We study the St¨uckelberg equation for a relativistic particle with two spin states S = 1 and S = 0 in the presence of an external uniform magnetic field. The particle is described by an 11-component wave function consisting of a scalar, a vector, and an antisymmetric tensor. On the solutions of the equation, the operators of energy, the third projection of the total angular momentum, and the third projection of the linear momentum along the direction of the magnetic field are diagonalized. After separation of variables, a system for 11 radial functions is obtained. Its solution is based on the use of the Fedorov-Gronsky method, in which all 11 radial functions are expressed in terms of three main functions. Exact solutions with cylindrical symmetry are constructed. Three series of energy levels are found.
St¨uckelberg particle, magnetic field, projective operators, Fedorov– Gronskiy method, exact solutions, bound states
Introduction
In the literature, the great interest can be noticed in
studying the Dirac-K¨ahler field [1,2]; also see [3–10]. This
field describes the complicated boson which contains the
fields 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 of
relativistic wave equations [16–20], the Dirac-K¨ahler field describes
a particle with a set of spin states: S = 1 and S = 0
(see [11–15]).
From this theory, by imposing additional constraints on
the 16 components of the Dirac-K¨ahler field, we can obtain
the usual theories for scalar and pseudoscalar particles, and
for vector and pseudovector particles:
(Φ, 0,Φa, 0, 0), S = 0;
(0, ˜Φ, 0, ˜Φa, 0), S = ˜0;
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(0, 0,Φa, 0,Φab), S = 1;
(0, 0, 0, ˜Φa,Φab), S = ˜1.
Also the system of equations describing the St¨uckelberg particle
is well known [1, 2]. In particular, this system can be obtained
from the Dirac-K¨ahler theory by imposing the following
constraints
(Φ, 0,Φa, 0,Φab), S = 0, 1;
(0, ˜Φ, 0, ˜Φa,Φab), S = ˜0,˜1.
There are possible two theories, corresponding to different
internal parities of the particle. In this paper the first variant
of the St¨uckelberg theory is studied.
We start with the known St¨uckelberg tensor system of 11
equations, which is transformed to a matrix form with the use
of the tetrad method [11, 12]. This equation is detailed in cylindric
system of coordinates and corresponding tetrad; wherein
we take into account the external uniform magnetic field. We
perform the separation of the variables and derive the system
of 11 equations in r variable. To resolve this system we
apply the method of Fedorov-Gronskiy [21], which is based on
projective operators referring to the third projection of the
11-dimensional spin matrix. According to this method we can
express all 11 functions through only three ones. We find 3 series
of physically interpretable energy levels, as solutions of
algebraic equations of order 1 and 3.
1. The basic equation
The 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 to
physical mass of the particle M is given by μ = −M. We
use the 11-dimensional wave function in the form
Φ = (Ψ;Ψ0,Ψ1,Ψ2,Ψ3;Ψ01,Ψ02,
Ψ03,Ψ23,Ψ31,Ψ12) = (H,H1,H2).
The above system can be presented in the matrix block form
DaGaH1 + 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 0
Δa 0 Ka
0 La 0
1
A, Φ =
0
@
H
H1
H2
1
A, (1)
where
Δ0 = (1, 0, 0, 0)t, Δ1 = (0, 1, 0, 0)t,
Δ2 = (0, 0, 1, 0)t, Δ3 = (0, 0, 0, 1)t,
K0 =
0
B@
0 0 0 0 0 0
−1 0 0 0 0 0
0 −1 0 0 0 0
0 0 −1 0 0 0
1
CA
,
K1 =
0
B@
−1 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 1
0 0 0 0 −1 0
1
CA
,
K2 =
0
B@
0 −1 0 0 0 0
0 0 0 0 0 −1
0 0 0 0 0 0
0 0 0 1 0 0
1
CA
,
K3 =
0
B@
0 0 −1 0 0 0
0 0 0 0 1 0
0 0 0 −1 0 0
0 0 0 0 0 0
1
CA
,
L0 =
0
BBBBB@
0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0
1
CCCCCA
,
L1 =
0
BBBBB@
−1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 −1
0 0 1 0
1
CCCCCA
,
L2 =
0
BBBBB@
0 0 0 0
−1 0 0 0
0 0 0 0
0 0 0 1
0 0 0 0
0 −1 0 0
1
CCCCCA
,
L3 =
0
BBBBB@
0 0 0 0
0 0 0 0
−1 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 0
1
CCCCCA
.
