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\begin{document}




\title{{\bf Spin 1/2 particle with anomalous magnetic moment in presence
 of external magnetic field, exact solutions}
}


\author{{\bf E.M. Ovsiyuk}\\ e.ovsiyuk@mail.ru; Mozyr State Pedagogical
University, Belarus\\
 {\bf V.V. Kisel}
 \\ Belarusian State University of Informatics and Radioelectronics, Minsk
  \\
 {\bf V.M. Red'kov} \\
  redkov@dragon.bas-net.by; B.I. Stepanov Institute of Physics}

\date{}

\maketitle




\begin{abstract}

We examine a generalize Dirac equation for spin 1/2 particle with
anomalous magnetic moment in presence of the external uniform
magnetic field. After separation of the variables, the problem is
reduced to a 4-order ordinary differential equation, which is
solved exactly with the use of the factorization method. A
generalized formula for Landau energy levels is found. Solutions
are expressed in terms of confluent hypergeometric functions.

\end{abstract}





\section{ Ordinary  Dirac equation in cylindrical coordinates, separation of the  variables}

 We use the known representation for  the  uniform  magnetic field:
$ {\bf A} = {1 \over 2} \; c {\bf B} \times {\bf r}, \; {\bf B} =
(0,0,B)$. After translating to cylindrical coordinates we get
\begin{eqnarray}
A_{t} = 0\; , \qquad A_{r} = 0 \; , \qquad A_{z} = 0 \; , \qquad
A_{\phi} = - Br^{2} /2  \; . \label{1b}
\end{eqnarray}

\noindent  a non-vanishing component of the electromagnetic tensor
is $ F_{\phi r} =
 B  r $.
We consider the Dirac equation in the magnetic field (\ref{1b}),
using the  tetrad formalism \cite{Book-2009} for  cylindrical
coordinates $x^{\alpha}= (t,r,\phi,z)$:
\begin{eqnarray}
dS^{2} =  dt^{2} - d r^{2} -
 r^{2} \; d \phi^{2}- dz^{2}\; , \qquad
  e_{(a)}^{\beta}(x) = \left |
\begin{array}{llll}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 &1/r& 0 \\
0 & 0 & 0 & 1
\end{array} \right | \; .
\label{2}
\end{eqnarray}
Generally covariant tetrad Dirac equation  \cite{Book-2009} is
\begin{eqnarray}
\left \{ \gamma^{c} [\; i \hbar \; (\;  e_{(c)}^{\beta}
\partial_{\beta}  + {1\over 2} \sigma^{ab} \gamma_{abc}  \; )\;
-\; {e \over c} A_{c}  \; ] \; - \;  mc  \right \} \Psi  = 0 \; ,
\label{5}
\end{eqnarray}

\noindent where  $\gamma_{abc}$  are the Ricci rotation symbols
rotation: $ \gamma_{bac} = - \gamma_{abc} = - e_{(b)\beta ; \alpha
} \; e_{(a)}^{\beta} e_{(c)}^{\alpha},$ $A_{a} = e_{(a)}^{\beta}
A_{\beta} $  is the tetrad components of the 4-vector $A_{\beta}
$; $\sigma^{ab} = {1/4} (\gamma^{a}\gamma^{b}- \gamma^{b}
\gamma^{a})$  are  generators for bispinor representation of the
Lorentz group.

We will use the shortening notation:  $ e /c \hbar \Longrightarrow
e, \; mc/ \hbar \Longrightarrow M$. The  Dirac equation takes the
form (let  $ \Psi = \varphi / \sqrt{r}$):
\begin{eqnarray}
\left [\;  i  \gamma^{0} { \partial  \over  \partial t}  +
  i \gamma^{1}  {\partial \over \partial r }  +  \gamma^{2} \;
 \left( {    i   \partial_{\phi}    \over  r}+{e B r\over 2}\right)
     +  i  \gamma^{3} { \partial \over  \partial z} -  M   \right ] \varphi =0\,.
\label{8}
\end{eqnarray}

\noindent We search solutions in the form
\begin{eqnarray}
\varphi = e^{-i\epsilon t} e^{im \phi} e^{ikz}
\left | \begin{array}{c}
f_{1}(r)\\ f_{2}(r)\\
f_{3}(r)\\
f_{4}(r)
\end{array} \right |, \quad
\left [   +\epsilon   \gamma^{0} + i \gamma^{1}  {\partial \over \partial r }  -  \gamma^{2} \;
 \mu (r)
      -k   \gamma^{3}  -  M   \right ]  \left | \begin{array}{r}
f_{1}\\
f_{2}\\
f_{3}\\
f_{4}
\end{array} \right |
 =0\, ,
\nonumber
 \label{9}
\end{eqnarray}


