi_{1}\cdot 2x+\varphi\cdot\{-(1/4)x^{2}+\nu x-l(l+1)\}=0$(1)$(\varphi_{a}=d^{\alpha}\varphi/dx^{\alpha} for \alpha>0, \varphi_{0}=\varphi, x\neq 0)$

has particular solutions of the forms (fractional dffferintegrated forms),

Group I;

1

$)$

2

$)$

3

$)$

4

$)$

$\varphi=x^{t}e^{x/2}(e^{-x}\cdot x^{\nu-(1+1)})_{t+v}\equiv\varphi_{[1]},$’$\varphi=x^{l}e^{x/2}(x^{v-(t+1)}\cdot e^{-x})_{l+}$。

$\equiv\varphi_{[2]}$$\varphi=x^{l}e^{-x/2}(e^{X}\cdot x^{-\nu-(l+1)})_{t-}$。

$\equiv\varphi_{[3]}$$\varphi=x^{t}e^{-x/2}(x^{-\nu-(l+1)}\cdot e^{x})_{l-}$。

$\equiv\varphi_{[4]}$(denote)

(2)

(3)
(4)
(5)

Group II;

1

$)$

2

$)$

3

$)$

4

$)$

$\varphi=x^{-(t+1)}e^{x/z}(e^{-x}\cdot x^{\nu+\mathfrak{l}})_{\nu-(l+1)}\equiv\varphi_{[5]}$(6)$\varphi=x^{-(l+1)}e^{x/2}(x^{v+t}\cdot e^{-x})_{v-(l+1)}\equiv\varphi_{[6]}$(7)$\varphi=x^{-(t+1)}e^{-x/2}(e^{x}\cdot x^{-\nu+t})_{-v-(l+1)}\equiv\varphi_{[7]}$(8)$\varphi=x^{-(t+1)}e^{-x/2}(x^{-\nu+t}\cdot e^{X})_{-v-(t+1)}\equiv\varphi_{[8]}$(9)Proof of Group I.
Set$\varphi=x^{\mu}\phi, (\phi=\phi(x))$

we have then$\varphi_{1}=\mu x^{\mu-1}\phi+x^{\mu}\phi_{1}$and$\varphi_{2}=\mu(\mu-1)x^{\mu-2}\phi+2\mu x^{\mu-1}\phi_{1}+x^{\mu}\phi_{2}$

(10)
(11)

(12)

56
Hence, yields$\phi_{2}\cdot x^{\mu+2}+\phi_{1}\cdot x^{\mu+1}(2\mu+2)+\phi\cdot\{-(1/4)x^{\mu+2}+vx^{\mu+1}+x^{\mu}(\mu^{2}+\mu-l^{2}-l)\}=0$(13)

from (1), applying (10), (11) and (12).
such that
Choose$\mu$(14)$\mu^{2}+\mu-l^{2}-l=0$

hence$\mu=\{\begin{array}{ll}l (15)-(l+1) (16)\end{array}$Case of $\mu=l,$
In this case we have

(I)

傷 x

1(21 2)$+\phi$$+$$+\phi$$\{v-(1/4)x\}=0$(17)

from (13). Next we set$\phi=e^{\lambda x}\psi (\psi=\psi(x))$(18)

we have then(19)$\phi_{1}=\lambda e^{\lambda x}\psi+e^{\lambda x}\psi_{1}$and$\phi_{2}=\lambda^{2}e^{\lambda x}\psi+2\lambda e^{\lambda x}\psi_{1}+e^{\lambdax}\psi_{2}$

.

(20)

Therefore, we obtain$\psi_{2}x+\psi_{1}\cdot(2\lambda x+2l+2)+\psi\cdot\{x(\lambda^{2}-\frac{1}{4})+2\lambda l+2\lambda+v\}=0$(21)

from (17), using (18), (19) and (20).
Choose$\lambda=\{\begin{array}{ll}1/2 (22)-1/2 (23)\end{array}$Case of $\lambda=1/2,$
In this case we have(i)$\psi_{2}\cdot x+\psi_{1}\cdot(x+21+2)+\psi\cdot(l+1+v)=0J$

from (21).

