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Недавно были развиты модели для частицы со спином 1/2 с двумя (или тремя) массовыми параметрами. Особенно- сти этих моделей заключаются в следующем. Для соответ- ствующих двух (или трех) биспиноров в отсутствие внеш- них полей выводятся отдельные уравнения дираковского типа, различающиеся массами. Однако при наличии внеш- них электромагнитных или гравитационных полей с нену- левым скаляром Риччи волновое уравнение для биспино- ров не распадается на отдельные уравнения, а возникает связанная система из двух (или трех) уравнений. В насто- ящей работе исследуется модель фермиона с тремя мас- совыми параметрами в присутствии внешнего однородно- го магнитного поля. После диагонализации матрицы сме- шивания получаются три отдельных уравнения для частиц с разными аномальными магнитными моментами; найдены их точные решения и получены обобщенные энергетиче- ские спектры.
фермион с тремя массовыми параметрами, магнитное по- ле, аномальный магнитный момент, точные решения, энер- гетический спектр
Introduction
In the context of existence of the similar neutrinos of different
masses, we examine a possibility within the theory of
relativistic wave equations to describe particles with several
mass parameters. In general, existence of more general wave
equations than commonly used ones is well known within the
Gel’fand-Yaglom formalism – see references [1-5].
In particular, models for a spin 1/2 particle with two and
three mass parameters were developed [6–15]. Specific features
of these models are as follows. For two (or three)
bispinors, in absence of external fields separate Dirac-like
equations are derived, they differ in masses. However, in
presence of external electromagnetic field or gravitational
field with non-vanishing Ricci scalar, the wave equation for
bispinors does not split into separated equations, instead a
quite definite mixing of two (or three) equations arises. It was
shown that generalized equations for Majorana particle with
several mass parameters exist as well. Such generalized Majorana
equations are not trivial if the Ricci scalar does not
vanish.
In the present paper, the model of a fermion with three
mass parameters is studied in presence of the external uniform
magnetics field. After applying the diagonalizing trans-
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77
formation, three separate equations are obtained effectively
for particles with different anomalous magnetic moments.
Their exact solutions and generalized energy spectra are
found.
1. General theory in presence of electromagnetic
and gravitational fields
We start with the system of equations for a fermion with
3 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 used
Y1 =
4c3
3M
c2(λ1 − c2), i = 1, 2, 3, L = c1 +
√c3
6
,
Φ(x) = L1Φ1(x) + L2Φ2(x) + L3Φ3(x),
Σ(x) = −ieFαβσαβ(x) +
1
4
R(x),
L1 =
−L|c4|2 − L|c3|2 + c22
− c2(λ2 + λ3) + λ2λ3
Lc2c3(λ1 − λ2)(λ1 − λ3)
,
L2 =
−L|c4|2 − L|c3|2 + c22
− c2(λ3 + λ1) + λ3λ1
Lc2c3(λ2 − λ3)(λ2 − λ1)
,
L3 =
−L|c4|2 − L|c3|2 + c22
− c2(λ1 + λ2) + λ1λ2
Lc2c3(λ3 − λ1)(λ3 − λ2)
.
With the notation |c4| = a, |c3| = b, the mixing matrix in
the equation (1) is specified by the relations
Y1L1 =
4
3M
(λ1 − c2)×
×
−L(a2 + b2) + c22
− c2(λ2 + λ3) + λ2λ3
L(λ1 − λ2)(λ1 − λ3)
,
Y1L2 =
4
3M
(λ1 − c2)×
×
−L(a2 + b2) + c22
− c2(λ3 + λ1) + λ3λ1
L(λ2 − λ3)(λ2 − λ1)
,
Y1L3 =
4
3M
(λ1 − c2)×
×
−L(a2 + b2) + c22
− c2(λ1 + λ2) + λ1λ2
L(λ3 − λ1)(λ3 − λ2)
,
Y2L1 =
4
3M
(λ2 − c2)×
×
−L(a2 + b2) + c22
− c2(λ2 + λ3) + λ2λ3
L(λ1 − λ2)(λ1 − λ3)
,
Y2L2 =
4
3M
(λ2 − c2)×
×
−L(a2 + b2) + c22
− c2(λ3 + λ1) + λ3λ1
L(λ2 − λ3)(λ2 − λ1)
,
Y2L3 =
4
3M
(λ2 − c2)×
×
−L(a2 + b2) + c22
− c2(λ1 + λ2) + λ1λ2
L(λ3 − λ1)(λ3 − λ2)
,
Y3L1 =
4
3M
(λ3 − c2)×
×
−L(a2 + b2) + c22
− c2(λ2 + λ3) + λ2λ3
L(λ1 − λ2)(λ1 − λ3)
,
Y3L2 =
4
3M
(λ3 − c2)×
×
−L(a2 + b2) + c22
− c2(λ3 + λ1) + λ3λ1
L(λ2 − λ3)(λ2 − λ1)
,
Y3L3 =
4
3M
(λ3 − c2)×
×
−L(a2 + b2) + c22− c2(λ1 + λ2) + λ1λ2
L(λ3 − λ1)(λ3 − λ2)
.
