IntroductionIn the context of existence of the similar neutrinos of differentmasses, we examine a possibility within the theory ofrelativistic wave equations to describe particles with severalmass parameters. In general, existence of more general waveequations than commonly used ones is well known within theGel’fand-Yaglom formalism – see references [1-5].In particular, models for a spin 1/2 particle with two andthree mass parameters were developed [6–15]. Specific featuresof these models are as follows. For two (or three)bispinors, in absence of external fields separate Dirac-likeequations are derived, they differ in masses. However, inpresence of external electromagnetic field or gravitationalfield with non-vanishing Ricci scalar, the wave equation forbispinors does not split into separated equations, instead aquite definite mixing of two (or three) equations arises. It wasshown that generalized equations for Majorana particle withseveral mass parameters exist as well. Such generalized Majoranaequations are not trivial if the Ricci scalar does notvanish.In the present paper, the model of a fermion with threemass parameters is studied in presence of the external uniformmagnetics field. After applying the diagonalizing trans-Известия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ru77formation, three separate equations are obtained effectivelyfor particles with different anomalous magnetic moments.Their exact solutions and generalized energy spectra arefound.1. General theory in presence of electromagneticand gravitational fieldsWe start with the system of equations for a fermion with3 mass parameters [12, 15]:iγα(x)[∂α + Γα(x) + ieAα(x)]Φ1(x)−−M1Φ1(x) + Y1Σ(x)Φ(x) = 0,iγα(x)[∂α + Γα(x)ieAα(x)]Φ2(x)−−M2Φ2(x) + Y2Σ(x)Φ(x) = 0,iγα(x)[∂α + Γα(x) + ieAα(x)]Φ3(x)−−M3Φ3(x) + Y3Σ(x)Φ(x) = 0, (1)where the notations are usedY1 =4c33Mc2(λ1 − c2), i = 1, 2, 3, L = c1 +√c36,Φ(x) = L1Φ1(x) + L2Φ2(x) + L3Φ3(x),Σ(x) = −ieFαβσαβ(x) +14R(x),L1 =−L|c4|2 − L|c3|2 + c22− c2(λ2 + λ3) + λ2λ3Lc2c3(λ1 − λ2)(λ1 − λ3),L2 =−L|c4|2 − L|c3|2 + c22− c2(λ3 + λ1) + λ3λ1Lc2c3(λ2 − λ3)(λ2 − λ1),L3 =−L|c4|2 − L|c3|2 + c22− c2(λ1 + λ2) + λ1λ2Lc2c3(λ3 − λ1)(λ3 − λ2).With the notation |c4| = a, |c3| = b, the mixing matrix inthe equation (1) is specified by the relationsY1L1 =43M(λ1 − c2)××−L(a2 + b2) + c22− c2(λ2 + λ3) + λ2λ3L(λ1 − λ2)(λ1 − λ3),Y1L2 =43M(λ1 − c2)××−L(a2 + b2) + c22− c2(λ3 + λ1) + λ3λ1L(λ2 − λ3)(λ2 − λ1),Y1L3 =43M(λ1 − c2)××−L(a2 + b2) + c22− c2(λ1 + λ2) + λ1λ2L(λ3 − λ1)(λ3 − λ2),Y2L1 =43M(λ2 − c2)××−L(a2 + b2) + c22− c2(λ2 + λ3) + λ2λ3L(λ1 − λ2)(λ1 − λ3),Y2L2 =43M(λ2 − c2)××−L(a2 + b2) + c22− c2(λ3 + λ1) + λ3λ1L(λ2 − λ3)(λ2 − λ1),Y2L3 =43M(λ2 − c2)××−L(a2 + b2) + c22− c2(λ1 + λ2) + λ1λ2L(λ3 − λ1)(λ3 − λ2),Y3L1 =43M(λ3 − c2)××−L(a2 + b2) + c22− c2(λ2 + λ3) + λ2λ3L(λ1 − λ2)(λ1 − λ3),Y3L2 =43M(λ3 − c2)××−L(a2 + b2) + c22− c2(λ3 + λ1) + λ3λ1L(λ2 − λ3)(λ2 − λ1),Y3L3 =43M(λ3 − c2)××−L(a2 + b2) + c22− c2(λ1 + λ2) + λ1λ2L(λ3 − λ1)(λ3 − λ2).We will consider eq. (1) in the cylindrical coordinates andtetraddS2 = dt2 − dr2 − r2dϕ2 − dz2,xα = (t, r, ϕ, z),eβ(a)(x) =0B@1 0 0 00 1 0 00 0 1/r 00 0 0 11CA.Ricci rotation coefficients are as follows: γab0 = 0, γab1 =0, γ122 = −γ212 = 1/r, γab3 = 0. The external magneticfield directed along the axis x3 is determined as followsAϕ = −12Br2, F12(x) = Frϕ = −Br,−ieFαβ(x) = −ieF12(x)γ1(x)γ2(x) = ieBγ1γ2 == ieB−iσ3 00 −iσ3= eBΣ.So, the main equation (1) takes the form (we simplify the notation,eB ⇒ B )iγ0 ∂∂t+ iγ1 ∂∂r+ iγ2r∂∂ϕ+ iBr22+ iγ3 ∂∂zΦ1−−MΦ1 + BΣY1(L1Φ1 + L2Φ2 + L3Φ3) = 0,78Известия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ruiγ0 ∂∂t+ iγ1 ∂∂r+ iγ2r∂∂ϕ+ iBr22+ iγ3 ∂∂zΦ2−−MΦ2 + BΣY2(L1Φ1 + L2Φ2 + L3Φ3) = 0,iγ0 ∂∂t+ iγ1 ∂∂r+ iγ2r∂∂ϕ+ iBr22+ iγ3 ∂∂zΦ3−−MΦ3 + BΣY3(L1Φ1 + L2Φ2 + L3Φ3) = 0.For three involved bispinors, we will use the following substitutionsΦ1 = e−iϵteimφeikz(f1, f2, f3, f4)t,Φ2 = e−iϵteimφeikz(g1, g2, g3, g4)t,Φ3 = e−iϵteimφeikz(h1, h2, h3, h4)t,where ()t stands for transpose. Then the previous systemreadsϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ1++(M −M1)Φ1 + BΣY1(L1Φ1 + L2Φ2 + L3Φ3) = 0,ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ2++(M −M2)Φ2 + BΣY2(L1Φ1 + L2Φ2 + L3Φ3) = 0,ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ3++(M −M3)Φ3 + BΣY3(L1Φ1 + L2Φ2 + L3Φ3) = 0,whereμ(r) = m +12Br2, Σ =0B@1 0 0 00 −1 0 00 0 1 00 0 0 −11CA.Using the shortening notations for elements of the mixing matrixZkj = (BYk)Lj = dkLj ,we present the system as followsb∂Φk + ZkjΦj = 0,b∂ = Σ−1ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −M.The mixing matrix should be reduced to a diagonal form bya linear transformationΦ = SΦ, b∂Φ = SZS−1Φ,SZS−1 =0@μ1 0 00 μ2 00 0 μ31A.As a result we getΣ−1 b∂0@Φ1Φ2Φ31A +0@μ1 0 00 μ2 00 0 μ31A0@Φ1Φ2Φ31A = 0,orϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ1+μ1ΣΦ1 = 0,ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ2+μ2ΣΦ2 = 0,ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ3+μ3ΣΦ3 = 0.