(24)

57
( $N$ -fractional calculus operator of order
Operatesides of (24), yields) to the both$N^{\alpha}$$\alpha$$N^{\alpha}(\psi_{2}\cdot x)+N^{\alpha}(\psi_{1}\cdot(x+21+2))+N^{\alpha}(\psi\cdot(l+1+\nu))=0,$$(\alpha\not\in Z_{0})$

(25)

hence
$\psi_{2+\alpha}\cdot x+\psi_{1+a}.(x+\alpha+2l+2)+\psi_{\alpha}.(\alpha+l+1+\nu)=0$(26)

from (25), by Lemma (iv), since(27)$N^{a}(\psi_{2}\cdot x)=(\psi_{2}\cdot x)_{\alpha}$

for example.
Choosesuch that$\alpha$$\alpha=-(t+1+\nu)\equiv p$

we have then(denote)$\psi_{2+p}\cdot x+\psi_{1+p}\cdot(x+p+2l+^{i}2)=0$

(28)
(29)

from (26), applying (28).
Therefore, we obtain$u_{]}\cdot x+u\cdot(x+p+2l+2)=0$(30)

from (29), setting$u=\psi_{1+p}.

(u_{-(?+p)}=\psi)$(31)

A particular solution to this variable separable form equation (30) is givenby$u=e^{-x}x^{-(p+2l+2)}$

We have then(32)$\psi=u_{-(1+p)}=(e^{-X}\cdot x^{-(p+2l+2)})_{-(1+p)}$(33)$\phi=e^{x/2}(e^{-X}\cdot x^{-(p+2/*2)})_{-(1+p)}$(34)

from (31), and hencefrom (18), applying (22) and (33).
Hence we obtain$\varphi=x^{l}e^{x/2}(e^{-x}\cdot x^{-(p+2l+2)})_{-(1+p)}$$=x^{t}e^{x\prime 2}(e^{-x}\cdot x^{v-(I+1)})_{+\nu}\equiv\varphi_{[1]}$from(10) using (15) and (34).

(35)

(2)

58
Next we obtain(3)$\varphi=x^{l}e^{x/2}(x^{\nu-(l*1)}.e^{-x})_{l+v}\equiv\varphi_{[2]}$changing the order of$e^{-X}$andin the parenth esis$x^{v-(l+1)}$$()_{l+t}$,in (2).

We have$\varphi_{[1|}=\varphi_{[2]}$when$(l+\nu)\in Z_{0}^{+}$

Case of $\lambda=-1/2,$
In this case we have(ii)$\psi_{2}.x+\psi_{1}.(-x+2l+2)+\psi.(\nu-l-1)=0$

,

(36)

from (21).
Therefore, in the same manner as above, we obtain(4)$\varphi=x^{l}e^{-x/2}(e^{X}\cdot x^{-\nu-(l+1)})_{-\nu}\equiv\varphi_{\mathfrak{l}3|}$from (36) and(5)$\varphi=x^{l-x/2}e(x^{-\nu-(l+1)}.e^{X})_{l-\nu}\equiv\varphi_{[4]}$changing the order of$e^{X}$$(\Pi)$

andin the parenthesis$x^{-\nu-(l+1)}$$()_{l-v}$

in (4).

Case of $\mu=-(l+1)$ ;Set $-(l+1)$ instead of$l$in the solutions$\varphi_{[1]},$$\varphi_{[2]},$$\varphi_{[.’]}$, and$\varphi_{|4)^{*}}$

we have then the solutions$\varphi_{[5]},$$\varphi_{[6]},$$\varphi_{|7)}$,

and$\varphi_{[8]},$respectively.
\S 3. The Familiar Forms of The Solutions in Section 2.

The familiar forms of the solutions that are obtained ones in section 2
are shown as follows.

