# The Exact Relativistic Scalar Quasibound States of The Dyonic Kerr-Sen Black
Hole: Quantized Energy, and Hawking Radiation

David Senjaya<EMAIL_ADDRESS>Piyabut Burikham<EMAIL_ADDRESS>High Energy Physics Theory Group, Department of Physics, Faculty of Science,
Chulalongkorn University, 254 Phayathai Road, Pathumwan, 10330, Bangkok,
Thailand  Tiberiu Harko<EMAIL_ADDRESS>Department of Physics,
Babes-Bolyai University,1 Mihail Kogalniceanu Street,400084, Cluj-
Napoca,Romania  Astronomical Observatory,19 Ciresilor Street,400487,Cluj-
Napoca,Romania

(August 28, 2024)

###### Abstract

We consider Klein-Gordon equation in the Dyonic Kerr-Sen black hole
background, which is the charged rotating axially symmetric solution of the
Einstein-Maxwell-Dilaton-Axion theory of gravity. The black hole incorporates
electric, magnetic, dilatonic and axionic charges and is constructed in 3+1
dimensional spacetime. We begin our investigations with the construction of
the scalar field’s governing equation, i.e., the covariant Klein-Gordon
equation. With the help of the ansatz of separation of variables, we
successfully separate the polar part, and find the exact solution in terms of
Spheroidal Harmonics, while the radial exact solution is obtained in terms of
the Confluent Heun function. The quantization of the quasibound state is done
by applying the polynomial condition of the Confluent Heun function that gives
rise to discrete complex-valued energy levels for massive scalar fields. The
real part is the scalar field relativistic quantized energy, while the
imaginary part represents the quasibound states’s decay. We present all of the
sixteen possible exact energy solutions for both massive and massless scalars.
We also present the investigation the Hawking radiation of the Dyonic Kerr-Sen
black hole’s apparent horizon, via the Sigurd-Sannan method by making use of
the obtained exact scalar wave functions. The radiation distribution function,
and the Hawking temperature are successfully obtained.

EMDA super gravity, Kerr-Sen black hole, Klein-Gordon equation, quasibound
states, Hawking radiation

##  I Introduction

Einstein’s theory of general relativity is a successful successor to the old
Newtonian theory of gravity. It describes the gravitational field as a
curvature effect in spacetime, determined by the mass of a body. The motion of
particles in the curved spacetime is governed by the geodesic equations, which
are directly related to the spacetime metric. Some of the gravitational
effects predicted by general relativity, such as the perihelion precession of
Mercury, the bending of light, the gravitational redshift of the distant
stars, the gravitational waves, and the existence of black holes, have been
confirmed observationally, or experimentally Hobson ; Will ; Abbott ; Abbott1
; EHTC .

However, despite its remarkable phenomenological success, Einstein’s theory
has also several problems, such as, for example, the existence of
singularities in its black hole and cosmological solutions. Also, dark matter
and dark energy are required to be added by hand inside the source term
($T_{\mu\nu}$), in order to explain the flat galaxy rotation curves, and the
accelerated expansion of the Universe. The need of modifying the Einstein’s
general relativity then gave birth to various extended theories of gravity,
such as the Kaluza-Klein theory, $f(R),f(T)$, $f\left(R,L_{m}\right)$,
$f(R,T)$ modified gravities, the Gauss-Bonnet theory, the Scalar-tensor-vector
gravity, Lovelock theory, etc Buchdahl ; Linder ; H1 ; H2 ; Moffat2006 ;
Moffat2007 ; Clifton ; papantonopoulos ; Petrov .

In this work, we will focus on the string inspired Einstein-Maxwell-dilaton-
axion (EMDA) theory of gravity, which belongs to the scalar-vector-tensor
supergravity family. The EMDA theory generalizes the Einstein-Maxwell theory
by introducing a coupling between the Maxwell electromagnetic field tensor
$F_{\mu\nu}$, the scalar dilaton field $\xi$ and the pseudo scalar axion field
$\phi$, and it can be constructed from the following four dimensional
effective action Sen ; Baner

$S_{EMDA}=\frac{1}{16\pi}\int\left[R-2\partial_{\mu}\xi\partial^{\mu}\xi\right.\\\
\left.-\frac{1}{3}H_{\rho\sigma\delta}H^{\rho\sigma\delta}+e^{-2\xi}F_{\alpha\beta}F^{\alpha\beta}\right]\sqrt{-g}d^{4}x,$
(1)

where $R$ is the Ricci scalar and $g$ is the metric tensor determinant. The
Maxwell electromagnetic field tensor $F_{\mu\nu}$ is defined as the partial
derivatives of the $U(1)$ gauge field $A_{\mu}$ as follows,

$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}.$ (2)

Moreover, $H^{\rho\sigma\delta}$ is the Kalb-Ramond field tensor that is
written in terms of the pseudo-scalar axion field $\phi$ according to,

$H_{\alpha\beta\delta}=\frac{1}{2}e^{4\xi}\varepsilon_{\alpha\beta\delta\gamma}\partial^{\gamma}\phi.$
(3)

Recently, Wu et al. Wu  did find a novel exact rotating black hole solution of
the EMDA gravity that depends on four charges, the i.e. electric, magnetic,
dilaton and axion charges, called the Dyonic Kerr-Sen black hole. The Dyonic
Kerr-Sen black hole is a generalization of the non-dyonic Kerr-Sen black hole,
found in Sen . The presence of the additional magnetic, dilaton and axion
fields in the theory also generalizes the electro-vacuum rotating black hole
solution of the Einstein-Maxwell theory of gravity, the Kerr-Newman black hole
solution Kerr . Due to its generality, in this work, the aforementioned Dyonic
Kerr-Sen solution becomes our topic of interest in order to investigate the
scalar quasibound states and the Hawking radiation, respectively.

The behavior of the relativistic scalar fields in curved spacetime is an
important subject of research, which may give some insights into the quantum
theory of gravity. In the presence of matter-charge distribution, the
surrounding spacetime will be curved, and the scalar field-gravity interaction
is then investigated by looking at the scalar field behavior in the curved
spacetime backgrounds. The behavior of the relativistic scalar fields in
curved spacetimes is mathematically described by the covariant Klein-Gordon
equation, in which the gravitational field is introduced via the metric tensor
$g_{\mu\nu}$. The investigation of the Klein-Gordon equation in the black hole
background may open up a new understanding of the quantum theory of gravity.
Instead of trying to consider the quantization of the general relativity, in
the present work we take the opposite approach, by ”gravitizing” the quantum
mechanics, to formulate it in line with the principles of the theory of
general relativity.

After the gravitational wave signal of a binary black hole merger was finally
detected for the first time on 14 September 2015 205 , quite recently, the
Hawking radiation of an optical black hole analog drori  was also finally
observed. This makes the investigation of the black hole spectroscopy a new
emerging field of major physical interest. The quasibound states, quasinormal
modes, and shadows of black holes are among the most interesting
characteristics of such astrophysical objects, and they appear in the
observationally measurable spectra that are released as the particles cross
into the black hole 100 .

