1. Introduction
The scattering states of the relativistic and nonrelativistic wave equation in recent times have received great attention in physics [1–9]. Scattering and bound state solutions of asymmetric Hulthen potential have been obtained by Arda and Sever [10]. Also, the bound state and scattering state of KleinGordon equation with effective mass formalism have been studied by Aydoğdu et al. [11]. Aydoğdu et al. [12] examined the scattering and bound states of massive Dirac equation with WoodsSaxon potential. Rojas and Vilalba [13] in their investigation had shown the relation between the bound state energy eigenvalues and transmission resonances for the KleinGordon particle with WoodsSaxon potential as the same for the Dirac particle. Hassanabadi et al. [14, 15] investigated the scattering state of relativistic spinless particles with Kinklike potential and scattering of WoodsSaxon potential in the framework of minimal length. The scattering states of relativistic and nonrelativistic particles with WoodsSaxon potential and mass dependent WoodsSaxon potential have been studied by Alpdoğan et al. [16] and Arda et al. [17]. Recently, Alpdoğan studied the scattering of massive Dirac particles with generalized asymmetric WoodsSaxon potential [18]. Yanar et al. studied the scattering and bound state of DuffinKemmerPetiau particles with qparameter hyperbolic PoschlTeller (qHPT) potential [19]. The qHPT potential is defined as follows:
(1)
V
x
=
4
λ
λ

1
cosh
q
2
α
x
=
4
λ
λ

1
·
θ

x
e
2
α
x
1
+
q
e
2
α
x
2
+
θ
x
e

2
α
x
1
+
q
e

2
α
x
2
,
where
θ
(
x
)
is the step function,
q
is the deformation parameter and
q
≠
0
,
λ
is the height of the potential,
λ
≠
0
,
λ
≠
1
, and
α
is the range of the potential barrier. This potential plays an important role in describing the interactions in molecular, atomic, and nuclear physics. One of the findings of physics is to understand the structures of nucleus, atoms, molecules, and other material objects. Thus, the sole aim of physicists is to create unique models that contain the physically motivated potential that will describe the interactions between particles. A few numbers of these potential models have been identified to describe the interaction in the nuclei and diatomic and polyatomic molecules [20, 21]. The applications of the PoschlTellerlike type potential for analyzing the bound states energy of
Λ
particle in hypernuclei in nuclear physics had been reported [19]. Furthermore, the qHPT potential may found much usefulness in molecular, atomic, and nuclear physics.
Understanding of our knowledge about fine scale systems has been gained by investigations into scattering and bounded state of such systems. Therefore, the scattering problem has become an interesting topic in relativistic or nonrelativistic quantum mechanics. In the case of nonrelativistic scattering problem, it has been shown that transmission and reflection coefficients take 1 and 0, respectively, as external potential has well behaved at infinity for the zero energy limit [22–24]. However, reflection coefficient goes to zero while transmission coefficient goes to unity in the zero energy limit when external potential supports a halfbound state. This situation was called as transmission resonance [25] which is escorted by fluctuations phenomena. The transmission resonance concept has been recently generalized to the relativistic case [26].
In this work, we attempt to study the solution of scattering state of the KleinGordon equation with equal vector and scalar (qHPT) potential. Our aim will be to calculate in detail the reflection (
R
) and transmission (
T
) coefficients and obtain the bound state solution of the qHPT potential using the equation of continuity of the wave function.
2. Scattering State Solutions of KleinGordon Equation for the qHPT Potential
The timeindependent KleinGordon equation with equal scalar
S
(
x
)
and vector
V
(
x
)
potentials can be written as [12]
(2)
d
2
ψ
x
d
x
2
+
E
2

m
2

2
E
+
m
V
x
ψ
x
=
0
,
ħ
=
c
=
1
,
where
E
is the relativistic energy of the particles and
m
is the mass of the particle. In order to obtain the scattering solution for qHPT potential, we consider both
x
<
0
and
x
>
0
at
E
>
m
. We intend to study the scattering of (2); we now seek the wave function for the case
x
<
0
. By substituting (1) into (2), we have
(3)
d
2
ψ
x
d
x
2
+
E
2