Here and below, t stands for transposition.
This matrix St¨uckelberg equation can be extended to the
Riemannian space-time in accordance with the known procedure.
To this end, for any given metric gαβ(x) we should
take the certain tetrad:
gαβ(x) → e(a)α(x), ηab = diag(1,−1,−1,−1),
then the equation (1) should have the structure
Γα(x)
∂
∂xα + Σα(x)
− μ
Ψ(x) = 0. (2)
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Local matrices Γα(x) and their blocks are determined with
the use of the tetrad
Γα(x) = eαa
(x)Γa =
0
@
0 −Gaeα (a) 0
Δaeα(
a) 0 Kaeα (a)
0 Laeα(
a) 0
1
A.
The connection Σα(x) is defined by the formulas
Jab =
0
@
0 0 0
0 Jab
1 0
0 0 Jab
2
1
A,
Σα(x) =
1
2
Jabeβ
(a)(x)e(b)β;α(x) =
=
0
@
0 0 0
0 (Σ1)α 0
0 0 (Σ2)α
1
A,
Σ1(x) =
1
2
Jab
(1)eβ
(a)(x)e(b)β;α(x),
Σ2(x) =
1
2
Jab
(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 coefficients
Γc
eα(
c)
∂
∂xα +
1
2
Jabγabc
− μ
Ψ(x) = 0. (3)
Recall that γ[ab]c = −γ[ba]c = e(b)ρσeρ
(a)eσ(
c). In detailed
form, Eq. (3) reads
−Gceα(
c)∂α − GcJab
(1)
1
2
γabc
H1 − μH = 0,
Δceα(
c)∂αH+
Kceα (c)∂α + KcJab
(2)
1
2
γabc
H2−μH1 = 0,
Lceα(
c)∂α + LcJab
(1)
1
2
γabc
H1 − μH2 = 0.
Let us consider the St¨uckelberg equation in presence of
the external uniform magnetic field. In cylindrical coordinates
with the use of the diagonal tetrad
xα = (t, r, ϕ, z), dS2 = dt2 − dr2 − r2dϕ2 − dz2,
Aϕ = −Br2
2
the above equation takes the form (let eB ⇒ B):
Γ0 ∂
∂t
+ Γ1 ∂
∂r
+ Γ2 ∂ϕ + iBr2/2 + J12
r
+
+Γ3 ∂
∂z
− μ
Ψ = 0.
In block form, it reads
−G0 ∂
∂t
− G1 ∂
∂r
− G2 1
r
∂
∂ϕ
+
iBr2
2
+ j12
1
−
−G3 ∂
∂z
H1 − μH = 0,
Δ0 ∂
∂t
+ Δ1 ∂
∂r
+
+ Δ2 1
r
∂ϕ +
iBr2
2
+ Δ3 ∂
∂z
H+
+
"
K0 ∂
∂t
+ K1 ∂
∂r
+ K2 ∂ϕ + iBr2
2 + j12
1
r
+
+ K3 ∂
∂z
H2 = μH1,
"
L0 ∂
∂t
+ L1 ∂
∂r
+ L2 ∂ϕ + iBr2
2 + j12
1
r
+
+ L3 ∂
∂z
H1 = μH2.
2. Cyclic basis
In the following, it will be convenient to apply the cyclic
basis (all quantities referring to it are marked by the overline).
In such a basis, the generators j12
1 and j12
2 are diagonal. The
necessary transformation ¯H1 = UH1 is determined by the
matrix U :
U =
0
BB@
1 0 0 0
0 −√1
2
√i
2
0
0 0 0 1
0 √1
2
√i
2
0
1
CCA
,
U
−1 =
0
BB@
1 0 0 0
0 −√1
2
0 √1
2
0 −√i
2
0 −√i
2
0
0 0 1 0
1
CCA
.
Correspondingly, the generators for tensor representation are
defined by the rule
¯ Jab
1 = UjabU
−1, ¯ Jab
2 = ¯j ab ⊗ I + I ⊗¯j ab.
