IntroductionIn the literature, the great interest can be noticed instudying the Dirac-K¨ahler field [1,2]; also see [3–10]. Thisfield describes the complicated boson which contains thefields with different parities and spins S = 1, S = 0;the complete wave function includes the scalar, pseudoscalar,vector, pseudovector, and antisymmetric tensor components(pseudo-quantities are marked by symbol of tilde):Φ, ˜Φ,Φa, ˜Φa,Φab. In the frames of the general theory ofrelativistic wave equations [16–20], the Dirac-K¨ahler field describesa particle with a set of spin states: S = 1 and S = 0(see [11–15]).From this theory, by imposing additional constraints onthe 16 components of the Dirac-K¨ahler field, we can obtainthe usual theories for scalar and pseudoscalar particles, andfor vector and pseudovector particles:(Φ, 0,Φa, 0, 0), S = 0;(0, ˜Φ, 0, ˜Φa, 0), S = ˜0;Известия Коми научного центра УрО РАН, серия «Физико-математические науки» № 5 (57), 2022www.izvestia.komisc.ru 69(0, 0,Φa, 0,Φab), S = 1;(0, 0, 0, ˜Φa,Φab), S = ˜1.Also the system of equations describing the St¨uckelberg particleis well known [1, 2]. In particular, this system can be obtainedfrom the Dirac-K¨ahler theory by imposing the followingconstraints(Φ, 0,Φa, 0,Φab), S = 0, 1;(0, ˜Φ, 0, ˜Φa,Φab), S = ˜0,˜1.There are possible two theories, corresponding to differentinternal parities of the particle. In this paper the first variantof the St¨uckelberg theory is studied.We start with the known St¨uckelberg tensor system of 11equations, which is transformed to a matrix form with the useof the tetrad method [11, 12]. This equation is detailed in cylindricsystem of coordinates and corresponding tetrad; whereinwe take into account the external uniform magnetic field. Weperform the separation of the variables and derive the systemof 11 equations in r variable. To resolve this system weapply the method of Fedorov-Gronskiy [21], which is based onprojective operators referring to the third projection of the11-dimensional spin matrix. According to this method we canexpress all 11 functions through only three ones. We find 3 seriesof physically interpretable energy levels, as solutions ofalgebraic equations of order 1 and 3.1. The basic equationThe initial St¨uckelberg system of equations is the following−DaΨa − μΨ = 0,DaΨ + DbΨab − μΨa = 0,DaΨb − DbΨa − μΨab = 0,where Da = ∂a +ieAa. The relation of the parameter μ tophysical mass of the particle M is given by μ = −M. Weuse the 11-dimensional wave function in the formΦ = (Ψ;Ψ0,Ψ1,Ψ2,Ψ3;Ψ01,Ψ02,Ψ03,Ψ23,Ψ31,Ψ12) = (H,H1,H2).The above system can be presented in the matrix block formDaGaH1 + μH = 0,ΔaDaH + KaDaH2 − μH1 = 0,DaLaH1 − μH2 = 0,or (note the minus sign in front of Ga)(−DaΓa − μ)Φ = 0,Γa =0@0 −Ga 0Δa 0 Ka0 La 01A, Φ =0@HH1H21A, (1)whereΔ0 = (1, 0, 0, 0)t, Δ1 = (0, 1, 0, 0)t,Δ2 = (0, 0, 1, 0)t, Δ3 = (0, 0, 0, 1)t,K0 =0B@0 0 0 0 0 0−1 0 0 0 0 00 −1 0 0 0 00 0 −1 0 0 01CA,K1 =0B@−1 0 0 0 0 00 0 0 0 0 00 0 0 0 0 10 0 0 0 −1 01CA,K2 =0B@0 −1 0 0 0 00 0 0 0 0 −10 0 0 0 0 00 0 0 1 0 01CA,K3 =0B@0 0 −1 0 0 00 0 0 0 1 00 0 0 −1 0 00 0 0 0 0 01CA,L0 =0BBBBB@0 1 0 00 0 1 00 0 0 10 0 0 00 0 0 00 0 0 01CCCCCA,L1 =0BBBBB@−1 0 0 00 0 0 00 0 0 00 0 0 00 0 0 −10 0 1 01CCCCCA,L2 =0BBBBB@0 0 0 0−1 0 0 00 0 0 00 0 0 10 0 0 00 −1 0 01CCCCCA,L3 =0BBBBB@0 0 0 00 0 0 0−1 0 0 00 0 −1 00 1 0 00 0 0 01CCCCCA.Here and below, t stands for transposition.This matrix St¨uckelberg equation can be extended to theRiemannian space-time in accordance with the known procedure.To this end, for any given metric gαβ(x) we shouldtake the certain tetrad:gαβ(x) → e(a)α(x), ηab = diag(1,−1,−1,−1),then the equation (1) should have the structureΓα(x)∂∂xα + Σα(x)− μΨ(x) = 0. (2)70Известия Коми научного центра УрО РАН, серия «Физико-математические науки» № 5 (57), 2022www.izvestia.komisc.ruLocal matrices Γα(x) and their blocks are determined withthe use of the tetradΓα(x) = eαa(x)Γa =0@0 −Gaeα (a) 0Δaeα(a) 0 Kaeα (a)0 Laeα(a) 01A.The connection Σα(x) is defined by the formulasJab =0@0 0 00 Jab1 00 0 Jab21A,Σα(x) =12Jabeβ(a)(x)e(b)β;α(x) ==0@0 0 00 (Σ1)α 00 0 (Σ2)α1A,Σ1(x) =12Jab(1)eβ(a)(x)e(b)β;α(x),Σ2(x) =12Jab(2)eβ(a)(x)e(b)β;α(x),where Jab(1) and Jab(2) designate generators for vector Ψk(x)and antisymmetric tensor Ψ[mn](x), respectively. Equation(2) may be presented with the use of the Ricci rotation coefficientsΓceα(c)∂∂xα +12Jabγabc− μΨ(x) = 0. (3)Recall that γ[ab]c = −γ[ba]c = e(b)ρσeρ(a)eσ(c). In detailedform, Eq. (3) reads−Gceα(c)∂α − GcJab(1)12γabcH1 − μH = 0,Δceα(c)∂αH+Kceα (c)∂α + KcJab(2)12γabcH2−μH1 = 0,Lceα(c)∂α + LcJab(1)12γabcH1 − μH2 = 0.Let us consider the St¨uckelberg equation in presence ofthe external uniform magnetic field. In cylindrical coordinateswith the use of the diagonal tetradxα = (t, r, ϕ, z), dS2 = dt2 − dr2 − r2dϕ2 − dz2,Aϕ = −Br22the above equation takes the form (let eB ⇒ B):Γ0 ∂∂t+ Γ1 ∂∂r+ Γ2 ∂ϕ + iBr2/2 + J12r++Γ3 ∂∂z− μΨ = 0.In block form, it reads−G0 ∂∂t− G1 ∂∂r− G2 1r∂∂ϕ+iBr22+ j121−−G3 ∂∂zH1 − μH = 0,Δ0 ∂∂t+ Δ1 ∂∂r++ Δ2 1r∂ϕ +iBr22+ Δ3 ∂∂zH++"K0 ∂∂t+ K1 ∂∂r+ K2 ∂ϕ + iBr22 + j121r++ K3 ∂∂zH2 = μH1,"L0 ∂∂t+ L1 ∂∂r+ L2 ∂ϕ + iBr22 + j121r++ L3 ∂∂zH1 = μH2.2. Cyclic basisIn the following, it will be convenient to apply the cyclicbasis (all quantities referring to it are marked by the overline).In such a basis, the generators j121 and j122 are diagonal. Thenecessary transformation ¯H1 = UH1 is determined by thematrix U :U =0BB@1 0 0 00 −√12√i200 0 0 10 √12√i201CCA,U−1 =0BB@1 0 0 00 −√120 √120 −√i20 −√i200 0 1 01CCA.Correspondingly, the generators for tensor representation aredefined by the rule¯ Jab1 = UjabU−1, ¯ Jab2 = ¯j ab ⊗ I + I ⊗¯j ab.Let us transform the generators to cyclic form¯j12 =0B@0 0 0 00 −i 0 00 0 0 00 0 0 i1CA,¯ J122 =0BBBBB@−1 · · · · ·· 0 · · · ·· · 1 · · ·· · · 1 · ·· · · · 0 ·· · · · · −11CCCCCA.