We will consider eq. (1) in the cylindrical coordinates and
tetrad
dS2 = dt2 − dr2 − r2dϕ2 − dz2,
xα = (t, r, ϕ, z),
eβ
(a)(x) =
0
B@
1 0 0 0
0 1 0 0
0 0 1/r 0
0 0 0 1
1
CA
.
Ricci rotation coefficients are as follows: γab0 = 0, γab1 =
0, γ122 = −γ212 = 1/r, γab3 = 0. The external magnetic
field directed along the axis x3 is determined as follows
Aϕ = −1
2
Br2, F12(x) = Frϕ = −Br,
−ieFαβ(x) = −ieF12(x)γ1(x)γ2(x) = ieBγ1γ2 =
= ieB
−iσ3 0
0 −iσ3
= eBΣ.
So, the main equation (1) takes the form (we simplify the notation,
eB ⇒ B )
iγ0 ∂
∂t
+ iγ1 ∂
∂r
+ i
γ2
r
∂
∂ϕ
+ i
Br2
2
+ iγ3 ∂
∂z
Φ1−
−MΦ1 + BΣY1(L1Φ1 + L2Φ2 + L3Φ3) = 0,
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iγ0 ∂
∂t
+ iγ1 ∂
∂r
+ i
γ2
r
∂
∂ϕ
+ i
Br2
2
+ iγ3 ∂
∂z
Φ2−
−MΦ2 + BΣY2(L1Φ1 + L2Φ2 + L3Φ3) = 0,
iγ0 ∂
∂t
+ iγ1 ∂
∂r
+ i
γ2
r
∂
∂ϕ
+ i
Br2
2
+ iγ3 ∂
∂z
Φ3−
−MΦ3 + BΣY3(L1Φ1 + L2Φ2 + L3Φ3) = 0.
For three involved bispinors, we will use the following substitutions
Φ1 = 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 system
reads
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ1+
+(M −M1)Φ1 + BΣY1(L1Φ1 + L2Φ2 + L3Φ3) = 0,
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ2+
+(M −M2)Φ2 + BΣY2(L1Φ1 + L2Φ2 + L3Φ3) = 0,
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ3+
+(M −M3)Φ3 + BΣY3(L1Φ1 + L2Φ2 + L3Φ3) = 0,
where
μ(r) = m +
1
2
Br2, Σ =
0
B@
1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1
1
CA
.
Using the shortening notations for elements of the mixing matrix
Zkj = (BYk)Lj = dkLj ,
we present the system as follows
b∂Φk + ZkjΦj = 0,
b∂ = Σ−1
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
.
The mixing matrix should be reduced to a diagonal form by
a linear transformation
Φ = SΦ, b∂Φ = SZS−1Φ,
SZS−1 =
0
@
μ1 0 0
0 μ2 0
0 0 μ3
1
A.
As a result we get
Σ−1 b∂
0
@
Φ1
Φ2
Φ3
1
A +
0
@
μ1 0 0
0 μ2 0
0 0 μ3
1
A
0
@
Φ1
Φ2
Φ3
1
A = 0,
or
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ1+μ1ΣΦ1 = 0,
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ2+μ2ΣΦ2 = 0,
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ3+μ3ΣΦ3 = 0.
(2)
Thus, after transformation, three separate equations are obtained
effectively for particles with different anomalous magnetic
moments.