\noindent where $ \mu (r) = m / r - e B r / 2$; further  we will
use the shortening notation $eB \Longrightarrow B$.
 When
choosing Dirac matrices in the spinor basis, we find equations for
the four functions $f_{a}(t,z)$:
\begin{eqnarray}
  ( {d \over d r }  +  \mu  )  f_{4}
   +   ik   f_{3}
+  i   ( \epsilon   f_{3}   -  M  f_{1} ) =0, \;
 ( { d \over d r }  -  \mu  ) f_{3}
-  ik  f_{4}
 +   i   (  \epsilon  f_{4}   - M  f_{2} ) =0,
 \nonumber
 \\
 ( { d \over d  r }  +    \mu )   f_{2}
+ ik  f_{1}
 - i     ( \epsilon f_{1} -M f_{3} ) =0,\;
  ( {d  \over d  r }  -   \mu  )  f_{1}
-  ik  f_{2}  - i    (   \epsilon f_{2} -M f_{4} ) =0 .
 \label{10}
\end{eqnarray}
The equations are consistent with the linear constraint
 $ f_{3} = A f_{1} , \;f_{4} = A f_{2} $,
if the following condition is imposed
\begin{eqnarray}
\epsilon -{M \over A} = - \epsilon + M A \qquad  \Longrightarrow \qquad
A = A_{1,2}= {\epsilon \pm \sqrt{\epsilon^{2} - M^{2}} \over M} \; .
\label{13}
\end{eqnarray}
As a result, the problem is reduced to the system of two equations
\begin{eqnarray}
\left  ( {d \over d r }  +    \mu \right  )  f_{2}
 + i     ( k - \epsilon  + M  A  )   f_{1} =0,\;
 \left  ( {d \over d  r }  -   \mu  \right )  f_{1}
+ i   ( -k  -  \epsilon   + M  A )   f_{2} =0 . \label{14}
\end{eqnarray}


In accordance with     (\ref{13}), we have two types of states:


\vspace{2mm}
 $
 AM =   \epsilon + \sqrt{\epsilon^{2} - M^{2}} \; ,  \qquad  (\; \sqrt{\epsilon^{2} - M^{2}} = p \; )
 $
 \begin{eqnarray}
 \left ( {d \over d r }  +   \; \mu \right )  \; f_{2}
+ i\;  (  k
 +    p    )\;   f_{1} =0\,,
 \;
 \left  ( {d  \over d  r }  -  \; \mu  \right ) \; f_{1}
- i\;  (  k  -  p )\; f_{2} =0\, ; \label{15}
\end{eqnarray}



$
 AM =   \epsilon - \sqrt{\epsilon^{2} - M^{2}} \; ,  \qquad  (\; \sqrt{\epsilon^{2} - M^{2}} = p \; )
$
 \begin{eqnarray}
 \left ( { d  \over d  r }  +   \; \mu \right )  \; f_{2}
+ i \;  ( k  -    p )   \;    f_{1} =0\,,
 \;
  \left ( { d \over d r }  -  \; \mu \right  ) \; f_{1}
-  i \; ( k  +  p  ) \;
f_{2} =0\, . \label{16}
\end{eqnarray}



For definiteness, we follow the variant  (\ref{15}).


\section{ Solving  equations for $r$-variable}


From   (\ref{15}) we obtain the second order equation for $R_{1}$
\begin{eqnarray}
{d^{2}R_{1}\over dr^{2}}+\left[{    m    \over  r^{2}}+{ B \over
2}-\left({    m    \over  r}-{ B r\over
2}\right)^{2}+\lambda^{2}\right]R_{1}=0\,. \label{17}
\end{eqnarray}
where
 $\lambda^{2} = \epsilon^{2} - m^{2} -k^{2}$.  Parameter $\lambda^{2}$
 describes the contribution  of the electron transversal motion to the total energy,
 this part of the energy is quantized.
Note that we  diagonalize  the operator
\begin{eqnarray}
-i {\partial \over  \partial \phi } \Psi = m  \; \Psi \; ,
\label{18}
\end{eqnarray}


\noindent which represents  the  third projection of the total
angular momentum of the Dirac particle in cylindrical tetrad
basis:
\begin{eqnarray}
\hat{J}_{3} \Psi_{Cart}  = ( -i {\partial \over  \partial \phi }
 + \Sigma_{3} )\; \Psi_{Cart} = m  \; \Psi = m\; \Psi_{Cart}
\label{19}
\end{eqnarray}


\noindent therefore for  $m$   are permitted only half-integer
values $ m = \pm 1 /2 , \; \pm 3 / 2, ...$.

We turn to eq. (\ref{17}) and introduce variable $ x = B r^{2} /
2$, equation becomes\footnote{Without loss of generality, we
assume that the parameter $B$ positive.}
\begin{eqnarray}
4x{d^{2}R_{1}\over dx^{2}}+2{dR_{1}\over dx}+\left( {m(1-m)\over
x}-x+1+2m+{2\lambda^{2}\over B}\right)R_{1}=0\,. \label{21}
\end{eqnarray}



\noindent We seek  solutions in the form $ R_{1} (x) = x^{A}
e^{-Cx} R (x) $. If $A, C$ are chosen according   $  A= m /  2\, ,
\;(1 - m)/ 2 \,, \;  C= \pm 1 /  2 $, the equation  for $R$ reads
$$
x{d^{2}R\over dx^{2}}\,+\left(2A+{1\over 2}-x\right){dR\over dx}-
\left(A-{m\over 2}-{\lambda^{2}\over 2eB}\right)R=0\, ,
$$

\noindent which is the confluent hypergeometric equation
\begin{eqnarray}
x \;  Y '' +( \gamma -x) Y'  - \alpha Y =0 \; ,
\qquad
\alpha= A-{m\over 2}-{\lambda^{2}\over 2B}\,,
\qquad
\gamma=2A+{1\over 2}\, .
\nonumber
\label{23}
\end{eqnarray}