Известия Коми научного центра УрО РАН, серия «Физико-математические науки» № 5 (57), 2022www.izvestia.komisc.ru 71We should also transform the main matrices Γa of the equationto the cyclic form. Starting with the formulas¯H= H, ¯H1 = C1H1, (C1 = U),¯H2 = U ⊗ UH2 = C2H2,we derive the rule0@0 −¯Ga 0¯Δa 0 ¯K a0 ¯La 01A ==0@0 −GaC−11 0C1Δa 0 C1KaC−120 C2LaC−11 01A.First, we find the matrices ¯Δa = C1Δa and ¯Ga =GaC−11 :¯Δ0 = (1, 0, 0, 0)t, ¯Δ1 =0,−√12, 0,√12t¯Δ2 =0,√i2, 0,√i2t, ¯Δ3 = (0, 0, 1, 0)t;¯G0 = (1, 0, 0, 0), ¯G1 =0,√12, 0,−√12,¯G2 =0,√i2, 0,√i2, ¯G3 = (0, 0,−1, 0).Having in mind the formula for C1, we can derive expressionsfor 6-dimensional transformation for C2:U ⊗ U ⇒ C2, ¯H2 = C2H2 ⇒C2 =0BBBBBBB@−√12√i20 0 0 00 0 1 0 0 0√12√i20 0 0 00 0 0 −√i2√1200 0 0 0 0 i0 0 0 √i2√1201CCCCCCCA,C−12 =0BBBBBBB@−√120 √120 0 0−√i20 −√i20 0 00 1 0 0 0 00 0 0 √i20 −√i20 0 0 √120 √120 0 0 0 −i 01CCCCCCCA.With the use of them we can obtain all other blocks in cyclicform:¯K0 =0B@0 0 0 0 0 0−1 0 0 0 0 00 −1 0 0 0 00 0 −1 0 0 01CA,¯K1 =0BBB@√120 −√120 0 00 0 0 0 √1200 0 0 −√120 −√120 0 0 0 √1201CCCA,¯K2 =0BBB@√i20 √i20 0 00 0 0 0 −√i200 0 0 √i20 −√i20 0 0 0 √i201CCCA,¯K3 =0B@0 −1 0 0 0 00 0 0 0 0 −10 0 0 0 0 00 0 0 1 0 01CA;¯L0 =0BBBBB@0 1 0 00 0 1 00 0 0 10 0 0 00 0 0 00 0 0 01CCCCCA,¯L1 =0BBBBBBB@√120 0 00 0 0 0−√120 0 00 0 −√1200 √120 √120 0 −√1201CCCCCCCA,¯L2 =0BBBBBBB@−√i20 0 00 0 0 0−√i20 0 00 0 −√i200 √i20 −√i20 0 √i201CCCCCCCA,¯L3 =0BBBBB@ 0000−1 0 0 00 0 0 00 0 0 10 0 0 00 −1 0 01CCCCCA.3. Separating the variablesWe apply the following substitution for the wave function(in cyclic basis)¯Ψ= e−iϵteimϕeikz0@¯H¯H1¯H21A, ¯H = h(r),¯H1 =0B@h0(r)h1(r)h2(r)h3(r)1CA, ¯H2 =Ei(r)Bi(r).After a simple calculation we derive the system of 11 equations.With the use of notationsam =ddr+m + Br2/2r, am+1 =ddr+m + 1 + Br2/2r,bm =ddr−m + Br2/2r, bm−1 =ddr−m − 1 + Br2/2r,72Известия Коми научного центра УрО РАН, серия «Физико-математические науки» № 5 (57), 2022www.izvestia.komisc.ruit reads−iϵh0 − ikh2 + bm−1h1 − am+1h3 = −μh,−iϵh − ikE2 + bm−1E1 − am+1E3 = μh0,−amh + am+1B2 − ikB3 + iϵE1 = μh1,ikh + iϵE2 − am+1B1 − bm−1B3 = μh2,bmh + bmB2 + ikB1 + iϵE3 = μh3,amh0 − iϵh1 = μE1, −ikh0 − iϵh2 = μE2,−bmh0 − iϵh3 = μE3, −bmh2 + ikh3 = μB1,bm−1h1 + am+1h3 = μB2, −ikh1 − amh2 = μB3.4. The Fedorov-Gronskiy methodWe will apply the Fedorov-Gronskiy method [21]. To thisend, let us consider the third