(2)Thus, after transformation, three separate equations are obtainedeffectively for particles with different anomalous magneticmoments.2. Solving the basic equationLet us briefly describe the procedure for solving the basicequation (2):ϵγ0 + iγ1 ∂∂r− γ2rσ(r) − kγ3 −M + ΓZ0B@f1f2f3f41CA= 0.Using the spinor basis for the Dirac matrices [3, 6], we obtainfour equations−iddr+ μf4 + (ϵ + k)f3 + (Γ −M)f1 = 0,−iddr− μf3 + (ϵ − k)f4 − (Γ +M)f2 = 0,iddr+ μf2 + (ϵ − k)f1 + (Γ −M)f3 = 0,iddr− μf1 + (ϵ + k)f2 − (Γ +M)f4 = 0. (3)Let ddr± μ(r) = D±. Equations (3) can be considered astwo linear subsystems and their solutions aref1 = +i(ϵ + k)D+f2 + (Γ −M)D+f4(Γ −M)2 − (ϵ2 − k2),f2 = +i(ϵ − k)D−f1 − (Γ +M)D−f3(Γ +M)2 − (ϵ2 − k2),f3 = −i(Γ −M)D+f2 + (ϵ − k)D+f4(Γ −M)2 − (ϵ2 − k2),f4 = −i−(Γ +M)D−f1 + (ϵ + k)D−f3(Γ +M)2 − (ϵ2 − k2).Eliminating the variables f1, f3 in the equations above, weobtain− Γ +MΓ −Mf2+ϵ − kΓ −Mf4 =D−D+f2(Γ −M)2 − (ϵ2 − k2)++ϵ − kΓ −M× D−D+f4(Γ −M)2 − (ϵ2 − k2), (4)Известия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ru79f2 − Γ +Mϵ + kf4 =D−D+f2(Γ −M)2 − (ϵ2 − k2)++Γ −Mϵ + k× D−D+f4(Γ −M)2 − (ϵ2 − k2).Subtract the second equation from the first equation and substitutethe resulting expression for f2 into (4), so we obtainthe fourth order equation for f4:−d4f4dr4 +e2B22r2 − eB(2m − 1)−−2(Γ2 −M2 − k2 + ϵ2) +2m(m + 1)r2d2f4dr2 ++e2B2r − 4m(m + 1)r3df4dr+"−e4B416r4++e2B24eB(2m − 1) + 2(Γ2 −M2 − k2 + ϵ2)r2−−eB(2m−1)(Γ2−M2−k2+ϵ2)−(Γ2+M2+k2−ϵ2)2++4Γ2M2 − e2B24(6m2 − 2m − 1)++m(m + 1)[eB(2m − 1) + 2(Γ2 −M2 − k2 + ϵ2)]r2−−m(m − 2)(m + 3)(m + 1)r4#f4 = 0. (5)Similarly, we can obtain the fourth order equation for thefunction f2. Equations for f2 and f4 turn out to be the same.To study the fourth order equation, we will use the factorizationmethod:b F4(r)f(r) = b f2(r)bg2(r)f(r) = 0,b f2(r) =d2dr2 + P0r2 + P1 +P2r2 ,bg2(r) =d2dr2 + Q0r2 + Q1 +Q2r2 .Computing the productb F4 =d2dr2 + P0r2 + P1 +P2r2××d2dr2 + Q0r2 + Q1 +Q2r2and equating the result to the operator (5), we find two solutionsfor sets of numerical coefficients:I P0 = −14B2e2, P2 = −m(m + 1),P1 = eBm − 12+ Γ2 −M2−− k2 + ϵ2 + 2Γ√ϵ2 − k2,II Q0 = −14B2e2, Q2 = −m(m + 1),Q1 = eBm − 12+ Γ2 −M2−− k2 + ϵ2 − 2Γ√ϵ2 − k2.The variant II differs only in sign at the parameter Γ. Thus, weare to solve two second-order differential equations:d2dr2− B2e2r24+ eBm − 12+ Γ2 −M2 − k2++ϵ2 + 2Γ√ϵ2 − k2 − m(m + 1)r2f = 0, (6)d2dr2− B2e2r24+ eBm − 12+ Γ2 −M2 − k2++ϵ2 − 2Γ√ϵ2 − k2 − m(m + 1)r2g = 0.They differ only in sign at the parameter Γ. Consider theequation (6). Let us make the change of the variable x =eBr2/2. Solutions are constructed in the form f = xaebxF.Taking into account the constraints a = −m/2, (m+ 1)/2and b = −1/2, we obtain the equation for F:xd2Fdx2 +12+ 2a − xdFdx−− 14eBheB(4a+1)−4Γ√ϵ2 − k2 −(2m−1)eB−−2(Γ2 −M2 − k2 + ϵ2)iF = 0.It is an equation of