2. Solving the basic equation
Let us briefly describe the procedure for solving the basic
equation (2):
ϵγ0 + iγ1 ∂
∂r
− γ2
r
σ(r) − kγ3 −M + ΓZ
0
B@
f1
f2
f3
f4
1
CA
= 0.
Using the spinor basis for the Dirac matrices [3, 6], we obtain
four equations
−i
d
dr
+ μ
f4 + (ϵ + k)f3 + (Γ −M)f1 = 0,
−i
d
dr
− μ
f3 + (ϵ − k)f4 − (Γ +M)f2 = 0,
i
d
dr
+ μ
f2 + (ϵ − k)f1 + (Γ −M)f3 = 0,
i
d
dr
− μ
f1 + (ϵ + k)f2 − (Γ +M)f4 = 0. (3)
Let d
dr
± μ(r) = D±. Equations (3) can be considered as
two linear subsystems and their solutions are
f1 = +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, we
obtain
− Γ +M
Γ −M
f2+
ϵ − k
Γ −M
f4 =
D−D+f2
(Γ −M)2 − (ϵ2 − k2)
+
+
ϵ − k
Γ −M
× D−D+f4
(Γ −M)2 − (ϵ2 − k2)
, (4)
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f2 − Γ +M
ϵ + k
f4 =
D−D+f2
(Γ −M)2 − (ϵ2 − k2)
+
+
Γ −M
ϵ + k
× D−D+f4
(Γ −M)2 − (ϵ2 − k2)
.
Subtract the second equation from the first equation and substitute
the resulting expression for f2 into (4), so we obtain
the fourth order equation for f4:
−d4f4
dr4 +
e2B2
2
r2 − eB(2m − 1)−
−2(Γ2 −M2 − k2 + ϵ2) +
2m(m + 1)
r2
d2f4
dr2 +
+
e2B2r − 4
m(m + 1)
r3
df4
dr
+
"
−e4B4
16
r4+
+
e2B2
4
eB(2m − 1) + 2(Γ2 −M2 − k2 + ϵ2)
r2−
−eB(2m−1)(Γ2−M2−k2+ϵ2)−(Γ2+M2+k2−ϵ2)2+
+4Γ2M2 − e2B2
4
(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 the
function f2. Equations for f2 and f4 turn out to be the same.
To study the fourth order equation, we will use the factorization
method:
b F4(r)f(r) = b f2(r)bg2(r)f(r) = 0,
b f2(r) =
d2
dr2 + P0r2 + P1 +
P2
r2 ,
bg2(r) =
d2
dr2 + Q0r2 + Q1 +
Q2
r2 .
Computing the product
b F4 =
d2
dr2 + P0r2 + P1 +
P2
r2
×
×
d2
dr2 + Q0r2 + Q1 +
Q2
r2
and equating the result to the operator (5), we find two solutions
for sets of numerical coefficients:
I P0 = −1
4
B2e2, P2 = −m(m + 1),
P1 = eB
m − 1
2
+ Γ2 −M2−
− k2 + ϵ2 + 2Γ
√
ϵ2 − k2,
II Q0 = −1
4
B2e2, Q2 = −m(m + 1),
Q1 = eB
m − 1
2
+ Γ2 −M2−
− k2 + ϵ2 − 2Γ
√
ϵ2 − k2.
The variant II differs only in sign at the parameter Γ. Thus, we
are to solve two second-order differential equations:
d2
dr2
− B2e2r2
4
+ eB
m − 1
2
+ Γ2 −M2 − k2+
+ϵ2 + 2Γ
√
ϵ2 − k2 − m(m + 1)
r2
f = 0, (6)
d2
dr2
− B2e2r2
4
+ eB
m − 1
2
+ Γ2 −M2 − k2+
+ϵ2 − 2Γ
√
ϵ2 − k2 − m(m + 1)
r2
g = 0.
They differ only in sign at the parameter Γ. Consider the
equation (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)/2
and b = −1/2, we obtain the equation for F:
x
d2F
dx2 +
1
2
+ 2a − x
dF
dx
−
− 1
4eB
h
eB(4a+1)−4Γ
√
ϵ2 − k2 −(2m−1)eB−
−2(Γ2 −M2 − k2 + ϵ2)
i
F = 0.