projection of 11-dimensional spinoperator Y = −i ¯ J12. We verify that it satisfies the minimalcubic equation, Y (Y − 1)(Y + 1) = 0. This permits us tointroduce three projective operatorsP1 =12Y (Y − 1), P2 =12Y (Y + 1), P3 = 1 − Y 2with the properties P20 = P0, P2+1 = P+1, P2−1 = P−1,P0 + P+1 + P−1 = 1.Therefore, the complete wave function may be decomposedinto the sum of three partsΨ = Ψ0 + Ψ+1 + Ψ−1,Ψσ = PσΨ, σ = 0, +1,−1.We can readily find an explicit form of them (according to theFedorov-Gronskiy method, each projective part should be determinedby only one function)Ψ1(r) = (0, 0, h1, 0, 0,E1, 0, 0, 0, 0,B3)tf1(r),Ψ2(r) = (0, 0, 0, 0, h3, 0, 0,E3,B1, 0, 0)tf2(r),Ψ3(r) = (h, h0, 0, h2, 0, 0,E2, 0, 0,B2, 0)tf3(r).Applying the projective operators to the system of 11 equations,Pi(A10×10Ψ) = 0, we obtainfor P1−amh + amB2 − ikB3 + iϵE1 = μh1,amh0 − iϵh1 = μE1,−ikh1 − amh2 = μB3;for P2bmh + bmB2 + ikB1 + iϵE3 = μh3,−bmh0 − iϵh3 = μE3,−bmh2 + ikh3 = μB1;for P3−iϵh0 − ikh2 + bm−1h1 − am+1h3 = μh,−iϵh − ikE2 + bm−1E1 − am+1E3 = μh0,ikh + iϵE2 − am+1B1 − bm−1B3 = μh2,−ikh0 − iϵh2 = μE2,bm−1h1 + am+1h3 = μB2.Besides, in accordance with the Fedorov-Gronskiy method,we impose the first order constraints which permit us totransform all differential equations into algebraic ones:for P1− amf3(r)h + amf3(r)B2 − ikf1(r)B3++ iϵf1(r)E1 = μf1(r)h1 ⇒ amf3 = C1f1,amf3(r)h0 − iϵf1(r)hi == μf1(r)E1 ⇒ amf3 = C1f1,− ikf1(r)h1 − amf3(r)h2 == μf1(r)B3 ⇒ amf3 = C1f1,for P2bmf3(r)h + bmf3(r)B2 + ikf2(r)B1++ iϵf2(r)E3 = μf2(r)h3 ⇒ bmf3 = C2f2,− bmf3(r)h0 − iϵf2(r)h3 == μf2(r)E3 ⇒ bmf3 = C2f2,− bmf3(r)h2 + ikf2(r)h3 == μf2(r)B1 ⇒ bmf3 = C2f2,for P3− iϵf3(r)h0 − ikf3(r)h2 + bm−1f1(r)h1−− bm−1f1(r)h3 = μf3(r)h ⇒ bm−1f1 = C3f3,− iϵf3(r)h − ikf3(r)E2 + bm−1f1(r)E1−− am+1f2(r)E3 = μf3(r)h0 ⇒⇒ bm−1f1 = C3f3, am+1f2 = C4f3,ikf3(r)h = iϵf3(r)E2 − am+1f2(r)B1−− bm−1f1(r)B3 = μf3(r)h2 ⇒⇒ bm−1f1 = C3f3, am+1f2 = C4f3,− ikf3(r)h0 − iϵf3(r)h2 = μf3(r)E2,bm−1f1(r)h1 + am+1f2(r)h3 = μf3(r)B2 ⇒⇒ bm−1f1 = C3f3, am+1f2 = C4f3.Thus, we have derived the algebraic equations−C1h + C1B2 − ikB3 + iϵE1 = μh1,C1h0 − iϵh1 = μE1, −ikh1 − C1h2 = μB3,C2h + C2B2 + ikB1 + iϵE3 = μh3,Известия Коми научного центра УрО РАН, серия «Физико-математические науки» № 5 (57), 2022www.izvestia.komisc.ru 73−C2h0 − iϵh3 = μE3, −C2h2 + ikh3 = μB1,−iϵh0 − ikh2 + C3h1 − C3h3 = μh,−iϵh − ikE2 + C3E1 − C4E3 = μh0,ikh + iϵE2 − C4B1 − C3B3 = μh2,−ikh0 − iϵh2 = μE2, C3h1 + C4h3 = μB2,and have the following constraintsbm−1f1(r) = C3f3(r), amf3(r) = C1f1(r),am+1f2(r) = C4f3(r), bmf3(r) = C2f2(r).