Spin-2 Particle in Coulomb Field: Non-Relativistic Approximation

Abstract

The primary objective of this paper is to derive a non-relativistic system of equations for a spin-2 particle in the presence of an external Coulomb field, solve these equations, and determine the corresponding energy spectra. We begin with the known radial system of 39 equations formulated for a free spin-2 particle and modify it to incorporate the effects of the Coulomb field. By eliminating the 28 components associated with vector and rank-3 tensor fields, we reduce the system to a set of 11 second-order equations related to scalar and symmetric tensor components. In accordance with parity constraints, this system naturally groups into two subsystems consisting of three and eight equations, respectively. To perform the non-relativistic approximation, we employ the method of projective operators constructed from the matrix ${\Gamma }^0$ of the original matrix equation. This approach allows us to derive two non-relativistic subsystems corresponding to the parity restrictions, comprising two and three coupled differential equations. Through a linear similarity transformation, we further decouple these into five independent equations with a Schrödinger-type non-relativistic structure, leading to explicit energy spectra. Special attention is given to the case of the minimal quantum number of total angular momentum, $j=0$, which requires separate consideration.

1. Introduction

The theory of massive and massless spin-2 fields, building on the foundational work of Pauli and Fierz [1], has long attracted significant attention [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]. Over the years, several key aspects of this theory have been explored, primarily within the framework of second-order differential equations. A central focus has been the additional constraints required to preserve the five independent degrees of freedom of a spin-2 particle—a problem that becomes even more intricate when extending the theory to curved Riemannian spacetimes.

Another important area of spin-2 theory is the investigation of anomalous solutions. An effective alternative approach for studying both massive and massless spin-2 fields involves formulating first-order systems. This method, rooted in the Gel’fand-Yaglom formalism [14], was first explored by Fedorov [3] and Regge [4]. Their work demonstrated that a complete description of a spin-2 particle requires a 39-component set of tensors.

In the present study, we focus on a specific problem within this 39-component framework: the non-relativistic approximation for a massive spin-2 particle in the presence of an external Coulomb field. Previously, using Cartesian coordinates, a Pauli-like equation was derived for a spin-2 particle interacting with electromagnetic fields [19]; see also [20].

The most significant novelty of this work is that we do not postulate a non-relativistic equation based on symmetry considerations. Instead, we apply a general procedure to derive non-relativistic systems directly from the relativistic equations. We employ a matrix formalism for first-order equations within the initial relativistic framework, combined with the method of projection operators (here applied to the spin-2 case). Another important point is that we develop this non-relativistic procedure directly at the level of the radial system of equations obtained after variable separation in the relativistic equation.

The choice of the Coulomb potential is not crucial for developing the general method; the procedure is applicable in any coordinate system and with any electromagnetic field. The use of spherical coordinates is also not essential—what matters is that variable separation has been performed in the relativistic equation. Our approach maintains a close connection between the relativistic wave function and its non-relativistic counterpart: in our method, the non-relativistic wave function is always defined through components of the relativistic function. For higher-spin particles, this relationship is nontrivial and cannot simply be guessed.

In addition, we aim to derive explicit forms of non-relativistic equations in the presence of arbitrary electromagnetic and gravitational fields with any symmetry—whether or not variable separation has been performed in the initial relativistic equation. This requires the use of generally covariant relativistic equations for particles of various spins, respecting the requirements of general relativity. Such problems cannot be solved correctly based solely on Galilean group theory, as is possible in flat space. These problems are far from trivial, especially for higher-spin particles.

The non-relativistic description of a spin-2 particle preserves certain relations to the relativistic description, because the non-relativistic equation is the result of an approximation, and the non-relativistic wave function is constructed from relativistic components according to strictly defined rules (which are given in this paper). Of course, there are no inverse formulas. Clearly, the non-relativistic equation cannot play a significant role in high-energy physics. However, low-energy spin-2 particles (for example, resonances) may arise in accelerator physics. Additionally, charged spin-2 objects may not be elementary particles at all; they may correspond to certain complex molecular-ion systems, and their interaction with environmental Coulomb forces is possible.

In Section 1, we present basic definitions and notations, including the structure of the 39-component wave function with spherical symmetry. In these solutions, the square and third projection of the total angular momentum are diagonalized. To separate the angular dependence of the wave function, we apply the Wigner D-functions.

In Section 2, we specify the known system of radial equations, originally derived for the case of a free spin-2 particle, and modify it to account for the presence of an external Coulomb field. Additionally, we present the constraints imposed by the diagonalization of the spatial reflection operator.

In Appendix A, we eliminate 28 variables associated with the four-vector and rank-2 antisymmetric tensor components, ${\Phi }_1$ and ${\Phi }_{a\left[bc\right]}$, respectively. This reduction yields a set of eleven second-order equations for the remaining variables corresponding to the scalar $\Phi$ and the symmetric tensor ${\Phi }_{\left(ab\right)}$. Considering parity restrictions, the system naturally divides into two subsystems comprising three and eight second-order radial equations, respectively.

In Appendix B, we apply a general method for performing the non-relativistic approximation, which distinguishes between large and small components of the wave function. The large and small components are determined using three projective operators, derived from the seventh-order minimal polynomial for the $39\times 39$ matrix ${\Gamma }^0$ of the initial matrix equation.

In Section 3, we analyze the subsystem of three second-order equations corresponding to states with parity $P={(-1)}^{j+1}$. By reducing these to two coupled radial equations with a non-relativistic structure, and applying a special linear transformation, we obtain two independent equations of the Schrödinger type in the presence of a Coulomb field.

In Section 4, we focus on the subsystem of eight second-order equations corresponding to states with parity $P={(-1)}^j$. These are reduced to three coupled radial equations with a non-relativistic structure. Applying the appropriate linear transformation, we obtain three independent equations of the Schrödinger type in the presence of a Coulomb field. As a result, we derive five distinct energy spectra corresponding to a non-relativistic spin-2 particle in the presence of a Coulomb field.

These five spectra correspond to the $2s+1=5$ degrees of freedom for a massive spin-2 particle, as expected. By analogy, spin-1/2 particles (Dirac case) produce two classes of solutions; spin-1 particles (Duffin-Kemmer case) yield three classes; and spin-3/2 particles produce four types. Our solutions are constructed using the standard quantum numbers $ϵ$, j, m, and P.

The case of the minimal value of the quantum number $j=0$ is a special scenario (addressed in Section 5), leading to a single Schrödinger-type equation with a simple energy spectrum.

2. Initial Equations

In [17,18], calculations were performed to separate variables in the matrix equation describing a spin-2 particle in spherical coordinates. Here, we briefly recall the essential points. The wave function consists of scalar (H), four-vector ($H_1$), symmetric tensor ($H_2$), and rank-3 antisymmetric tensor ($H_3$) components, resulting in a total of 39 components. For solutions with spherical symmetry, the following operators are diagonalized:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill {\overrightarrow{J}}^{2}H=j(j+1)H,J_{3}H=mH;{\overrightarrow{J}}_{\left(1\right)}^2{H}_1=j(j+1)H_1,J_3^{\left(1\right)}{H}_1=m{H}_1;\\ \hfill {\overrightarrow{J}}_{\left(2\right)}^2{H}_2=j(j+1)H_2,J_3^{\left(2\right)}{H}_2=m{H}_2;{\overrightarrow{J}}_{\left(3\right)}^2{H}_3=j(j+1)H_3,J_3^{\left(3\right)}{H}_2=m{H}_3,\end{array}\end{array}$$

(1)

where j is the quantum number of the squared angular momentum, and m is the quantum number of the third projection of the total angular momentum (see in [21]).

To separate the variables, we employ the Wigner D-functions technique, specifically $D_{-m,-s_3}^j(\varphi ,\theta ,0)=D_{-s_3}$, as detailed in [21]; the well-known recurrence relations are utilized [22]:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill {\partial }_{\theta }{D}_0=+{\displaystyle \frac{1}{2}}a{D}_{-1}-{\displaystyle \frac{1}{2}}a{D}_{+1},{\displaystyle \frac{-m}{sin\theta }}{D}_0=-{\displaystyle \frac{1}{2}}a{D}_{-1}-{\displaystyle \frac{1}{2}}a{D}_{+1},\\ \hfill {\partial }_{\theta }{D}_{+1}=+{\displaystyle \frac{1}{2}}a{D}_0-{\displaystyle \frac{1}{2}}b{D}_{+2},{\displaystyle \frac{-m-cos\theta }{sin\theta }}{D}_{+1}=-{\displaystyle \frac{1}{2}}a{D}_0-{\displaystyle \frac{1}{2}}b{D}_{+2},\\ \hfill {\partial }_{\theta }{D}_{-1}=+{\displaystyle \frac{1}{2}}b{D}_{-2}-{\displaystyle \frac{1}{2}}a{D}_0,{\displaystyle \frac{-m+cos\theta }{sin\theta }}{D}_{-1}=-{\displaystyle \frac{1}{2}}b{D}_{-2}-{\displaystyle \frac{1}{2}}a{D}_0,\\ \hfill {\partial }_{\theta }{D}_{+2}=+{\displaystyle \frac{1}{2}}b{D}_{+1}-{\displaystyle \frac{1}{2}}c{D}_{+3},{\displaystyle \frac{-m-2cos\theta }{sin\theta }}{D}_{+2}=-{\displaystyle \frac{1}{2}}b{D}_{+1}-{\displaystyle \frac{1}{2}}c{D}_{+3},\\ \hfill {\partial }_{\theta }{D}_{-2}=+{\displaystyle \frac{1}{2}}c{D}_{-3}-{\displaystyle \frac{1}{2}}b{D}_{-1},{\displaystyle \frac{-m+2cos\theta }{sin\theta }}{D}_{-2}=-{\displaystyle \frac{1}{2}}c{D}_{-3}-{\displaystyle \frac{1}{2}}b;D_{-1},\end{array}\end{array}$$

(2)

where $\sqrt{j(j+1)}=a,\sqrt{(j-1)(j+2)}=b,\sqrt{(j-2)(j+3)}=c.$

Thus, we use the following substitutions, while omitting the common multiplier $e^{-iϵt}$ (for more details see [21]):

$$H=h\left(r\right)D_0,H_1=\left|\begin{array}{c}{h}_0\left(r\right)D_0\hfill \\ h_1\left(r\right)D_{-1}\hfill \\ h_2\left(r\right)D_0\hfill \\ h_3\left(r\right)D_{+1}\hfill \end{array}\right|,H_2=\left|\begin{array}{c}{f}_1\left(r\right)D_{-2}\hfill \\ f_2\left(r\right)D_0\hfill \\ f_3\left(r\right)D_{+2}\hfill \\ c_1\left(r\right)D_{+1}\hfill \\ c_2\left(r\right)D_0\hfill \\ c_3\left(r\right)D_{-1}\hfill \\ d_1\left(r\right)D_{-1}\hfill \\ d_2\left(r\right)D_0\hfill \\ d_3\left(r\right)D_{+1}\hfill \\ f_0\left(r\right)D_0\hfill \end{array}\right|,$$

$$\begin{array}{c}\hfill H_3,{\phi }_0=\left|\begin{array}{c}{E}_{10}\left(r\right)D_{-1}\hfill \\ E_{20}\left(r\right)D_0\hfill \\ E_{30}\left(r\right)D_{+1}\hfill \\ B_{10}\left(r\right)D_{+1}\hfill \\ B_{20}\left(r\right)D_0\hfill \\ B_{30}\left(r\right)D_{-1}\hfill \end{array}\right|,{\phi }_1=\left|\begin{array}{c}{E}_{11}\left(r\right)D_{-2}\hfill \\ E_{21}\left(r\right)D_{-1}\hfill \\ E_{31}\left(r\right)D_0\hfill \\ B_{11}\left(r\right)D_0\hfill \\ B_{21}\left(r\right)D_{-1}\hfill \\ B_{31}\left(r\right)D_{-2}\hfill \end{array}\right|,{\phi }_2=\left|\begin{array}{c}{E}_{12}\left(r\right)D_{-1}\hfill \\ E_{22}\left(r\right)D_0\hfill \\ E_{32}\left(r\right)D_{+1}\hfill \\ B_{12}\left(r\right)D_{+1}\hfill \\ B_{22}\left(r\right)D_0\hfill \\ B_{32}\left(r\right)D_{-1}\hfill \end{array}\right|,{\phi }_3=\left|\begin{array}{c}{E}_{13}\left(r\right)D_0\hfill \\ E_{23}\left(r\right)D_{+1}\hfill \\ E_{33}\left(r\right)D_{+2}\hfill \\ B_{13}\left(r\right)D_{+2}\hfill \\ B_{23}\left(r\right)D_{+1}\hfill \\ B_{33}\left(r\right)D_0\hfill \end{array}\right|.\end{array}$$

(3)

We employ the so-called cyclic basis [21], in which the third projection of spin is represented as a diagonal matrix.