the confluent hypergeometric type withparametersγ = 2a +12,a = − 14eBheB(4a+1)−4Γ√ϵ2 − k2−(2m−1)eB−−2(Γ2 −M2 − k2 + ϵ2)i.To construct solutions corresponding to bound states, oneshould use the positive values of the parameter a (for definiteness,we assume that eB > 0):a = −m2(m < 0); a =m + 12> 0 (m ≥ 0).The polynomial conditions α = −n (let ϵ2 − k2 = λ )provides us with the quantization rule for energy valuesa +12− m2+M2 − Γ22eB+ n =Γ√λeB+λ2eB.Hence, using the notation M2 + 2eB(a + 1/2 − m/2 +n) = N, we obtainλ = (√N − Γ)2 > 0. (7)80Известия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ruFrom (7) we find the formula for the energy valuesϵ2 − k2 ="sM2 + 2eBa +1 − m2+ n− Γ#2.Depending on the value of a , we have two expressions forN:m < 0, a = −m2,ϵ2 − k2 =hpM2 + 2eB(1/2 − m + n) − Γi2;m ≥ 0, a =m + 12,ϵ2 − k2 =hpM2 + 2eB(1 + n) − Γi2.Thus, we obtain two series of energiesI : λ = (√N − Γ)2; II : λ = (√N + Γ)2.Let us consider a special case of an electrically neutralparticle with the magnetic moment (neutron). The transitionto the case of a neutral particle can be carried out with thehelp of simple formal changes, e → 0, λ → ∞, eλ → Λ,so thatΓ = λeℏBℏmc⇒ Γ = ΛBmc2 .No additional calculations are needed. We obtain two secondorderequations:d2dr2 + (√ϵ2 − k2 + Γ)2 −M2 − m(m + 1)r2f = 0,d2dr2 + (√ϵ2 − k2 − Γ)2 −M2 − m(m + 1)r2f = 0.(8)General solutions of equations (8) have the formf(r) =√r(Jm+1/2(x) + Ym+1/2(x)),x =r√ϵ2 − k2 + Γ2−M2r;g(r) =√r(Jm+1/2(y) + Ym+1/2(y)),y =r√ϵ2 − k2 − Γ2−M2r.From the form of these equations, we can conclude that themagnetic moment manifests itself in an external magneticfield in a quite definite way: in fact, everything reduces to solutionswith cylindrical symmetry for an ordinary free particlewith spin 1/2, but with a certain replacementϵ2 −M2 ⇒√ϵ2 − k2 ± Γ2−M2.The main manifestation of the magnetic moment of a neutralparticle is the modification (spatial scaling) of the wavefunctions in directions transverse to the magnetic field. Apparently,such a modification of the transverse structure ofa neutron beam can be observed experimentally, for example,in neutron Bessel beams. Obviously, the transition to thesituation of an electrically neutral particle can also be carriedout in the case of a particle with three mass parameters.3. Diagonalization of the mixing matrixLet us turn back to the system of three equationsΣ−1ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ1++(d1L1Φ1 + d1L2Φ2 + d1L3Φ3 + (M −M1)Φ1) = 0,Σ−1ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ2++(d2L1Φ1 + d2L2Φ2 + d2L3Φ3 + (M −M2)Φ2) = 