\noindent
To obtain solutions that vanish at the origin
$r\rightarrow 0$ and infinity $r \rightarrow \infty$, we  must take  positive values for $A$ and $C$:
\begin{eqnarray}
C = + 1 /2 \; , \qquad  A = \left \{ \begin{array}{l}
m = +1 / 2, + 3 /  2, ..., \;\; A = m /  2 \; ,\\[3mm]
m = -1 / 2, - 3 /  2, ..., \;\; A = (1 -m)/  2\; .
\end{array} \right.
\nonumber
\label{24}
\end{eqnarray}



To obtain  polynomials, we impose the know restriction
$
\alpha = - n, \; n=0,1,2, ...
$
This leads to the following rule for quantization of  the parameter $\lambda^{2}$:
\begin{eqnarray}
{\lambda^{2} \over 2eB} = A - {m\over 2}  + n \; .
\nonumber
\end{eqnarray}


 \noindent Depending on the sign of the quantum number $m$ we get two formulas
  for  $\lambda^{2} = \epsilon^{2} - M^{2} - k^{2}$:
\begin{eqnarray}
m > 0 \; , \qquad \qquad
\lambda^{2} =  2eB\;  n  \;, \qquad n = 0, 1,2, .... ;
\nonumber
\\
m < 0 \; , \qquad
\lambda^{2}  =  2eB\;\left  ( n  - m + {1 \over 2} \right  ), \quad  n = 0,1, 2, ...
\label{25}
\end{eqnarray}



\section{ Accounting of the anomalous magnetic moment }





The Dirac equation for a particle with spin $1/2$ with anomalous magnetic moment in the Riemannian space-time (using the tetrad formalism) can be represented in the form \cite{Kisel-Toc-Red-2002, Kisel-Toc-Red-2002'}
\begin{eqnarray}
\left \{  \gamma^{c}  [\; i  (  e_{(c)}^{\beta}
\partial_{\beta}  + {1\over 2} \sigma^{ab} \gamma_{abc}  \; )\;
-\; {e \over \hbar c} A_{c}  \; ]
-i        \lambda   { 2e \over Mc^{2} } \sigma^{\alpha \beta} (x) F_{\alpha \beta} (x) -
       {Mc \over \hbar } \right \} \Psi  = 0 \; . \label{7'}
\end{eqnarray}
Dimensions of  the parameters in the equation are
$$
[{Mc \over \hbar } ] = l^{-1}, \quad [ {e \over \hbar c}  A ] =
l^{-1}, \quad [ {e \over \hbar c} F ] = l^{-2}, \quad [ {e F \over
Mc^{2} } ] = l^{-1} \; ;
$$
free parameter   $\lambda$ is  dimensionless.
Consider this equation in the uniform magnetic field. In view of
$$
\sigma^{\alpha \beta} (x) F_{\alpha \beta} (x) = 2\sigma^{\phi r } F_{\phi r } =
i
  \gamma^{2} \gamma^{1} B = i B \Sigma_{3}
 $$


\noindent instead of (\ref{8}) we get a more general equation
\begin{eqnarray}
\left [\;  i  \gamma^{0} { \partial  \over  \partial t}  +
  i \gamma^{1}  {\partial \over \partial r }  +  \gamma^{2} \;
 \left( {    i   \partial_{\phi}    \over  r}+{e B r\over 2}\right)
     +  i  \gamma^{3} { \partial \over  \partial z}
     +       \lambda    { 2e B \over Mc^{2} }  \Sigma_{3}
     - {M c \over \hbar }   \right ] \varphi =0\,.
\label{8'}
\end{eqnarray}



\noindent Substitution for  the wave function is the same as above
\begin{eqnarray}
\left [   \epsilon   \gamma^{0} + i \gamma^{1}  {\partial \over
\partial r }  -  \gamma^{2} \;
 \mu (r)
      -k   \gamma^{3}  +  \Gamma \Sigma_{3} -  M   \right ]  \left | \begin{array}{c}
f_{1}(r)\\
f_{2}(r)\\
f_{3}(r)\\
f_{4} (r)
\end{array} \right |
 =0\,,
 \label{9'}
\end{eqnarray}


\noindent
where we use the notation
\begin{eqnarray}
     {m    \over  r} - {e B r \over  2 } \Longrightarrow  \mu (r)\;, \quad
    \lambda  { 2e B \over Mc^{2} } \Longrightarrow \Gamma \; ,
     \qquad {M c \over \hbar } \Longrightarrow M \;,
     \quad
     {\epsilon \over \hbar c } \Longrightarrow \epsilon\;  .
 \label{9''}
\end{eqnarray}

Further we get four radial equations
\begin{eqnarray}
  -i \left ({d \over dr}  + \mu \right )f_{4} +(\epsilon+  k )f_{3} + (\Gamma -M )f_{1}
  =0,
  \nonumber
  \\
  -i \left ( {d \over dr}  -  \mu \right  )f_{3} + (\epsilon -k )f_{4} - (\Gamma +M) f_{2} =0,
\nonumber\\
 +i \left ({d \over dr}  + \mu \right  ) f_{2} + (\epsilon -k) f_{1} + (\Gamma -M ) f_{3} =0,
\nonumber
\\
  +i \left ({d \over dr}  - \mu \right ) f_{1} +( \epsilon + k) f_{2} -(\Gamma +M)  f_{4}  =0  .
\nonumber
\label{26}
\end{eqnarray}