3. Separation of the Variables

We start with the radial system. The presence of the external Coulomb field is incorporated by the substitution $ϵ⟹ϵ+{\displaystyle \frac{\alpha }{r}}$ (the prime denotes the derivative $d/dr$):

$$H,-{\displaystyle \frac{\sqrt{2}a}{2r}}\left(h_1+h_3\right)-i(ϵ+{\displaystyle \frac{\alpha }{r}})h_0-h_2^{\prime }-{\displaystyle \frac{2}{2}}{h}_2-Mh=0;$$

$$H_1,\sqrt{2}a\left(d_1+d_3\right)+2r\left(d_2^{\prime }+i(ϵ+{\displaystyle \frac{\alpha }{r}})f_0-3M{h}_0\right)+4d_2-3ir(ϵ+{\displaystyle \frac{\alpha }{r}})h=0,$$

$${\displaystyle \frac{2a{c}_2-3ah+2b{f}_1}{6\sqrt{2}r}}+{\displaystyle \frac{1}{3}}(c_3^{\prime }+{\displaystyle \frac{3c_3}{r}}+i(ϵ+{\displaystyle \frac{\alpha }{r}})d_1)-M{h}_1=0,$$

$$\sqrt{2}a\left(c_1+c_3\right)+4c_2+r(2i(ϵ+{\displaystyle \frac{\alpha }{r}})d_2+2f_2^{\prime }+3h^{\prime }-6M{h}_2)+4f_2=0,$$

$${\displaystyle \frac{2a{c}_2-3ah+2b{f}_3}{6\sqrt{2}r}}+{\displaystyle \frac{1}{3}}(c_1^{\prime }+{\displaystyle \frac{3c_1}{r}}+i(ϵ+{\displaystyle \frac{\alpha }{r}})d_3)-M{h}_3=0;$$

$H_2$

$${\displaystyle \frac{B_{31}}{r}}-i(ϵ+{\displaystyle \frac{\alpha }{r}})E_{11}-{\displaystyle \frac{b\left(B_{21}+2h_1\right)}{\sqrt{2}r}}+B_{31}^{\prime }-M{f}_1=0,$$

$$4B_{11}-4B_{33}+4\left(E_{20}-2h_2\right)+\sqrt{2}a\left(3B_{12}+B_{21}-B_{23}-3B_{32}+E_{10}+E_{30}-2\left(h_1+h_3\right)\right)$$

$$+2r\left(-i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(E_{13}+3E_{22}+E_{31}+2h_0\right)-B_{11}^{\prime }+B_{33}^{\prime }+E_{20}^{\prime }+6h_2^{\prime }\right)-8Mr{f}_2=0,$$

$$-2B_{13}+\sqrt{2}b\left(B_{23}-2h_3\right)-2r\left(i(ϵ+{\displaystyle \frac{\alpha }{r}})E_{33}+M{f}_3+B_{13}^{\prime }\right)=0,$$

$$\sqrt{2}b{B}_{13}+2B_{23}+\sqrt{2}a\left(B_{22}-B_{33}-2h_2\right)-4h_3$$

$$-2r\left(2M{c}_1+i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(E_{23}+E_{32}\right)+B_{12}^{\prime }-2h_3^{\prime }\right)=0,$$

$$\sqrt{2}a\left(B_{12}+B_{21}-B_{23}-B_{32}-E_{10}-E_{30}-2\left(h_1+h_3\right)\right)-2(2\left(E_{20}+2h_2\right)$$

$$+r\left(4M{c}_2+i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(E_{13}+E_{22}+E_{31}-2h_0\right)+B_{11}^{\prime }-B_{33}^{\prime }+E_{20}^{\prime }-2h_2^{\prime }\right))=0,$$

$$-4Mr{c}_3+\sqrt{2}\left(a{B}_{11}-b{B}_{31}-a\left(B_{22}+2h_2\right)\right)$$

$$-2\left(B_{21}+ir(ϵ+{\displaystyle \frac{\alpha }{r}})\left(E_{12}+E_{21}\right)+2h_1-r\left(B_{32}^{\prime }+2h_1^{\prime }\right)\right)=0,$$

$$-4Mr{d}_1+\sqrt{2}\left(b{E}_{11}+a\left(-B_{20}+E_{31}-2h_0\right)\right)$$

$$+2\left(B_{30}+E_{12}+2E_{21}+r\left(-i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(E_{10}+2h_1\right)+B_{30}^{\prime }+E_{21}^{\prime }\right)\right)=0,$$

$$\sqrt{2}a\left(B_{10}-B_{30}+E_{12}+E_{32}\right)$$

$$+2\left(E_{13}+2E_{22}+E_{31}+r\left(-2M{d}_2-i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(E_{20}+2h_2\right)+E_{22}^{\prime }+2h_0^{\prime }\right)\right)=0,$$

$$-2B_{10}+4E_{23}+2E_{32}+\sqrt{2}b{E}_{33}$$

$$+\sqrt{2}a\left(B_{20}+E_{13}-2h_0\right)-2r\left(2M{d}_3+i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(E_{30}+2h_3\right)+B_{10}^{\prime }-E_{23}^{\prime }\right)=0,$$

$$4B_{11}+\sqrt{2}a\left(B_{12}-B_{21}+B_{23}-B_{32}+3E_{10}+3E_{30}+2\left(h_1+h_3\right)\right)+2\left(-2B_{33}+6E_{20}+4h_2\right.$$

$$\left.+r\left(-4M{f}_0+i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(E_{13}-E_{22}+E_{31}-6h_0\right)+B_{11}^{\prime }-B_{33}^{\prime }+3E_{20}^{\prime }+2h_2^{\prime }\right)\right)=0,$$

$H_3$

$${\displaystyle \frac{6c_3+\sqrt{2}\left(a{c}_2+3a{f}_0+b{f}_1\right)}{r}}+2\left(-2i(ϵ+{\displaystyle \frac{\alpha }{r}})d_1-3M{E}_{10}+c_3^{\prime }\right)=0,$$

$$4c_2+\sqrt{2}a\left(c_1+c_3\right)+4f_2+2r\left(-2i(ϵ+{\displaystyle \frac{\alpha }{r}})d_2-3\left(M{E}_{20}+f_0^{\prime }\right)+f_2^{\prime }\right)=0,$$

$${\displaystyle \frac{6c_1+\sqrt{2}\left(a{c}_2+3a{f}_0+b{f}_3\right)}{r}}+2\left(-2i(ϵ+{\displaystyle \frac{\alpha }{r}})d_3-3M{E}_{30}+c_1^{\prime }\right)=0,$$

$${\displaystyle \frac{a{d}_2}{\sqrt{2}r}}+{\displaystyle \frac{d_3}{r}}+d_3^{\prime }-M{B}_{10}=0,{\displaystyle \frac{a\left(d_3-d_1\right)}{\sqrt{2}r}}-M{B}_{20}=0,2d_1+\sqrt{2}a{d}_2+2r{d}_1^{\prime }+2rM{B}_{30}=0,$$

$${\displaystyle \frac{b{d}_1}{\sqrt{2}r}}-i(ϵ+{\displaystyle \frac{\alpha }{r}})f_1-M{E}_{11}=0,-i(ϵ+{\displaystyle \frac{\alpha }{r}})c_3-d_1^{\prime }-M{E}_{21}=0,$$

$$2d_2+\sqrt{2}a\left(2d_1-d_3\right)-2r\left(3M{E}_{31}+i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(3c_2+f_0\right)+d_2^{\prime }\right)=0,$$

$$\sqrt{2}a\left(c_1-2c_3\right)-2\left(c_2+f_2\right)+2r\left(3M{B}_{11}+i(ϵ+{\displaystyle \frac{\alpha }{r}})d_2-3c_2^{\prime }+f_2^{\prime }\right)=0,$$

$$+i(ϵ+{\displaystyle \frac{\alpha }{r}})d_1+{\displaystyle \frac{\sqrt{2}\left(2a{c}_2-b{f}_1\right)}{r}}+c_3^{\prime }-3M{B}_{21}=0,$$

$$\sqrt{2}b{c}_3+2f_1+2r\left(M{B}_{31}+f_1^{\prime }\right)=0,-i(ϵ+{\displaystyle \frac{\alpha }{r}})c_3+{\displaystyle \frac{d_1}{r}}+{\displaystyle \frac{a{d}_2}{\sqrt{2}r}}-M{E}_{12}=0,$$

$$4d_2+\sqrt{2}a\left(d_1+d_3\right)-6Mr{E}_{22}+2ir(ϵ+{\displaystyle \frac{\alpha }{r}})\left(f_0-3f_2\right)-4r{d}_2^{\prime }=0,$$

$$-i(ϵ+{\displaystyle \frac{\alpha }{r}})c_1+{\displaystyle \frac{a{d}_2}{\sqrt{2}r}}+{\displaystyle \frac{d_3}{r}}-M{E}_{32}=0,$$

$$6Mr{B}_{12}-6c_1+2ir(ϵ+{\displaystyle \frac{\alpha }{r}})d_3+\sqrt{2}\left(a{c}_2-3a{f}_2+b{f}_3\right)-4r{c}_1^{\prime }=0,$$

$${\displaystyle \frac{a\left(c_1-c_3\right)}{\sqrt{2}r}}-M{B}_{22}=0,$$

$$-6Mr{B}_{32}-6c_3+2ir(ϵ+{\displaystyle \frac{\alpha }{r}})d_1+\sqrt{2}\left(a{c}_2+b{f}_1-3a{f}_2\right)-4r{c}_3^{\prime }=0,$$

$$-2d_2+\sqrt{2}a\left(d_1-2d_3\right)+2r\left(3M{E}_{13}+i(ϵ+{\displaystyle \frac{\alpha }{r}})\left(3c_2+f_0\right)+d_2^{\prime }\right)=0,$$

$$-i(ϵ+{\displaystyle \frac{\alpha }{r}})c_1-M{E}_{23}-d_3^{\prime }=0,{\displaystyle \frac{b{d}_3}{\sqrt{2}r}}-M{E}_{33}-i(ϵ+{\displaystyle \frac{\alpha }{r}})f_3=0,$$

$$-M{B}_{13}+{\displaystyle \frac{b{c}_1}{\sqrt{2}r}}+{\displaystyle \frac{f_3}{r}}+f_3^{\prime }=0,-3M{B}_{23}-i(ϵ+{\displaystyle \frac{\alpha }{r}})d_3+{\displaystyle \frac{\sqrt{2}\left(b{f}_3-2a{c}_2\right)}{r}}-c_1^{\prime }=0,$$

$$\sqrt{2}a\left(c_3-2c_1\right)-2\left(c_2+f_2\right)+2r\left(-3M{B}_{33}+i(ϵ+{\displaystyle \frac{\alpha }{r}})d_2-3c_2^{\prime }+f_2^{\prime }\right)=0.$$

To simplify the problem, we can take into account the parity restrictions. In fact, we started with the substitution given in Section 1, then we applied the 39-component presentation for the relevant space reflection operator (in accordance with the tensors involved in the complete wave function); after that, we studied the eigen-value problem for the reflection operator, $\widehat{P}\Psi =P\Psi$; and with the use of the known behavior of Wigner D-functions under the space reflection, the problem was reduced to simple algebraic restrictions of radial variables. Therefore, the parity restrictions given in this paper are the results of solving these algebraic constraints. This procedure is correct; moreover, it was checked by direct calculations: the parity constraints turn outed to be consistent with the complete system of 39 radial equations. For more detail, see [18].

The resulting parity restrictions are as follows:

$$\begin{array}{c}\hfill P={(-1)}^{j+1},\begin{array}{c}\hfill h=0;h_0=0,h_2=0,h_3=-h_1;\\ \hfill f_0=0,f_2=0,c_2=0,d_2=0,f_3=-f_1,c_3=-c_1,d_3=-d_1;\\ \hfill {\overline{E}}_{30}=-{\overline{E}}_{10},{\overline{E}}_{20}=0,{\overline{B}}_{30}=+{\overline{B}}_{10};\\ \hfill {\overline{E}}_{32}=-{\overline{E}}_{12},{\overline{E}}_{22}=0,{\overline{B}}_{32}=+{\overline{B}}_{12};\\ \hfill {\overline{E}}_{13}=-{\overline{E}}_{31},{\overline{E}}_{23}=-{\overline{E}}_{21},{\overline{E}}_{33}=-{\overline{E}}_{11},\\ \hfill {\overline{B}}_{13}=+{\overline{B}}_{31},{\overline{B}}_{23}=+{\overline{B}}_{21},{\overline{B}}_{33}=+{\overline{B}}_{11};\end{array}\end{array}$$

(4)

$$\begin{array}{c}\hfill P={(-1)}^j,\begin{array}{c}\hfill h_3=+h_1;f_3=+f_1,c_3=+c_1,d_3=+d_1;\\ \hfill {\overline{B}}_{30}=-{\overline{B}}_{10},{\overline{B}}_{20}=0,{\overline{E}}_{30}=+{\overline{E}}_{10};\\ \hfill {\overline{B}}_{32}=-{\overline{B}}_{12},{\overline{B}}_{22}=0,{\overline{E}}_{32}=+{\overline{E}}_{12};\\ \hfill {\overline{B}}_{13}=-{\overline{B}}_{31},{\overline{B}}_{23}=-{\overline{B}}_{21},{\overline{B}}_{33}=-{\overline{B}}_{11},\\ \hfill {\overline{E}}_{13}=+{\overline{E}}_{31},{\overline{E}}_{23}=+{\overline{E}}_{21},{\overline{E}}_{33}=+{\overline{E}}_{11};\end{array}\end{array}$$

(5)

There are two special cases where the initial substitutions should be simplified from the very beginning.