It is an equation of the confluent hypergeometric type with
parameters
γ = 2a +
1
2
,
a = − 1
4eB
h
eB(4a+1)−4Γ
√
ϵ2 − k2−(2m−1)eB−
−2(Γ2 −M2 − k2 + ϵ2)
i
.
To construct solutions corresponding to bound states, one
should use the positive values of the parameter a (for definiteness,
we assume that eB > 0):
a = −m
2
(m < 0); a =
m + 1
2
> 0 (m ≥ 0).
The polynomial conditions α = −n (let ϵ2 − k2 = λ )
provides us with the quantization rule for energy values
a +
1
2
− m
2
+
M2 − Γ2
2eB
+ n =
Γ
√
λ
eB
+
λ
2eB
.
Hence, using the notation M2 + 2eB(a + 1/2 − m/2 +
n) = N, we obtain
λ = (
√
N − Γ)2 > 0. (7)
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From (7) we find the formula for the energy values
ϵ2 − k2 =
"s
M2 + 2eB
a +
1 − m
2
+ n
− Γ
#2
.
Depending on the value of a , we have two expressions forN:
m < 0, a = −m
2
,
ϵ2 − k2 =
hp
M2 + 2eB(1/2 − m + n) − Γ
i2
;
m ≥ 0, a =
m + 1
2
,
ϵ2 − k2 =
hp
M2 + 2eB(1 + n) − Γ
i2
.
Thus, we obtain two series of energies
I : λ = (
√
N − Γ)2; II : λ = (
√
N + Γ)2.
Let us consider a special case of an electrically neutral
particle with the magnetic moment (neutron). The transition
to the case of a neutral particle can be carried out with the
help of simple formal changes, e → 0, λ → ∞, eλ → Λ,
so that
Γ = λ
e
ℏ
Bℏ
mc
⇒ Γ = Λ
B
mc2 .
No additional calculations are needed. We obtain two secondorder
equations:
d2
dr2 + (
√
ϵ2 − k2 + Γ)2 −M2 − m(m + 1)
r2
f = 0,
d2
dr2 + (
√
ϵ2 − k2 − Γ)2 −M2 − m(m + 1)
r2
f = 0.
(8)
General solutions of equations (8) have the form
f(r) =
√
r(Jm+1/2(x) + Ym+1/2(x)),
x =
r√
ϵ2 − k2 + Γ
2
−M2r;
g(r) =
√
r(Jm+1/2(y) + Ym+1/2(y)),
y =
r√
ϵ2 − k2 − Γ
2
−M2r.
From the form of these equations, we can conclude that the
magnetic moment manifests itself in an external magnetic
field in a quite definite way: in fact, everything reduces to solutions
with cylindrical symmetry for an ordinary free particle
with spin 1/2, but with a certain replacement
ϵ2 −M2 ⇒
√
ϵ2 − k2 ± Γ
2
−M2.
The main manifestation of the magnetic moment of a neutral
particle is the modification (spatial scaling) of the wave
functions in directions transverse to the magnetic field. Apparently,
such a modification of the transverse structure of
a neutron beam can be observed experimentally, for example,
in neutron Bessel beams. Obviously, the transition to the
situation of an electrically neutral particle can also be carried
out in the case of a particle with three mass parameters.
3. Diagonalization of the mixing matrix
Let us turn back to the system of three equations
Σ−1
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ1+
+(d1L1Φ1 + d1L2Φ2 + d1L3Φ3 + (M −M1)Φ1) = 0,
Σ−1
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ2+
+(d2L1Φ1 + d2L2Φ2 + d2L3Φ3 + (M −M2)Φ2) = 0,
Σ−1
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
Φ3+
+(d3L1Φ1 + d3L2Φ2 + d3L3Φ3 + (M −M3)Φ3) = 0,
or briefly
b∂ = Σ−1
ϵγ0 + iγ1 ∂
∂r
− γ2
r
μ(r) − kγ3 −M
,
b∂Φk + TkjΦj = 0.