(4)From Eqs. (4) we derive the second order equations for separatefunctions:bm−1amf3 = C1C3f3, ambm−1f1 = C1C3f1,am+1bmf3 = C2C4f3, bmam+1f2 = C2C4f2.Evidently, within these pairs we can setC3 = C1, C4 = C2.Therefore, the differential constraints and second order equationstake on the formbm−1f1(r) = C1f3, amf3 = C1f1,am+1f2(r) = C2f3, bmf3 = C2f2;[bm−1am − C21 ]f3 = 0, [ambm−1 − C21 ]f1 = 0,[am+1bm − C22 ]f3 = 0, [bmam+1 − C22 ]f2 = 0.(5)In explicit form, Eqs. (5) readd2dr2 +1rddr− B2r24− m2r2−−Bm + B − C21f3 = 0,d2dr2 +1rddr− B2r24− (m − 1)2r2−−Bm − C21f1 = 0,d2dr2 +1rddr− B2r24− m2r2−−Bm − B − C22f3 = 0,d2dr2 +1rddr− B2r24− (m + 1)2r2−−Bm − C22f2 = 0.So we get the following identity C22 = C21− 2B and onlythree different equations:(1)d2dr2 +1rddr− B2r24− m2r2−− Bm + B − C21f3 = 0,(2)d2dr2 +1rddr− B2r24−−(m − 1)2r2− Bm − C21f1 = 0,(3)d2dr2 +1rddr− B2r24−−(m + 1)2r2− Bm − C21 + 2Bf2 = 0.Let B − C21 = X, then these equations are written in amore symmetrical form(1)d2dr2 +1rddr− B2r24− m2r2− Bm + Xf3 = 0,(2)d2dr2 +1rddr− B2r24−−(m − 1)2r2− B(m + 1) + Xf1 = 0,(3)d2dr2 +1rddr− B2r24−−(m + 1)2r2− B(m − 1) + Xf2 = 0.With the new variable x = Br22 , they take on the form(1)d2dx2 +1xddx− 14− (m/2)2x2 ++1x−m2+X2Bf3 = 0,(2)d2dx2 +1xddx− 14− [(m − 1)/2]2x2 ++1x−m + 12+X2Bf1 = 0,(3)d2dx2 +1xddx− 14− [(m + 1)/2]2x2 ++1x−m − 12+X2Bf2 = 0.It is sufficient to consider only the first equation in detail. Letus search for solutions in the form f3(x) = xAeCxF(x),then we readily obtainxF′′+(2A+1+2Cx)F′+A2 − (m/2)2x+ 2AC++C − m2+X2B+ xC2 − 14F = 0.Let us impose restrictionsA2 − (m/2)2 = 0 ⇒ A = ±|m/2|,74Известия Коми научного центра УрО РАН, серия «Физико-математические науки» № 5 (57), 2022www.izvestia.komisc.ruC2 − 14= 0 ⇒ C + ±12.In order to have the equations referring to bound the states,we should assumeA = ±|m/2|, C = −12.This results in the equation of a confluent hypergeometrictypexF′′+(|m|+1−x)F′−|m| + m2+12− X2BF = 0with parametersa =|m| + m2+12− X2B,c = |m| + 1, F = Φ(a, c, x).The polynomial condition a = −n1 leads to(3) ⇒ X = +2B|m| + m2+12+ n1> 0,n1 = 0, 1, 2, . . . .The following solutions correspond to this spectrum(3) f3(x) = x+|m|2 x−x/2F1(x),F1(x) = Φ(−n1, |m| + 1, x).Two other equations give similar results. Thus, we have(3) f3(x) = x+|m|2 x−x/2F1(x),F3(x) = Φ(−n1, |m| + 1, x),X = 2B|m| + m2+12+ n1> B,n3 = 0, 1, 2, . . .(6)(1) f1(x) = x+|m−1|2 x−x/2F2(x),F1(x) = Φ(−n2, |m − 1| + 1, x),X = 2B|m − 1| + m + 12+12+ n2> B,n3 = 0, 1, 2, . . .(7)(2) f2(x) = x+|m+1|2 x−x/2F3(x),F2(x) = Φ(−n3, |m + 1| + 1, x),X = 2B|m + 1| + m − 12+12+ n3> B,n2 = 0, 1, 2, . . .