m
2

8
λ
λ

1
E
+
m
e
2
α
x
1
+
q
e
2
α
x
2
ψ
x
=
0
.
To solve (3), we used a new variable defined as
z
=

q
e
2
α
x
which yields
(4)
z
1

z
d
2
ψ
d
z
2
+
1

z
d
ψ
d
z
+
1
z
1

z
γ
1
z
2
+
γ
2
z
+
γ
ψ
z
=
0
,
where
(5)
γ
1
=
1
4
α
2
E
2

m
2
,
γ
2
=
2
α
2
E
+
m
λ
λ

1
q
+
m
2

E
2
4
,
γ
3
=
1
4
α
2
E
2

m
2
.
Defining the wave function in (4) as
ψ
(
z
)
=
z
μ
(
1

z
)
ν
φ
(
z
)
, then (4) turns into the hypergeometric differential equation [27]
(6)
z
1

z
d
2
φ
d
z
2
+
1
+
2
μ

1
+
2
μ
+
2
ν
z
d
φ
d
z

μ
+
ν
+
δ
μ
+
ν

δ
φ
z
=
0
,
where
μ
,
ν
, and
δ
are defined as follows:
(7)
μ
=
i
k
2
α
,
ν
=
1
2
±
1
2
1

8
E
+
m
λ
λ

1
α
2
q
,
δ
=
i
k
2
α
,
k
2
=
E
2

m
2
.
Regarding properties of hypergeometric functions the minus sign should be chosen for
ν
. The solutions of (6) can be written in terms of the hypergeometric function as follows:
(8)
φ
z
<
0
=
A
F
2
1
μ
+
ν
+
δ
,
μ
+
ν

δ
,
1
+
2
μ
,
z
+
B
z
2

2
μ
F
1

μ
+
ν
+
δ
,

μ
+
ν

δ
,
1

2
μ
,
z
.
The general solutions for
x
<
0
are given as
(9)
ψ
z
<
0
=
A
z
μ
1

z
ν
·
F
2
1
μ
+
ν
+
δ
,
μ
+
ν

δ
,
1
+
2
μ
,
z
+
B
z

μ
1

z
ν
·
F
2
1

μ
+
ν
+
δ
,

μ
+
ν

δ
,
1

2
μ
,
z
.
Next, we investigate the solution for
x
>
0
. Again substituting (1) into (2) yields
(10)
d
2
ψ
x
d
x
2
+
E
2

m
2

8
λ
λ

1
E
+
m
e

2
α
x
1
+
q
e

2
α
x
2
ψ
x
=
0
.
Defining the new variable
z
~
=

q
e

2
α
x
and redefining the ansatz for the wave function as
ψ
(
z
~
)
=
z
~
μ
~
1

z
~
ν
~
f
(
z
~
)
, the solutions of (10) become
(11)
ψ
z
~
>
0
=
C
z
~
μ
~
1

z
~
ν
~
·
F
2
1
μ
~
+
ν
~
+
δ
~
,
μ
~
+
ν
~

δ
~
,
1
+
2
μ
~
,
z
~
+
D
z
~

μ
~
1

z
~
ν
~
·
F
2
1

μ
~
+
ν
~
+
δ
~
,

μ
~
+
ν
~

δ
~
,
1

2
μ
~
,
z
~
,
where
(12)
μ
~
=
i
k
2
α
,
ν
~
=
1
2
±
1
2
1

8
E
+
m
λ
λ

1
q
α
2
,
δ
~
=
i
k
2
ħ
,
k
2
=
E
2

m
2
.
Here we also should choose the minus sign for
ν
~
. We seek for the physical results of the problem under investigation; therefore in order to get this physical result, the solutions obtained have to be used with appropriate boundary conditions as
x
→

∞
and
x
→
+
∞
. Applying the asymptotic solution to (9) for
x
→

∞
, as
z
L
→
0
and
1

z
ν
→
1
, becomes
(13)
ψ
L
x
⟶

∞
=
A

1
q
i
k
/
2
α
e
i
k
+
B

1
q

i
k
/
2
α
e

i
k
.
In order to find a plane wave traveling from left to right, we set
C
=
0
in (11) and the asymptotic behavior of the right solution becomes
(14)
ψ
R
x
⟶
∞
=
D