The first case occurs when $j=0$; then, the initial general substitution must be modified according to the following constraints (we allow for the fact that the multiples before the Wigner functions $D_{-m,\sigma }^0$ should vanish identically if $m\ne 0,m\ne 0$), as follows ($j=0,a=0,b=\sqrt{-2}$):

$${\overline{H}}_1{\overline{h}}_1=0,{\overline{h}}_3=0,$$

$${\overline{H}}_2{\overline{f}}_1=0,{\overline{f}}_3=0,{\overline{c}}_1=0,{\overline{c}}_3=0,{\overline{d}}_1=0,{\overline{d}}_3=0,$$

$$\begin{array}{c}\hfill {\overline{H}}_3\begin{array}{c}\hfill {\overline{E}}_{10}=0{\overline{E}}_{11}=0{\overline{E}}_{12}=0{\overline{E}}_{23}=0\\ \hfill {\overline{E}}_{30}=0{\overline{E}}_{21}=0{\overline{E}}_{32}=0{\overline{E}}_{33}=0\\ \hfill {\overline{B}}_{10}=0{\overline{B}}_{21}=0{\overline{B}}_{12}=0{\overline{B}}_{13}=0\\ \hfill {\overline{B}}_{30}=0{\overline{B}}_{31}=0{\overline{B}}_{32}=0{\overline{B}}_{23}=0.\end{array}\end{array}$$

(6)

The second special case is

$$\begin{array}{c}\hfill j=1,a=\sqrt{2},b=0;{\overline{f}}_1,{\overline{f}}_3=0,{\overline{E}}_{11}=0,{\overline{B}}_{31}=0,{\overline{E}}_{33}=0,{\overline{B}}_{13}=0.\end{array}$$

(7)

Further, we should make the temporary change in notation $\left(ϵ+{\displaystyle \frac{\alpha }{r}}\right)⟹D_0$ to account for the presence of the external Coulomb potential (see Appendix A). In addition, we should eliminate the variables related to the vector and rank-3 tensor components involved in the description of the spin-2 particle. This reduction yields eleven second-order equations for the radial variables associated with the scalar and symmetric tensor components (see Appendix A). Finally, we should impose the constraints due to spatial parity conservation, producing subsystems of three and eight equations, respectively (see Appendix A).

To perform the non-relativistic approximation in the resulting second-order radial equations, we need to identify the large and small components of the wave function. To this end, we apply the known method based on projective operators associated with the matrix ${\Gamma }_{39\times 39}^0$ of the initial first-order equation. In this way, we determine all the independent large and small variables associated with the scalar and symmetric tensor components (see Appendix B).

4. Non-Relativistic Approximation: Parity $\mathit{P}={(-\mathbf{1})}^{\mathit{j}+\mathbf{1}}$

Consider the states with parity $P={(-1)}^{j+1}$. For this case, the relevant components are decomposed into large and small parts as shown below (see Appendix B):

$$\begin{array}{c}\hfill f_1=(L_1+S_1),c_1=(L_4+S_4),d_1=S_9.\end{array}$$

(8)

The final goal is to derive a system of equations for the large variables $L_1$ and $L_4$ by eliminating all small variables. Taking (8) into account, we obtain three radial equations in the following form (note the notations $D_0=ϵ+{\displaystyle \frac{\alpha }{r}}$ and $D_r=d/dr$):

$${\displaystyle \frac{\sqrt{2}b}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_4+S_4)+{\displaystyle \frac{\sqrt{2}b}{M^2{r}^2}}(L_4+S_4)-{\displaystyle \frac{\sqrt{2}b}{Mr}}{\displaystyle \frac{1}{M}}i{D}_0{s}_9-{\displaystyle \frac{1}{M^2}}{D}_0^2(L_1+S_1)$$

$$-{\displaystyle \frac{2}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_1+S_1)-{\displaystyle \frac{1}{M^2}}{D}_r^2(L_1+S_1)-(L_1+S_1)=0,$$

$${\displaystyle \frac{3a^2+b^2-6}{M^2{r}^2}}(L_4+S_4)-4(L_4+S_4)-{\displaystyle \frac{4}{M^2}}{D}_0^2(L_4+S_4)+{\displaystyle \frac{4i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_9$$

$$-{\displaystyle \frac{2i}{M^2}}{D}_0{D}_r{s}_9-{\displaystyle \frac{2i}{M^2}}{D}_r{D}_0{s}_9-{\displaystyle \frac{2\sqrt{2}b}{rm}}{\displaystyle \frac{1}{M}}{D}_r(L_1+S_1)=0,$$

$${\displaystyle \frac{3a^2+b^2+2}{M^2{r}^2}}{s}_9-4s_9-{\displaystyle \frac{8}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{s}_9-{\displaystyle \frac{4}{M^2}}{D}_r^2{s}_9+{\displaystyle \frac{12i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0(L_4+S_4)$$

$$+{\displaystyle \frac{2i}{M^2}}{D}_0{D}_r(L_4+S_4)+{\displaystyle \frac{2i}{M^2}}{D}_r{D}_0(L_4+S_4)-{\displaystyle \frac{i2\sqrt{2}b}{rm}}{\displaystyle \frac{1}{M}}{D}_0(L_1+S_1)=0.$$

When performing the non-relativistic approximation, we separate the rest energy by making the formal substitution $D_0⟹M+i{D}_0$ (for more detail see [19,20]), where M is the mass parameter. Additionally, we should assume the following smallness orders for the wave function components and derivative operators:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill L\sim 1,S_k\sim x,s_k\sim x,{\displaystyle \frac{1}{M}}{D}_r\sim x,{\displaystyle \frac{1}{Mr}}\sim x,\\ \hfill {\displaystyle \frac{1}{M}}i{D}_0⟹{\displaystyle \frac{1}{M}}(M+i{D}_0)={\displaystyle \frac{1}{M}}\left[M+(E+{\displaystyle \frac{\alpha }{r}})\right]\sim (1+x^2),\\ \hfill {\displaystyle \frac{1}{M^2}}{D}_0^2=\left(-1-{\displaystyle \frac{2}{M}}(E+{\displaystyle \frac{\alpha }{r}})-{\displaystyle \frac{1}{M^2}}{(E+{\displaystyle \frac{\alpha }{r}})}^2\right)\sim \left(-1-x^2-x^4\right).\end{array}\end{array}$$

(9)

In order to derive the non-relativistic equations, we retain only the terms of orders $x^0$, x, and $x^2$, and neglect the higher-order terms (note that at order $x^0$ we should obtain identities). As a result, we derive equations of orders x and $x^2$:

x

$$-{\displaystyle \frac{2\sqrt{2}b{L}_1}{Mr}}+{\displaystyle \frac{4D_r{L}_4}{M}}+{\displaystyle \frac{12L_4}{Mr}}-4s_9=0,$$

$x^2$

$${\displaystyle \frac{\sqrt{2}b{D}_r{L}_4}{M^{2}r}}+{\displaystyle \frac{\sqrt{2}b{L}_4}{M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}b{s}_9}{Mr}}+{\displaystyle \frac{2}{M}}(E+\alpha /r)L_1-{\displaystyle \frac{D_r^2{L}_1}{M^2}}-{\displaystyle \frac{2D_r{L}_1}{M^{2}r}}=0,$$

$${\displaystyle \frac{3a^2{L}_4}{M^2{r}^2}}+{\displaystyle \frac{b^2{L}_4}{M^2{r}^2}}-{\displaystyle \frac{2\sqrt{2}b{D}_r{L}_1}{M^{2}r}}+4{\displaystyle \frac{2}{M}}(E+\alpha /r)L_4-{\displaystyle \frac{4D_r{s}_9}{M}}-{\displaystyle \frac{6L_4}{M^2{r}^2}}+{\displaystyle \frac{4s_9}{Mr}}=0,$$

$$-{\displaystyle \frac{2\sqrt{2}b{S}_1}{Mr}}+{\displaystyle \frac{4D_r{S}_4}{M}}+{\displaystyle \frac{12S_4}{Mr}}=0\left(\mathrm{we\; do\; not\; need\; this\; constraint\; on\; small\; variables}\right).$$

Expressing $S_9$ from the first equation and substituting the result into the two remaining equations, we obtain two equations for the large components.

$$(E+\alpha /r)L_1+{\displaystyle \frac{1}{2M}}\left[-D_r^2{L}_1-{\displaystyle \frac{2D_r{L}_1}{r}}+{\displaystyle \frac{b^2{L}_1}{r^2}}-{\displaystyle \frac{2\sqrt{2}b{L}_4}{r^2}}\right]=0,$$

$$4(E+\alpha /r)L_4+{\displaystyle \frac{1}{2M}}\left[-4D_r^2{L}_4-{\displaystyle \frac{8D_r{L}_4}{r}}+{\displaystyle \frac{3a^2{L}_4}{r^2}}+{\displaystyle \frac{b^2{L}_4}{r^2}}-{\displaystyle \frac{4\sqrt{2}b{L}_1}{r^2}}+{\displaystyle \frac{18L_4}{r^2}}\right]=0.$$

With the use of a new variable, $4L_4={\overline{L}}_4$, we obtain

$$(E+\alpha /r)L_1+{\displaystyle \frac{1}{2M}}\left[-(D_r^2{L}_1+{\displaystyle \frac{2D_r{L}_1}{r}})+{\displaystyle \frac{1}{r^2}}\left(b^2{L}_1-{\displaystyle \frac{b}{\sqrt{2}}}{\overline{L}}_4\right)\right]=0,$$

$$(E+\alpha /r){\overline{L}}_4+{\displaystyle \frac{1}{2M}}[-(D_r^2{\overline{L}}_4+{\displaystyle \frac{2D_r{\overline{L}}_4}{r}})+{\displaystyle \frac{1}{r^2}}\left(-4\sqrt{2}b{L}_1+({\displaystyle \frac{3a^2}{4}}+{\displaystyle \frac{b^2}{4}}+{\displaystyle \frac{18}{4}}){\overline{L}}_4\right]=0.$$

Using the abbreviated notation

$$\begin{array}{c}\hfill \Delta =r^2\left[{\displaystyle \frac{d^2}{d{r}^2}}+2{\displaystyle \frac{d}{dr}}-2M(E+\alpha /r)\right],\end{array}$$

(10)

we transform the system to the form

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \Delta L_1=b^2{L}_1-{\displaystyle \frac{b}{\sqrt{2}}}{\overline{L}}_4,\Delta L_2=-4\sqrt{2}b{L}_1+{\displaystyle \frac{3a^2+b^2+18}{4}}{\overline{L}}_4=0.\end{array}\end{array}$$

(11)

Taking into account the following identities:

$$a=\sqrt{j(j+1)},b=\sqrt{(j-1)(j+2)},{\displaystyle \frac{3a^2+b^2+18}{4}}=j^2+j+4=c,$$

we obtain the matrix representation of the system:

$$\Delta \left|\begin{array}{c}{L}_1\\ {\overline{L}}_4\end{array}\right|=\left|\begin{array}{cc}{b}^2& -{\displaystyle \frac{b}{\sqrt{2}}}\\ -4\sqrt{2}b& c\end{array}\right|\left|\begin{array}{c}{L}_1\\ {\overline{L}}_4\end{array}\right|,\Delta F=AF$$

Let us find the linear transformation that reduces the system to a diagonal form:

$$F^{\prime }=SF,S\Delta S^{-1}{F}^{\prime }=SA{S}^{-1}{F}^{\prime },\Delta F^{\prime }=\left(SA{S}^{-1}\right)F^{\prime },\left(SA{S}^{-1}\right)=A^{\prime }=\left|\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right|.$$

This leads to

$$\left|\begin{array}{cc}{s}_{11}& s_{12}\\ s_{21}& s_{22}\end{array}\right|\left|\begin{array}{cc}{b}^2& -{\displaystyle \frac{b}{\sqrt{2}}}\\ -4\sqrt{2}b& c\end{array}\right|=\left|\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right|\left|\begin{array}{cc}{s}_{11}& s_{12}\\ s_{21}& s_{22}\end{array}\right|$$

From this, it follows that

$$b^2{s}_{11}-4\sqrt{2}b{s}_{12}={\lambda }_1{s}_{11},-{\displaystyle \frac{b}{\sqrt{2}}}{s}_{11}+c{s}_{12}={\lambda }_1{s}_{12},\left|\begin{array}{cc}(b^2-\lambda )& -4\sqrt{2}b\\ -b/\sqrt{2}& (c-\lambda )\end{array}\right|\left|\begin{array}{c}{s}_{11}\\ s_{12}\end{array}\right|=0,$$

$$b^2{s}_{21}-4\sqrt{2}b{s}_{22}={\lambda }_2{s}_{21},-{\displaystyle \frac{b}{\sqrt{2}}}{s}_{21}+c{s}_{22}={\lambda }_2{s}_{22},\left|\begin{array}{cc}(b^2-\lambda ))& -4\sqrt{2}b\\ -b/\sqrt{2}& (c-\lambda )\end{array}\right|\left|\begin{array}{c}{s}_{21}\\ s_{22}\end{array}\right|=0.$$

From the condition of the vanishing determinant, by using the substitution $j(j+1)=x$, we get

$${\lambda }^2-2\lambda (x+1)+x^2-2x=0,$$

and we arrive at the solutions

$${\lambda }_1=x+1+\sqrt{4{(j+1/2)}^2}=(j+1)(j+2),(b^2-{\lambda }_1)s_{11}-4\sqrt{2}b{s}_{12}=0,s_{11}=1,$$

$${\lambda }_2=x+1-\sqrt{4{(j+1/2)}^2}=(j-1)j,(b^2-{\lambda }_2)s_{21}-4\sqrt{2}b{s}_{22}=0,s_{22}=1,$$

In this way, we obtain two independent equations:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill (\Delta -{\lambda }_1)F_1=0,\left[{\displaystyle \frac{d^2}{d{r}^2}}+2{\displaystyle \frac{d}{dr}}-2M(E+{\displaystyle \frac{\alpha }{r}})-{\displaystyle \frac{(j+1)(j+2)}{r^2}}\right]F_1=0,\\ \hfill \Delta F_2={\lambda }_2{F}_2,\left[{\displaystyle \frac{d^2}{d{r}^2}}+2{\displaystyle \frac{d}{dr}}-2M(E+{\displaystyle \frac{\alpha }{r}})\right)-{\displaystyle \frac{(j-1)j}{r^2}}]F_2=0.\end{array}\end{array}$$

(12)

The solutions of these equations and energy spectra are well-known [23], so for states with this parity, we have found two types of solutions ant two relevant energy spectra.