Let us transform the generators to cyclic form
¯j
12 =
0
B@
0 0 0 0
0 −i 0 0
0 0 0 0
0 0 0 i
1
CA
,
¯ J12
2 =
0
BBBBB@
−1 · · · · ·
· 0 · · · ·
· · 1 · · ·
· · · 1 · ·
· · · · 0 ·
· · · · · −1
1
CCCCCA
.
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We should also transform the main matrices Γa of the equation
to the cyclic form. Starting with the formulas
¯H
= H, ¯H1 = C1H1, (C1 = U),
¯H
2 = U ⊗ UH2 = C2H2,
we derive the rule
0
@
0 −¯Ga 0
¯Δ
a 0 ¯K a
0 ¯La 0
1
A =
=
0
@
0 −GaC
−1
1 0
C1Δa 0 C1KaC
−1
2
0 C2LaC
−1
1 0
1
A.
First, we find the matrices ¯Δa = C1Δa and ¯Ga =
GaC
−1
1 :
¯Δ
0 = (1, 0, 0, 0)t, ¯Δ1 =
0,−√1
2
, 0,
√1
2
t
¯Δ
2 =
0,
√i
2
, 0,
√i
2
t
, ¯Δ3 = (0, 0, 1, 0)t;
¯G
0 = (1, 0, 0, 0), ¯G1 =
0,
√1
2
, 0,−√1
2
,
¯G
2 =
0,
√i
2
, 0,
√i
2
, ¯G3 = (0, 0,−1, 0).
Having in mind the formula for C1, we can derive expressions
for 6-dimensional transformation for C2:
U ⊗ U ⇒ C2, ¯H2 = C2H2 ⇒
C2 =
0
BBBBBBB@
−√1
2
√i
2
0 0 0 0
0 0 1 0 0 0
√1
2
√i
2
0 0 0 0
0 0 0 −√i
2
√1
2
0
0 0 0 0 0 i
0 0 0 √i
2
√1
2
0
1
CCCCCCCA
,
C
−1
2 =
0
BBBBBBB@
−√1
2
0 √1
2
0 0 0
−√i
2
0 −√i
2
0 0 0
0 1 0 0 0 0
0 0 0 √i
2
0 −√i
2
0 0 0 √1
2
0 √1
2
0 0 0 0 −i 0
1
CCCCCCCA
.
With the use of them we can obtain all other blocks in cyclic
form:
¯K
0 =
0
B@
0 0 0 0 0 0
−1 0 0 0 0 0
0 −1 0 0 0 0
0 0 −1 0 0 0
1
CA
,
¯K
1 =
0
BBB@
√1
2
0 −√1
2
0 0 0
0 0 0 0 √1
2
0
0 0 0 −√1
2
0 −√1
2
0 0 0 0 √1
2
0
1
CCCA
,
¯K
2 =
0
BBB
@
√i
2
0 √i
2
0 0 0
0 0 0 0 −√i
2
0
0 0 0 √i
2
0 −√i
2
0 0 0 0 √i
2
0
1
CCCA
,
¯K
3 =
0
B@
0 −1 0 0 0 0
0 0 0 0 0 −1
0 0 0 0 0 0
0 0 0 1 0 0
1
CA
;
¯L
0 =
0
BBBBB@
0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0
1
CCCCCA
,
¯L
1 =
0
BBBBBBB@
√1
2
0 0 0
0 0 0 0
−√1
2
0 0 0
0 0 −√1
2
0
0 √1
2
0 √1
2
0 0 −√1
2
0
1
CCCCCCCA
,
¯L
2 =
0
BBBBBBB@
−√i
2
0 0 0
0 0 0 0
−√i
2
0 0 0
0 0 −√i
2
0
0 √i
2
0 −√i
2
0 0 √i
2
0
1
CCCCCCCA
,
¯L3 =
0
BBBBB@ 0
0
0
0
−1 0 0 0
0 0 0 0
0 0 0 1
0 0 0 0
0 −1 0 0
1
CCCCCA
.