The quasibound states are relativistic quasistationary resonances outside the
black hole’s event horizon, localized in the black hole’s finite gravitational
potential well. In contrast to the electron’s effective potential in the
hydrogen atoms, scalar quasibound states are leaking and crossing into the
black hole, causing the spectrum to be complex valued, similar to the case of
damped oscillations, where the real part is the scalar field’s quantized
energy, and the imaginary part determines the stability of the system. It is
possible, in principle, to extract the information about the central black
hole, as well as to validate some modified theories of gravity, by measuring
the frequency of the quasibound states 100 . Analogously to atomic transitions
emitting photons, transitions of axions around black holes between different
energy levels emit gravitons 32a .

However, due to the complexity of the equations involved, especially the
radial equation, analytical methods were used less often, and only for some
very special problems Anal1 ; Anal2 ; Anal3 . The vast majority of these
studies made use of numerical techniques such as asymptotical analysis, WKB,
and continued fraction to investigate the specific task at hand. However, very
recently, Vier22 ; senjaya1 ; senjaya2 ; senjaya3 ; senjaya4 ; senjaya5 ;
senjaya6 ; senjaya7  have successfully worked out, and presented some novel
exact scalar quasibound states solutions in various black hole backgrounds,
and obtained the radial exact solutions of the Klein-Gordon equation in terms
of the Confluent Heun and the General Heun functions. The polynomial condition
of the Heun functions leads to the quasibound states’ energy quantization.

In this work, we present in detail the analytical derivations of both massive
and massless scalar quasibound states of the exact solutions in the Dyonic
Kerr-Sen black hole background. The exact wave function is found in terms of
the harmonic function as the temporal solution, the Spheroidal harmonics as
the angular solution, and the Confluent Heun functions as the radial
solutions. The expression of the relativistic quantized energy levels is
obtained after applying the polynomial condition of the Confluent Heun
function. We also investigate the wave function of the quasibound states in
the two extreme regions, i.e., close to the black hole horizon, and at
infinity, respectively, and find out that they behave like an ingoing wave,
close to the black hole horizon, and vanishing far away from the horizon. In
the last Section, using exact radial solutions, the Hawking radiation of the
black hole’s apparent horizon is investigated, and the Hawking temperature is
obtained.

##  II The Dyonic Kerr-Sen Metric

In the present Section we briefly review the basic properties of the Dyonic
Kerr-Sen metric, we obtain the singular points of the line element, and we
calculate the inverse of the metric tensor coefficients.

###  II.1 The Metric

The line element of rotating black hole’s exterior can be written in either
the Cartesian (Kerr-Schild) coordinates, the Boyer-Lindquist (Schwarzschild-
like) coordinates, or the Eddington-Finkelstein (Tortoise) coordinates. In
this work, the Dyonic Kerr-Sen spacetime in the Boyer-Lindquist coordinate
system is adopted. The line element is given as follows Wu ; Jana ,

$ds^{2}=-\left[1-\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}\right]c^{2}dt^{2}\\\
-2\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}a{{\sin}^{{\rm 2}}\theta\
}d\phi cdt+\frac{\rho^{2}}{\Delta}dr^{2}+\rho^{2}d\theta^{2}\\\
+\left[r\left(r-2d\right)-k^{2}+a^{2}+\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}a^{2}{{\sin}^{{\rm
2}}\theta\ }\right]\times\\\ {{\sin}^{{\rm 2}}\theta\ }d\phi^{2},$ (4)

where,

$\displaystyle\rho^{2}=r\left(r-2d\right)-k^{2}+a^{2}{{\cos}^{{\rm 2}}\theta\
},$ (5)
$\displaystyle\Delta=r\left(r-2d\right)-r_{s}\left(r-d\right)-k^{2}+a^{2}+r^{2}_{D}.$
(6)

We have also introduced the notations,

$\displaystyle
r^{2}_{D}=\frac{G\left(Q^{2}_{E}+Q^{2}_{M}\right)}{4\pi\epsilon_{0}c^{4}}=Q^{2}+P^{2},$
(7) $\displaystyle k=2\frac{PQ}{r_{s}},$ (8) $\displaystyle
d=\frac{P^{2}-Q^{2}}{r_{s}},$ (9) $\displaystyle
d^{2}+k^{2}={\left(\frac{P^{2}+Q^{2}}{r_{s}}\right)}^{2}\leftrightarrow
r_{s}\sqrt{d^{2}+k^{2}}=r^{2}_{D},$ (10) $\displaystyle a=\frac{J}{M},$ (11)

where $M,J,Q,P,d,k$ are respectively the black hole’s mass, spin, electric,
magnetic/dyonic, dilaton and axion charge.

The Boyer-Lindquist coordinates are connected with the Cartesian coordinates
by the following transformation,

$\displaystyle
x=\sqrt{r\left(r-2d\right)+a^{2}+r^{2}_{D}-k^{2}+dr_{s}}{\sin\theta{\cos\phi\
}\ },$ (12) $\displaystyle
y=\sqrt{r\left(r-2d\right)+a^{2}+r^{2}_{D}-k^{2}+dr_{s}}{\sin\theta{\sin\phi\
}\ },$ (13) $\displaystyle z=\sqrt{r\left(r-2d\right)}\;{\cos\theta\ }.$ (14)

The Dyonic Kerr-Sen geometry is singular when $\rho^{2}=0$ and
$\Delta{\rm=0}$. The first condition,

$\rho^{2}=0=r\left(r-2d\right)-k^{2}+a^{2}{{\cos}^{{\rm 2}}\theta\ },$ (15)

is satisfied at $\theta=\frac{\pi}{2}$. By using the Cartesian-Boyer-Lindquist
coordinates relationship,

$x^{2}+y^{2}=r\left(r-2d\right)+a^{2}+r^{2}_{D}-k^{2}+r_{s}d,$ (16)

we obtain,

$r\left(r-2d\right)-k^{2}=0\leftrightarrow
x^{2}+y^{2}=a^{2}+r^{2}_{D}+r_{s}d.$ (17)

Here we obtain a ring singularity on the Cartesian $x-y$ plane, with radius
$\sqrt{a^{2}+r^{2}_{D}+r_{s}d}$, measured from the coordinate’s origin. The
second singularity, $\Delta{\rm=0,}$ is related to the one way surfaces, i.e.,
the black hole’s horizons, determined by the condition $g_{rr}=\infty$. The
positions of the horizons can be obtained by solving the following quadratic
equation,

$\displaystyle\Delta{\rm=0=}r\left(r-2d\right)-r_{s}\left(r-d\right)-k^{2}+a^{2}+r^{2}_{D},$
(18) $\displaystyle
r_{\pm}=\frac{r_{s}}{2}+d\pm\sqrt{{\left(\frac{r_{s}}{2}\right)}^{2}+d^{2}+k^{2}-\left(a^{2}+r^{2}_{D}\right)},$
(19)

where $r_{+}$ and $r_{-}$ are the outer and inner horizons, respectively.

It is important to mention that by taking $d=0$ in the expression of $k$, we
reobtain the non-dilatonic, axionic dyonic black hole, and the non-axionic,
dilatonic dyonic black hole solutions. By taking $P$ equal to zero, we obtain
the Kerr-Newman black hole, which is the most general electro-vacuum black
hole solution of the Einstein-Maxwell theory. By taking $P,a$ equal to zero,
we reobtain the static spherically symmetric Reissner-Nordström black hole. By
taking as zero all of the Dyonic Kerr-Sen black hole’s parameters, except the
mass, we obtain the basic Schwazschild black hole.