59
Corollary 1. WehaveGroup I.

1

$)$

2

$)$

$(|-1/x|<1)$ ,$\varphi_{[1]}=(e^{i\pi})^{l+\nu}e^{-x/2}x^{v-1}{}_{2}F_{0}(-l-v\cdot,l+1-\nu;-1/x)$$\varphi_{[2]1}=(e^{-i\pi})^{l+\nu}e^{-x/2}x^{-(t+1)}\frac{\Gamma(2t+1)}{\Gamma(l+1-\nu)}F_{1}(-l-v;-2t;x),$$(])$$(|x|<1),$ $(2)$$(|\Gamma(2l+1-k)/\Gamma(l+1-\nu)|<\infty)$

3

$)$

4

$)$

(3)

$(|1/x|<1)$ ,$\varphi_{[3]}=e^{X/2}x^{-(1+\nu)}{}_{2}F_{0}(-t+\nu,1+1+\nu;1/x)$$\varphi_{[4]}=(e^{-i\pi})^{l-\nu}e^{x/2}x^{-(t+1)}\frac{\Gamma(21+1)}{\Gamma(l+1+\nu)}{}_{1}F_{1}(\nu-l;-2l;-x)$,(4)$(|-x|<1)$ ,$(|\Gamma(2l+1-k)/\Gamma(\nu+l+1)|<\infty)$

Group II.
1

$)$

2

$)$

$\varphi_{[5]}=(e^{i\pi})^{v-(l+1)}e^{-x/2}x^{v-1}{}_{2}F_{0}(l+1-v, -l-\nu;-1/x)$,$(|-1/x|<1),$$(5)$$(|x|<1)$ ,(6)$\varphi_{[6]}=(e^{-i\pi})^{-(t+1)+v}e^{-x/2}x^{t}\frac{\Gamma(-2l-1)}{\Gamma(-l-\nu)}{}_{1}F_{1}(l+1-\nu;2l+2;x),$$(|\Gamma(-2l-1-k)/\Gamma(-\nu-1)|<\infty)$

3

$)$

4

$)$

$\varphi_{[7]}=e^{x/2}x^{-(1+\nu)}{}_{2}F_{0}(t+1+\nu, \nu-1;1/x)$(7)$(|1/x|<1)$ ,$\varphi_{[8]}=(e^{i\pi})^{t+1+v}e^{x/2}x^{l}\frac{\Gamma(-2l-1)}{\Gamma(\nu-l)}{}_{1}F_{1}(l+1+\nu;2l+2;-x),$(8)$(|-x|<1)$ ,$(|\Gamma(-2l-1-k)/\Gamma(\nu-l)|<\infty)$

Where$pqF(\cdots\cdots)$

are thegeneralized Gauss hypergeometricfunctions.

Proof of Group I.
We have1$)$$\varphi_{[1]}=x^{l}e^{x/2}(e^{-x}\cdot x^{v-(l+1)})_{l+v}$$=x^{t}e^{x/2} \sum_{k-0}^{\infty}\frac{\Gamma(l+\nu+1)}{k!\Gamma(l+\nu+1-k)}(e^{-x})_{l+\nu-k}(x^{v-(l+1)})_{k}$(\S 2. (2))(by Lemma (iv))$= x^{1}e^{x/2}\sum_{k\Rightarrow 0}^{\infty}\frac{\Gamma(l+v+1)}{k!\Gamma(l+\nu+1-k)}(ei\pi)le\{e^{-i\pi k}\frac{\Gamma(k+l+1-v)}{\Gamma(l+1-\nu)}x^{v-(l+1)-k}\}$

(9)

(10)

60
(11)

$=(e^{i\pi})^{l+v}e^{-x/2}x^{\nu-1} \sum_{k-0}^{\infty}\frac{[-l-v]_{k}[t+1-vL}{k!}(-x)^{-k}$$=(e^{i\pi})^{l+v}e^{-x/2}x^{\nu-1}{}_{2}F_{0}(-l-\nu, l+1-\nu;-1/x)$,$(|-1/x|<1)$ ,