To find the transformation matrix S, which will diagonalize
the mixing matrix, we are to study the following equation
ST = T0S, or in detail
0
@
s11 s12 s13
s21 s22 s23
s31 s32 s33
1
A×
×
0
@
M −M1 + d1L1 d1L2
d2L1 M −M2 + d2L2
d3L1 d3L2
d1L3
d2L3
M −M3 + d3L3
1
A =
=
0
@
μ1 0 0
0 μ2 0
0 0 μ3
1
A
0
@
s11 s12 s13
s21 s22 s23
s31 s32 s33
1
A.
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.
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81
Here we have three eigenvalue problems
0
@
M −M1 + d1L1 − μ1 d2L1 d3L1
d1L2 M −M2 + d2L2 − μ1 d3L2
d1L3 d2L3 M −M3 + d3L3 − μ1
1
A
0
@
s11
s12
s13
1
A = 0,
0
@
M −M1 + d1L1 − μ2 d2L1 d3L1
d1L2 M −M2 + d2L2 − μ2 d3L2
d1L3 d2L3 M −M3 + d3L3 − μ2
1
A
0
@
s21
s22
s23
1
A = 0,
0
@
M −M1 + d1L1 − μ3 d2L1 d3L1
d1L2 M −M2 + d2L2 − μ3 d3L2
d1L3 d2L3 M −M3 + d3L3 − μ3
1
A
0
@
s31
s32
s33
1
A = 0.
Note that the three rows of the matrix S may be found only up to arbitrary multipliers. The condition for the existence of
solutions for these three systems is the vanishing of the determinant
det
0
@
M −M1 + d1L1 − μ d2L1 d3L1
d1L2 M −M2 + d2L2 − μ d3L2
d1L3 d2L3 M −M3 + d3L3 − μ
1
A = 0.
Let us get the explicit form of the third order equation for
parameter 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 transformed
into the following form
−μ3 +
"
M
3 − 1
λ1
− 1
λ2
− 1
λ3
+ L1d1 + L2d2+
+L3d3
#
μ2 +
−ML1d1
2 − 1
λ2
− 1
λ3
−
−ML2d2
2 − 1
λ1
− 1
λ3
−ML3d3
2 − 1
λ1
− 1
λ2
−
−3M2 + 2M2
1
λ1
+
1
λ2
+
1
λ3
−
−M2
1
λ1
1
λ2
+
1
λ1
1
λ3
+
1
λ2
1
λ3
μ+
+
M2L1d1
1 − 1
λ2
1 − 1
λ3
+
+M2L2d2
1 − 1
λ3
1 − 1
λ1
+
+M2L3d3
1 − 1
λ1
1 − 1
λ2
+
+ M3
1 − 1
λ1
1 − 1
λ2
1 − 1
λ2
= 0. (9)
Let us detail the explicit form of the elements of the mixing
matrix. First, we take into account that the parametrization
of possible values of the mass parametersMi =
M
λi
can be
simplified. Recall that the roots λi are solutions of the characteristic
equation
λ3−(c1+c2)λ2+(c1c2+a2+b2)λ−(c1a2+c2b2) = 0,
where c3 = b > 0, c4 = a > 0. Taking into account the
identities
λ1 + λ2 = c1 + c2 − λ3, λ1λ2 =
c1a2 + c2b2
λ3
,
we find expressions for λ1, λ2 in terms of λ3 :
λ1 =
c1 + c2 − λ3
2
−
−
s
c1 + c2 − λ3
2
2
− c1a2 + c2b2
λ3
,
λ2 =
c1 + c2 − λ3
2
+
+
s
c1 + c2 − λ3
2
2
− c1a2 + c2b2
λ3
.
The value of the root λ3 may be arbitrary, because the physically
meaningful parameter is M3 = M/λ3 (at an arbitrary
M). The simplest expression for λ3 is obtained when
c1 = c2 = 1. In this case, the cubic equation is simplified
λ3 − 2λ2 + (1 + k)λ − k = 0, k = a2 + b2.
Its roots are given by the formulas
λ3 = 1, λ1 =
1
2
− 1
2
√
1 − 4k,
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λ2 =
1
2
+
1
2
√
1 − 4k, k ∈
0,
1
4
.
Accordingly, the mass parameters Mi are specified by the
formulas
M3 = M, M1 =
2M
1 −
√
1 − 4k
,
M2 =
2M
1 +
√
1 − 4k
, k ∈
0,
1
4
.