5. Non-Relativistic Approximation: Parity $\mathit{P}={(-\mathbf{1})}^{\mathit{j}}$

Consider the case of parity $P={(-1)}^j$. The decompositions of the relevant components into large and small parts are as follows (see Appendix B):

$$\begin{array}{c}\hfill \begin{array}{c}\hfill h=S_1,f_1=(L_1+S_1),c_1=(L_4+S_4),d_1=S_9,\\ \hfill f_2=(L_2+S_2),c_2={\displaystyle \frac{1}{2}}(L_2+S_2),d_2=S_{10},f_0=S_{12}.\end{array}\end{array}$$

(13)

Explicitly, we obtain the following eight radial equations (coinciding equations are listed by pairs):

$${\displaystyle \frac{2a^2}{3M^2{r}^2}}{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{a^2}{M^2{r}^2}}{s}_1+{\displaystyle \frac{i4\sqrt{2}a}{3rm}}{\displaystyle \frac{1}{M}}{D}_0{s}_9+{\displaystyle \frac{2ab}{3M^2{r}^2}}(L_1+S_1)$$

$$+{\displaystyle \frac{4\sqrt{2}a}{3Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_4+S_4)+{\displaystyle \frac{4\sqrt{2}a}{3M^2{r}^2}}(L_4+S_4)$$

$$+{\displaystyle \frac{4\sqrt{2}a}{3M^2{r}^2}}(L_4+S_4)-{\displaystyle \frac{2}{3Mr}}{\displaystyle \frac{1}{M}}{D}_0^2{f}_0+{\displaystyle \frac{1}{M^2}}{D}_0^2{s}_1$$

$$+{\displaystyle \frac{8i}{3Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_{10}+{\displaystyle \frac{2i}{3M^2}}{D}_0{D}_r{s}_{10}+{\displaystyle \frac{2i}{3M^2}}{D}_r{D}_0{s}_{10}$$

$$+{\displaystyle \frac{4}{3Mr}}{\displaystyle \frac{1}{M}}{D}_r{c}_2+{\displaystyle \frac{4}{3M^2{r}^2}}{\displaystyle \frac{1}{2}}(L_2+S_2)+{\displaystyle \frac{8}{3Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_2+S_2)+{\displaystyle \frac{2}{3M^2}}{D}_r^2(L_2+S_2)$$

$$+{\displaystyle \frac{4}{3M^2{r}^2}}(L_2+S_2)+{\displaystyle \frac{1}{M^2}}{D}_r^2{s}_1+{\displaystyle \frac{2}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{s}_1+2s_1=0,$$

$${\displaystyle \frac{ab}{2M^2{r}^2}}{s}_1-{\displaystyle \frac{\sqrt{2}b}{M^2{r}^2}}(L_4+S_4)-{\displaystyle \frac{ab}{M^2{r}^2}}{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{2ib}{\sqrt{2}Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_9-(L_1+S_1)$$

$$-{\displaystyle \frac{1}{M^2}}{D}_0^2(L_1+S_1)-{\displaystyle \frac{\sqrt{2}b}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_4+S_4)-{\displaystyle \frac{2}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_1+S_1)-{\displaystyle \frac{1}{M^2}}{D}_r^2(L_1+S_1)=0,$$

$${\displaystyle \frac{2a^2}{M^2{r}^2}}{s}_1+{\displaystyle \frac{2a^2}{M^2{r}^2}}{f}_0+{\displaystyle \frac{6a^2}{M^2{r}^2}}(L_2+S_2)-{\displaystyle \frac{8i\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_9-{\displaystyle \frac{4ba}{M^2{r}^2}}(L_1+S_1)$$

$$+{\displaystyle \frac{8\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_4+S_4)-{\displaystyle \frac{2}{M^2}}{D}_0^2{s}_1$$

$$-{\displaystyle \frac{4}{M^2}}{D}_0^2{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{8}{M^2{r}^2}}{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{16i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_{10}$$

$$+{\displaystyle \frac{2}{M^2}}{D}_0^2{f}_0-8(L_2+S_2)-{\displaystyle \frac{6}{M^2}}{D}_0^2(L_2+S_2)-{\displaystyle \frac{8}{M^2{r}^2}}(L_2+S_2)$$

$$-{\displaystyle \frac{4}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{s}_1+{\displaystyle \frac{16}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{\displaystyle \frac{1}{2}}(L_2+S_2)+{\displaystyle \frac{8i}{M^2}}{D}_0{D}_r{s}_{10}$$

$$-{\displaystyle \frac{4}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{f}_0+{\displaystyle \frac{4}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_2+S_2)+{\displaystyle \frac{6}{M^2}}{D}_r^2{s}_1-{\displaystyle \frac{4}{M^2}}{D}_r^2{\displaystyle \frac{1}{2}}(L_2+S_2)$$

$$-{\displaystyle \frac{2}{M^2}}{D}_r^2{f}_0+{\displaystyle \frac{6}{M^2}}{D}_r^2(L_2+S_2)=0,$$

$${\displaystyle \frac{b^2-6-a^2}{M^2{r}^2}}(L_4+S_4)+{\displaystyle \frac{2\sqrt{2}a}{M^2{r}^2}}{s}_1-{\displaystyle \frac{4\sqrt{2}a}{M^2{r}^2}}{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{2i\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_{10}$$

$$-{\displaystyle \frac{2\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{s}_1+{\displaystyle \frac{2\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{2\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_2+S_2)$$

$$-4(L_4+S_4)-{\displaystyle \frac{4}{M^2}}{D}_0^2(L_4+S_4)-{\displaystyle \frac{4i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_9+{\displaystyle \frac{2i}{M^2}}{D}_0{D}_r{s}_9$$

$$+{\displaystyle \frac{2i}{M^2}}{D}_r{D}_0{s}_9+{\displaystyle \frac{2\sqrt{2}b}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_1+S_1)=0,$$

$${\displaystyle \frac{2a^2}{M^2{r}^2}}{s}_1-{\displaystyle \frac{2a^2}{M^2{r}^2}}{f}_0+{\displaystyle \frac{2a^2}{M^2{r}^2}}(L_2+S_2)-{\displaystyle \frac{8\sqrt{2}a}{M^2{r}^2}}(L_4+S_4)-{\displaystyle \frac{4ba}{M^2{r}^2}}(L_1+S_1)$$

$$+{\displaystyle \frac{2}{M^2}}{D}_0^2{s}_1-8{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{4}{M^2}}{D}_0^2{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{8}{M^2{r}^2}}{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{2}{M^2}}{D}_0^2{f}_0$$

$$-{\displaystyle \frac{2}{M^2}}{D}_0^2(L_2+S_2)-{\displaystyle \frac{8}{M^2{r}^2}}(L_2+S_2)-{\displaystyle \frac{4}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{s}_1+{\displaystyle \frac{4i}{M^2}}{D}_0{D}_r{s}_{10}$$

$$+{\displaystyle \frac{4i}{M^2}}{D}_r{D}_0{s}_{10}+{\displaystyle \frac{4}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{f}_0-{\displaystyle \frac{4}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_2+S_2)$$

$$+{\displaystyle \frac{2}{M^2}}{D}_r^2{s}_1-{\displaystyle \frac{4}{M^2}}{D}_r^2{\displaystyle \frac{1}{2}}(L_2+S_2)+{\displaystyle \frac{2}{M^2}}{D}_r^2{f}_0+{\displaystyle \frac{2}{M^2}}{D}_r^2(L_2=S_2)=0,$$

$${\displaystyle \frac{b^2+2-a^2}{M^2{r}^2}}{s}_9+{\displaystyle \frac{2i\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_1-{\displaystyle \frac{2i\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{2i\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{f}_0-{\displaystyle \frac{2\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{s}_{10}$$

$$-{\displaystyle \frac{12i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0(L_4+S_4)-4s_9-{\displaystyle \frac{2i\sqrt{2}b}{Mr}}{\displaystyle \frac{1}{M}}{D}_0(L_1+S_1)$$

$$-{\displaystyle \frac{2i}{M^2}}{D}_0{D}_r(L_4+S_4)-{\displaystyle \frac{2i}{M^2}}{D}_r{D}_0(L_4+S_4)-{\displaystyle \frac{8}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{d}_1-{\displaystyle \frac{4}{M^2}}{D}_r^2{s}_9=0,$$

$${\displaystyle \frac{4a^2}{M^2{r}^2}}{s}_{10}-{\displaystyle \frac{4i\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_0(L_4+S_4)+{\displaystyle \frac{4\sqrt{2}a}{M^2{r}^2}}{s}_9+{\displaystyle \frac{4\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{s}_9$$

$$-{\displaystyle \frac{8i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{\displaystyle \frac{1}{2}}(L_2+S_2)-4s_{10}-{\displaystyle \frac{8i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0(L_2+S_2)-{\displaystyle \frac{2i}{M^2}}{D}_0{D}_r{s}_1-{\displaystyle \frac{2i}{M^2}}{D}_r{R}_0{s}_1$$

$$+{\displaystyle \frac{2i}{M^2}}{D}_0{D}_r{f}_0+{\displaystyle \frac{2i}{M^2}}{D}_r{D}_0{f}_0-{\displaystyle \frac{2i}{M^2}}{D}_0{D}_r(L_2+S_2)-{\displaystyle \frac{2i}{M^2}}{D}_r{D}_0(L_2+S_2)=0,$$

$$-{\displaystyle \frac{2a^2}{M^2{r}^2}}{s}_1+{\displaystyle \frac{6a^2}{M^2{r}^2}}{f}_0+{\displaystyle \frac{2a^2}{M^2{r}^2}}(L_2+S_2)+{\displaystyle \frac{16\sqrt{2}a}{M^2{r}^2}}(L_4+S_4)-{\displaystyle \frac{8i\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_9$$

$$+{\displaystyle \frac{4ba}{M^2{r}^2}}(L_1+S_1)+{\displaystyle \frac{8\sqrt{2}a}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_4+S_4)-{\displaystyle \frac{6}{M^2}}{D}_0^2{s}_1+{\displaystyle \frac{4}{M^2}}{D}_0^2{\displaystyle \frac{1}{2}}(L_2+S_2)+{\displaystyle \frac{8}{M^2{r}^2}}{\displaystyle \frac{1}{2}}(L_2+S_2)$$

$$-{\displaystyle \frac{8i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_{10}-{\displaystyle \frac{8i}{Mr}}{\displaystyle \frac{1}{M}}{D}_0{s}_{10}-8f_0+{\displaystyle \frac{6}{M^2}}{D}_0^2{f}_0$$

$$-{\displaystyle \frac{2}{M^2}}{D}_0^2(L_2+S_2)+{\displaystyle \frac{8}{M^2{r}^2}}(L_2+S_2)+{\displaystyle \frac{4}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{s}_1+{\displaystyle \frac{16}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{\displaystyle \frac{1}{2}}(L_2+S_2)$$

$$-{\displaystyle \frac{4i}{M^2}}{D}_0{D}_r{s}_{10}-{\displaystyle \frac{4i}{M^2}}{D}_r{D}_0{s}_{10}-{\displaystyle \frac{12}{Mr}}{\displaystyle \frac{1}{M}}{D}_r{f}_0+{\displaystyle \frac{12}{Mr}}{\displaystyle \frac{1}{M}}{D}_r(L_2+S_2)$$

$$+{\displaystyle \frac{2}{M^2}}{D}_r^2{s}_1+{\displaystyle \frac{4}{M^2}}{D}_r^2{\displaystyle \frac{1}{2}}(L_2+S_2)-{\displaystyle \frac{6}{M^2}}{D}_r^2{f}_0+{\displaystyle \frac{2}{M^2}}{D}_r^2(L_2+S_2)=0.$$