In order to find the inverse of the metric of the Dyonic Kerr-Sen spacetime,
as a first step we express the line element (4) as follows,

$\displaystyle
g_{\mu\nu}=\left(\begin{array}[]{cccc}-\left[1-\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}\right]&0&0&g_{c\phi}\\\
0&\frac{\rho^{2}}{\Delta}&0&0\\\ 0&0&\rho^{2}&0\\\
g_{c\phi}&0&0&g_{\phi\phi}\end{array}\right),$ (24)

where,

$g_{c\phi}=-\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}a{\sin}^{{\rm
2}}\theta,$ (25) $g_{\phi\phi}=\Bigg{[}r\left(r-2d\right)-k^{2}+a^{2}\\\
\left.+\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}a^{2}{{\sin}^{{\rm
2}}\theta\ }\right]{{\sin}^{{\rm 2}}\theta\ }.$ (26)

Since we need to calculate the determinant of the metric tensor, let us first
modify the metric tensor to be a block matrix as follows,

$\displaystyle g_{\mu\nu}$
$\displaystyle=\left(\begin{array}[]{cccc}\frac{\rho^{2}}{\Delta}&0&0&0\\\
0&\rho^{2}&0&0\\\
0&0&-\left[1-\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}\right]&g_{c\phi}\\\
0&0&g_{c\phi}&g_{\phi\phi}\end{array}\right)$ (31)
$\displaystyle=\left(\begin{array}[]{cc}\tilde{A}&0\\\
0&\tilde{B}\end{array}\right).$ (34)

The determinant of the metric tensor can then be calculated as follows,

${\det\left(g_{\mu\nu}\right)\
}=g=\det\left(\tilde{A}\right)\det\left(\tilde{B}\right),$ (35)

and after some lines of straightforward algebra, we obtain,

$g=\rho^{4}{{\sin}^{{\rm 2}}\theta\ }.$ (36)

###  II.2 The Metric Inverse

Now we will proceed to calculate the inverse of the metric tensor. Using the
metric in block matrix form, the inverse is calculated block per block as
follows,

$\displaystyle g_{\mu\nu}=\left(\begin{array}[]{cc}\tilde{A}&0\\\
0&\tilde{B}\end{array}\right)\to
g^{\mu\nu}=\left(\begin{array}[]{cc}{\tilde{A}}^{-1}&0\\\
0&{\tilde{B}}^{-1}\end{array}\right),$ (41)
$\displaystyle{\tilde{A}}^{-1}=\left(\begin{array}[]{cc}\frac{\Delta}{\rho^{2}}&0\\\
0&\frac{1}{\rho^{2}}\end{array}\right),$ (44)
$\displaystyle{\tilde{B}}^{-1}=\frac{1}{\left|\tilde{B}\right|}\left(\begin{array}[]{cc}g_{\phi\phi}&-g_{c\phi}\\\
-g_{c\phi}&-\left[1-\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}\right]\end{array}\right),$
(47)
$\displaystyle\left|\tilde{B}\right|=\frac{-g}{\left(\frac{\rho^{4}}{\Delta}\right)}=-\Delta{{\sin}^{2}\theta\
}.$ (48)

After finding the inverse of each block, the metric tensor’s inverse is then
obtained as follows,

$\displaystyle
g^{\mu\nu}=\left(\begin{array}[]{cccc}-\frac{1}{\Delta}f_{\phi\phi}&0&0&\frac{g_{c\phi}}{\Delta\sin^{2}\theta}\\\
0&\frac{\Delta}{\rho^{2}}&0&0\\\ 0&0&\frac{1}{\rho^{2}}&0\\\
\frac{g_{c\phi}}{\Delta\sin^{2}\theta}&0&0&{\frac{1}{\Delta{\sin}^{{\rm
2}}\theta}\left(1-\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}\right)}\end{array}\right),$
(53) $\displaystyle
f_{\phi\phi}=r\left(r-2d\right)-k^{2}+a^{2}+\frac{r_{s}\left(r-d\right)-r^{2}_{D}}{\rho^{2}}a^{2}{{\sin}^{{\rm
2}}\theta\ }.$ (54)

##  III The Klein-Gordon Equation, its solutions, and the quantization of the
quasibound states

In the present Section we formulate first the Klein-Gordon equation in the
Dyonic Kerr-Sen black hole geometry, and then we obtain its exact solutions.
The quantization of the quasibound states is also realized by using the
properties of the Confluent Heun function.

###  III.1 The Klein-Gordon equation

We aim to investigate the massive and massless scalar fields in the curved
spacetime generated by a Dyonic Kerr-Sen black hole, which is the solution of
the EMDA theory in the low energy limit of the heterotic string field theory.
We will first construct, and then search for the exact solutions of the
governing relativistic matter wave equation. We begin with writing the generic
form of the Klein-Gordon in a curved spacetime background as follows,

$-{\hbar}^{2}\nabla_{\mu}\nabla^{\mu}\psi+m^{2}c^{2}=0,$ (55)

where $\nabla_{\mu}$ is the covariant derivative with respect to the metric.

For the covariant derivatives, we obtain,
$\nabla_{\mu}\nabla^{\mu}\psi={\nabla}_{\mu}{\partial}^{\mu}\psi={\partial}_{\mu}{\partial}^{\mu}\psi+{\Gamma}^{\mu}_{\mu\nu}{\partial}^{\nu}\psi.$
With the use of the identity
${\Gamma}^{\alpha}_{\alpha\beta}=\left(1/\sqrt{-g}\right)\left(\partial\sqrt{-g}/\partial
x^{\beta}\right)$, we can express the d’Alembert operator
$\nabla_{\mu}\nabla^{\mu}\psi$ as
$\nabla_{\mu}\nabla^{\mu}\psi=\left(1/\sqrt{-g}\right){\partial}_{\mu}\left(\sqrt{-g}g^{\mu\nu}{\partial}_{\nu}\psi\right)$.

Now, the Klein-Gordon equation can be expressed in terms of partial
derivatives and the inverse metric tensor components as follows,

$\displaystyle\left\\{-{\hbar}^{2}\left[\frac{1}{\sqrt{-g}}\left({\partial}_{\mu}\sqrt{-g}g^{\mu\nu}{\partial}_{\nu}\right)\right]+m^{2}c^{2}\right\\}\psi=0,$
(56)

where $m$ is the rest mass of the scalar.

As the metric determinant and the metric inverse have already been obtained,
the Laplace-Beltrami operator of the Klein-Gordon equation can be found
component by component as follows,