Let us try to impose a linear constraint (see the  case of ordinary Dirac particle):
$
f_{3} = A f_{1} ,  \; f_{4} = A f_{2},
$
equations take the form
%$$
%  -i \left ({d \over dr}  + \mu \right )A f_{2} +(\epsilon+  k ) A f_{1} + (\Gamma -M )f_{1}  =0\;,
%$$
%$$
%  -i \left ( {d \over dr}  -  \mu  \right )A f_{1} + (\epsilon -k )A f_{2} - (\Gamma +M) f_{2} =0\;,
%$$
%$$
% +i \left ({d \over dr}  + \mu\right  ) f_{2} + (\epsilon -k) f_{1} + (\Gamma -M ) A f_{1} =0\;,
%$$
%$$
%  +i \left  ({d \over dr}  - \mu \right ) f_{1} +( \epsilon + k) f_{2} -(\Gamma +M)  A f_{2}  =0 \; .
%$$
%или
\begin{eqnarray}
  -i  \left ({d \over dr}  + \mu \right ) f_{2} + \left [ \epsilon+  k  + {(\Gamma -M )\over A} \right ] f_{1}  =0\;,
\nonumber
\\
  -i \left ( {d \over dr}  -  \mu \right  ) f_{1} + \left [ \epsilon -k  - {(\Gamma +M)\over A} \right ]  f_{2} =0\;,
\nonumber
\\
 +i \left ({d \over dr}  + \mu \right) f_{2} + \left [ \epsilon -k  + (\Gamma -M ) A \right ] f_{1} =0\;,
\nonumber
\\
  +i \left ({d \over dr}  - \mu \right ) f_{1} +\left [  \epsilon + k   -  (\Gamma +M)   A \right ] f_{2}  =0 \; .
\label{28}
\end{eqnarray}






In the system (\ref{28}) equations 1 and 3, as well as equations 2 and 4 are the same, only if the ratio
$$
 \epsilon+  k  + {(\Gamma -M )\over A}  =-
\left [ \epsilon -k  + (\Gamma -M ) A \right ],
$$
$$
 \epsilon -k  - {(\Gamma +M)\over A}=-
\left [  \epsilon + k   -  (\Gamma +M)   A \right ] ;
$$
they can be rewritten as:
\begin{eqnarray}
 \epsilon  + {(\Gamma -M )\over A}  =
- \epsilon   - (\Gamma -M ) A \quad \Longrightarrow \quad 2\epsilon = (M-\Gamma) (A + {1 \over A})\; ,
\nonumber
\\
 \epsilon  - {(\Gamma +M)\over A} =-
  \epsilon    +  (\Gamma +M)   A
  \quad \Longrightarrow \quad 2\epsilon = (M+\Gamma) (A + {1 \over A})\; .
\label{29}
\end{eqnarray}


Obviously, this system is not consistent.
Thus, we are to examine the system of four equation
in a different way.



\section{Solving  the radial  equations}

We turn again to the  system of four equations.
They can be represented in the form of two linear systems
\begin{eqnarray}
 (\Gamma -M )f_{1}    +(\epsilon+  k )f_{3}   = i \left ({d \over dr}  + \mu  \right )f_{4} =  i D_{+} f_{4} \;,
\nonumber
\\
  + (\epsilon -k) f_{1} + (\Gamma -M ) f_{3} =- i \left ({d \over dr}  + \mu\right  ) f_{2} = - i D_{+} f_{2}\;;
\label{6.3}
\\
 - (\Gamma +M) f_{2}   + (\epsilon -k )f_{4}  = i \left ( {d \over dr}  -  \mu \right  )f_{3} = iD_{-} f_{3}\;,
\nonumber
\\
   +( \epsilon + k) f_{2} -(\Gamma +M)  f_{4}  = -i \left ({d \over dr}  - \mu  \right  ) f_{1} =- iD_{-} f_{1} \; .
\label{6.4}
\end{eqnarray}
Their solutions are as follows:
\begin{eqnarray}
f_{1} = + i
{ ( \epsilon + k) D_{+} f_{2} + (\Gamma -M)D_{+} f_{4}  \over (\Gamma -M)^{2} - (\epsilon^{2} - k^{2}) } ,
\quad
f_{3} = -i{
(\Gamma -M)  D_{+} f_{2} + ( \epsilon - k) D_{+} f_{4} \over (\Gamma -M)^{2} - (\epsilon^{2} - k^{2}) } ;
\label{6.5}
\end{eqnarray}
\begin{eqnarray}
f_{2} = + i {
( \epsilon - k) D_{-} f_{1} - (\Gamma + M)D_{-} f_{3}  \over (\Gamma +M)^{2} - (\epsilon^{2} - k^{2}) } ,
\quad
f_{4} = -i {
 -(\Gamma + M)  D_{-} f_{1} + ( \epsilon + k) D_{-} f_{3} \over (\Gamma + M)^{2} - (\epsilon^{2} - k^{2}) } .
\label{6.6}
\end{eqnarray}
We note two identities
\begin{eqnarray}
D_{+}D_{-} = ({d \over dr}  + \mu ) ({d \over dr}  - \mu ) ={d^{2} \over dr^{2}} -\mu' -\mu^{2} \; ,
\nonumber
\\
D_{-}D_{+} = ({d \over dr}  - \mu ) ({d \over dr}  + \mu ) ={d^{2} \over dr^{2}} +\mu' -\mu^{2}\; .
\label{6.7}
\end{eqnarray}