Now we take into consideration the smallness orders in accordance with the rules in (9); as a result, we obtain

$$x^3(-{\displaystyle \frac{a^2{s}_1}{M^2{r}^2}}+{\displaystyle \frac{a^2{S}_2}{3M^2{r}^2}}+{\displaystyle \frac{2ab{S}_1}{3M^2{r}^2}}+{\displaystyle \frac{4\sqrt{2}a{D}_r{S}_4}{3M^{2}r}}+{\displaystyle \frac{8\sqrt{2}a{S}_4}{3M^2{r}^2}}-{\displaystyle \frac{D_0^2{s}_1}{M^2}}$$

$$+{\displaystyle \frac{D_r^2{s}_1}{M^2}}+{\displaystyle \frac{2D_r^2{S}_2}{3M^2}}+{\displaystyle \frac{2D_r{s}_1}{M^{2}r}}+{\displaystyle \frac{10D_r{S}_2}{3M^{2}r}}+{\displaystyle \frac{2S_2}{M^2{r}^2}})$$

$$+x^2({\displaystyle \frac{a^2{L}_2}{3M^2{r}^2}}+{\displaystyle \frac{2ab{L}_1}{3M^2{r}^2}}+{\displaystyle \frac{4\sqrt{2}a{D}_r{L}_4}{3M^{2}r}}+{\displaystyle \frac{8\sqrt{2}a{L}_4}{3M^2{r}^2}}+{\displaystyle \frac{4\sqrt{2}a{s}_9}{3Mr}}$$

$$+{\displaystyle \frac{2D_r^2{L}_2}{3M^2}}+{\displaystyle \frac{10D_r{L}_2}{3M^{2}r}}+{\displaystyle \frac{4D_r{s}_{10}}{3M}}+{\displaystyle \frac{2L_2}{M^2{r}^2}}+{\displaystyle \frac{8s_{10}}{3Mr}}+{\displaystyle \frac{2s_{12}}{3Mr}})$$

$$+x^4\left({\displaystyle \frac{4i\sqrt{2}a{D}_0{s}_9}{3M^{2}r}}+{\displaystyle \frac{2D_0^2{s}_{12}}{3M^{2}r}}+{\displaystyle \frac{8i{D}_0{s}_{10}}{3M^{2}r}}+{\displaystyle \frac{2i{D}_0{D}_r{s}_{10}}{3M^2}}+{\displaystyle \frac{2i{D}_r{D}_0{s}_{10}}{3M^2}}\right)$$

$$+{\displaystyle \frac{2D_0^2{s}_{12}{x}^6}{3M^{2}r}}-{\displaystyle \frac{D_0^2{s}_1{x}^5}{M^2}}+s_{1}x=0,$$

$$x^2\left(-{\displaystyle \frac{ab{L}_2}{2M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}b{D}_r{L}_4}{M^{2}r}}-{\displaystyle \frac{\sqrt{2}b{L}_4}{M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}b{s}_9}{Mr}}+{\displaystyle \frac{D_0^2{L}_1}{M^2}}-{\displaystyle \frac{D_r^2{L}_1}{M^2}}-{\displaystyle \frac{2D_r{L}_1}{M^{2}r}}\right)$$

$$+x^3\left({\displaystyle \frac{ab{s}_1}{2M^2{r}^2}}-{\displaystyle \frac{ab{S}_2}{2M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}b{D}_r{S}_4}{M^{2}r}}-{\displaystyle \frac{\sqrt{2}b{S}_4}{M^2{r}^2}}+{\displaystyle \frac{D_0^2{S}_1}{M^2}}-{\displaystyle \frac{D_r^2{S}_1}{M^2}}-{\displaystyle \frac{2D_r{S}_1}{M^{2}r}}\right)$$

$$+x^4\left({\displaystyle \frac{D_0^2{L}_1}{M^2}}-{\displaystyle \frac{i\sqrt{2}b{D}_0{s}_9}{M^{2}r}}\right)+{\displaystyle \frac{D_0^2{S}_1{x}^5}{M^2}}=0,$$

$$x^2({\displaystyle \frac{6a^2{L}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{L}_4}{M^{2}r}}-{\displaystyle \frac{8\sqrt{2}a{s}_9}{Mr}}+{\displaystyle \frac{8D_0^2{L}_2}{M^2}}+{\displaystyle \frac{4D_r^2{L}_2}{M^2}}$$

$$+{\displaystyle \frac{12D_r{L}_2}{M^{2}r}}+{\displaystyle \frac{8D_r{s}_{10}}{M}}-{\displaystyle \frac{12L_2}{M^2{r}^2}}-{\displaystyle \frac{16s_{10}}{Mr}})$$

$$+x^3({\displaystyle \frac{2a^2{s}_1}{M^2{r}^2}}+{\displaystyle \frac{2a^2{s}_{12}}{M^2{r}^2}}+{\displaystyle \frac{6a^2{S}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{S}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{S}_4}{M^{2}r}}+{\displaystyle \frac{2D_0^2{s}_1}{M^2}}-{\displaystyle \frac{2D_0^2{s}_{12}}{M^2}}$$

$$+{\displaystyle \frac{8D_0^2{S}_2}{M^2}}+{\displaystyle \frac{6D_r^2{s}_1}{M^2}}-{\displaystyle \frac{2D_r^2{s}_{12}}{M^2}}+{\displaystyle \frac{4D_r^2{S}_2}{M^2}}-{\displaystyle \frac{4D_r{s}_1}{M^{2}r}}-{\displaystyle \frac{4D_r{s}_{12}}{M^{2}r}}+{\displaystyle \frac{12D_r{S}_2}{M^{2}r}}-{\displaystyle \frac{12S_2}{M^2{r}^2}})$$

$$+x^4\left(-{\displaystyle \frac{8i\sqrt{2}a{D}_0{s}_9}{M^{2}r}}+{\displaystyle \frac{8D_0^2{L}_2}{M^2}}-{\displaystyle \frac{16i{D}_0{s}_{10}}{M^{2}r}}+{\displaystyle \frac{8i{D}_0{D}_r{s}_{10}}{M^2}}\right)$$

$$+x^5\left({\displaystyle \frac{2D_0^2{s}_1}{M^2}}-{\displaystyle \frac{2D_0^2{s}_{12}}{M^2}}+{\displaystyle \frac{8D_0^2{S}_2}{M^2}}\right)+\left(2s_1-2s_{12}\right)x=0,$$

$$x^2(-{\displaystyle \frac{a^2{L}_4}{M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}a{D}_r{L}_2}{M^{2}r}}-{\displaystyle \frac{2\sqrt{2}a{L}_2}{M^2{r}^2}}-{\displaystyle \frac{2\sqrt{2}a{s}_{10}}{Mr}}+{\displaystyle \frac{b^2{L}_4}{M^2{r}^2}}+{\displaystyle \frac{2\sqrt{2}b{D}_r{L}_1}{M^{2}r}}$$

$$+{\displaystyle \frac{4D_0^2{L}_4}{M^2}}+{\displaystyle \frac{4D_r{s}_9}{M}}-{\displaystyle \frac{6L_4}{M^2{r}^2}}-{\displaystyle \frac{4s_9}{Mr}})$$

$$+x^3(-{\displaystyle \frac{a^2{S}_4}{M^2{r}^2}}-{\displaystyle \frac{2\sqrt{2}a{D}_r{s}_1}{M^{2}r}}-{\displaystyle \frac{\sqrt{2}a{D}_r{S}_2}{M^{2}r}}+{\displaystyle \frac{2\sqrt{2}a{s}_1}{M^2{r}^2}}-{\displaystyle \frac{2\sqrt{2}a{S}_2}{M^2{r}^2}}+{\displaystyle \frac{b^2{S}_4}{M^2{r}^2}}$$

$$+{\displaystyle \frac{2\sqrt{2}b{D}_r{S}_1}{M^{2}r}}+{\displaystyle \frac{4D_0^2{S}_4}{M^2}}-{\displaystyle \frac{6S_4}{M^2{r}^2}})$$

$$+x^4\left(-{\displaystyle \frac{2i\sqrt{2}a{D}_0{s}_{10}}{M^{2}r}}+{\displaystyle \frac{4D_0^2{L}_4}{M^2}}-{\displaystyle \frac{4i{D}_0{s}_9}{M^{2}r}}+{\displaystyle \frac{2i{D}_0{D}_r{s}_9}{M^2}}+{\displaystyle \frac{2i{D}_r{D}_0{s}_9}{M^2}}\right)+{\displaystyle \frac{4D_0^2{S}_4{x}^5}{M^2}}=0,$$

$$x^3({\displaystyle \frac{2a^2{s}_1}{M^2{r}^2}}-{\displaystyle \frac{2a^2{s}_{12}}{M^2{r}^2}}+{\displaystyle \frac{2a^2{S}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{S}_1}{M^2{r}^2}}-{\displaystyle \frac{8\sqrt{2}a{S}_4}{M^2{r}^2}}-{\displaystyle \frac{2D_0^2{s}_1}{M^2}}+{\displaystyle \frac{2D_0^2{s}_{12}}{M^2}}+{\displaystyle \frac{4D_0^2{S}_2}{M^2}}$$

$$+{\displaystyle \frac{2D_r^2{s}_1}{M^2}}+{\displaystyle \frac{2D_r^2{s}_{12}}{M^2}}-{\displaystyle \frac{4D_r{s}_1}{M^{2}r}}+{\displaystyle \frac{4D_r{s}_{12}}{M^{2}r}}-{\displaystyle \frac{4D_r{S}_2}{M^{2}r}}-{\displaystyle \frac{12S_2}{M^2{r}^2}})$$

$$+x^2\left({\displaystyle \frac{2a^2{L}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}-{\displaystyle \frac{8\sqrt{2}a{L}_4}{M^2{r}^2}}+{\displaystyle \frac{4D_0^2{L}_2}{M^2}}-{\displaystyle \frac{4D_r{L}_2}{M^{2}r}}+{\displaystyle \frac{8D_r{s}_{10}}{M}}-{\displaystyle \frac{12L_2}{M^2{r}^2}}\right)$$

$$+x^4\left({\displaystyle \frac{4D_0^2{L}_2}{M^2}}+{\displaystyle \frac{4i{D}_0{D}_r{s}_{10}}{M^2}}+{\displaystyle \frac{4i{D}_r{D}_0{s}_{10}}{M^2}}\right)$$

$$+x^5\left(-{\displaystyle \frac{2D_0^2{s}_1}{M^2}}+{\displaystyle \frac{2D_0^2{s}_{12}}{M^2}}+{\displaystyle \frac{4D_0^2{S}_2}{M^2}}\right)\left(2s_{12}-2s_1\right)x=0,$$

$$x^3(-{\displaystyle \frac{a^2{s}_9}{M^2{r}^2}}-{\displaystyle \frac{i\sqrt{2}a{D}_0{L}_2}{M^{2}r}}-{\displaystyle \frac{2\sqrt{2}a{D}_r{s}_{10}}{M^{2}r}}+{\displaystyle \frac{b^2{s}_9}{M^2{r}^2}}-{\displaystyle \frac{2i\sqrt{2}b{D}_0{L}_1}{M^{2}r}}-{\displaystyle \frac{2i{D}_0{D}_r{L}_4}{M^2}}$$

$$-{\displaystyle \frac{12i{D}_0{L}_4}{M^{2}r}}-{\displaystyle \frac{2i{D}_0{D}_r{L}_4}{M^2}}-{\displaystyle \frac{4D_r^2{s}_9}{M^2}}-{\displaystyle \frac{8D_r{s}_9}{M^{2}r}}+{\displaystyle \frac{2s_9}{M^2{r}^2}})$$

$$+x^4({\displaystyle \frac{2i\sqrt{2}a{D}_0{s}_1}{M^{2}r}}-{\displaystyle \frac{2i\sqrt{2}a{D}_0{s}_{12}}{M^{2}r}}-{\displaystyle \frac{i\sqrt{2}a{D}_0{S}_2}{M^{2}r}}-{\displaystyle \frac{2i\sqrt{2}b{D}_0{S}_1}{M^{2}r}}$$

$$-{\displaystyle \frac{2i{D}_0{D}_r{S}_4}{M^2}}-{\displaystyle \frac{12i{D}_0{S}_4}{M^{2}r}}-{\displaystyle \frac{2i{D}_0{D}_r{S}_4}{M^2}})$$

$$+x\left(-{\displaystyle \frac{\sqrt{2}a{L}_2}{Mr}}-{\displaystyle \frac{2\sqrt{2}b{L}_1}{Mr}}-{\displaystyle \frac{4D_r{L}_4}{M}}-{\displaystyle \frac{12L_4}{Mr}}-4s_9\right)$$

$$+x^2\left({\displaystyle \frac{2\sqrt{2}a{s}_1}{Mr}}-{\displaystyle \frac{2\sqrt{2}a{s}_{12}}{Mr}}-{\displaystyle \frac{\sqrt{2}a{S}_2}{Mr}}-{\displaystyle \frac{2\sqrt{2}b{S}_1}{Mr}}-{\displaystyle \frac{4D_r{S}_4}{M}}-{\displaystyle \frac{12S_4}{Mr}}\right)=0,$$

$$x^3\left({\displaystyle \frac{4a^2{s}_{10}}{M^2{r}^2}}-{\displaystyle \frac{4i\sqrt{2}a{D}_0{L}_4}{M^{2}r}}+{\displaystyle \frac{4\sqrt{2}a{D}_r{s}_9}{M^{2}r}}+{\displaystyle \frac{4\sqrt{2}a{s}_9}{M^2{r}^2}}-{\displaystyle \frac{12i{D}_0{L}_2}{M^{2}r}}-{\displaystyle \frac{2i{D}_0{D}_r{L}_2}{M^2}}-{\displaystyle \frac{2i{D}_r{D}_0{L}_2}{M^2}}\right)$$

$$+x^4(-{\displaystyle \frac{4i\sqrt{2}a{D}_0{S}_4}{M^{2}r}}-{\displaystyle \frac{12i{D}_0{S}_2}{M^{2}r}}-{\displaystyle \frac{2i{D}_0{D}_r{s}_1}{M^2}}+{\displaystyle \frac{2i{D}_0{D}_r{s}_{12}}{M^2}}$$

$$-{\displaystyle \frac{2i{D}_0{D}_r{S}_2}{M^2}}-{\displaystyle \frac{2i{D}_r{D}_0{s}_1}{M^2}}+{\displaystyle \frac{2i{D}_r{D}_0{s}_{12}}{M^2}}-{\displaystyle \frac{2i{D}_r{D}_0{S}_2}{M^2}})$$