q

i
k
/
2
α
ħ
c
e
i
k
/
ħ
c
.
Now in order to give explicit expressions for the coefficients, we use the continuity conditions of the wave function and its derivative defined as
(15)
ψ
L
x
=
0
=
ψ
R
x
=
0
,
ψ
L
′
x
=
0
=
ψ
R
′
x
=
0
,
where the prime denotes differential with respect to
x
. Applying these conditions on the wave function and matching the wave function at
x
=
0
, we get
(16)
A
Q
1
U
1
+
B
Q
2
U
2
=
D
Q
3
U
3
(17)
A
Q
4
U
4
+
Q
5
U
5
+
Q
6
S
1
U
6
+
B
Q
7
U
7
+
Q
8
U
8
+
Q
9
S
2
U
9
=
D
Q
10
U
10
+
Q
11
U
11
+
Q
12
S
3
U
12
,
where the following abbreviations have been used:
(18)
Q
1
=

q
μ
1
+
q
ν
,
Q
2
=

q

μ
1
+
q
ν
,
Q
3
=

q

μ
~
1
+
q
ν
~
,
Q
4
=

2
α
μ

q
μ
1
+
q
ν
,
Q
5
=
2
α
ν

q
μ
+
1
1
+
q
ν

1
,
Q
6
=

2
α

q
μ
+
1
1
+
q
ν
,
Q
7
=
2
α
μ

q

μ
1
+
q
ν
,
Q
8
=
2
α
ν

q

μ
+
1
1
+
q
ν

1
,
Q
9
=

2
α

q

μ
+
1
1
+
q
ν
,
Q
10
=

2
α
μ
~

q

μ
~
1
+
q
ν
~
,
Q
11
=

2
α
ν
~

q

μ
~
+
1
1
+
q
ν
~

1
,
Q
12
=
2
α

q

μ
~
+
1
1
+
q
ν
~
,
S
1
=
μ
+
ν
+
δ
μ
+
ν

δ
1
+
2
μ
,
S
2
=

μ
+
ν
+
δ

μ
+
ν

δ
1

2
μ
,
S
3
=

μ
~
+
ν
+
δ

μ
~
+
ν

δ
1

2
μ
~
,
U
1
=
F
2
1
μ
+
ν
+
δ
,
μ
+
ν

δ
,
1
+
2
μ
,

q
,
U
2
=
F
2
1

μ
+
ν
+
δ
,

μ
+
ν

δ
,
1

2
μ
,

q
,
U
3
=
F
2
1

μ
~
+
ν
~
+
δ
~
,
μ
~
+
ν
~

δ
~
,
1

2
μ
~
,

q
,
U
4
=
F
2
1
μ
+
ν
+
δ
,
μ
+
ν

δ
,
1
+
2
μ
,

q
,
U
5
=
F
2
1
μ
+
ν
+
δ
,
μ
+
ν

δ
;
1
+
2
μ
,

q
,
U
6
=
F
2
1
μ
+
ν
+
δ
+
1
,
μ
+
ν

δ
+
1
;
2
+
2
μ
,

q
,
U
7
=
F
2
1

μ
+
ν
+
δ
,

μ
+
ν

δ
,
1

2
μ
,

q
,
U
8
=
F
2
1

μ
+
ν
+
δ
,

μ
+
ν

δ
,
1

2
μ
,

q
,
U
9
=
F
2
1

μ
+
ν
+
δ
+
1
,

μ
+
ν

δ
+
1,2

2
μ
,

q
,
U
10
=
F
2
1

μ
~
+
ν
~
+
δ
~
,

μ
~
+
ν
~

δ
~
,
1

2
μ
~
,

q
,
U
11
=
F
2
1

μ
~
+
ν
~
+
δ
~
,

μ
~
+
ν
~

δ
~
,
1

2
μ
~
,

q
,
U
12
=
F
2
1

μ
~
+
ν
~
+
δ
~
+
1
,

μ
~
+
ν
~

δ
~
+
1,2

2
μ
~
,

q
.
And the property of hypergeometric function
d
/
d
s
F
2
1
a
,
b
,
c
,
s
=
a
b
/
c
F
2
1
a
+
1
,
b
+
1
,
c
+
1
,
s
[28] has been used in obtaining (17). The onedimensional current density for the KleinGordon equation in terms of the relativistic units (
ħ
=
c
=
1
) is given by [29]
(19)
J
x
=
i
2
m
ψ
∗
∇
x
ψ

ψ
∇
x
ψ
∗
.
The reflection and transmission coefficients are defined in terms of current density as follows:
(20)
R
=
j
ref
j
inc
=
B
A
2
,
T
=
j
trans
j
inc
=
D
A
2
.
Now using these equations and after a tedious algebra, we obtain the reflection and transmission coefficients as
(21)
R
=
B
A
2
=
Q
4
U
4
+
Q
5
U
5
+
Q
6
S
1
U
6