3. Separating the variables
We apply the following substitution for the wave function
(in cyclic basis)
¯Ψ
= e
−iϵteimϕeikz
0
@
¯H¯H
1
¯H
2
1
A, ¯H = h(r),
¯H
1 =
0
B@
h0(r)
h1(r)
h2(r)
h3(r)
1
CA
, ¯H2 =
Ei(r)
Bi(r)
.
After a simple calculation we derive the system of 11 equations.
With the use of notations
am =
d
dr
+
m + Br2/2
r
, am+1 =
d
dr
+
m + 1 + Br2/2
r
,
bm =
d
dr
−m + Br2/2
r
, bm−1 =
d
dr
−m − 1 + Br2/2
r
,
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it 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 method
We will apply the Fedorov-Gronskiy method [21]. To this
end, let us consider the third projection of 11-dimensional spin
operator Y = −i ¯ J12. We verify that it satisfies the minimal
cubic equation, Y (Y − 1)(Y + 1) = 0. This permits us to
introduce three projective operators
P1 =
1
2
Y (Y − 1), P2 =
1
2
Y (Y + 1), P3 = 1 − Y 2
with the properties P2
0 = P0, P2
+1 = P+1, P2−
1 = P−1,
P0 + P+1 + P−1 = 1.
Therefore, the complete wave function may be decomposed
into the sum of three parts
Ψ = Ψ0 + Ψ+1 + Ψ−1,
Ψσ = PσΨ, σ = 0, +1,−1.
We can readily find an explicit form of them (according to the
Fedorov-Gronskiy method, each projective part should be determined
by 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 obtain
for P1
−amh + amB2 − ikB3 + iϵE1 = μh1,
amh0 − iϵh1 = μE1,
−ikh1 − amh2 = μB3;
for P2
bmh + 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 to
transform 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 P2
bmf3(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,
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−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 constraints
bm−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 separate
functions:
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 equations
take on the form
bm−1f1(r) = C1f3, amf3 = C1f1,
am+1f2(r) = C2f3, bmf3 = C2f2;
[bm−1am − C2
1 ]f3 = 0, [ambm−1 − C2
1 ]f1 = 0,
[am+1bm − C2
2 ]f3 = 0, [bmam+1 − C2
2 ]f2 = 0.
(5)
In explicit form, Eqs. (5) read
d2
dr2 +
1
r
d
dr
− B2r2
4
− m2
r2
−
−Bm + B − C2
1
f3 = 0,
d2
dr2 +
1
r
d
dr
− B2r2
4
− (m − 1)2
r2
−
−Bm − C2
1
f1 = 0,
d2
dr2 +
1
r
d
dr
− B2r2
4
− m2
r2
−
−Bm − B − C2
2
f3 = 0,
d2
dr2 +
1
r
d
dr
− B2r2
4
− (m + 1)2
r2
−
−Bm − C2
2
f2 = 0.
So we get the following identity C2
2 = C2
1
− 2B and only
three different equations:
(1)
d2
dr2 +
1
r
d
dr
− B2r2
4
− m2
r2
−
− Bm + B − C2
1
f3 = 0,
(2)
d2
dr2 +
1
r
d
dr
− B2r2
4
−
−(m − 1)2
r2
− Bm − C2
1
f1 = 0,
(3)
d2
dr2 +
1
r
d
dr
− B2r2
4
−
−(m + 1)2
r2
− Bm − C2
1 + 2B
f2 = 0.
Let B − C2
1 = X, then these equations are written in a
more symmetrical form
(1)
d2
dr2 +
1
r
d
dr
− B2r2
4
− m2
r2
− Bm + X
f3 = 0,
(2)
d2
dr2 +
1
r
d
dr
− B2r2
4
−
−(m − 1)2
r2
− B(m + 1) + X
f1 = 0,
(3)
d2
dr2 +
1
r
d
dr
− B2r2
4
−
−(m + 1)2
r2
− B(m − 1) + X
f2 = 0.
With the new variable x = Br2
2 , they take on the form
(1)
d2
dx2 +
1
x
d
dx
− 1
4
− (m/2)2
x2 +
+
1
x
−m
2
+
X
2B
f3 = 0,
(2)
d2
dx2 +
1
x
d
dx
− 1
4
− [(m − 1)/2]2
x2 +
+
1
x
−m + 1
2
+
X
2B
f1 = 0,
(3)
d2
dx2 +
1
x
d
dx
− 1
4
− [(m + 1)/2]2
x2 +
+
1
x
−m − 1
2
+
X
2B
f2 = 0.