Let us construct tables of values for λi and λ
−1
i =
Mi/M, i = 1, 2, (λ3 = λ
−1
3 = 1), depending on the
parameter k:
Table 1
Parameter values λi = 1
2
1 ∓
√
1 − 4k
, i = 1, 2 depending on k
Таблица 1
Значения параметров λi = 1
2
1 ∓
√
1 − 4k
, i = 1, 2 в зависимости от k
k 0.24 0.22 0.209 0.207 0.205 0.203 0.20 0.16 0.12 0.08 0.04 0.01
λ1 0.4 0.327 0.298 0.293 0.288 0.283 0.276 0.2 0.139 0.088 0.042 0.01
λ2 0.6 0.673 0.702 0.707 0.712 0.717 0.724 0.8 0.861 0.912 0.958 0.989
Table 2
Parameter values λ
−1
i = 2
1 ∓
√
1 − 4k
−1, i = 1, 2 depending on k
Таблица 2
Значения параметров λ
−1
i = 2
1 ∓
√
1 − 4k
−1, i = 1, 2 в зависимости от k
k 0.24 0.22 0.209 0.207 0.205 0.203 0.20 0.16 0.12 0.08 0.04 0.01
1/λ1 2.5 3.058 3.356 3.413 3.472 3.534 3.623 5 7.194 11.364 23.810 100
1/λ2 1.667 1.486 1.425 1.414 1.404 1.395 1.381 1.250 1.161 1.096 1.044 1.011
Let us detail the coefficients Li in the expression for
Φ(x):
Φ(x) = L1Φ1(x) + L2Φ2(x) + L3Φ3(x),
L1 =
−2k
b
1
(λ1 − λ2)(λ1 − λ3)
+
+
1
2Lb
1 − 2(λ2 + λ3) + 4λ2λ3
(λ1 − λ2)(λ1 − λ3)
,
L2 =
−2k
b
1
(λ2 − λ3)(λ2 − λ1)
+
+
1
2Lb
1 − 2(λ3 + λ1) + 4λ3λ1
(λ2 − λ3)(λ2 − λ1)
,
L3 =
−2k
b
1
(λ3 − λ1)(λ3 − λ2)
+
+
1
2Lb
1 − 2(λ1 + λ2) + 4λ1λ2
(λ3 − λ1)(λ3 − λ2)
,
or (we take into account that λ3 = 1)
L1 =
1
b
1
(λ1 − 1)(λ1 − λ2)
−2k +
1
2L
(−1 + 2λ2)
,
L2 =
1
b
1
(λ2 − 1)(λ2 − λ1)
×
×
−2k +
1
2L
(−1 + 2λ1)
,
L3 =
1
b
1
(1 − λ1)(1 − λ2)
×
×
−2k +
1
2L
(1 − 2λ1 − 2λ2 + 4λ1λ2)
, (10)
where
L = 1 +
√b
6
, (a2 + b2) = k <
1
4
, 0 < 2b < 1.
Let us detail the coefficients di:
d1 =
4Bb
6M
λ1 − 1
2
, d2 =
4Bb
6M
λ2 − 1
2
,
d3 =
4Bb
6M
λ3 − 1
2
. (11)
Combinations diLj appear in the mixing matrix, so the parameter
b in the denominators in (10) will be canceled with the
parameter b in (11). Also, since the dimension of the quantity
B isM2 (inverse square meter), we can introduce substitutions
4B = 6rM2 ⇒ di = M
4B
6M2 b
λi − 1
2
=
= Mrb
λi − 1
2
= MDi,
where the quantitiesDi are dimensionless. With this in mind,
the cubic equation (9) is transformed to the form
−μ3 +M
"
3 − 1
λ1
− 1
λ2
− 1
λ3
+ L1D1 + L2D2+
+L3D3
#
μ2 +M2
"
−L1D1
2 − 1
λ2
− 1
λ3
−
−L2D2
2 − 1
λ1
− 1
λ3
− L3D3
2 − 1
λ1
−
− 1
λ2
− 3 + 2
1
λ1
+
1
λ2
+
1
λ3
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83
−
1
λ1
1
λ2
+
1
λ1
1
λ3
+
1
λ2
1
λ3
#
μ+
+M3
"
L1D1
1 − 1
λ2
1 − 1
λ3
+
+L2D2
1 − 1
λ3
1 − 1
λ1
+
+L3D3
1 − 1
λ1
1 − 1
λ2
+
+
1 − 1
λ1
1 − 1
λ2
1 − 1
λ3
#
= 0.