$\displaystyle\frac{1}{\sqrt{-g}}{\partial}_{0}\sqrt{-g}g^{00}{\partial}_{0}{\rm=-}\frac{{\rm
1}}{\Delta\rho^{2}}\left\\{{\left[r\left(r-2d\right)-k^{2}+a^{2}\right]}^{2}-\Delta
a^{2}{{\sin}^{{\rm 2}}\theta\ }\right\\}{\partial}^{2}_{ct},$ (57)
$\displaystyle\frac{1}{\sqrt{-g}}{\partial}_{0}\sqrt{-g}g^{03}{\partial}_{3}{\rm=-}\frac{\left[r\left(r-2d\right)-k^{2}+a^{2}-\Delta\right]a}{\Delta\rho^{2}}{\partial}_{ct}{\partial}_{\phi},$
(58) $\displaystyle\frac{1}{\sqrt{-g}}{\partial}_{{\rm
3}}\sqrt{-g}g^{30}{\partial}_{0}{\rm=-}\frac{\left[r\left(r-2d\right)-k^{2}+a^{2}-\Delta\right]a}{\Delta\rho^{2}}{\partial}_{ct}{\partial}_{\phi},$
(59) $\displaystyle\frac{1}{\sqrt{-g}}{\partial}_{{\rm
1}}\sqrt{-g}g^{11}{\partial}_{1}{\rm=}\frac{1}{\rho^{2}}{\partial}_{r}\left(\Delta{\partial}_{r}\right),$
(60) $\displaystyle\frac{1}{\sqrt{-g}}{\partial}_{{\rm
2}}\sqrt{-g}g^{22}{\partial}_{2}{\rm=}\frac{1}{\rho^{2}{\sin\theta\
}}{\partial}_{\theta}\left({\sin\theta\ }{\partial}_{\theta}\right),$ (61)
$\displaystyle\frac{1}{\sqrt{-g}}{\partial}_{{\rm
3}}\sqrt{-g}g^{33}{\partial}_{3}{\rm=}\frac{\Delta-a^{2}{{\sin}^{{\rm
2}}\theta\ }}{\Delta{{\sin}^{2}\theta\ }\rho^{2}}{\partial}^{2}_{\theta}$ (62)

By combining all components, we obtain the explicit Klein-Gordon equation in
the Dyonic Kerr-Sen black hole background as follows,

$\left[{\rm-}\frac{{\rm
1}}{\Delta\rho^{2}}\left\\{{\left[r\left(r-2d\right)-k^{2}+a^{2}\right]}^{2}-\Delta
a^{2}{{\sin}^{{\rm 2}}\theta\ }\right\\}{\partial}^{2}_{ct}\right.\\\
\left.-2\frac{\left[r\left(r-2d\right)-k^{2}+a^{2}-\Delta\right]a}{\Delta\rho^{2}}{\partial}_{ct}{\partial}_{\phi}\right.\\\
\left.+\frac{1}{\rho^{2}}{\partial}_{r}\left(\Delta{\partial}_{r}\right)+\frac{1}{\rho^{2}{\sin\theta\
}}{\partial}_{\theta}\left({\sin\theta\ }{\partial}_{\theta}\right)\right.\\\
\left.+\frac{\Delta-a^{2}{{\sin}^{{\rm 2}}\theta\ }}{\Delta{{\sin}^{2}\theta\
}\rho^{2}}{\partial}^{2}_{\phi}\right]\psi-\frac{m^{2}c^{2}}{{\hbar}^{2}}\psi=0.$
(63)

The presence of temporal and azimuthal symmetry enables us to apply the
separation ansatz 35 ,

$\displaystyle\psi\left(t,r,\theta,\phi\right)=e^{i\frac{E}{c}ct-
im_{l}\phi}R\left(r\right)T\left(\theta\right).$ (64)

###  III.2 The Polar Wave Equation

Let us define first the following dimensionless variables,

$\displaystyle\Omega=\frac{Er_{s}}{\hbar c},\Omega_{0}=\frac{E_{0}r_{s}}{\hbar
c},$ (65)

where $E_{0}=mc^{2}$ is the scalar’s rest energy.

Substituting the separation ansatz into the Eq. (63), after multiplying the
entire equation by $r^{2}/\psi\left(t,r,\theta,\phi\right)$, we obtain,

$\left\\{{\rm-}\frac{{\rm
1}}{\Delta\rho^{2}}\left\\{{\left(r\left(r-2d\right)-k^{2}+a^{2}\right)}^{2}-\Delta
a^{2}{{\sin}^{{\rm 2}}\theta\ }\right\\}\times\right.\\\
\left.\left(-\frac{E^{2}}{\hbar^{2}c^{2}}\right)-2\frac{\left(r\left(r-2d\right)-k^{2}+a^{2}-\Delta\right)a}{\Delta\rho^{2}}\left(\frac{Em_{l}}{\hbar
c}\right)\right.\\\
\left.+\frac{1}{R\rho^{2}}{\partial}_{r}\left(\Delta{\partial}_{r}R\right)+\frac{1}{T\rho^{2}{\sin\theta\
}}{\partial}_{\theta}\left({\sin\theta\ }{\partial}_{\theta}T\right)\right.\\\
\left.+\frac{\Delta-a^{2}{{\sin}^{{\rm 2}}\theta\ }}{\Delta{{\sin}^{2}\theta\
}\rho^{2}}\left(-m^{2}_{l}\right)\right\\}-\frac{m^{2}c^{2}}{{\hbar}^{2}}=0.$
(66)

By multiplying the entire wave equation by $\rho^{2}$, and after using the
trigonometric identity ${{\sin}^{{\rm 2}}\theta\ }=1-{{\cos}^{2}\theta\ }$, we
obtain the following radial-polar equation,

$\left[\frac{1}{T{\sin\theta\ }}{\partial}_{\theta}\left({\sin\theta\
}{\partial}_{\theta}T\right)-\frac{m^{2}_{l}}{{{\sin}^{2}\theta\
}}-\left(\frac{\Omega^{2}_{0}a^{2}}{r^{2}_{s}}-\frac{\Omega^{2}a^{2}}{r^{2}_{s}}\right){{\cos}^{2}\theta\
}\right]\\\
+\left[\frac{1}{R}{\partial}_{r}\left(\Delta{\partial}_{r}R\right)+\frac{\Omega^{2}}{r^{2}_{s}}\frac{{\left(r\left(r-2d\right)-k^{2}+a^{2}\right)}^{2}}{\Delta}-\frac{\Omega^{2}a^{2}}{r^{2}_{s}}\right.\\\
\left.-2\frac{\left(r\left(r-2d\right)-k^{2}+a^{2}-\Delta\right)a}{\Delta}\left(\frac{\Omega
m_{l}}{r_{s}}\right)+\frac{m^{2}_{l}a^{2}}{\Delta}\right.\\\
\left.-\frac{\Omega^{2}_{0}}{r^{2}_{s}}\left(r\left(r-2d\right)-k^{2}\right)\right]=0.$
(67)

The polar part can be separated as follows,

$\frac{1}{T{\sin\theta\ }}{\partial}_{\theta}\left({\sin\theta\
}{\partial}_{\theta}T\right)-\frac{m^{2}_{l}}{{{\sin}^{2}\theta\ }}\\\
-\left(\frac{\Omega^{2}_{0}a^{2}}{r^{2}_{s}}-\frac{\Omega^{2}a^{2}}{r^{2}_{s}}\right){{\cos}^{2}\theta\
}+\lambda^{m_{l}}_{l}=0.$ (68)

In the case of $a=k=0$, the separation constant is set to be
$\lambda^{m_{l}}_{l}=l\left(l+1\right)$, and the polar wave solution is
obtained in terms of the Legendre polynomial, $P^{m_{l}}_{l}(\cos\theta)$.
But, in the case of a rotating black hole with a non-zero axion charge, the
polar solution is obtained in terms of the Spheroidal function,
$S^{m_{l}}_{l}\left(c_{a},{\cos\theta\ }\right)$, as follows NIST ,

$T(\theta)=S^{m_{l}}_{l}\left(c_{a},{\cos\theta\
}\right)=\sum^{\infty}_{r=-\infty}{d^{\;lm_{l}}_{r}\left(c_{a}\right)P^{m_{l}}_{l+r}\left({\cos\theta\
}\right)},$ (69)

where,

$c_{a}=\frac{\Omega^{2}_{0}a^{2}}{r^{2}_{s}}-\frac{\Omega^{2}a^{2}}{r^{2}_{s}}.$
(70)