Q
1
U
1
/
Q
3
U
3
Q
10
U
10
+
Q
11
U
11
+
Q
12
S
3
U
12
Q
2
U
2
/
Q
3
U
3
U
10
Q
10
+
Q
11
U
11
+
Q
12
S
3
U
12

Q
7
U
7
+
Q
8
U
8
+
Q
9
S
2
U
9
2
,
T
=
D
A
2
=
Q
2
U
2
/
Q
3
U
3
Q
4
U
4
+
Q
5
U
5
+
Q
6
S
1
U
6

Q
1
U
1
/
Q
3
U
3
Q
7
U
7
+
Q
8
U
8
+
Q
9
S
2
U
9
Q
2
U
2
/
Q
3
U
3
U
10
Q
10
+
Q
11
U
11
+
Q
12
S
3
U
12

Q
7
U
7
+
Q
8
U
8
+
Q
9
S
2
U
9
2
.
We have depicted in Figure 5 that the sum of reflection and transmission coefficients approaches unity.
3. Bound State Solutions of KleinGordon Equation for the qHPT Potential
In order to find the bound state solutions for the KleinGordon particle with qHPT potential, we map
8
λ
(
λ

1
)
→

V
0
.
3.1. Bound State Solutions in the Negative Region <inlineformula>
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M63">
<mml:mo stretchy="false">(</mml:mo>
<mml:mi>x</mml:mi>
<mml:mo><</mml:mo>
<mml:mn mathvariant="normal">0</mml:mn>
<mml:mo stretchy="false">)</mml:mo></mml:math>
</inlineformula>
The bound state solutions can be found in this region by changing the variable
z
=

q
e
2
α
x
and taking into consideration
8
λ
(
λ

1
)
→

V
0
and (3) becomes
(22)
z
1

z
d
2
ψ
d
z
2
+
1

z
d
ψ
d
z
+
1
z
1

z
β
1
z
2
+
β
2
z
+
β
ψ
z
=
0
,
where
(23)
β
1
=
1
4
α
2
E
2

m
2
,
β
2
=

1
2
α
2
V
0
E
+
m
q
+
E
2

m
2
,
β
3
=
1
4
α
2
E
2

m
2
.
Taking a wave function of the form
ψ
(
z
)
=
z
μ
1
(
1

z
)
ν
1
φ
(
z
)
, then (22) turns into the hypergeometric differential equation [27]
(24)
z
1

z
d
2
φ
d
z
2
+
1
+
2
μ
1

1
+
2
μ
1
+
2
ν
1
z
d
φ
d
z

μ
1
+
ν
1
+
δ
1
μ
1
+
ν
1

δ
1
φ
z
=
0
with
μ
1
,
ν
1
, and
δ
1
defined as follows:
(25)
μ
1
=
i
k
2
α
,
ν
1
=
1
2
±
1
2
1
+
E
+
m
V
0
α
2
q
,
δ
1
=
i
k
2
α
,
k
2
=
E
2

m
2
.
We choose minus sign for
ν
1
in equation (25). The general solutions for
x
<
0
are given as
(26)
ψ
z
<
0
=
A
1
z
μ
1
1

z
ν
1
·
F
2
1
μ
1
+
ν
1
+
δ
1
,
μ
1
+
ν
1

δ
1
,
1
+
2
μ
1
,
z
+
B
1
z

μ
1
1

z
ν
1
·
F
2
1

μ
1
+
ν
1
+
δ
1
,

μ
1
+
ν
1

δ
1
,
1

2
μ
1
,
z
.
3.2. Bound State Solutions in the Positive Region <inlineformula>
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M77">
<mml:mo stretchy="false">(</mml:mo>
<mml:mi>x</mml:mi>
<mml:mo>></mml:mo>
<mml:mn mathvariant="normal">0</mml:mn>
<mml:mo stretchy="false">)</mml:mo></mml:math>
</inlineformula>
In the positive region, we defined the variable
z
~
=

q
e

2
α
x
with
8
λ
(
λ

1
)
→

V
0
and taking the wave function
ψ
(
z
~
)
=
z
~
μ
~
1
1

z
~
ν
~
1
φ
(
z
~
)
and following the same procedures as the case of negative region, we obtain the solutions for the positive region as follows:
(27)
ψ
z
~
>
0
=
C
1
z
~
μ
~
1
1