The roots of this equation can be found in the form μi =
MΔi, where Δi are dimensionless quantities; then the
equation for Δ1,Δ2,Δ3 takes the form
−Δ3+
"
3− 1
λ1
− 1
λ2
− 1
λ3
+L1D1+L2D2+L3D3
#
Δ2+
+
"
−L1D1
2 − 1
λ2
− 1
λ3
−L2D2
2 − 1
λ1
− 1
λ3
−
−L3D3
2 − 1
λ1
− 1
λ2
− 3 + 2
1
λ1
+
1
λ2
+
1
λ3
−
−
1
λ1
1
λ2
+
1
λ1
1
λ3
+
1
λ2
1
λ3
#
Δ+
+
"
L1D1
1 − 1
λ2
1 − 1
λ3
+
+L2D2
1 − 1
λ3
1 − 1
λ1
+
+L3D3
1 − 1
λ1
1 − 1
λ2
+
+
1 − 1
λ1
1 − 1
λ2
1 − 1
λ3
#
= 0.
Allow for that λ3 = 1, then we get
Δ3 +
"
2 − 1
λ1
− 1
λ2
+ L1D1 + L2D2 + L3D3
#
Δ2+
+
"
L1D1
1 − 1
λ2
+ L2D2
1 − 1
λ1
+
+L3D3
2 − 1
λ1
− 1
λ2
+ 3 − 2
1
λ1
+
1
λ2
+ 1
+
+
1
λ1
1
λ2
+
1
λ1
+
1
λ2
#
Δ−
−L3D3
1 − 1
λ1
1 − 1
λ2
= 0.
Since the parameter b enters as a multiplier intoD1,D2,D3
and the denominators of the expressions for the coefficients
L1,L2,L3, then in combinations
L1D1 + L2D2 + L3D3, L1D1, L2D2, L3D3
this parameter is reduced. With this in mind, we have the
identities
L1 =
1
(λ1 − 1)(λ1 − λ2)
−2k +
1
2L
(2λ2 − 1)
,
L2 =
1
(λ2 − 1)(λ2 − λ1)
−2k +
1
2L
(2λ1 − 1)
,
L3 =
1
(1 − λ1)(1 − λ2)
×
×
−2k +
1
2L
(1 − 2λ1 − 2λ2 + 4λ1λ2)
,
D1 =
r
2
(2λ1 − 1), D2 =
r
2
(2λ2 − 1), D3 =
r
2
,
where
λ1 =
1
2
1 −
√
1 − 4k
, λ2 =
1
2
1 +
√
1 − 4k
,
k ∈
1,
1
4
, L = 1 +
√1
6
, (a2 + b2) = k <
1
4
.
Also, we should take into account the identities
2λ1 − 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 expressions
L1 =
2
(1 +
√
1 − 4k)
√
1 − 4k
−2k +
1
2L
√
1 − 4k
,
L2 =
2
(1 +
√
1 − 4k)
√
1 − 4k
2k +
1
2L
√
1 − 4k
,
L3 =
1
k
−2k +
1
2L
(4k − 1)
,
D1 = −r
2
√
1 − 4k, D2 = +
r
2
√
1 − 4k, D3 =
r
2
,
L = 1 +
√b
6
, 0 < b <
1
2
.
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 properties
of the roots λ1, λ2)
A = 2 − 1
k
+ L1D1 + L2D2 + L3D3, C = L3D3,
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B = −L1D1
1 −
√
1 − 4k
1 +
√
1 − 4k
−
− L2D2
1 +
√
1 − 4k
1 −
√
1 − 4k
+ L3D3
2 − 1
k
+ 1.