###  III.3 The Radial Wave Equation

After solving the polar part, we are left with the following radial equation,

${\partial}_{r}\left(\Delta{\partial}_{r}R\right)+\left[\frac{\Omega^{2}}{r^{2}_{s}}\frac{{\left(r\left(r-2d\right)-k^{2}+a^{2}\right)}^{2}}{\Delta}-\frac{\Omega^{2}a^{2}}{r^{2}_{s}}\right.\\\
\left.-2\frac{\left(r\left(r-2d\right)-k^{2}+a^{2}-\Delta\right)a}{\Delta}\left(\frac{\Omega
m_{l}}{r_{s}}\right)+\frac{m^{2}_{l}a^{2}}{\Delta}\right.\\\
\left.-\frac{\Omega^{2}_{0}}{r^{2}_{s}}\left(r\left(r-2d\right)-k^{2}\right)-\lambda^{m_{l}}_{l}\right]R{\rm=0}.$
(71)

The radial equation needs a careful treatment. First, remember that the
condition ${\Delta}=0$ leads to two solutions, i.e., $r_{\pm}$. Hence,
$\Delta$ can be rewritten in the factorized form as follows,

$\displaystyle{\Delta}=(r-r_{-})(r-r_{+}),$ (72)

giving

$\partial_{r}({\Delta}\partial_{r}R)=(r-r_{-})\partial_{r}R\\\
+(r-r_{+})\partial_{r}R+(r-r_{-})(r-r_{+})\partial^{2}_{r}R,$ (73)

and

$\displaystyle\frac{r_{+}-r_{-}}{{\Delta}}=\frac{\delta_{r}}{{\Delta}}=\frac{1}{r-r_{+}}-\frac{1}{r-r_{-}},$
(74)

respectively. Rearranging Eq. (71), we obtain,

${\partial}_{r}\left(\Delta{\partial}_{r}R\right)+\left[\frac{1}{\Delta}{\left\\{\frac{\Omega}{r_{s}}\left(r\left(r-2d\right)-k^{2}+a^{2}\right)-m_{l}a\right\\}}^{2}\right.\\\
\left.-\left\\{\frac{\Omega^{2}_{0}}{r^{2}_{s}}\left(r\left(r-2d\right)-k^{2}+a^{2}\right)+\frac{\Omega^{2}}{r^{2}_{s}}a^{2}-\frac{\Omega^{2}_{0}}{r^{2}_{s}}a^{2}\right.\right.\\\
\left.\left.-2\frac{\Omega
m_{l}}{r_{s}}a+\lambda^{m_{l}}_{l}\right\\}\right]R{\rm=0},$ (75)

where the following constant, $K^{m_{l}}_{l}$, has been defined as,

$\displaystyle
K^{m_{l}}_{l}=\frac{\Omega^{2}}{r^{2}_{s}}a^{2}-\frac{\Omega^{2}_{0}}{r^{2}_{s}}a^{2}-2\frac{\Omega
m_{l}}{r_{s}}a+\lambda^{m_{l}}_{l}.$ (76)

Working out the double radial differentiation in the first term of the
equation (75), followed by multiplying the entire equation by
$\delta_{r}^{2}$, we obtain the following equation,

$\delta^{2}_{r}{\partial}^{{\rm
2}}_{r}R+\left\\{\frac{1}{r-r_{+}}+\frac{1}{r-r_{-}}\right\\}\delta^{2}_{r}{\partial}_{r}R+\left[{\left(\frac{1}{r-r_{+}}-\frac{1}{r-r_{-}}\right)}^{2}{\left\\{\frac{\Omega}{r_{s}}\left(r\left(r-2d\right)-k^{2}+a^{2}\right)-m_{l}a\right\\}}^{2}\right.\\\
\left.-\delta_{r}\left(\frac{1}{r-r_{+}}-\frac{1}{r-r_{-}}\right)\left\\{\frac{\Omega^{2}_{0}}{r^{2}_{s}}\left(r\left(r-2d\right)-k^{2}+a^{2}\right)+K^{m_{l}}_{l}\right\\}\right]R=0.$
(77)

The region of interest of the investigation is outside the outer horizon, i.e.
$r_{+}\leq r<\infty$. Hence, let us define the following new radial variable,

$\displaystyle\delta_{r}y=r-r_{+}\to\delta_{r}dy=dr,$ (78) $\displaystyle
r-r_{-}=\delta_{r}\left(y+1\ \right).$ (79)

The domain is now shifted to $0\leq y<\infty$. In terms of $y$, the radial
equation becomes,

${\partial}^{{\rm
2}}_{y}R+\left\\{\frac{1}{y}+\frac{1}{y+1}\right\\}{\partial}_{y}R+\left[{{F.T.}^{2}}+S.T.\right]R{\rm=0}.$
(80)

where,

$F.T.=\frac{1}{\delta_{r}}\left(\frac{1}{y}-\frac{1}{y+1}\right)\times\\\
\left\\{\frac{\Omega}{r_{s}}\left({\left(\delta_{r}y+r_{+}\right)}^{2}-2d\left(\delta_{r}y+r_{+}\right)-k^{2}+a^{2}\right)-m_{l}a\right\\},$
(81)

and,

$S.T.=-\left(\frac{1}{y}-\frac{1}{y+1}\right)\times\\\
\left\\{\frac{\Omega^{2}_{0}}{r^{2}_{s}}\left({\left(\delta_{r}y+r_{+}\right)}^{2}-2d\left(\delta_{r}y+r_{+}\right)-k^{2}+a^{2}\right)+K^{m_{l}}_{l}\right\\}.$
(82)

Let us consider now the term $F.T.$, which can be rewritten as

$F.T.=\frac{1}{\delta_{r}}\frac{1}{y\left(y+1\right)}\left\\{\frac{\Omega}{r_{s}}\left[\delta^{2}_{r}y^{2}+2y\delta_{r}(r_{+}-2d)\right]\right.\\\
\left.+\underbrace{\frac{\Omega}{r_{s}}\left[r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right]-m_{l}a}_{K_{1}}\right\\}.$
(83)

We need to decompose the terms such as $\frac{y}{y+1}$ by using the fractional
decomposition, $\frac{y}{y+1}=1-\frac{1}{y+1}$. Hence we continue as follows,

$F.T.=\frac{\Omega}{r_{s}}\delta_{r}+\frac{1}{y}\frac{K_{1}}{\delta_{r}}\\\
+\frac{1}{y+1}\left\\{\underbrace{\frac{\Omega}{r_{s}}\left(r_{+}+r_{-}{-4d}\right)-\frac{K_{1}}{\delta_{r}}}_{{K_{3}}}\right\\}.$
(84)

After some algebraic transformations we obtain,

${F.T.^{2}}=\frac{\Omega^{2}}{r^{2}_{s}}\delta^{2}_{r}+\frac{K^{2}_{1}}{\delta^{2}_{r}y^{2}}+\frac{K^{2}_{3}}{{\left(y+1\right)}^{2}}+\frac{2\Omega
K_{1}}{r_{s}y}\\\
+\frac{2\Omega\delta_{r}K_{3}}{r_{s}\left(y+1\right)}+\frac{2K_{1}K_{3}}{\delta_{r}}\left(\frac{1}{y}-\frac{1}{y+1}\right).$
(85)