Substituting expression (\ref{6.5})  into equation (\ref{6.4}), we get
\begin{eqnarray}
 - (\Gamma +M) f_{2}   + (\epsilon -k )f_{4}  =  { (\Gamma -M)   D_{-}D_{+} f_{2}\over (\Gamma -M)^{2} - (\epsilon^{2} - k^{2}) }+
{ ( \epsilon - k)  D_{-}D_{+} f_{4} \over (\Gamma -M)^{2} - (\epsilon^{2} - k^{2}) }\;,
\nonumber
\\
   +( \epsilon + k) f_{2} -(\Gamma +M)  f_{4}  =   {( \epsilon + k) D_{-}D_{+} f_{2} \over (\Gamma -M)^{2} - (\epsilon^{2} - k^{2}) }+
{(\Gamma -M)D_{-}D_{+} f_{4}  \over (\Gamma -M)^{2} - (\epsilon^{2} - k^{2}) }\,.
\label{6.8}
\end{eqnarray}
By combining equations  (\ref{6.8}), we obtain
\begin{eqnarray}
  f_{2}  = { 1 \over 2\Gamma\,(\epsilon+k) }
   \left ( \Gamma^{2}-M^{2} -k^{2}+\epsilon^{2}+
   {d^{2} \over dr^{2}} +\mu' -\mu^{2} \right  )   f_{4} \,,
\label{6.10}
\end{eqnarray}



\noindent and then we derive  the 4-th order equation   for the function $f_{4}$
$$
-{\frac {d^{4}f_{4}}{d{r}^{4}}} + \left[{{B}^{2}\over 2}\,{r}^{2}- B\,(2\,m-1)-2\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2
})+2\,{\frac {m \left( m+1 \right) }{{r}^{2}}} \right] {\frac {d^{2}f_{4}}{d
{r}^{2}}}+
$$
$$
+ \left[ {B}^{2}r-4\,{\frac {m
 \left( m+1 \right) }{{r}^{3}}} \right] {\frac {df_{4}}{dr}} +
 $$
 $$+ \left[-{{B}^{4}\over 16}\,{r}^{4}+ { {B}^{2}\over 4} \left( B\,(2m-1)+2\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+
{\epsilon}^{2}) \right)
 {r}^{2}
-\right.
$$
$$- B\,(2m-1)\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2})-  ({\Gamma}^{2}+{M}^{2}+{k}^{2}-{\epsilon}^{2})^{2}+4\,\Gamma^{2}M^{2}
-{ {B}^{2}\over 4}\, \left( 6\,{m}^{2}-2\,m-1 \right)
+
$$
 $$\left. + {\frac {  m\,(m+1)\left(B\,(2m-1)+2\,({\Gamma}^{2}-{M}
^{2}-{k}^{2}+{\epsilon}^{2})\right)  }{{r}^{2}}}-{\frac {m \left( m-2 \right)  \left( m+3 \right)  \left( m+1 \right)
}{{r}^{4}}} \right] f_{4}
=0\,.
$$
\begin{eqnarray}
\label{6.11}
\end{eqnarray}



Similarly, we can find  the  expression for $f_{4}$
\begin{eqnarray}
    f_{4}
=  {    1 \over 2\Gamma\,(\epsilon-k) } \left ( \Gamma^{2} -M^{2}-k^{2}+\epsilon^{2} +
{d^{2} \over dr^{2}} +\mu' -\mu^{2}\right )
 f_{2}
\,,
\label{6.21}
\end{eqnarray}
\noindent and then derive the  4-th order equation  for the function $f_{2}$:
$$
{\frac {d^{4}f_{2}}{d{r}^{4}}} + \left[-{ {B}^{2}\over 2}\,{r}^{2}+B\,(2\,m-1)+2\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2
})-2\,{\frac {m \left( m+1 \right) }{{r}^{2}}} \right] {\frac {d^{2}f_{2}}{d
{r}^{2}}}+
$$
$$
+ \left[ - {B}^{2}r+4\,{\frac {m
 \left( m+1 \right) }{{r}^{3}}} \right] {\frac {df_{2}}{dr}} +
 $$
 $$+ \left[{ {B}^{4}\over 16}\,{r}^{4}- { {B}^{2}\over 4} \left( B\,(2m-1)+2\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+
{\epsilon}^{2}) \right)
 {r}^{2}
+\right.
$$
$$+B\,(2m-1)\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2})+ ({\Gamma}^{2}+{M}^{2}+{k}^{2}-{\epsilon}^{2})^{2}-4\,\Gamma^{2}M^{2}
+{{e}^{2}{B}^{2}\over 4}\, \left( 6\,{m}^{2}-2\,m-1 \right)
-
$$
 $$\left. - {\frac {  m\,(m+1)\left(B\,(2m-1)+2\,({\Gamma}^{2}-{M}
^{2}-{k}^{2}+{\epsilon}^{2})\right)  }{{r}^{2}}}+{\frac {m \left( m-2 \right)  \left( m+3 \right)  \left( m+1 \right)
}{{r}^{4}}} \right] f_{2}
=0\,.
$$
\begin{eqnarray}
\label{6.22}
\end{eqnarray}