0,Σ−1ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −MΦ3++(d3L1Φ1 + d3L2Φ2 + d3L3Φ3 + (M −M3)Φ3) = 0,or brieflyb∂ = Σ−1ϵγ0 + iγ1 ∂∂r− γ2rμ(r) − kγ3 −M,b∂Φk + TkjΦj = 0.To find the transformation matrix S, which will diagonalizethe mixing matrix, we are to study the following equationST = T0S, or in detail0@s11 s12 s13s21 s22 s23s31 s32 s331A××0@M −M1 + d1L1 d1L2d2L1 M −M2 + d2L2d3L1 d3L2d1L3d2L3M −M3 + d3L31A ==0@μ1 0 00 μ2 00 0 μ31A0@s11 s12 s13s21 s22 s23s31 s32 s331A.Hence we obtain three linear subsystems:(M −M1 + d1L1)s11 + d2L1s12 + d3L1s13 = μ1s11,d1L2s11 + (M −M2 + d2L2)s12 + d3L2s13 = μ1s12,d1L3s11 + d2L3s12 + (M −M3 + d3L3)s13 = μ1s13;(M −M1 + d1L1)s11 + d2L1s12 + d3L1s13 = μ1s11,d1L2s11 + (M −M2 + d2L2)s12 + d3L2s13 = μ1s12,d1L3s11 + d2L3s12 + (M −M3 + d3L3)s13 = μ1s13;(M −M1 + d1L1)s31 + d2L1s32 + d3L1s33 = μ3s31,d1L2s31 + (M −M2 + d2L2)s32 + d3L2s33 = μ3s32,d1L3s31 + d2L3s32 + (M −M3 + d3L3)s33 = μ3s33.Известия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ru81Here we have three eigenvalue problems0@M −M1 + d1L1 − μ1 d2L1 d3L1d1L2 M −M2 + d2L2 − μ1 d3L2d1L3 d2L3 M −M3 + d3L3 − μ11A0@s11s12s131A = 0,0@M −M1 + d1L1 − μ2 d2L1 d3L1d1L2 M −M2 + d2L2 − μ2 d3L2d1L3 d2L3 M −M3 + d3L3 − μ21A0@s21s22s231A = 0,0@M −M1 + d1L1 − μ3 d2L1 d3L1d1L2 M −M2 + d2L2 − μ3 d3L2d1L3 d2L3 M −M3 + d3L3 − μ31A0@s31s32s331A = 0.Note that the three rows of the matrix S may be found only up to arbitrary multipliers. The condition for the existence ofsolutions for these three systems is the vanishing of the determinantdet0@M −M1 + d1L1 − μ d2L1 d3L1d1L2 M −M2 + d2L2 − μ d3L2d1L3 d2L3 M −M3 + d3L3 − μ1A = 0.Let us get the explicit form of the third order equation forparameter μ:−μ3+(3M−M1−M2−M3+d1L1+d2L2+d3L3)μ2++−d1L1(2M−M2−M3)−d2L2(2M−M1−M3)−−d3L3(2M−M2−M1)−3M2+2(M1+M2+M3)M−−(M1M2 +M1M3 +M2M3)μ++d1L1(M−M2)(M−M3)+d2L2(M−M3)(M−M1)++d3L3(M −M1)(M −M2)++(M −M1)(M −M2)(M −M3) = 0.Given the relation Mi = M/λi, the equation can be transformedinto the following form−μ3 +"M3 − 1λ1− 1λ2− 1λ3+ L1d1 + L2d2++L3d3#μ2 +−ML1d12 − 1λ2− 1λ3−−ML2d22 − 1λ1− 1λ3−ML3d32 − 1λ1− 1λ2−−3M2 + 2M21λ1+1λ2+1λ3−−M21λ11λ2+1λ11λ3+1λ21λ3μ++M2L1d11 − 1λ21 − 1λ3++M2L2d21 − 1λ31 − 1λ1++M2L3d31 − 1λ11 − 1λ2++ M31 − 1λ11 − 1λ21 − 1λ2= 0. (9)Let us detail the explicit form of the elements of the mixingmatrix. First, we take into account that the parametrizationof possible values of the mass parametersMi =Mλican besimplified. Recall that the roots λi are solutions of the characteristicequationλ3−(c1+c2)λ2+(c1c2+a2+b2)λ−(c1a2+c2b2) = 0,where c3 = b > 0, c4 = a > 0. Taking into account theidentitiesλ1 + λ2 = c1 + c2 − λ3, λ1λ2 =c1a2 + c2b2λ3,we find expressions for λ1, λ2 in terms of λ3 :λ1 =c1 + c2 − λ32−−sc1 + c2 − λ322− c1a2 + c2b2λ3,λ2 =c1 + c2 − λ32++sc1 + c2 − λ322− c1a2 + c2b2λ3.The value of the root λ3 may be arbitrary, because the physicallymeaningful parameter is M3 = M/λ3 (at an arbitraryM). The simplest expression for λ3 is obtained whenc1 = c2 = 1. In this case, the cubic equation is simplifiedλ3 − 2λ2 + (1 + k)λ − k = 0, k = a2 + b2.Its roots are given by the formulasλ3 = 1, λ1 =12− 12√1 − 4k,82Известия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ruλ2 =12+12√1 − 4k, k ∈0,14.Accordingly, the mass parameters Mi are specified by theformulasM3 = M, M1 =2M1 −√1 − 4k,M2 =2M1 +√1 − 4k, k ∈0,14.Let us construct tables of values for λi and λ−1i =Mi/M, i = 1, 2, (λ3 = λ−13 = 1), depending on theparameter k:Table 1Parameter values λi = 12􀀀1 ∓√1 − 4k, i = 1, 2 depending on kТаблица 1Значения параметров λi = 12􀀀1 ∓√1 − 4k, i = 1, 2 в зависимости от kk 0.24 0.22 0.209 0.207 0.205 0.203 0.20 0.16 0.12 0.08 0.04 0.01λ1 0.4 0.327 0.298 0.293 0.288 0.283 0.276 0.2 0.139 0.088 0.042 0.01λ2 0.6 0.673 0.702 0.707 0.712 0.717 0.724 0.8 0.861 0.912 0.958 0.989Table 2Parameter values λ−1i = 2􀀀1 ∓√1 − 4k−1, i = 1, 2 depending on kТаблица 2Значения параметров λ−1i = 2􀀀1 ∓√1 − 4k−1, i = 1, 2 в зависимости от kk 0.24 0.22 0.209 0.207 0.205 0.203 0.20 0.16 0.12 0.08 0.04 0.011/λ1 2.5 3.058 3.356 3.413 3.472 3.534 3.623 5 7.194 11.364 23.810 1001/λ2 1.667 1.486 1.425 1.414 1.404 1.395 1.381 1.250 1.161 1.096 1.044 1.011Let us detail the coefficients Li in the expression forΦ(x):Φ(x) = L1Φ1(x) + L2Φ2(x) + L3Φ3(x),L1 =−2kb1(λ1 − λ2)(λ1 − λ3)++12Lb1 − 2(λ2 + λ3) + 4λ2λ3(λ1 − λ2)(λ1 − λ3),L2 =−2kb1(λ2 − λ3)(λ2 − λ1)++12Lb1 − 2(λ3 + λ1) + 4λ3λ1(λ2 − λ3)(λ2 − λ1),L3 =−2kb1(λ3 − λ1)(λ3 − λ2)++12Lb1 − 2(λ1 + λ2) + 4λ1λ2(λ3 − λ1)(λ3 − λ2),or (we take into account that λ3 = 1)L1 =1b1(λ1 − 1)(λ1 − λ2)−2k +12L(−1 + 2λ2),L2 =1b1(λ2 − 1)(λ2 − λ1)××−2k +12L(−1 + 2λ1),L3 =1b1(1 − λ1)(1 − λ2)××−2k +12L(1 − 2λ1 − 2λ2 + 4λ1λ2), (10)whereL = 1 +√b6, (a2 + b2) = k <14, 0 < 2b < 1.Let us detail the coefficients di:d1 =4Bb6Mλ1 − 12, d2 =4Bb6Mλ2 − 12,d3 =4Bb6Mλ3 − 12. (11)Combinations diLj appear in the mixing matrix, so the parameterb in the denominators in (10) will be canceled with theparameter b in (11). Also, since the dimension of the quantityB isM2 (inverse square meter), we can introduce substitutions4B = 6rM2 ⇒ di = M4B6M2 bλi − 12== Mrbλi − 12= MDi,where the quantitiesDi are dimensionless. With this in mind,the cubic equation (9) is transformed to the form−μ3 +M"3 − 1λ1− 1λ2− 1λ3+ L1D1 + L2D2++L3D3#μ2 +M2"−L1D12 − 1λ2− 1λ3−−L2D22 − 1λ1− 1λ3− L3D32 − 1λ1−− 1λ2− 3 + 21λ1+1λ2+1λ3−Известия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ru83−1λ11λ2+1λ11λ3+1λ21λ3#μ++M3"L1D11 − 1λ21 − 1λ3++L2D21 − 1λ31 − 1λ1++L3D31 − 1λ11 − 1λ2++1 − 1λ11 − 1λ21 − 1λ3#= 0.The roots of this equation can be found in the form μi =MΔi, where Δi are dimensionless quantities; then theequation for Δ1,Δ2,Δ3 takes the form−Δ3+"3− 1λ1− 1λ2− 1λ3+L1D1+L2D2+L3D3#Δ2++"−L1D12 − 1λ2− 1λ3−L2D22 − 1λ1− 1λ3−−L3D32 − 1λ1− 1λ2− 3 + 21λ1+1λ2+1λ3−−1λ11λ2+1λ11λ3+1λ21λ3#Δ++"L1D11 − 1λ21 − 1λ3++L2D21 − 1λ31 − 1λ1++L3D31 − 1λ11 − 1λ2++1 − 1λ11 − 1λ21 − 1λ3#= 0.Allow for that λ3 = 1, then we getΔ3 +"2 − 1λ1− 1λ2+ L1D1 + L2D2 + L3D3#Δ2++"L1D11 − 1λ2+ L2D21 − 1λ1++L3D32 − 1λ1− 1λ2+ 3 − 21λ1+1λ2+ 1++1λ11λ2+1λ1+1λ2#Δ−−L3D31 − 1λ11 − 1λ2= 0.Since the parameter b enters as a multiplier intoD1,D2,D3and the denominators of the expressions for the coefficientsL1,L2,L3, then in combinationsL1D1 + L2D2 + L3D3, L1D1, L2D2, L3D3this parameter is reduced. With this in mind, we have theidentitiesL1 =1(λ1 − 1)(λ1 − λ2)−2k +12L(2λ2 − 1),L2 =1(λ2 − 1)(λ2 − λ1)−2k +12L(2λ1 − 1),L3 =1(1 − λ1)(1 − λ2)××−2k +12L(1 − 2λ1 − 2λ2 + 4λ1λ2),D1 =r2(2λ1 − 1), D2 =r2(2λ2 − 1), D3 =r2,whereλ1 =121 −√1 − 4k, λ2 =121 +√1 − 4k,k ∈1,14, L = 1 +√16, (a2 + b2) = k <14.Also, we should take into account the identities2λ1 − 1 = −√1 − 4k, 2λ2 − 1 = +√1 − 4k,λ1 − λ2 = −√1 − 4k, λ2 − λ1 = +√1 − 4k,λ1 + λ2 = 1, λ1 − 1 = −λ2, λ2 − 1 = −λ1,1 − 2λ1 − 2λ2 + 4λ1λ2 = 4k − 1, λ1λ2 = k,then we get simpler expressionsL1 =2(1 +√1 − 4k)√1 − 4k−2k +12L√1 − 4k,L2 =2(1 +√1 − 4k)√1 − 4k2k +12L√1 − 4k,L3 =1k−2k +12L(4k − 1),D1 = −r2√1 − 4k, D2 = +r2√1 − 4k, D3 =r2,L = 1 +√b6, 0 < b <12.In this way, we arrive at the following cubic equation for Δ:Δ3 + AΔ2 + BΔ + C = 0,where the coefficients are (take into account the propertiesof the roots λ1, λ2)A = 2 − 1k+ L1D1 + L2D2 + L3D3, C = L3D3,84Известия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ruB = −L1D11 −√1 − 4k1 +√1 − 4k−− L2D21 +√1 − 4k1 −√1 − 4k+ L3D32 − 1k+ 1.Using identitiesL1D1 =2k − l√1 − 4k1 +√1 − 4kr,L2D2 =2k + l√1 − 4k1 −√1 − 4kr,L3D3 = −1 + l1 − 4k2kr,we transform the cubic equation to a simpler formΔ3 +2k − 1kΔ2 +1 +1 − l1 − 4k2krΔ++1 + l1 − 4k2kr = 0.Allowing for the expression for the parameter l:l =12L=√62(√6 + b),we obtainΔ3 +2k − 1kΔ2++"1 + 1 −√62(√6 + b)1 − 4k2k!r#Δ++ 1 +√62(√6 + b1 − 4k2k!r = 0, (12)where0 < k <14, 0 < b < 2.For definiteness, we set b1 = 0, b2 = 1, b3 = 2. The parameterr was introduced by the relation 4B = 6rM2from physical considerations, we will assume that the dimensionlessparameter