Now let us proceed to the calculation of $S.T.$,

$S.T.=-\frac{1}{y\left(y+1\right)}\left\\{\frac{\Omega^{2}_{0}}{r^{2}_{s}}\left(\delta^{2}_{r}y^{2}+2\delta_{r}y\left(r_{+}-2d\right)\right)\
\right.\\\
\left.+\underbrace{\frac{\Omega^{2}_{0}}{r^{2}_{s}}\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)+K^{m_{l}}_{l}}_{K_{2}}\right\\}.$
(86)

Again, applying the fractional decomposition, $\frac{y}{y+1}=1-\frac{1}{y+1}$,
we find,

$S.T.=-\frac{\Omega^{2}_{0}}{r^{2}_{s}}\delta^{2}_{r}-\frac{K_{2}}{y}\\\
+\frac{1}{y+1}\left[\underbrace{\frac{\Omega^{2}_{0}}{r^{2}_{s}}\delta^{2}_{r}-2\delta_{r}(r_{+}-2d)\frac{\Omega^{2}_{0}}{r^{2}_{s}}+K_{2}}_{K_{4}}\right].$
(87)

Up to this step, $F.T.^{2}$ and $S.T.$ have successfully been expressed in
terms of $\frac{1}{y+1}$ and $\frac{1}{y}$. Let us rewrite the fully
decomposed radial equation,

${\partial}^{{\rm
2}}_{y}R+\left[\frac{1}{y}+\frac{1}{y+1}\right]\partial_{y}R+\left[\left(\frac{\Omega^{2}}{r^{2}_{s}}\delta^{2}_{r}-\frac{\Omega^{2}_{0}}{r^{2}_{s}}\delta^{2}_{r}\right)\right.\\\
\left.+\frac{1}{y}\left(\frac{2\Omega
K_{1}}{r_{s}}+\frac{2K_{1}K_{3}}{\delta_{r}}-K_{2}\right)\right.\\\
\left.+\frac{1}{y+1}\left(\frac{2\Omega\delta_{r}K_{3}}{r_{s}}-\frac{2K_{1}K_{3}}{\delta_{r}}+K_{4}\right)\right.\\\
\left.+\frac{1}{y^{2}}\left(\frac{K^{2}_{1}}{\delta^{2}_{r}}\right)+\frac{1}{(y+1)^{2}}K^{2}_{3}\right]y=0.$
(88)

By redefining the radial variable $x=-y$, we obtain,

${\partial}^{{\rm
2}}_{x}R+\left[\frac{1}{x}+\frac{1}{x-1}\right]\partial_{x}R+\left[\left(\frac{\Omega^{2}}{r^{2}_{s}}\delta^{2}_{r}-\frac{\Omega^{2}_{0}}{r^{2}_{s}}\delta^{2}_{r}\right)\right.\\\
\left.-\frac{1}{x}\left(\frac{2\Omega
K_{1}}{r_{s}}+\frac{2K_{1}K_{3}}{\delta_{r}}-K_{2}\right)\right.\\\
\left.-\frac{1}{x-1}\left(\frac{2\Omega\delta_{r}K_{3}}{r_{s}}-\frac{2K_{1}K_{3}}{\delta_{r}}+K_{4}\right)\right.\\\
\left.+\frac{1}{x^{2}}\left(\frac{K^{2}_{1}}{\delta^{2}_{r}}\right)+\frac{1}{(x-1)^{2}}K^{2}_{3}\right]y=0.$
(89)

The radial equation above is in the natural general form of the asymmetrical
Confluent Heun equation (162). Direct comparison between (89) and (162) allows
us to identify $A_{1,2,3,4,5}$ as follows,

$\displaystyle A_{1}=-\left(\frac{2\Omega
K_{1}}{r_{s}}+\frac{2K_{1}K_{3}}{\delta_{r}}-K_{2}\right),$ (90)
$\displaystyle
A_{2}=-\left(\frac{2\Omega\delta_{r}K_{3}}{r_{s}}-\frac{2K_{1}K_{3}}{\delta_{r}}+K_{4}\right),$
(91) $\displaystyle A_{3}=\frac{K^{2}_{1}}{\delta^{2}_{r}},$ (92)
$\displaystyle A_{4}=K^{2}_{3},$ (93) $\displaystyle
A_{5}=\frac{\Omega^{2}}{r^{2}_{s}}\delta^{2}_{r}-\frac{\Omega^{2}_{0}}{r^{2}_{s}}\delta^{2}_{r}.$
(94)

Following the results of Appendix A, the exact solutions of the radial
equation (89) are obtained in terms of the Confluent Heun functions as
follows,

$y=e^{i\sqrt{A_{5}}}x^{i\sqrt{A_{3}}}(x-1)^{i\sqrt{A_{4}}}\left[A\operatorname{HeunC}\left(\alpha,\beta,\gamma,\delta,\eta,x\right)\right.\\\
\left.+Bx^{-\beta}\operatorname{HeunC}\left(\alpha,-\beta,\gamma,\delta,\eta,x\right)\right],$
(95)

where according to (A.16-A.18), we can express $i\sqrt{A_{3,4,5}}$ in terms of
$\alpha,\beta,\gamma$ as follows,

$\displaystyle\pm i\sqrt{A_{3}}=\frac{1}{2}\beta,$ (96) $\displaystyle\pm
i\sqrt{A_{4}}=\frac{1}{2}\gamma,$ (97) $\displaystyle\pm
i\sqrt{A_{5}}=\frac{1}{2}\alpha.$ (98)

The Confluent Heun function’s parameters, following (A.15)-(A.19), are then
explicitly obtained as follows,

$\displaystyle\alpha=\pm
2i\frac{\delta_{r}}{r_{s}}\sqrt{\Omega^{2}-\Omega^{2}_{0}},$ (99)
$\displaystyle\beta=\pm\frac{{\rm
2}i}{\delta_{r}}\left[\frac{\Omega}{r_{s}}\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)-m_{l}a\right],$
(100)
$\displaystyle\gamma=\pm\frac{2i}{\delta_{r}}\left[\frac{\Omega}{r_{s}}\left(r_{-}\left(r_{-}-2d\right)-k^{2}+a^{2}+2d\delta_{r}\right)-m_{l}a\right],$
(101)
$\displaystyle\delta=-\frac{\delta_{r}}{r^{2}_{s}}\left(r_{+}+r_{-}-4d\right)\left[2\Omega^{2}-\Omega^{2}_{0}\right],$
(102)
$\eta=-\frac{2}{\delta^{2}_{r}}\left(\frac{\Omega}{r_{s}}\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)-m_{l}a\right)\times\\\
\left(\frac{\Omega}{r_{s}}\left(-r_{+}\left(r_{+}-2d\right)+2r_{-}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)-m_{l}a\right)\\\
-\frac{\Omega^{2}_{0}}{r^{2}_{s}}\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\
\right)-K^{m_{l}}_{l}.$ (103)

Finally, we can present the complete exact solution of the scalar field’s wave
function in Dyonic Kerr-Sen black hole background as follows,