We note that the equation for $f_{2}$ and  $f_{4}$ are the same.
 Therefore it is sufficient to consider one of them.
 To study the arising equations   (\ref{6.11})  and (\ref{6.22})   we
 will use the factorization method:
$$  \hat{F}_{4} (r) \;  f (r)  =
\hat{f}_{2}  (r) \hat{g}_{2}(r)   \; f  (r) = 0\, , $$
\begin{eqnarray}
 \hat{f}_{2} (r) =
 {d^{2} \over dr^{2}} +
  P_{0}  r^{2} +  P_{1}  + {P_{2} \over r^{2}} \,,\quad
    \hat{g}_{2} (r) =
 {d^{2} \over dr^{2}} +
Q_{0}  r^{2} +  Q_{1}  + {Q_{2} \over r^{2}}  \,.
 \label{factor}
 \end{eqnarray}


\noindent Calculating the operator $\hat{F}_{4} $ and comparing with
(\ref{6.22}), we find two sets of numerical coefficients:
$$
1) \qquad \qquad \qquad P_{0}=-{1\over 4}\,{B}^{2}\,, \quad P_{{2}}=-m\,(m+1)\,,
$$
$$
P_{1}=B\,\left(m-{1\over 2}\right)+{\Gamma}^{2}-{M}^{2}-
{k}^{2}+{\epsilon}^{2}+2\,\Gamma\sqrt {{\epsilon}^{2}-{k}^{2}}\,,
$$
$$
Q_{0}=-{1\over 4}\,{B}^{2}\,, \quad Q_{{2}}=-m\,(m+1)\, ,
$$
$$
Q_{1}=B\,\left(m-{1\over 2}\right)+{\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2}-2\,\Gamma\sqrt {{\epsilon}^{2}-
{k}^{2}}\,;
$$


\noindent
and
$$
 2) \qquad \qquad \qquad P_{0}=-{1\over 4}\,{B}^{2}\,,\quad P_{{2}}=-m\,(m+1)\,,
 $$
 $$
 P_{1}=B\,\left(m-{1\over 2}\right)+{\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2}-2\,\Gamma\sqrt {{\epsilon}^{2}-
 {k}^{2}}\,,
 $$
  $$
 Q_{0}=-{1\over 4}\,{B}^{2}\,, \quad Q_{{2}}=-m\,(m+1)\;,
 $$
 $$
 Q_{1}=B\,\left(m-{1\over 2}\right)+{\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2}+2\,\Gamma\sqrt {{\epsilon}^{2}-
 {k}^{2}}\,.
 $$

  Thus,  we are to solve two equations (they differ only in the sign of $\Gamma$)
  $$
 \left({d^{2} \over dr^{2}}
 -{ B^{2}  r^{2} \over 4}  +  B\,\left(m-{1\over 2}\right)+{\Gamma}^{2}-{M}^{2}-{k}^{2}+
 {\epsilon}^{2}+2\,\Gamma\sqrt {{\epsilon}^{2}-{k}^{2}}  - {m\,(m+1) \over r^{2}} \right)\,f=0\,,
$$
 \begin{eqnarray}
 \label{factor-3}
 \end{eqnarray}
  $$
  \left({d^{2} \over dr^{2}}
  -{ {B}^{2} r^{2} \over 4} +  B\,\left(m-{1\over 2}\right)+{\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2}-2\,\Gamma\sqrt {{\epsilon}^{2}-{k}^{2}}  - {m\,(m+1) \over r^{2}} \right)\,g=0\,.
 $$
\begin{eqnarray}
 \label{factor-4}
 \end{eqnarray}

 \noindent Consider the first equation (\ref{factor-3}). We turn it to the variable $x=Br^{2}/2$:
 $$
x\, {d^{2}f\over dx^{2}} +{1\over 2}\,{df\over dx}+
$$
$$
+ \left[ -{x\over 4}+{\frac {4\,\Gamma\sqrt {\epsilon^{2}-k^
{2}}+ ( 2\,m-1 )\, B+2\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2})}{4B}}-{1\over 4}\,{\frac {m ( m+1) }
{x}} \right] f
=0\,.
 $$

\noindent We will build  solutions in the form:
$$
f=x^{a}e^{bx}F\,,
$$
$$
x\, {d^{2}F\over dx^{2}} +\left({1\over 2}+2a+2bx\right)\,{dF\over dx}+
\left[ \left(b^{2}-{1\over 4}\right)\,x+\right.
$$
$$
+{\frac {2\,Bb ( 4\,a+1 )+4\,\Gamma\sqrt {\epsilon^{2}-k^
{2}}+ ( 2\,m-1 )\, B+2\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2})}{4B}}+
$$
$$
\left.+{1\over 4}\,{\frac {(2a+m)\, ( 2a-m-1) }
{x}} \right] F
=0\,.
 $$

 \noindent If $a,\,b$ are chosen as
 \begin{eqnarray}
 a=-{m\over 2},\;{1\over 2}+{m\over 2}, \quad b=-{1\over 2}\,,
  \label{factor-5}
 \end{eqnarray}

 \noindent the equation is simplified
$$
x\, {d^{2}F\over dx^{2}} +\left({1\over 2}+2a-x\right)\,{dF\over dx}-
$$
$$
-{\frac {B ( 4\,a+1 )-4\,\Gamma\sqrt {\epsilon^{2}-k^
{2}}- ( 2\,m-1 )\, B-2\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2})}{4B}} \, F
=0\,,
 $$