(1)

since
$(k\in Z_{0}^{+})$$\Gamma(\lambda-k)=(-1)^{-k}\frac{\Gamma(\lambda)\Gamma(1-\lambda)}{\Gamma(k+1-\lambda)}=(-1)^{-k}\frac{\Gamma(\lambda)}{[1-\lambda]_{k}},$

(12)

(13)

$(e^{\lambda x})_{t+\nu-k}=\lambda^{t+\nu-k}e^{\lambda x}$(14)$(x^{\nu-(t+1)})_{k}=e^{-i\pi k} \frac{\Gamma(k+1+1-\nu)}{\Gamma(l+1-\nu)}x^{\nu-(l*1)-k}$and$[\lambda]_{k}=\lambda(\lambda+1)\cdots(\lambda+k-1)=\Gamma(\lambda+k)/\Gamma(\lambda)$with$[\lambda\iota=1.$(Notation of Pohhammer).

2$)$(\S 2. (3))$\varphi_{[2]}=x^{t}e^{x/2}(x^{v-(l+1)}.e^{-x})_{l+\nu}$(15)$=x^{l}e^{x/2} \sum_{k-0}^{\infty}\frac{\Gamma(l+v+1)}{k!\Gamma(l+v+1-k)}(x^{v-(l+1)})_{l+\nu-k}(e^{-x})_{k}$$=x^{l}e^{x/2} \sum_{k-0}^{\infty}\frac{(-1)^{k}[-l-v]_{k}}{k!}\{e^{-i\pi(l+\nu-k)}\Gamma(2l+1-k)_{x^{-2l-1+k}\}(e^{i\pi})^{k}e^{-x}}$(16)$\Gamma(l+1-v)$$=x^{-t-1}(e^{-i\pi})^{t+\nu}e^{-x/2} \sum_{k\cdot 0}^{\infty}\frac{(-1)^{k}[-l-v]_{k}(-1)^{-k}\Gamma(2l+1)}{k!\Gamma(l+1-v)[-2l]_{k}}x^{k}$(18)$=(e^{-i\pi})^{t+\nu} \frac{\Gamma(2l+1)}{\Gamma(l+1-\nu)}e^{-x/2}x^{-l-1}\sum_{k-0}^{\infty}\frac{[-l-v]_{k}}{k![-2l]_{k}}x^{k}$,$=(e^{-i\pi})^{l+\nu}e^{-x/2}x^{-l-1} \frac{\Gamma(2l+1)}{\Gamma(l+1-\nu)}{}_{1}F_{1}(-l-v;-2l;x)$(17)$(|x|<1)$ ,

(2)

61
3

$)$

(\S 2. (4))$\varphi_{[3]}=x^{l}e^{-x/2}(e^{X}.x^{-v-(l+1)})_{t-v}$(19)$=x^{t-x/2}e \sum_{k=0}^{\infty}\frac{\Gamma(l-v+1)}{k!\Gamma(t-\nu+1-k)}(e^{x})_{t-v-k}(x^{-v-(f+1)})_{k}$(20)$=x^{I}e^{x/2} \sum_{k=0}^{\infty}\frac{(-1)^{k}[\nu-IL}{k!}\{e^{-i_{d}\tau k}\frac{\Gamma(v+t+1+k)}{\Gamma(\nu+l+1)}x^{-\nu-(t+1)-k}\}$(21)$=x^{-\nu-1}e^{x/2} \sum_{k=0}^{\infty}\frac{[\nu-l]_{k}[\nu+l+1]_{k}}{k!}x^{-k}$$=x^{-v-1}e^{x/2}{}_{2}F_{0}(v-1, \nu+l+1;1/x)$ ,4$)$$(|1/x|<1)$ .