(8)The quantity X in all three cases (6)–(8) should be thesame which assumes existence of some correlations withinn − 1, n2, n3. Bellow we will apply the simplest quantizationruleX = 2BN > 0, N =|m| + m2+12+ n,N =12,32, · · · .5. Solving the algebraic systemLet us turn to the algebraic equations−iϵh0 − ikh2 + C3h1 − C3h3 = −μh,−iϵh − ikE2 + C3E1 − C4E3 = μh0,−C1h + C1B2 − ikB3 + iϵE1 = μh1,ikh + iϵE2 − C4B1 − C3b3 = μh2,C2h + C2B2 + ikB1 + iϵE3 = μh3,C1h0 − iϵh1 = μE1, −ikh0 − iϵh2 = μE2,−C2h0 − iϵh3 = μE3, −C2h2 + ikh3 = μB1,C3h1 + C4h3 = μB2, −ikh1 − C1h2 = μB3.Recall thatC1 = C3 =√X − B, C2 = C4 =√X + B.We can present the above system in the matrix formAΨ = 0.As its determinant vanishes we get the equationμ3(k2+μ2−2X −ϵ2)h−2B2(5k2+μ2−2X −5ϵ2)++B(−k2−μ2+2X+ϵ2)(2pX2 − B2−k2−μ2+ϵ2)−−(k2 + μ2 − 2X − ϵ2)2pX2 − B2 − k2++μ2 + X + ϵ2i= 0.This equation is factorized, P8 = P2P6 :k2 + μ2 − 2X − ϵ2 = 0 ⇒ ϵ2 − μ2 = k2 − 2X.The second equation with the use of the quantityW = ϵ2 −k2 reads as follows−W3 +W2−pX2 − B2 + B + μ2 − 5X++Wh2BpX2 − B2 − μ2 + X−−(2X − μ2)2pX2 − B2 + μ2 + 4X+ 10B2i++(2X − μ2)hB2pX2 − B2 − μ2−−(2X − μ2)pX2 − B2 + μ2 + X+ 2B2i= 0.With dimensionless variablesWμ2⇒ w,Xμ2⇒ x,Bμ2⇒ b,μμ⇒ 1,Известия Коми научного центра УрО РАН, серия «Физико-математические науки» № 5 (57), 2022www.izvestia.komisc.ru 75it takes on the formw3 − w2−px2 − b2 + b + 1 − 5x−−wh2bpx2 − b2 − 1 + x−−(2x − 1)2px2 − b2 + 1 + 4x+ 10b2i++(1 − 2x)hb2px2 − b2 − 1++(1 − 2x)px2 − b2 + 1 + x+ 2b2i= 0.Its analytical solutions are found straightforwardly, but theyare helpless. By this reason, let us study its solutions numerically.Recall that x = 2bN. First let us consider the simplecase w = 1−2x = 1−4bN, 1−2x > 0. For several typicalexamples we find the roots forw (physically interpretableare only positive ones):b = 0.01(0.98, 0.94, 0.9, 0.86, 0.82, 0.78, 0.74, 0.7)t,b = 0.05(0.9, 0.7, 0.5, 0.3, 0.1,−0.1,−0.3,−0.5)t,b = 0.1(0.8, 0.4, 0.,−0.4,−0.8,−1.2,−1.6,−2)t.Now let us examine three roots of the third order equation:b = 0.0010BBBBBBBBB@−1 0.996002 1.−1.00483 0.992007 0.995996−1.0089 0.988011 0.991992−1.01293 0.984015 0.987988−1.01695 0.980019 0.983984−1.02096 0.976023 0.97998−1.02496 0.972027 0.975976−1.02897 0.968031 0.9719721CCCCCCCCCA,b = 0.010BBBBBBBBB@−1.0002 0.960204 1.−1.04858 0.920701 0.959595−1.0893 0.88113 0.919181−1.1296 0.841564 0.878757−1.16977 0.802008 0.838324−1.20988 0.762461 0.797879−1.24996 0.722926 0.757423−1.29002 0.683403 0.7169551CCCCCCCCCA,b = 0.050BBBBBBBBB@−1.00554 0.805539−1.25012 0.619508−1.45486 0.433263−1.65743 0.249873−1.85931 0.0735313−2.06088 −0.0934233 − 0.0312522i−226227 −0.2929 − 0.0548385i−2.46357 −0.49238 − 0.0709556i10.789190.5766510.3611490.138565−0.0934233 + 0.0312522i−0.2929 + 0.0548385i−0.49238 + 0.0709556i1CCCCCCCCCA.In the second case, we can see only two positive roots. Intotal, we have 3 physically interpretable roots and the correspondingenergy series.ConclusionFor better understanding of the problem, we will considerthe nonrelativistic approximation for this model in a separatepaper.