 \noindent and it is the equation for a confluent hypergeometric function with parameters:
 \begin{eqnarray}
  \alpha={\frac {B ( 4\,a+1 )-4\,\Gamma\sqrt {\epsilon^{2}-k^{2}}- ( 2\,m-1 )\,
 eB-2\,({\Gamma}^{2}-{M}^{2}-{k}^{2}+{\epsilon}^{2})}{4B}} \,, \;
 \gamma={1\over 2}+2a\,.
  \label{factor-6}
 \end{eqnarray}


To get  solutions that meet the bound states, we  should use the positive value
 of the parameter $a$ and the negative values of the parameter $b$ (for definiteness assume that $B >0$):
\begin{eqnarray}
a=-{m\over 2}\,, \quad  ( m<0 ) \; ; \quad\quad  a = {m\over 2}
+{1\over 2} >0 \quad (  m \geq 0) \, . \label{II}
 \end{eqnarray}

Conditions of terminating the hypergeometric series to  polynomials  $\alpha = -n$ (introduce the notation
$\epsilon^{2}-k^{2} = \lambda$):
\begin{eqnarray}
{B ( 4\,a+1 )-4\,\Gamma\sqrt \lambda - ( 2\,m-1 )\,
 B- 2\,({\Gamma}^{2}-{M}^{2})   - 2\lambda  \over 4B}=-n
\label{factor-8}
 \end{eqnarray}
gives quantization rule for the energy values:
\begin{eqnarray}
 a+{1 \over 2} - { m\over2}+ {M^{2} - \Gamma^{2}   \over 2B}
  +n={\Gamma\sqrt \lambda\over B}  + {\lambda  \over 2B}  \,,
\nonumber
 \end{eqnarray}

\noindent we obtain
\begin{eqnarray}
(\sqrt{\lambda} + \Gamma )^{2} = N, \; N = M^{2}  + 2B  ( a +
{1 \over 2} -{m \over 2}  +n  )\,\;\Longrightarrow \;\;
\lambda  = (  \sqrt{N} - \Gamma   )^{2} > 0 \; .
\label{rule-1}
 \end{eqnarray}

\noindent From (\ref{rule-1}) we find the formula for the allowed values of $\lambda$
\begin{eqnarray}
\epsilon ^{2}     - k^{2} =
\left (    \sqrt{ M^{2} + 2B  ( a+ {1 \over 2} - { m \over 2 } +n  ) } - \Gamma \right  ) ^{2}
.
\label{rule-2}
\end{eqnarray}



Depending on the sign of  $m$,  we obtain two formulas:
\begin{eqnarray}
m<0, \quad a=-{m\over 2}\,,\quad \epsilon^{2}     -  k^{2} = \left
(    \sqrt { M^{2} + 2B
 ( {1 \over 2} -  m +n ) } - \Gamma \right )  ^{2}  ;
\label{energy-1}
\end{eqnarray}
\begin{eqnarray}
 m \geq 0, \quad a = {m\over 2} +{1\over 2}\; ,
\quad \epsilon^{2}  - k^{2}  = \left (  \sqrt { M^{2} + 2B
 ( 1 +n  ) }  - \Gamma \right  ) ^{2}  .
\label{energy-2}
\end{eqnarray}


This, we have two possibilities for quantization
\begin{eqnarray}
I, \quad \lambda  = (  \sqrt{N} - \Gamma   )^{2}   \;, \qquad
II, \quad \lambda  = (  \sqrt{N} + \Gamma   )^{2}   \; .
\end{eqnarray}
that is the particle with anomalous magnetic moment has two series
of energy levels, formally differing in sign of the parameter
$\Gamma$.





\section{Further analysis of solutions}







Let us consider the function $f_{2}(r) $ as the primary.
Obtained above ratios allow us to calculate other three functions.
We start from the explicit form of the function $f_{2}$, the solution of equation (\ref{factor-3}):
\begin{eqnarray}
f_{2} = x^{a} e^{-x/2} F (  \alpha  , \gamma , x) \; ,
\label{7.1}
\end{eqnarray}
where the parameters are given by
\begin{eqnarray}
\alpha=
a+{1 \over 2} - { m\over 2}+ {M^{2} - \Gamma^{2}   \over 2B}
  - {\Gamma\sqrt \lambda\over B}  - {\lambda  \over 2B} = -n, \quad\quad
 \gamma={1\over 2}+2a\,.
 \label{7.2}
\end{eqnarray}

The function $f_{4}$ can be found according to the following relationship:
$$
    f_{4} =
{    1 \over 2\Gamma\,(\epsilon-k) } \left ( \Gamma^{2} -M^{2}-k^{2}+\epsilon^{2} +
{d^{2} \over dr^{2}}  - {m(m+1) \over r^{2}} +{B (2m -1)\over 2} -{ B^{2} \over 4} r^{2}\right )  f_{2} \; ;
$$
from where  after translating to the variable $x$ we find
\begin{eqnarray}
f_{4} =
{    2B  \over 2\Gamma\,(\epsilon-k) } \left (
 x {d^{2} \over dx^{2}} + {1 \over 2} {d \over dx}
+{ \lambda \over 2B}   - {M^{2} -\Gamma^{2}  \over 2B} +  {m \over 2} -{1\over 4}
  - { m(m+1) \over 4 x  } -{ 1 \over 4} x \right )  f_{2}\; .
\label{7.3}
\end{eqnarray}