$\psi=e^{i{\frac{E}{\hbar
c}}ct}S_{l}^{m_{l}}\left(\theta,\phi\right)e^{-\frac{1}{2}\alpha\left(\frac{r-r_{+}}{\delta_{r}}\right)}{\left(\frac{r-r_{-}}{\delta_{r}}\right)}^{\frac{1}{2}\gamma}\\\
\times\left[A{\left(\frac{r-r_{+}}{\delta_{r}}\right)}^{\frac{1}{2}\beta}\operatorname{HeunC}\left(\alpha,\beta,\gamma,\delta,\eta,-\frac{r-r_{+}}{\delta_{r}}\right)\right.\\\
\left.+B{\left(\frac{r-r_{+}}{\delta_{r}}\right)}^{-\frac{1}{2}\beta}\operatorname{HeunC}\left(\alpha,-\beta,\gamma,\delta,\eta,-\frac{r-r_{+}}{\delta_{r}}\right)\right],$
(104)

where,

$S_{l}^{m_{l}}\left(\theta,\phi\right)=e^{im_{l}\phi}S_{l}^{m_{l}}\left(\frac{\Omega^{2}_{0}a^{2}}{r^{2}_{s}}-\frac{\Omega^{2}a^{2}}{r^{2}_{s}},\cos\theta\right).$
(105)

###  III.4 QBS Energy Quantization

The radial quantization condition is obtained from the degree of the
interpolating radial function. The radial quantum number $n$ is defined as the
number of zeros of the radial wave. The Confluent Heun function will have $n$
zeros if it is a polynomial function with degree $n$, and this condition is
fulfilled when (see A),

$\displaystyle\frac{\delta}{\alpha}+\frac{\beta+\gamma}{2}=-n,$ (106)

where we have redefined $n=n_{r}+1=1,2,...$.

####  III.4.1 The Quasistationary Modes

In this Subsection, we will present the energy quantization expression (106)
for all possible quasistationary modes, which corresponds to combinations of
positive and negative solutions of $\alpha,\beta,\gamma$. We will use the
notation $X_{\pm}$, where $X$ could be $\alpha,\beta,\gamma$ and $\pm$
corresponds to the positive or negative solution of $X$.

1\. For the case with $\alpha_{+},\beta_{+},\gamma_{+}$ the quantization
condition (106) can be rewritten explicitly as follows,

$-\frac{{\left(r_{+}+r_{-}-4d\right)}\left[2\Omega^{2}-\Omega^{2}_{0}\right]}{2r_{s}\sqrt{\Omega^{2}_{0}-\Omega^{2}}}+\frac{i}{\delta_{r}}\left\\{\frac{\Omega}{r_{s}}\left(r_{+}^{2}\right.\right.\\\
\left.\left.+r_{-}\left(r_{-}-4d\right)+2\left(a^{2}-k^{2}\right)\right)-2m_{l}a\right\\}=-n.$
(107)

2\. For the case with $\alpha_{+},\beta_{+},\gamma_{-}$, we obtain,

$-\frac{{\left(r_{+}+r_{-}-4d\right)}\left[-\Omega^{2}_{0}+2\Omega^{2}+2i\Omega\sqrt{\Omega^{2}_{0}-\Omega^{2}}\right]}{2r_{s}\sqrt{\Omega^{2}_{0}-\Omega^{2}}}=-n.$
(108)

3\. For the case with $\alpha_{+},\beta_{-},\gamma_{+}$ , we obtain,

$-\frac{{\left(r_{+}+r_{-}-4d\right)}\left[-\Omega^{2}_{0}+2\Omega^{2}-2i\Omega\sqrt{\Omega^{2}_{0}-\Omega^{2}}\right]}{2r_{s}\sqrt{\Omega^{2}_{0}-\Omega^{2}}}=-n.$
(109)

4\. For the case with $\alpha_{+},\beta_{-},\gamma_{-}$, we obtain,

$-\frac{{\left(r_{+}+r_{-}-4d\right)}\left[2\Omega^{2}-\Omega^{2}_{0}\right]}{2r_{s}\sqrt{\Omega^{2}_{0}-\Omega^{2}}}-\frac{i}{\delta_{r}}\left\\{\frac{\Omega}{r_{s}}\left(r_{+}^{2}\right.\right.\\\
\left.\left.+r_{-}\left(r_{-}-4d\right)+2\left(a^{2}-k^{2}\right)\right)-2m_{l}a\right\\}=-n.$
(110)

5\. For the case with $\alpha_{-},\beta_{+},\gamma_{+}$, we obtain,

$\frac{{\left(r_{+}+r_{-}-4d\right)}\left[2\Omega^{2}-\Omega^{2}_{0}\right]}{2r_{s}\sqrt{\Omega^{2}_{0}-\Omega^{2}}}+\frac{i}{\delta_{r}}\left\\{\frac{\Omega}{r_{s}}\left(r_{+}^{2}\right.\right.\\\
\left.\left.+r_{-}\left(r_{-}-4d\right)+2\left(a^{2}-k^{2}\right)\right)-2m_{l}a\right\\}=-n.$
(111)

6\. For the case with $\alpha_{-},\beta_{+},\gamma_{-}$ , we obtain,

$\frac{{\left(r_{+}+r_{-}-4d\right)}\left[-\Omega^{2}_{0}+2\Omega^{2}-2i\Omega\sqrt{\Omega^{2}_{0}-\Omega^{2}}\right]}{2r_{s}\sqrt{\Omega^{2}_{0}-\Omega^{2}}}=-n.$
(112)

7\. For the case with $\alpha_{-},\beta_{-},\gamma_{+}$ , we obtain,

$\frac{{\left(r_{+}+r_{-}-4d\right)}\left[-\Omega^{2}_{0}+2\Omega^{2}+2i\Omega\sqrt{\Omega^{2}_{0}-\Omega^{2}}\right]}{2r_{s}\sqrt{\Omega^{2}_{0}-\Omega^{2}}}=-n.$
(113)

8\. For the case with $\alpha_{-},\beta_{-},\gamma_{-}$, we obtain,

$\frac{{\left(r_{+}+r_{-}-4d\right)}\left[2\Omega^{2}-\Omega^{2}_{0}\right]}{2r_{s}\sqrt{\Omega^{2}_{0}-\Omega^{2}}}-\frac{i}{\delta_{r}}\left\\{\frac{\Omega}{r_{s}}\left(r_{+}^{2}\right.\right.\\\
\left.\left.+r_{-}\left(r_{-}-4d\right)+2\left(a^{2}-k^{2}\right)\right)-2m_{l}a\right\\}=-n.$
(114)

Note that the scalar’s energy levels in a rotating black hole spacetime
possess hyperfine splitting, where the energy depends on the azimuthal quantum
number $m_{l}$ and is directly coupled to the black hole’s spin parameter $a$.
This is analogous to the Zeeman effect, which occurs when a hydrogen atom is
immersed in a magnetic environment. The existence of the term can be
understood as an interaction between the orbiting scalar field, with magnetic
state $m_{l}$, and the black hole’s angular momentum $a$.