$$+x\left(-{\displaystyle \frac{4\sqrt{2}a{L}_4}{Mr}}-{\displaystyle \frac{4D_r{L}_2}{M}}-{\displaystyle \frac{12L_2}{Mr}}-4s_{10}\right)$$

$$+x^2\left(-{\displaystyle \frac{4\sqrt{2}a{S}_4}{Mr}}-{\displaystyle \frac{4D_r{s}_1}{M}}+{\displaystyle \frac{4D_r{s}_{12}}{M}}-{\displaystyle \frac{4D_r{S}_2}{M}}-{\displaystyle \frac{12S_2}{Mr}}\right)=0,$$

$$x^3(-{\displaystyle \frac{2a^2{s}_1}{M^2{r}^2}}+{\displaystyle \frac{6a^2{s}_{12}}{M^2{r}^2}}+{\displaystyle \frac{2a^2{S}_2}{M^2{r}^2}}+{\displaystyle \frac{4ab{S}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{S}_4}{M^{2}r}}+{\displaystyle \frac{16\sqrt{2}a{S}_4}{M^2{r}^2}}+{\displaystyle \frac{6D_0^2{s}_1}{M^2}}-{\displaystyle \frac{6D_0^2{s}_{12}}{M^2}}$$

$$+{\displaystyle \frac{2D_r^2{s}_1}{M^2}}-{\displaystyle \frac{6D_r^2{s}_{12}}{M^2}}+{\displaystyle \frac{4D_r^2{S}_2}{M^2}}+{\displaystyle \frac{4D_r{s}_1}{M^{2}r}}-{\displaystyle \frac{12D_r{s}_{12}}{M^{2}r}}+{\displaystyle \frac{20D_r{S}_2}{M^{2}r}}+{\displaystyle \frac{12S_2}{M^2{r}^2}})$$

$$+x^2({\displaystyle \frac{2a^2{L}_2}{M^2{r}^2}}+{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{L}_4}{M^{2}r}}+{\displaystyle \frac{16\sqrt{2}a{L}_4}{M^2{r}^2}}-{\displaystyle \frac{8\sqrt{2}a{s}_9}{Mr}}+{\displaystyle \frac{4D_r^2{L}_2}{M^2}}$$

$$+{\displaystyle \frac{20D_r{L}_2}{M^{2}r}}-{\displaystyle \frac{8D_r{s}_{10}}{M}}+{\displaystyle \frac{12L_2}{M^2{r}^2}}-{\displaystyle \frac{16s_{10}}{Mr}})$$

$$+x^4\left(-{\displaystyle \frac{8i\sqrt{2}a{D}_0{s}_9}{M^{2}r}}-{\displaystyle \frac{16i{D}_0{s}_{10}}{M^{2}r}}-{\displaystyle \frac{4i{D}_0{D}_r{s}_{10}}{M^2}}-{\displaystyle \frac{4i{D}_r{D}_0{s}_{10}}{M^2}}\right)$$

$$+x^5\left({\displaystyle \frac{6D_0^2{s}_1}{M^2}}-{\displaystyle \frac{6D_0^2{s}_{12}}{M^2}}\right)+\left(6s_1-14s_{12}\right)x=0.$$

We now retain only the equations of orders x and $x^2$:

x

$$s_1=0,2s_1-2s_{12}=0,2s_{12}-2s_1=0,$$

$$-{\displaystyle \frac{\sqrt{2}a{L}_2}{Mr}}-{\displaystyle \frac{2\sqrt{2}b{L}_1}{Mr}}-{\displaystyle \frac{4D_r{L}_4}{M}}-{\displaystyle \frac{12L_4}{Mr}}-4s_9=0,$$

$$-{\displaystyle \frac{4\sqrt{2}a{L}_4}{Mr}}-{\displaystyle \frac{4D_r{L}_2}{M}}-{\displaystyle \frac{12L_2}{Mr}}-4s_{10}=0,6s_1-14s_{12}=0;$$

$x^2$

$${\displaystyle \frac{a^2{L}_2}{3M^2{r}^2}}+{\displaystyle \frac{2ab{L}_1}{3M^2{r}^2}}+{\displaystyle \frac{4\sqrt{2}a{D}_r{L}_4}{3M^{2}r}}+{\displaystyle \frac{8\sqrt{2}a{L}_4}{3M^2{r}^2}}+{\displaystyle \frac{4\sqrt{2}a{s}_9}{3Mr}}+{\displaystyle \frac{2D_r^2{L}_2}{3M^2}}$$

$$+{\displaystyle \frac{10D_r{L}_2}{3M^{2}r}}+{\displaystyle \frac{4D_r{s}_{10}}{3M}}+{\displaystyle \frac{2L_2}{M^2{r}^2}}+{\displaystyle \frac{8s_{10}}{3Mr}}+{\displaystyle \frac{2s_{12}}{3Mr}}=0,$$

$$-{\displaystyle \frac{ab{L}_2}{2M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}b{D}_r{L}_4}{M^{2}r}}-{\displaystyle \frac{\sqrt{2}b{L}_4}{M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}b{s}_9}{Mr}}+{\displaystyle \frac{D_0^2{L}_1}{M^2}}-{\displaystyle \frac{D_r^2{L}_1}{M^2}}-{\displaystyle \frac{2D_r{L}_1}{M^{2}r}}=0,$$

$${\displaystyle \frac{6a^2{L}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{L}_4}{M^{2}r}}-{\displaystyle \frac{8\sqrt{2}a{s}_9}{Mr}}+{\displaystyle \frac{8D_0^2{L}_2}{M^2}}+{\displaystyle \frac{4D_r^2{L}_2}{M^2}}+{\displaystyle \frac{12D_r{L}_2}{M^{2}r}}$$

$$+{\displaystyle \frac{8D_r{s}_{10}}{M}}-{\displaystyle \frac{12L_2}{M^2{r}^2}}-{\displaystyle \frac{16s_{10}}{Mr}}=0,$$

$$-{\displaystyle \frac{a^2{L}_4}{M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}a{D}_r{L}_2}{M^{2}r}}-{\displaystyle \frac{2\sqrt{2}a{L}_2}{M^2{r}^2}}-{\displaystyle \frac{2\sqrt{2}a{s}_{10}}{Mr}}+{\displaystyle \frac{b^2{L}_4}{M^2{r}^2}}+{\displaystyle \frac{2\sqrt{2}b{D}_r{L}_1}{M^{2}r}}$$

$$+{\displaystyle \frac{4D_0^2{L}_4}{M^2}}+{\displaystyle \frac{4D_r{s}_9}{M}}-{\displaystyle \frac{6L_4}{M^2{r}^2}}-{\displaystyle \frac{4s_9}{Mr}}=0,$$

$${\displaystyle \frac{2a^2{L}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}-{\displaystyle \frac{8\sqrt{2}a{L}_4}{M^2{r}^2}}+{\displaystyle \frac{4D_0^2{L}_2}{M^2}}-{\displaystyle \frac{4D_r{L}_2}{M^{2}r}}+{\displaystyle \frac{8D_r{s}_{10}}{M}}-{\displaystyle \frac{12L_2}{M^2{r}^2}}=0,$$

$${\displaystyle \frac{2\sqrt{2}a{s}_1}{Mr}}-{\displaystyle \frac{2\sqrt{2}a{s}_{12}}{Mr}}-{\displaystyle \frac{\sqrt{2}a{S}_2}{Mr}}-{\displaystyle \frac{2\sqrt{2}b{S}_1}{Mr}}-{\displaystyle \frac{4D_r{S}_4}{M}}-{\displaystyle \frac{12S_4}{Mr}}=0,$$

$$-{\displaystyle \frac{4\sqrt{2}a{S}_4}{Mr}}-{\displaystyle \frac{4D_r{s}_1}{M}}+{\displaystyle \frac{4D_r{s}_{12}}{M}}-{\displaystyle \frac{4D_r{S}_2}{M}}-{\displaystyle \frac{12S_2}{Mr}}=0,$$

$${\displaystyle \frac{2a^2{L}_2}{M^2{r}^2}}+{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{L}_4}{M^{2}r}}+{\displaystyle \frac{16\sqrt{2}a{L}_4}{M^2{r}^2}}-{\displaystyle \frac{8\sqrt{2}a{s}_9}{Mr}}+{\displaystyle \frac{4D_r^2{L}_2}{M^2}}$$

$$+{\displaystyle \frac{20D_r{L}_2}{M^{2}r}}-{\displaystyle \frac{8D_r{s}_{10}}{M}}+{\displaystyle \frac{12L_2}{M^2{r}^2}}-{\displaystyle \frac{16s_{10}}{Mr}}=0.$$

Whence, it follows that

x

$$s_1=0,2s_1-2s_{12}=0,2s_{12}-2s_1=0,$$

$$-{\displaystyle \frac{\sqrt{2}a{L}_2}{Mr}}-{\displaystyle \frac{2\sqrt{2}b{L}_1}{Mr}}-{\displaystyle \frac{4D_r{L}_4}{M}}-{\displaystyle \frac{12L_4}{Mr}}-4s_9=0,$$

$$-{\displaystyle \frac{4\sqrt{2}a{L}_4}{Mr}}-{\displaystyle \frac{4D_r{L}_2}{M}}-{\displaystyle \frac{12L_2}{Mr}}-4s_{10}=0,6s_1-14s_{12}=0;$$

$x^2$

$${\displaystyle \frac{a^2{L}_2}{3M^2{r}^2}}+{\displaystyle \frac{2ab{L}_1}{3M^2{r}^2}}+{\displaystyle \frac{4\sqrt{2}a{D}_r{L}_4}{3M^{2}r}}+{\displaystyle \frac{8\sqrt{2}a{L}_4}{3M^2{r}^2}}+{\displaystyle \frac{4\sqrt{2}a{s}_9}{3Mr}}+{\displaystyle \frac{2D_r^2{L}_2}{3M^2}}$$

$$+{\displaystyle \frac{10D_r{L}_2}{3M^{2}r}}+{\displaystyle \frac{4D_r{s}_{10}}{3M}}+{\displaystyle \frac{2L_2}{M^2{r}^2}}+{\displaystyle \frac{8s_{10}}{3Mr}}+{\displaystyle \frac{2s_{12}}{3Mr}}=0,$$

$$-{\displaystyle \frac{ab{L}_2}{2M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}b{D}_r{L}_4}{M^{2}r}}-{\displaystyle \frac{\sqrt{2}b{L}_4}{M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}b{s}_9}{Mr}}+{\displaystyle \frac{2}{M}}(E+\alpha /r)L_1-{\displaystyle \frac{D_r^2{L}_1}{M^2}}-{\displaystyle \frac{2D_r{L}_1}{M^{2}r}}=0,$$

$${\displaystyle \frac{6a^2{L}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{L}_4}{M^{2}r}}-{\displaystyle \frac{8\sqrt{2}a{s}_9}{Mr}}+8{\displaystyle \frac{2}{M}}(E+\alpha /r)L_2+{\displaystyle \frac{4D_r^2{L}_2}{M^2}}$$

$$+{\displaystyle \frac{12D_r{L}_2}{M^{2}r}}+{\displaystyle \frac{8D_r{s}_{10}}{M}}-{\displaystyle \frac{12L_2}{M^2{r}^2}}-{\displaystyle \frac{16s_{10}}{Mr}}=0,$$

$$-{\displaystyle \frac{a^2{L}_4}{M^2{r}^2}}-{\displaystyle \frac{\sqrt{2}a{D}_r{L}_2}{M^{2}r}}-{\displaystyle \frac{2\sqrt{2}a{L}_2}{M^2{r}^2}}-{\displaystyle \frac{2\sqrt{2}a{s}_{10}}{Mr}}+{\displaystyle \frac{b^2{L}_4}{M^2{r}^2}}+{\displaystyle \frac{2\sqrt{2}b{D}_r{L}_1}{M^{2}r}}$$

$$+4{\displaystyle \frac{2}{M}}(E+\alpha /r)L_4+{\displaystyle \frac{4D_r{s}_9}{M}}-{\displaystyle \frac{6L_4}{M^2{r}^2}}-{\displaystyle \frac{4s_9}{Mr}}=0,$$

$${\displaystyle \frac{2a^2{L}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}-{\displaystyle \frac{8\sqrt{2}a{L}_4}{M^2{r}^2}}+4{\displaystyle \frac{2}{M}}(E+\alpha /r)L_2-{\displaystyle \frac{4D_r{L}_2}{M^{2}r}}+{\displaystyle \frac{8D_r{s}_{10}}{M}}-{\displaystyle \frac{12L_2}{M^2{r}^2}}=0,$$

$${\displaystyle \frac{2\sqrt{2}a{s}_1}{Mr}}-{\displaystyle \frac{2\sqrt{2}a{s}_{12}}{Mr}}-{\displaystyle \frac{\sqrt{2}a{S}_2}{Mr}}-{\displaystyle \frac{2\sqrt{2}b{S}_1}{Mr}}-{\displaystyle \frac{4D_r{S}_4}{M}}-{\displaystyle \frac{12S_4}{Mr}}=0,$$

$$-{\displaystyle \frac{4\sqrt{2}a{S}_4}{Mr}}-{\displaystyle \frac{4D_r{s}_1}{M}}+{\displaystyle \frac{4D_r{s}_{12}}{M}}-{\displaystyle \frac{4D_r{S}_2}{M}}-{\displaystyle \frac{12S_2}{Mr}}=0,$$

$${\displaystyle \frac{2a^2{L}_2}{M^2{r}^2}}+{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{L}_4}{M^{2}r}}+{\displaystyle \frac{16\sqrt{2}a{L}_4}{M^2{r}^2}}-{\displaystyle \frac{8\sqrt{2}a{s}_9}{Mr}}+{\displaystyle \frac{4D_r^2{L}_2}{M^2}}$$

$$+{\displaystyle \frac{20D_r{L}_2}{M^{2}r}}-{\displaystyle \frac{8D_r{s}_{10}}{M}}+{\displaystyle \frac{12L_2}{M^2{r}^2}}-{\displaystyle \frac{16s_{10}}{Mr}}=0.$$