z
~
ν
~
1
·
F
2
1
μ
~
1
+
ν
~
1
+
δ
~
1
,
μ
~
1
+
ν
~
1

δ
~
1
,
1
+
2
μ
~
1
,
z
~
+
D
1
z

μ
~
1
1

z
~
ν
~
1
·
F
2
1

μ
~
1
+
ν
~
1
+
δ
~
1
,

μ
~
1
+
ν
~
1

δ
~
1
,
1

2
μ
~
1
,
z
~
,
where
(28)
μ
~
1
=
i
k
2
α
,
ν
~
1
=
1
2
±
1
2
1
+
2
E
+
m
c
2
V
0
α
2
q
,
δ
~
1
=
i
k
2
α
,
k
2
=
E
2

m
2
.
And as mentioned above the minus sing should be chosen. In order to find the equation for the energy eigenvalues, we set
B
=
D
=
0
and used the conditions of continuity for the wave function as
ψ
L
(
x
=
0
)
=
ψ
R
(
x
=
0
)
,
ψ
L
′
(
x
=
0
)
=
ψ
R
′
(
x
=
0
)
and we get
(29)
A
1
a
1
K
1
=
C
1
b
1
K
2
,
A
1
a
2
K
3
+
a
3
K
4
+
a
5
K
5
=
C
1
b
2
K
6
+
b
3
K
7
+
b
4
K
8
,
where
(30)
a
1
=

q
μ
1
1
+
q
ν
1
,
K
1
=
F
2
1
μ
1
+
ν
1
+
δ
1
,
μ
1
+
ν
1

δ
1
,
1
+
2
μ
1
;

q
,
b
1
=

q
μ
~
1
1
+
q
ν
~
1
,
K
2
=
F
2
1
μ
~
1
+
ν
~
1
+
δ
~
1
,
μ
~
1
+
ν
~
1

δ
~
1
,
1
+
2
μ
~
1
;

q
,
a
2
=
2
α
μ
1

q
μ
1
1
+
q
ν
1
,
a
3
=
2
α
ν

q
μ
1
+
1
1
+
q
ν
1

1
,
a
4
=
2
α

q
μ
1
+
1
1
+
q
ν
1
,
K
3
=
F
2
1
μ
1
+
ν
1
+
δ
1
,
μ
1
+
ν
1

δ
1
,
1
+
2
μ
1
;

q
,
K
4
=
F
2
1
μ
1
+
ν
1
+
δ
1
,
μ
1
+
ν
1

δ
1
,
1
+
2
μ
1
;

q
,
b
2
=

2
α
μ
~
1

q
μ
~
1
1
+
q
ν
~
1
,
b
3
=
2
α
ν
~
1

q
μ
~
1
+
1
1
+
q
ν
~
1

1
,
b
4
=

2
α

q
μ
~
1
+
1
1
+
q
ν
~
1
,
K
5
=
μ
1
+
ν
1
+
δ
1
μ
1
+
ν
1

δ
1
1
+
2
μ
1
F
2
1
μ
1
+
ν
1
+
δ
1
+
1
,
μ
1
+
ν
1

δ
1
+
1,2
+
2
μ
1
;

q
,
K
6
=
F
2
1
μ
~
1
+
ν
~
1
+
δ
~
1
,
μ
~
1
+
ν
~
1

δ
~
1
,
1
+
2
μ
~
1
;

q
,
K
7
=
F
2
1
μ
~
1
+
ν
~
1
+
δ
~
1
,
μ
~
1
+
ν
~
1

δ
~
1
,
1
+
2
μ
~
1
;

q
,
K
8
=
μ
~
1
+
ν
~
1
+
δ
~
1
μ
~
1
+
ν
~
1

δ
~
1
1
+
2
μ
~
1
F
2
1
μ
~
1
+
ν
~
1
+
δ
~
1
+
1
,
μ
~
1
+
ν
~
1

δ
~
1
+
1,2
+
2
μ
~
1
;

q
.
Equation (29) admits a solution if and only if its determinant is zero [30]. This thus provides a condition for getting the energy eigenvalues as
(31)
b
1
a
2
K
2
K
3
+
b
1
a
3
K
2
K
4
+
b
1
a
4
K
2
K
5
=
a
1
b
2
K
1
K
6
+
a
1
b
3
K
1
K
7
+
a
1
b
4
K
1
K
8
.
Energy equation (31) is a complicated transcendental equation and can only be solved numerically.