r is small. Below we will follow severalsituations r = 10−5, r = 10−3, r = 1. The last value|r| = 1 corresponds to a very strong magnetic field. Numericalstudy showed that dependence of the roots Δi uponparameter b ∈ (0, 2) is very insignificant. By this reason belowwe take the value b = 0. Let us construct tables of valuesfor the roots Δ1,Δ2,Δ3 equations (12) (see tables 3–5).Table 3The values of the rootsΔi, i = 1, 2, 3 of equation (12) for b = 0, r = 10−5Таблица 3Значения корнейΔi, i = 1, 2, 3 уравнения (12) для b = 0, r = 10−5k 0.24 0.22 0.208 0.204 0.20 0.16 0.10 0.06 0.02 0.01Δ1 0.667 0.485 0.419 0.400 0.382 0.250 0.127 0.069 0.021 0.010Δ2 1.500 2.060 2.389 2.502 2.618 3.999 7.873 14.598 47.979 97.990Δ3 -0.00001 -0.00001 -0.00001 -0.00001 -0.00001 -0.00002 -0.00002 -0.00004 -0.00012 -0.00024Table 4The values of the rootsΔi, i = 1, 2, 3 of equation (12) for b = 0, r = 10−3Таблица 4Значения корнейΔi, i = 1, 2, 3 уравнения (12) для b = 0, r = 10−3k 0.24 0.22 0.208 0.204 0.20 0.16 0.12 0.08 0.04 0.01Δ1 0.670 0.487 0.420 0.401 0.384 0.252 0.164 0.099 0.049 0.021Δ2 1.489 2.059 2.388 2.502 2.617 3.999 6.171 10.404 22.957 97.990Δ3 -0.001 -0.001 -0.001 -0.001 -0.001 -0.002 -0.002 -0.003 -0.005 -0.012Finally, with an even greater increase in the parameter r, the roots become complex-valuedTable 5The values of the rootsΔi, i = 1, 2, 3 of equation (12) for b = 0, r = 1Таблица 5Значения корнейΔi, i = 1, 2, 3 уравнения (12) для b = 0, r = 1k 0.24 0.209 0.206 0.203 0.18 0.12 0.06 0.01Δ1 1.265+1.129i 1.588+0.726i 1.625+0.656i 1.662+0.574i 1.173 0.695 0.503 0.406Δ2 -0.362 -0.392 -0.395 -0.398 2.804 6.128 14.727 98.221Δ3 1.265-1.129i 1.588-0.726i 1.625-0.656i 1.662-0.574i -0.422 -0.489 -0.653 -0.627This means that at such magnetic field, the model becomesnon-interpretable.ConclusionsIn the present paper, the model of a fermion with threemass parameters is studied in presence of the externaluniform magnetic field. After diagonalizing transformation,three separate equations are obtained effectively for particleswith different anomalous magnetic moments. Their exactsolutions and generalized energy spectra are found. Afterdiagonalizing the mixing matrix, we reduce the problemfor three separated Dirac–like equations for particles withdifferent anomalous magnetic moment. It is shown that forИзвестия Коми научного центра Уральского отделения Российской академии наук № 4 (62), 2023Серия «Физико-математические науки»www.izvestia.komisc.ru85a very strong magnetic field, the model becomes non-interpretable,because the effective anomalous moments turns outto be complex-valued.