Expressing the small components from the equations of order x in terms of the large components, and substituting the results into the equations of order $x^2$, we derive

$$\begin{array}{c}\hfill -{\displaystyle \frac{a^2{L}_2}{3M^2{r}^2}}-{\displaystyle \frac{2ab{L}_1}{3M^2{r}^2}}-{\displaystyle \frac{4\sqrt{2}a{D}_r{L}_4}{3M^{2}r}}-{\displaystyle \frac{4\sqrt{2}a{L}_4}{M^2{r}^2}}-{\displaystyle \frac{2D_r^2{L}_2}{3M^2}}-{\displaystyle \frac{10D_r{L}_2}{3M^{2}r}}-{\displaystyle \frac{6L_2}{M^2{r}^2}}=0,\end{array}$$

(14)

$$\begin{array}{c}\hfill {\displaystyle \frac{b^2{L}_1}{M^2{r}^2}}+{\displaystyle \frac{2\sqrt{2}b{L}_4}{M^2{r}^2}}-{\displaystyle \frac{D_r^2{L}_1}{M^2}}-{\displaystyle \frac{2D_r{L}_1}{M^{2}r}}+{\displaystyle \frac{2\alpha L_1}{Mr}}+{\displaystyle \frac{2L_{1}E}{M}}=0,\end{array}$$

(15)

$$\begin{array}{c}\hfill {\displaystyle \frac{10a^2{L}_2}{M^2{r}^2}}+{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}+{\displaystyle \frac{8\sqrt{2}a{D}_r{L}_4}{M^{2}r}}+{\displaystyle \frac{40\sqrt{2}a{L}_4}{M^2{r}^2}}-{\displaystyle \frac{4D_r^2{L}_2}{M^2}}+{\displaystyle \frac{4D_r{L}_2}{M^{2}r}}\\ \hfill +{\displaystyle \frac{36L_2}{M^2{r}^2}}+{\displaystyle \frac{16\alpha L_2}{Mr}}+{\displaystyle \frac{16L_{2}E}{M}}=0,\end{array}$$

(16)

$$\begin{array}{c}\hfill {\displaystyle \frac{3a^2{L}_4}{M^2{r}^2}}+{\displaystyle \frac{5\sqrt{2}a{L}_2}{M^2{r}^2}}+{\displaystyle \frac{b^2{L}_4}{M^2{r}^2}}+{\displaystyle \frac{2\sqrt{2}b{L}_1}{M^2{r}^2}}-{\displaystyle \frac{4D_r^2{L}_4}{M^2}}-{\displaystyle \frac{8D_r{L}_4}{M^{2}r}}+{\displaystyle \frac{6L_4}{M^2{r}^2}}+{\displaystyle \frac{8\alpha L_4}{Mr}}+{\displaystyle \frac{8L_{4}E}{M}}=0,\end{array}$$

(17)

$$\begin{array}{c}\hfill {\displaystyle \frac{2a^2{L}_2}{M^2{r}^2}}-{\displaystyle \frac{4ab{L}_1}{M^2{r}^2}}-{\displaystyle \frac{8\sqrt{2}a{D}_r{L}_4}{M^{2}r}}-{\displaystyle \frac{8\sqrt{2}a{L}_4}{M^2{r}^2}}-{\displaystyle \frac{8D_r^2{L}_2}{M^2}}-{\displaystyle \frac{28D_r{L}_2}{M^{2}r}}-{\displaystyle \frac{12L_2}{M^2{r}^2}}+{\displaystyle \frac{8\alpha L_2}{Mr}}+{\displaystyle \frac{8L_{2}E}{M}}=0,\end{array}$$

(18)

$$\begin{array}{c}\hfill -{\displaystyle \frac{\sqrt{2}a{S}_2}{Mr}}-{\displaystyle \frac{2\sqrt{2}b{S}_1}{Mr}}-{\displaystyle \frac{4D_r{S}_4}{M}}-{\displaystyle \frac{12S_4}{Mr}}=0,\mathrm{not\; needed}\\ \hfill -{\displaystyle \frac{4\sqrt{2}a{S}_4}{Mr}}-{\displaystyle \frac{4D_r{S}_2}{M}}-{\displaystyle \frac{12S_2}{Mr}}=0,\mathrm{not\; needed}\end{array}$$

$$\begin{array}{c}\hfill {\displaystyle \frac{6a^2{L}_2}{M^2{r}^2}}+{\displaystyle \frac{12ab{L}_1}{M^2{r}^2}}+{\displaystyle \frac{24\sqrt{2}a{D}_r{L}_4}{M^{2}r}}+{\displaystyle \frac{56\sqrt{2}a{L}_4}{M^2{r}^2}}+{\displaystyle \frac{12D_r^2{L}_2}{M^2}}+{\displaystyle \frac{60D_r{L}_2}{M^{2}r}}+{\displaystyle \frac{60L_2}{M^2{r}^2}}=0.\end{array}$$

(19)

Thus, we arrive at the equations for $L_1,L_2,L_4$:

$$\begin{array}{c}\hfill {\displaystyle \frac{2}{3}}{D}_r^2{L}_2+{\displaystyle \frac{10}{3r}}{D}_r{L}_2+{\displaystyle \frac{4\sqrt{2}a}{3r}}{D}_r{L}_4+{\displaystyle \frac{4\sqrt{2}a}{r^2}}{L}_4+{\displaystyle \frac{2ab}{3r^2}}{L}_1+{\displaystyle \frac{a^2+18}{3r^2}}{L}_2=0,\end{array}$$

(20)

$$\begin{array}{c}\hfill 12D_r^2{L}_2+{\displaystyle \frac{60}{r}}{D}_r{L}_2+{\displaystyle \frac{6a^2+60}{r^2}}{L}_2+{\displaystyle \frac{24\sqrt{2}a}{r}}{D}_r{L}_4+{\displaystyle \frac{56\sqrt{2}a}{r^2}}{L}_4+{\displaystyle \frac{12ab}{r^2}}{L}_1=0,\end{array}$$

(21)

$$\begin{array}{c}\hfill -4D_r^2{L}_2+{\displaystyle \frac{4D_r}{r}}{L}_2+16M(E+{\displaystyle \frac{\alpha }{r}})L_2+{\displaystyle \frac{10a^2+36}{r^2}}{L}_2\\ \hfill +{\displaystyle \frac{8\sqrt{2}a}{r}}{D}_r{L}_4+{\displaystyle \frac{40\sqrt{2}a}{r^2}}{L}_4+{\displaystyle \frac{4ab{L}_1}{r^2}}=0,\end{array}$$

(22)

$$\begin{array}{c}\hfill 8D_r^2{L}_2+{\displaystyle \frac{28}{r}}{D}_r{L}_2-8M(E+{\displaystyle \frac{\alpha }{r}})L_2-{\displaystyle \frac{2a^2-12}{r^2}}{L}_2\\ \hfill +{\displaystyle \frac{8\sqrt{2}a}{r}}{D}_r{L}_4+{\displaystyle \frac{8\sqrt{2}a}{r^2}}{L}_4+{\displaystyle \frac{4ab{L}_1}{r^2}}=0,\end{array}$$

(23)

$$\begin{array}{c}\hfill \left[D_r^2{L}_1+{\displaystyle \frac{2}{r}}{D}_r{L}_1-2M(E+{\displaystyle \frac{\alpha }{r}})L_1\right]-{\displaystyle \frac{b^2}{r^2}}{L}_1-{\displaystyle \frac{2\sqrt{2}b}{r^2}}{L}_4=0,\end{array}$$

(24)

$$D_r^2{L}_4+{\displaystyle \frac{8}{r}}{D}_r{L}_4-8M(E+{\displaystyle \frac{\alpha }{r}})L_4-{\displaystyle \frac{3a^2+b^2+6}{r^2}}{L}_4-{\displaystyle \frac{5\sqrt{2}a}{r^2}}{L}_2-{\displaystyle \frac{2\sqrt{2}b}{r^2}}{L}_1=0,$$

let $4L_4={\overline{L}}_4$, then the last equation reads as follows:

$$\begin{array}{c}\hfill \left[D_r^2{\overline{L}}_4+{\displaystyle \frac{2}{r}}{D}_r{\overline{L}}_4-2M(E+{\displaystyle \frac{\alpha }{r}}){\overline{L}}_4\right]-{\displaystyle \frac{3a^2+b^2+6}{4r^2}}{\overline{L}}_4-{\displaystyle \frac{5\sqrt{2}a}{r^2}}{L}_2-{\displaystyle \frac{2\sqrt{2}b}{r^2}}{L}_1=0.\end{array}$$

(25)

Equations (24) and (25) exhibit the desired non-relativistic structure:

$$\left[D_r^2+{\displaystyle \frac{2}{r}}{D}_r-2M(E+{\displaystyle \frac{\alpha }{r}})\right]L_1=\Delta L_1,\left[D_r^2+{\displaystyle \frac{2}{r}}{D}_r-2M(E+{\displaystyle \frac{\alpha }{r}})\right]{\overline{L}}_4=\Delta {\overline{L}}_4.$$

From Equation (22), we express the variable $L_1$ and substitute the result into Equation (23). This leads to

$$\left[D_r^2+{\displaystyle \frac{2}{r}}{D}_r-2M(E+{\displaystyle \frac{\alpha }{r}})\right]L_2-{\displaystyle \frac{a^2{L}_2}{r^2}}-{\displaystyle \frac{6L_2}{r^2}}-{\displaystyle \frac{4\sqrt{2}a{L}_4}{r^2}}=0.$$

Let us now consider Equation (20) and express the variable ${\displaystyle \frac{1}{r}}{D}_r{L}_4$ from it. Substituting this result into Equation (21), we obtain the identity $0\equiv 0$. Therefore, we have derived three equations with the desired structure (remembering that $4L_4={\overline{L}}_4$):

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left[D_r^2+{\displaystyle \frac{2}{r}}{D}_r-2M(E+{\displaystyle \frac{\alpha }{r}})\right]L_1-{\displaystyle \frac{b^2}{r^2}}{L}_1-{\displaystyle \frac{\sqrt{2}b}{2r^2}}{\overline{L}}_4=0,\\ \hfill \left[D_r^2+{\displaystyle \frac{2}{r}}{D}_r-2M(E+{\displaystyle \frac{\alpha }{r}})\right]L_2-{\displaystyle \frac{a^2+6}{r^2}}{L}_2-{\displaystyle \frac{\sqrt{2}a}{r^2}}{\overline{L}}_4=0,\\ \hfill \left[D_r^2+{\displaystyle \frac{2}{r}}{D}_r-2M(E+{\displaystyle \frac{\alpha }{r}})\right]{\overline{L}}_4-{\displaystyle \frac{3a^2+b^2+6}{4r^2}}{\overline{L}}_4-{\displaystyle \frac{5\sqrt{2}a}{r^2}}{L}_2-{\displaystyle \frac{2\sqrt{2}b}{r^2}}{L}_1=0.\end{array}\end{array}$$

(26)

In matrix form, this becomes

$$\Delta =r^2\left[D_r^2+{\displaystyle \frac{2}{r}}{D}_r-2M(E+{\displaystyle \frac{\alpha }{r}})\right],L=\left|\begin{array}{c}{L}_1\\ L_2\\ {\overline{L}}_4\end{array}\right|,$$

$$\Delta L=AL,A=\left|\begin{array}{ccc}{b}^2& 0& {\displaystyle \frac{\sqrt{2}b}{2}}\\ 0& a^2+6& \sqrt{2}a\\ 2\sqrt{2}b& 5\sqrt{2}a& {\displaystyle \frac{3a^2+b^2+6}{4}}\end{array}\right|,a=\sqrt{j(j+1)},b=\sqrt{(j-1)(j+2)},$$

$$F=SL=\left|\begin{array}{c}{F}_1\\ F_2\\ F_3\end{array}\right|,\Delta F=SA{S}^{-1}F,S=\left|\begin{array}{ccc}{s}_{11}& s_{12}& s_{13}\\ s_{21}& s_{22}& s_{23}\\ s_{31}& s_{32}& s_{33}\end{array}\right|,SA{S}^{-1}=A^{\prime }=\left|\begin{array}{ccc}{\lambda }_1& 0& 0\\ 0& {\lambda }_2& 0\\ 0& 0& {\lambda }_3\end{array}\right|.$$

Thus, we obtain $SA=A^{\prime }S$; here, we have a linear system with the same structure: ($i=1,2,3$):

$$s_{i1}\left(b^2-{\lambda }_i\right)+2\sqrt{2}b{s}_{i3}=0,$$

$$s_{i2}\left(a^2-{\lambda }_i+6\right)+5\sqrt{2}a{s}_{i3}=0,$$

$${\displaystyle \frac{b{s}_{i1}}{\sqrt{2}}}+\sqrt{2}a{s}_{i2}+{\displaystyle \frac{1}{4}}{s}_{i3}\left(3a^2+b^2+6-4{\lambda }_i\right)=0.$$