It is sufficient to consider only the first equation in detail. Let
us search for solutions in the form f3(x) = xAeCxF(x),
then we readily obtain
xF
′′
+(2A+1+2Cx)F
′
+
A2 − (m/2)2
x
+ 2AC+
+C − m
2
+
X
2B
+ x
C2 − 1
4
F = 0.
Let us impose restrictions
A2 − (m/2)2 = 0 ⇒ A = ±|m/2|,
74
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C2 − 1
4
= 0 ⇒ C + ±1
2
.
In order to have the equations referring to bound the states,
we should assume
A = ±|m/2|, C = −1
2
.
This results in the equation of a confluent hypergeometric
type
xF
′′
+(|m|+1−x)F
′−
|m| + m
2
+
1
2
− X
2B
F = 0
with parameters
a =
|m| + m
2
+
1
2
− X
2B
,
c = |m| + 1, F = Φ(a, c, x).
The polynomial condition a = −n1 leads to
(3) ⇒ X = +2B
|m| + m
2
+
1
2
+ 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| + m
2
+
1
2
+ 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 + 1
2
+
1
2
+ 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 − 1
2
+
1
2
+ n3
> B,
n2 = 0, 1, 2, . . .
(8)
The quantity X in all three cases (6)–(8) should be the
same which assumes existence of some correlations within
n − 1, n2, n3. Bellow we will apply the simplest quantization
rule
X = 2BN > 0, N =
|m| + m
2
+
1
2
+ n
,
N =
1
2
,
3
2
, · · · .
5. Solving the algebraic system
Let 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 that
C1 = 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 equation
μ3(k2+μ2−2X −ϵ2)
h
−2B2(5k2+μ2−2X −5ϵ2)+
+B(−k2−μ2+2X+ϵ2)(2
p
X2 − B2−k2−μ2+ϵ2)−
−(k2 + μ2 − 2X − ϵ2)2
p
X2 − B2 − k2+
+μ2 + X + ϵ2
i
= 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
−
p
X2 − B2 + B + μ2 − 5X
+
+W
h
2B
p
X2 − B2 − μ2 + X
−
−(2X − μ2)
2
p
X2 − B2 + μ2 + 4X
+ 10B2
i
+
+(2X − μ2)
h
B
2
p
X2 − B2 − μ2
−
−(2X − μ2)
p
X2 − B2 + μ2 + X
+ 2B2
i
= 0.
With dimensionless variables
W
μ2
⇒ w,
X
μ2
⇒ x,
B
μ2
⇒ b,
μ
μ
⇒ 1,
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www.izvestia.komisc.ru 75
it takes on the form
w3 − w2
−
p
x2 − b2 + b + 1 − 5x
−
−w
h
2b
p
x2 − b2 − 1 + x
−
−(2x − 1)
2
p
x2 − b2 + 1 + 4x
+ 10b2
i
+
+(1 − 2x)
h
b
2
p
x2 − b2 − 1
+
+(1 − 2x)
p
x2 − b2 + 1 + x
+ 2b2
i
= 0.
Its analytical solutions are found straightforwardly, but they
are helpless. By this reason, let us study its solutions numerically.
Recall that x = 2bN. First let us consider the simple
case w = 1−2x = 1−4bN, 1−2x > 0. For several typical
examples we find the roots forw (physically interpretable
are 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 = 0.001
0
BBBBBBBBB@
−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 0.971972
1
CCCCCCCCCA
,
b = 0.01
0
BBBBBBBBB@
−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 0.716955
1
CCCCCCCCCA
,
b = 0.05
0
BBBBBBBBB@
−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 − 0.0709556i
1
0.78919
0.576651
0.361149
0.138565
−0.0934233 + 0.0312522i
−0.2929 + 0.0548385i
−0.49238 + 0.0709556i
1
CCCCCCCCCA
.
In the second case, we can see only two positive roots. In
total, we have 3 physically interpretable roots and the corresponding
energy series.
Conclusion
For better understanding of the problem, we will consider
the nonrelativistic approximation for this model in a separate
paper.
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