Given the identities
$$
+{\lambda \over 2 B} -{M^{2} - \Gamma^{2} \over 2B} + {m \over 2}  -{1 \over 4}    =
n  + a  + {1\over 4} - {\Gamma \sqrt{\lambda} \over B}
$$
the above relation can be written as
\begin{eqnarray}
f_{4} =
{    2B  \over 2\Gamma\,(\epsilon-k) } \left (
 x {d^{2} \over dx^{2}} + {1 \over 2} {d \over dx}
+ n  + a  + {1\over 4} - {\Gamma \sqrt{\lambda} \over B}
  - { m(m+1) \over 4 x  } -{ 1 \over 4} x \right )  f_{2}\,.
\label{7.4}
\end{eqnarray}




Depending on the values of the parameter $m$, we have two different
cases:
$$
A) \quad m < 0, \qquad a = -{m \over 2}, \quad f_{2} = x^{-m /2 } e^{-x/2} F (  -n  , -m +{1\over 2}  ,\, x)\,,
$$
$$
\alpha=
m  +{1 \over 2} + {M^{2} - \Gamma^{2}   \over 2B}
  - {\Gamma\sqrt \lambda\over B}  - {\lambda  \over 2B} = -n, \quad\quad
 \gamma= -m + {1\over 2}\,,
 $$
$$
f_{4} =
{    2B  \over 2\Gamma\,(\epsilon-k) } \left (
 x {d^{2} \over dx^{2}} + {1 \over 2} {d \over dx}
+ n  -{m \over 2}   + {1\over 4} - {\Gamma \sqrt{\lambda} \over B}
  - { m(m+1) \over 4 x  } -{ x \over 4}  \right )  f_{2}\,,
$$
\begin{eqnarray}
\sqrt{\lambda} =   \sqrt { M^{2} + 2B
 ( {1 \over 2} -  m +n ) } - \Gamma \,;
 \label{7.5}
\end{eqnarray}
$$
B) \quad m \geq   0, \qquad a = {m \over 2} +{1 \over 2} , \quad f_{2} = x^{(m+1)/2 } e^{-x/2} F (  -n   ,
 m +{3 \over 2 } ,\, x)\,,
$$
$$
\alpha=
1 + {M^{2} - \Gamma^{2}   \over 2B}
  - {\Gamma\sqrt \lambda\over B}  - {\lambda  \over 2B} = -n, \quad\quad
 \gamma= m + {3\over 2} \,,
 $$
$$
f_{4} =
{    2B  \over 2\Gamma\,(\epsilon-k) } \left (
 x {d^{2} \over dx^{2}} + {1 \over 2} {d \over dx}
+ n  + {m+1 \over 2}  + {1\over 4} - {\Gamma \sqrt{\lambda} \over B}
  - { m(m+1) \over 4 x  } -{ x \over 4}  \right )  f_{2}\,,
$$
\begin{eqnarray}
\sqrt{\lambda} =  \sqrt { M^{2} + 2B
 ( 1 +n  ) }  - \Gamma \, .
 \label{7.6}
\end{eqnarray}



Using the relations (\ref{7.5})  and  (\ref{7.6}) we find the explicit expression for the function $f_{4}$:


 for the variant $A$,
\begin{eqnarray}
m<0, \qquad f_{4} = -{\sqrt{\lambda}\over \epsilon-k}\,x^{-m /2 } e^{-x/2} F (  -n  , -m +{1\over 2}  , x)=
-{\sqrt{\lambda}\over \epsilon-k}\, f_{2}(x) \,,
 \label{7.7}
\end{eqnarray}


\noindent for the variant $B$,
\begin{eqnarray}
m\geq 0, \qquad f_{4} =-{\sqrt{\lambda}\over \epsilon-k}\, x^{(m+1)/2 } e^{-x/2} F (  -n   ,
 m +{3 \over 2 } , x)= -{\sqrt{\lambda}\over \epsilon-k} f_{2}(x)\,.
 \label{7.8}
\end{eqnarray}





Changing  in these formulas the parameter $\Gamma$ on $-\Gamma$,
we obtain the relations describing the second series of states.
It is not difficult to calculate the explicit form of the other two functions
$f_{1}$,  $f_{3}$; corresponding calculated are made, we will not detail on this.


\begin{thebibliography}{xxx}





\bibitem{Book-2009}
 V.M. Red'kov Fields of particles in Riemannian space and the Lorentz group. --
 Belarusian Science: Minsk, 2009. --  486 pages.




\bibitem{Kisel-Toc-Red-2002}
  V.V. Kisel,  N.G. Tokarevskaya,  V.M  Red'kov.
    Petrash theory for particles with spin 1/2 in curved space-time.
     Preprint  737. --   Institute of Physics of NAS of Belarus: Minsk, 2002.


\bibitem{Kisel-Toc-Red-2002'}
 A.A.  Bogush,   V.V. Kisel,  N.G. Tokarevskaya,  V.M. Red'kov
   Petrasha theory for particles with spin 1/2 in curved space-time
    // Vesti of NASB. Ser. fiz.-mat. nauk. 2002.  no  1. P. 63--68.



\end{thebibliography}

\end{document}