By setting the determinant of the last system to zero, we derive

$$det\left|\begin{array}{ccc}\left(b^2-\lambda \right)& 0& 2\sqrt{2}b\\ 0& \left(a^2-\lambda +6\right)& 5\sqrt{2}a\\ {\displaystyle \frac{b}{\sqrt{2}}}& \sqrt{2}a& {\displaystyle \frac{1}{4}}\left(3a^2+b^2-4\lambda +6\right)\end{array}\right|=0.$$

From this, it follows that

$${\lambda }^3-\left(3j^2+3j+5\right){\lambda }^2+\left(3j^4+6j^3+j^2-2j-4\right)\lambda$$

$$-(j-2)(j-1)(j+2)(j+3)\left(j^2+j+1\right)=0,{\lambda }_1{\lambda }_2{\lambda }_3>0.$$

Using the notation

$$A=\sqrt[3]{6\sqrt{3}\sqrt{-\left(\left(j^2+j+1\right)\left(9j\left(48(j+2)j^2+j-47\right)-847\right)\right)}-378j(j+1)-377},$$

the solutions are given by

$${\lambda }_1=-{\displaystyle \frac{1}{3}}A+{\displaystyle \frac{-36j(j+1)-37}{3A}}+j^2+{\displaystyle \frac{5}{3}},$$

$${\lambda }_2=+{\displaystyle \frac{1}{6}}\left(1-i\sqrt{3}\right)A+{\displaystyle \frac{\left(1+i\sqrt{3}\right)(36j(j+1)+37)}{6A}}+j^2+j+{\displaystyle \frac{5}{3}},$$

$${\lambda }_3={\displaystyle \frac{1}{6}}\left(1+i\sqrt{3}\right)A+{\displaystyle \frac{\left(1-i\sqrt{3}\right)(36j(j+1)+37)}{6A}}+j^2+j+{\displaystyle \frac{5}{3}}.$$

Let us now study the roots numerically for fixed values of $j=1,2,\dots ,10$:

$$\begin{array}{cccc}j=0\hfill & -0.5-1.32288i\hfill & -0.5+1.32288i\hfill & 6\hfill \\ j=1\hfill & 0\hfill & 0.376525\hfill & 10.6235\hfill \\ j=2\hfill & 0\hfill & 5.15571\hfill & 17.8443\hfill \\ j=3\hfill & 2.54917\hfill & 11.2483\hfill & 27.2025\hfill \\ j=4\hfill & 7.10818\hfill & 19.283\hfill & 38.6088\hfill \\ j=5\hfill & 13.6632\hfill & 29.3\hfill & 52.0368\hfill \\ j=6\hfill & 22.2141\hfill & 41.3096\hfill & 67.4763\hfill \\ j=7\hfill & 32.7618\hfill & 55.3156\hfill & 84.9226\hfill \\ j=8\hfill & 45.3072\hfill & 71.3196\hfill & 104.373\hfill \\ j=9\hfill & 59.8508\hfill & 89.3223\hfill & 125.827\hfill \\ j=10\hfill & 76.3931\hfill & 109.324\hfill & 149.283\hfill \end{array}$$

The first line of the table is meaningless. However, it is important to recall that the case $j=0$ is special, as the initial general substitution must be modified. This case should be examined separately.

Instead of the parameter $\lambda$, let us introduce a more convenient parameter, $\lambda =L(L+1)$. With this substitution, the main algebraic equation becomes

$$(j-2)(j-1)(j+2)(j+3)\left(j^2+j+1\right)$$

$$+\left(4-j\left(3(j+2)j^2+j-2\right)\right)L+(9-j(j+1)(3j(j+1)-5))L^2$$

$$+(6j(j+1)+9)L^3+(3j(j+1)+2)L^4-3L^5-L^6=0.$$

From here, the numerical analysis yields the following:

$$\begin{array}{ccccccc}j=1& -3.7975& -1.29153& -1.0& 0.0& 0.291533& 2.7975\\ j=2& -4.75374& -2.82502& -1.0& 0.0& 1.82502& 3.75374\\ j=3& -5.73952& -3.89092& -2.17307& 1.17307& 2.89092& 4.73952\\ j=4& -6.73368& -4.91962& -3.2126& 2.2126& 3.91962& 5.73368\\ j=5& -7.73096& -5.93599& -4.23004& 3.23004& 4.93599& 6.73096\\ j=6& -8.7296& -6.94668& -5.23963& 4.23963& 5.94668& 7.7296\\ j=7& -9.7289& -7.95423& -6.24559& 5.24559& 6.95423& 8.7289\\ j=8& -10.7286& -8.95988& -7.2496& 6.2496& 7.95988& 9.72855\\ j=9& -11.7284& -9.96427& -8.25247& 7.25247& 8.96427& 10.7284\\ j=10& -12.7284& -10.9678& -9.2546& 8.2546& 9.96778& 11.7284\end{array}$$

The last three columns yield positive values for $L_1,L_2,L_3$. Therefore, after performing the linear transformation, we obtain three separate equations for states with parity $P={(-1)}^j$:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left[{\displaystyle \frac{d^2}{d{r}^2}}+{\displaystyle \frac{2}{r}}{\displaystyle \frac{d}{dr}}-2M(E+{\displaystyle \frac{\alpha }{r}})-L_1(L_1+1)\right]F_1=0,\\ \hfill \left[{\displaystyle \frac{d^2}{d{r}^2}}+{\displaystyle \frac{2}{r}}{\displaystyle \frac{d}{dr}}-2M(E+{\displaystyle \frac{\alpha }{r}})-L_2(L_2+1)\right]F_2=0,\\ \hfill \left[{\displaystyle \frac{d^2}{d{r}^2}}+{\displaystyle \frac{2}{r}}{\displaystyle \frac{d}{dr}}-2M(E+{\displaystyle \frac{\alpha }{r}})-L_3(L_3+1)\right]F_3=0.\end{array}\end{array}$$

(27)

The solutions to these equations are well-known [23]. It should be noted that $L_1,L_2,L_3$ are not integers.

6. States with $\mathit{j}=\mathbf{0}$

For the case $j=0$, we present only the final result. It turns out that solutions with parity $P={(-1)}^{j+1}$ do not exist. For states with parity $P={(-1)}^j$, the five independent components are divided into large (L) and small (S) parts as follows:

$$\begin{array}{c}\hfill h=S_1,f_2=L_2+S_2,c_2={\displaystyle \frac{L_2+S_2}{2}},d_2=S_{10},f_0=S_{12}.\end{array}$$

(28)

For states with $j=0$ in the non-relativistic approximation, only one large variable exists: $L_2$.

After performing the required calculations (similar to those described above), we arrive at a second-order equation for the function $L_2\left(r\right)=f\left(r\right)$:

$$\begin{array}{c}\hfill \left[{\displaystyle \frac{d^2}{d{r}^2}}+{\displaystyle \frac{4}{r}}{\displaystyle \frac{d}{dr}}+{\displaystyle \frac{2ME}{3}}+{\displaystyle \frac{2M\alpha }{3}}{\displaystyle \frac{1}{r}}\right]L_2=0.\end{array}$$

(29)

It is solved in terms of confluent hypergeometric functions; the relevant energy spectrum is

$$\begin{array}{c}\hfill E_n=-{\displaystyle \frac{M{\alpha }^2}{6}}{\displaystyle \frac{1}{{(2+n)}^2}},n=0,1,2,\dots .\end{array}$$

(30)

7. Conclusions

In this study, we discussed the non-relativistic energy spectra for a spin-2 particle with an anomalous magnetic moment in an external Coulomb field, starting from the 39-component radial system of equations. By introducing the Coulomb interaction via the substitution $ϵ\to ϵ+\alpha /r$ and eliminating components associated with the vector and antisymmetric tensor parts, we obtained a tractable formulation of the problem. The use of parity constraints led to a natural separation of the system into subsystems of three and eight equations for states with parities $P={(-1)}^{j+1}$ and $P={(-1)}^j$, respectively.

For the subsystem corresponding to $P={(-1)}^{j+1}$, we analyzed three second-order equations and reduced them to two coupled radial equations with a non-relativistic structure. By applying a special linear transformation, we obtained two independent Schrodinger-type equations in the presence of a Coulomb field.

For the subsystem corresponding to $P={(-1)}^j$, we analyzed eight second-order equations and reduced them to three coupled radial equations with a non-relativistic structure. By applying an appropriate linear transformation, we obtained three independent Schrodinger-type equations in the presence of a Coulomb field. Consequently, we obtained five distinct energy spectra corresponding to the non-relativistic spin-2 particle in the presence of a Coulomb field, arising from the general case of nonzero j.

For the special case $j=0$, we presented only the final result. States with parity $P={(-1)}^{j+1}$ do not exist in this case. For states with parity $P={(-1)}^j$, the five independent components of the wave function are divided into large (L) and small (S) components, which correspond to a single Schrodinger equation with a hydrogen-like spectrum.

These findings highlight the effectiveness of the projective operator method in analyzing high-spin systems, and extend the Pauli–Fierz framework by incorporating Coulomb-type interactions. The analytical energy spectra derived here deepen our understanding of spin-2 dynamics and are expected to inform future studies in quantum field theory and gravitational analog models, where spin-2 particles represent composite or emergent excitations.

Future research could include relativistic corrections, interactions with additional external fields, or generalizations to curved spacetime geometries. Potential applications range from fundamental particle physics to low-energy condensed-matter systems, where the behavior of emergent high-spin states remains an open frontier.

Regarding the problem of formulating a Pauli-like equation for a spin-2 particle in curved spacetime, we must begin with the generally covariant matrix equation, generalized according to the Weyl–Fock–Ivanenko tetrad method [17]. The non-relativistic approximation is feasible only for spacetime models with the following metric structure:

$$\begin{array}{c}\hfill d{S}^2={\left(d{x}^0\right)}^2+g_{ij}\left(x\right)d{x}^{i}d{x}^j,e_{\left(a\right)\alpha }\left(x\right)=\left|\begin{array}{cc}1& 0\\ 0& e_{\left(i\right)k}\left(x\right)\end{array}\right|.\end{array}$$

(31)

Accordingly, the components of the relevant connection take the following form:

$$\begin{array}{c}\hfill {\Sigma }_0={\displaystyle \frac{1}{2}}{J}^{ik}{e}_{\left(i\right)}^M\left({\nabla }_0{e}_{\left(k\right)M}\right),{\Sigma }_l={\displaystyle \frac{1}{2}}{J}^{ik}{e}_{\left(i\right)}^M\left({\nabla }_l{e}_{\left(k\right)M}\right),\end{array}$$

(32)

where contributions from the generators $J^{0k}$ vanish identically. With these simplifications, the generally covariant matrix equation for a spin-2 particle in the first-order form [17] becomes

$$\begin{array}{c}\hfill \left(i{\Gamma }^0{\overline{D}}_0+i{\Gamma }^1{D}_1+i{\Gamma }^2{D}_2+{\Gamma }^3{D}_3-M\right)\Psi =0,\end{array}$$

(33)

where the derivative operators are defined as

$$\begin{array}{c}\hfill \begin{array}{c}\hfill D_0=({\partial }_0+ie{A}_0)+J^{23}{\gamma }_{\left[23\right]0}+J^{31}{\gamma }_{\left[31\right]0}+J^{12}{\gamma }_{\left[12\right]0},\\ \hfill D_1=e_{\left(1\right)}^k({\partial }_k+ie{A}_k)+J^{23}{\gamma }_{\left[23\right]1}+J^{31}{\gamma }_{\left[31\right]1}+J^{12}{\gamma }_{\left[12\right]1},\\ \hfill D_2=e_{\left(2\right)}^k({\partial }_k+ie{A}_k)+J^{23}{\gamma }_{\left[23\right]2}+J^{31}{\gamma }_{\left[31\right]2}+J^{12}{\gamma }_{\left[12\right]2},\\ \hfill D_3=e_{\left(3\right)}^k({\partial }_k+ie{A}_k)+J^{23}{\gamma }_{\left[23\right]3}+J^{31}{\gamma }_{\left[31\right]3}+J^{12}{\gamma }_{\left[12\right]3},\end{array}\end{array}$$

(34)

The generators have dimensions of $39\times 39$, and the Ricci rotation coefficients ${\gamma }_{\left[ij\right]a}$ are used. To distinguish the large and small constituents of the wave function, we employ projective operators constructed from the minimal polynomial of order 7 for the matrix ${\Gamma }_{39\times 39}^0$.

During the non-relativistic approximation, the rest energy is separated by substituting $D_0\to M+i{D}_0$, where $M>0$. The wave function components and derivative operators are assumed to satisfy the following smallness orders:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill L\sim 1,S_k\sim x,s_k\sim x,{\displaystyle \frac{1}{M}}{D}_j\sim x,\\ \hfill {\displaystyle \frac{1}{M}}i{D}_0\sim x^2,{\gamma }_{\left[ij\right]0}\sim x^2,{\displaystyle \frac{1}{M}}{D}_j\sim x,{\displaystyle \frac{1}{M}}{\gamma }_{\left[ij\right]j}\sim x;\end{array}\end{array}$$

(35)

noting the smallness orders for the Ricci rotation coefficients. These observations set the groundwork for deriving a Pauli-like equation for a spin-2 particle in curved spacetime within the non-relativistic approximation. This will be the subject of future work.