Россия
Исследуется уравнение Штюкельберга для релятивист- ской частицы с двумя спиновыми состояниями S = 1 и S=0 в присутствии внешнего однородного магнитного по- ля. Частица описывается 11-компонентной волновой функ- цией, состоящей из скаляра, вектора и антисимметричного тензора. На решениях уравнения диагонализируются опе- раторы энергии, третьей проекции полного углового мо- мента и третьей проекции линейного момента вдоль на- правления магнитного поля. После разделения перемен- ных получена система для 11 радиальных функций. Ее ре- шение основано на использовании метода Федорова-Грон- ского, в рамках которого все 11 радиальных функций выра- жаются через три основные функции. Построены точные решения с цилиндрической симметрией. Найдены три се- рии уровней энергии.
частица Штюкельберга, магнитное поле, проективные опе- раторы, метод Федорова–Гронского, точные решения, свя- занные состояния
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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