(3)(\S 2. (5))$\varphi[4]^{=x^{I-x/2}}e(x^{-v-(l+\ddagger)}\cdot e^{-x})_{\check{l}-v}$(22)$=x^{t-x/2}e \sum_{k=0}^{\infty}\frac{\Gamma(t-\nu+1)}{k!\Gamma(t-v+1-k)}(x^{-v-(l+1)})_{1-\nu-k}(e^{x})_{k}$$=x^{t-x/2}e \sum_{k\Leftrightarrow 0}^{\infty}\frac{(-1)^{k}[\nu-l]_{k}}{k!}\{e^{-i\pi(1-v-k\rangle}\frac{\Gamma(2l+1-k)}{\Gamma(v+l+1)}x^{-2t-1+k}\}e^{X}$(23)$=(e- \pi)xe\sum_{k=0}^{\infty}\frac{(-1)^{k}[\nu-l]_{k}(-1)^{-k}\Gamma(2l+1)}{k!\Gamma(\nu+t+1)[-2l]_{k}}(-x)^{k}$(24)$=(e^{-i\pi})^{l-v} x^{-l-1}e^{x/2}\frac{\Gamma(2l+1)}{\Gamma(\nu+t+1)}{}_{1}F_{1}(\nu-l;-2l;-x) , (|-x|<1)$.

(4)

Proof of Group II.
We can obtain the solutions in Group II from the ones

Group I, setting $-(l+1)$ instead of$t$, respectively. (cf.

$\varphi_{[1]}\sim\varphi_{[4]}$\S 2. )

in

62
\S 4. Commentary[I]

In the quantum mechanics, it is requested that$l\leq(n-1)\in Z_{0}^{\star};(l=0,1,2,\cdots, n-1)$

$n$

$t$

(1)

,

,

The principal quantum number,,

Orbital angular momentum quantum number.

$(n=v)$

Moreover, for the relationship (1), we have
(i)

(2)

$((l+n)\in Z^{+})$$\varphi_{[1]}=\varphi_{[2]}=x^{l}e^{x/2}(x^{n-(l+1)}.e^{-x})_{l+n}$$=(e^{-i\pi})^{l+n}e^{-x/2}x^{-l-1} \frac{\Gamma(21+1)}{\Gamma(l+1-tl)}{}_{1}F_{1}(-l-\cdot n;-2l;x)$$(|x.|<1)$ ,

,

(3)

and
$(ii)$$\varphi_{r5J}=\varphi_{[6]}=x^{-\sigma+1)}e^{x/2}(x^{n+t}\cdot e^{-x})_{n-(t+1)}$$=(e^{-i\pi})^{n-l-1}x^{t}e$- $x$$((n-1-1)\in Z_{0}^{+})$

(4)

,

(5)

/2 $\Gamma(-2l-1)$$\overline{\Gamma(-n-l)}{}_{1}F_{1}(1+l-r\iota;2l+2;x)$$(|x|<1)$ ,respectively.
We have the below classically.
$(D=d/dx)$$\varphi_{(11}=\varphi_{[2]}=x^{l}e^{x/2}D^{t+n}(x^{n-(i+1)}\cdot e^{-x})$

(6)

and
$\varphi_{[5]}=\varphi_{[6]}=x^{-(l+1)}e^{x/2}D^{n-(l+1)}(x^{n+l}\cdot e^{-x})$(7)

from (2) and (4), respectively.
Nexf we have(iii)$\varphi_{[3]}=\varphi_{[4]}=xte-x/2(e^{x}.x^{-n-(t+1)})_{l-n}$$=x^{t}e^{-\kappa/z}(e^{x}.x^{-n-(t+1)})_{-m}$$((n-1)=m\in Z^{+})$$=x^{t}e^{-x/2} \int\cdots\int e^{X}x^{-(n+I\perp 1)}(d\kappa)^{m}$

and

(8)
(9)
(10)

63
(iv)

(11)