Using identities
L1D1 =
2k − l
√
1 − 4k
1 +
√
1 − 4k
r,
L2D2 =
2k + l
√
1 − 4k
1 −
√
1 − 4k
r,
L3D3 = −
1 + l
1 − 4k
2k
r,
we transform the cubic equation to a simpler form
Δ3 +
2k − 1
k
Δ2 +
1 +
1 − l
1 − 4k
2k
r
Δ+
+
1 + l
1 − 4k
2k
r = 0.
Allowing for the expression for the parameter l:
l =
1
2L
=
√
6
2(
√
6 + b)
,
we obtain
Δ3 +
2k − 1
k
Δ2+
+
"
1 +
1 −
√
6
2(
√
6 + b)
1 − 4k
2k
!
r
#
Δ+
+
1 +
√
6
2(
√
6 + b
1 − 4k
2k
!
r = 0, (12)
where
0 < k <
1
4
, 0 < b < 2.
For definiteness, we set b1 = 0, b2 = 1, b3 = 2. The parameter
r was introduced by the relation 4B = 6rM2
from physical considerations, we will assume that the dimensionless
parameter r is small. Below we will follow several
situations r = 10−5, r = 10−3, r = 1. The last value
|r| = 1 corresponds to a very strong magnetic field. Numerical
study showed that dependence of the roots Δi upon
parameter b ∈ (0, 2) is very insignificant. By this reason below
we take the value b = 0. Let us construct tables of values
for the roots Δ1,Δ2,Δ3 equations (12) (see tables 3–5).
Table 3
The values of the rootsΔi, i = 1, 2, 3 of equation (12) for b = 0, r = 10−5
Таблица 3
Значения корнейΔi, i = 1, 2, 3 уравнения (12) для b = 0, r = 10−5
k 0.24 0.22 0.208 0.204 0.20 0.16 0.10 0.06 0.02 0.01
Δ1 0.667 0.485 0.419 0.400 0.382 0.250 0.127 0.069 0.021 0.010
Δ2 1.500 2.060 2.389 2.502 2.618 3.999 7.873 14.598 47.979 97.990
Δ3 -0.00001 -0.00001 -0.00001 -0.00001 -0.00001 -0.00002 -0.00002 -0.00004 -0.00012 -0.00024
Table 4
The values of the rootsΔi, i = 1, 2, 3 of equation (12) for b = 0, r = 10−3
Таблица 4
Значения корнейΔi, i = 1, 2, 3 уравнения (12) для b = 0, r = 10−3
k 0.24 0.22 0.208 0.204 0.20 0.16 0.12 0.08 0.04 0.01
Δ1 0.670 0.487 0.420 0.401 0.384 0.252 0.164 0.099 0.049 0.021
Δ2 1.489 2.059 2.388 2.502 2.617 3.999 6.171 10.404 22.957 97.990
Δ3 -0.001 -0.001 -0.001 -0.001 -0.001 -0.002 -0.002 -0.003 -0.005 -0.012
Finally, with an even greater increase in the parameter r, the roots become complex-valued
Table 5
The values of the rootsΔi, i = 1, 2, 3 of equation (12) for b = 0, r = 1
Таблица 5
Значения корнейΔi, i = 1, 2, 3 уравнения (12) для b = 0, r = 1
k 0.24 0.209 0.206 0.203 0.18 0.12 0.06 0.01
Δ1 1.265+1.129i 1.588+0.726i 1.625+0.656i 1.662+0.574i 1.173 0.695 0.503 0.406
Δ2 -0.362 -0.392 -0.395 -0.398 2.804 6.128 14.727 98.221
Δ3 1.265-1.129i 1.588-0.726i 1.625-0.656i 1.662-0.574i -0.422 -0.489 -0.653 -0.627
This means that at such magnetic field, the model becomes
non-interpretable.
Conclusions
In the present paper, the model of a fermion with three
mass parameters is studied in presence of the external
uniform magnetic field. After diagonalizing transformation,
three separate equations are obtained effectively for particles
with different anomalous magnetic moments. Their exact
solutions and generalized energy spectra are found. After
diagonalizing the mixing matrix, we reduce the problem
for three separated Dirac–like equations for particles with
different anomalous magnetic moment. It is shown that for
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85
a very strong magnetic field, the model becomes non-interpretable,
because the effective anomalous moments turns out
to be complex-valued.
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