$\varphi_{[7]}=\varphi_{[8]}=x^{-(l+1)}e^{-x/2}(e^{x}.x^{-n+l})_{-n-(t+1)}$$=x^{-(t+1)}e^{-x/2}(e^{x}.x^{-n\sim t})_{-m}$$((n+l+1)=m\in Z^{+})$$=x^{-(t\sim I)}e^{-x/2} \int\cdot\cdot\int e^{X}x^{-n+I}(dx)^{m}$

(12)

,

(13)

omitting the additional arbitraly constants of the integrations.
Note. We have $(for m, n\in Z_{0}^{+})$$L_{n}(x)=e^{X}D^{n}(e^{-X}x^{n})$$L_{n}^{m}(x)=D^{m}L_{n}(x)$$[\Pi]$,$\dot{f}$Laguerre function.

Associated Laguerre function.

We have$\varphi_{2}\cdot x^{2}+\varphi_{1}\cdot 2x+\varphi\cdot\{-(1/4)x^{2}+nx-n(n-1)\}=0$

,

(14)

from \S 2. (1), setting $\nu=n$ and $l=n-1$
And when $n=l+1$ we have$\varphi=\varphi_{[5]}=\varphi_{[6]}=x^{-n}e^{x/2}(e^{-x}\prime x^{2n-1})_{0}=x^{n-1}e^{-x/2}$(15)

from (4). The function shown by (15) satisfies equation (14) drearly.
$[m]$ The homogeneous Fukuhara equation is given by$\varphi_{2}+\varphi_{1}\cdot(a+b/x)+\varphi\cdot(?+q/x+r/x^{2})=0, (x\neq 0)$(16)

that is,$\varphi_{2}\cdot x^{2}+\varphi_{1}\cdot(ax^{2}+bx)+\varphi\cdot(px^{2}+qx+r)=0$

.

(17)

Therefore, we have$\varphi_{2}\cdot x^{2}+\varphi_{1}\cdot 2x+\varphi\cdot\{(-1/4)x^{2}+vx-l(l+1)\}=0$(\S 2. (1))

from (17), setting$a=0,b=2,$ $p=-1/4,$ $q=\nu$ and $r=-l(l+1)$(18)

Namely, the radial Schrodinger equation of the Hydrogen atom is a
special one of the homogeneous Fukuhara equation (16).
(cf. Fractional Calculus. Vol. 5, (1996) pp.136-142, by K. Nishimoto)$\cdot$64[lV]. Usually the solutions to the equation so-called special differential

equations such as the above ones are obtained by means of Frobenius.
Then compare the our solving manner (Method of NFCO) with thatof Frobenius and others.
[V] The equation \S 2. (1) is shown in Chapter 5 in the volume of Ian
N. Sneddon (Special Functions of Mathematical Physics and Chemistry(1961); Oliver and Boyd, Edingburgh.) in the form$\frac{d^{2}R}{d\kappa^{2}}+\frac{2}{x}\cdot\frac{dR}{\ }- \{\frac{1}{4}-\frac{v}{x}+\frac{1(1+1)}{x^{2}}\}R=0, (x\neq 0)$(19)

and a solution is given by

瓦，t $(x)=e^{-(1/2)x}x^{l}L_{n+l}^{2t+1}(x)$[VI] Indeed we have (see the Note above)
$R=$

$R=$

$(h\geq l+1)$

.

(20)

尺，t $(x)=e^{-(1/2\mu}x^{f}d^{1+1}(e^{x}D^{n+t}(e^{-x}x^{n+t}))$ ,(21)

hence we obtain$R=R_{7,0}(x)=e^{-(1/2)x}D^{1}(e^{X}D^{1}(e^{-x}x))=-e^{-(1/2)x}$

,

(22)

for $l=0$ and $n=1$
And we obtain, when $l=0$ and $\nu=n=1,$$\varphi_{[1]}=\varphi_{[2]}=x^{t}e^{x/2}(e^{-x}.x^{v-\sigma+I)})_{t+\nu}=e^{X/2}(e^{-x}1)_{1}$(23)$=-e^{-x/2}$

(24)

65
References

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