Now, let us investigate the energy levels in the limit $Er_{s}\to 0$. For
modes with $\beta_{+},\gamma_{+}$ and $\beta_{-},\gamma_{-}$, we obtain,

$\displaystyle E_{n}=E_{0}\sqrt{1-{\left[\frac{\frac{E_{0}r_{s}}{\hbar
c}\frac{{\left(r_{+}+r_{-}-4d\right)}}{r_{s}}}{2\left(i\frac{2m_{l}a}{\delta_{r}}+n\right)}\right]}^{2}},$
(115) $\displaystyle
E_{n}-E_{0}\approx-\frac{E_{0}}{2}{\left[\frac{\frac{E_{0}r_{s}}{\hbar
c}\frac{{\left(r_{+}+r_{-}-4d\right)}}{r_{s}}}{2\left(i\frac{2m_{l}a}{\delta_{r}}+n\right)}\right]}^{2}.$
(116)

while for the cases with $\beta_{-},\gamma_{+}$ and $\beta_{+},\gamma_{-}$, we
obtain,

$\displaystyle E_{n}=E_{0}\sqrt{1-{\left[\frac{\frac{E_{0}r_{s}}{\hbar
c}\frac{{\left(r_{+}+r_{-}-4d\right)}}{r_{s}}}{2n}\right]}^{2}},$ (117)
$\displaystyle
E_{n}-E_{0}\approx-\frac{E_{0}}{2}{\left[\frac{\frac{E_{0}r_{s}}{\hbar
c}{\frac{{\left(r_{+}+r_{-}-4d\right)}}{r_{s}}}}{2n}\right]}^{2}.$ (118)

In contrast to the quasibound states around static black holes, where the
energy levels are real valued in the small black hole limit Daniel ; Anal1 ;
Anal2 ; Anal3 ; Anal10 ; David-Senjaya , for a rotating black hole, there are
modes where complex valued energy levels are obtained even in a small black
hole limit. The non zero decay of the quasibound state even in the small black
hole limit in the rotating black hole backgrounds is expected, since,
classically, an astrophysical rotating black hole with
$a=0.998\frac{r_{s}}{2}$ has $32\%$ efficiency in converting the mass accreted
from the disk to radiation. This value is five times greater than that of the
non rotating Schwarzschild black hole Hobson . Considering the non-rotating
case, by taking as zero the black hole’s angular momentum, $a=0$, we obtain
the expression of the energy levels of the real valued quasi-bound states of
the static axionic dyonic dilatonic Reissner-Nordstrom black hole.

Now, let us consider a massless scalar field in the Dyonic Kerr-Sen black
hole’s gravitational potential. Setting $\Omega_{0}=0$, we can derive the
energy expressions for each modes as follows,

1\. For the case with $\alpha_{+},\beta_{+},\gamma_{+}$, we obtain,

$\displaystyle\Omega_{1}=\frac{r_{s}\left(in(r_{+}-r_{-})+2m_{l}a\right)}{2\left(r_{+}(r_{+}-2d)+a^{2}-k^{2}\right)},$
(119)
$\displaystyle\Omega_{2}=\frac{r_{s}\left(in(r_{+}-r_{-})+2m_{l}a\right)}{2\left(r_{-}(r_{-}-4d)+2r_{+}d+a^{2}-k^{2}\right)}.$
(120)

2\. For the case with $\alpha_{+},\beta_{+},\gamma_{-}$ , we obtain,

$\Omega=\frac{inr_{s}}{r_{+}+r_{-}-4d}.$ (121)

3\. For the case with $\alpha_{+},\beta_{-},\gamma_{+}$ , we obtain,

$\Omega=-\frac{inr_{s}}{r_{+}+r_{-}-4d}.$ (122)

4\. For the case with $\alpha_{+},\beta_{-},\gamma_{-}$, we obtain,

$\displaystyle\Omega_{1}=\frac{r_{s}\left(-in(r_{+}-r_{-})+2m_{l}a\right)}{2\left(r_{+}(r_{+}-2d)+a^{2}-k^{2}\right)},$
(123)
$\displaystyle\Omega_{2}=\frac{r_{s}\left(-in(r_{+}-r_{-})+2m_{l}a\right)}{2\left(r_{-}(r_{-}-4d)+2r_{+}d+a^{2}-k^{2}\right)}.$
(124)

5\. For the case with $\alpha_{-},\beta_{+},\gamma_{+}$, we obtain,

$\displaystyle\Omega_{1}=\frac{r_{s}\left(in(r_{+}-r_{-})+2m_{l}a\right)}{2\left(r_{+}(r_{+}-2d)+a^{2}-k^{2}\right)},$
(125)
$\displaystyle\Omega_{2}=\frac{r_{s}\left(in(r_{+}-r_{-})+2m_{l}a\right)}{2\left(r_{-}(r_{-}-4d)+2r_{+}d+a^{2}-k^{2}\right)}.$
(126)

6\. For the case with $\alpha_{-},\beta_{+},\gamma_{-}$ , we obtain,

$\Omega=\frac{inr_{s}}{r_{+}+r_{-}-4d}.$ (127)

7\. For the case with $\alpha_{-},\beta_{-},\gamma_{+}$ , we obtain,

$\Omega=-\frac{inr_{s}}{r_{+}+r_{-}-4d}.$ (128)

8\. For the case with $\alpha_{-},\beta_{-},\gamma_{-}$, we obtain,

$\displaystyle\Omega_{1}=\frac{r_{s}\left(-in(r_{+}-r_{-})+2m_{l}a\right)}{2\left(r_{+}(r_{+}-2d)+a^{2}-k^{2}\right)},$
(129)
$\displaystyle\Omega_{2}=\frac{r_{s}\left(-in(r_{+}-r_{-})+2m_{l}a\right)}{2\left(r_{-}(r_{-}-4d)+2r_{+}d+a^{2}-k^{2}\right)}.$
(130)

Notice that the cases with $\beta_{-},\gamma_{+}$ and $\beta_{+},\gamma_{-}$
have purely imaginary energy levels, indicating a more rapid absorption of the
field by the Dyonic Kerr-Sen black hole. There is also an apparent degeneracy
in $\alpha$, since the solutions with $\alpha_{+}$ are the same as the
solutions with $\alpha_{-}$.

##  IV The Horizon’s Hawking Radiation

In the previous Section, we have presented the detailed derivations for
obtaining the complete expressions of the polar and radial waves in terms of
the Spheriodal Harmonics and of the Confluent Heun functions. In this Section,
we focus on the Hawking radiation from the apparent horizon of the Dyonic
Kerr-Sen black hole. Following the Sigurd - Sannan method Sannan:1988eh , we
can use the exact radial wave solutions to derive the Hawking radiation
distribution function, and search for the expression of the Hawking
temperature of the black hole’s apparent horizon $r_{+}$. Approaching the
exterior event horizon $r\to r_{+}$, we can approximate $\operatorname{HeunC}$
as follows,

$\operatorname{HeunC}(0)=\operatorname{HeunC}^{\prime}(0)\approx 1,$ (131)

also the exponential,
$e^{-\frac{1}{2}\alpha\left(\frac{r-r_{+}}{\delta_{r}}\right)}\approx 1$,

$\displaystyle
R={\left(\frac{r_{+}-r_{-}}{\delta_{r}}\right)}^{\frac{1}{2}\gamma}\left[B{\left(\frac{r-r_{+}}{\delta_{r}}\right)}^{-\frac{1}{2}\beta}+A{\left(\frac{r-r_{+}}{\delta_{r}}\right)}^{\frac{1}{2}\beta}\right],$
(132) $\displaystyle\beta=\frac{{\rm
2}i}{\delta_{r}}\left[\frac{\Omega}{r_{s}}\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)-{m_{l}a}\right].$
(133)

The radial wave consists of two independent parts,

$\displaystyle
R=\left\\{\begin{array}[]{cc}\psi_{+in}=A{\left(\frac{r_{+}-r_{-}}{\delta_{r}}\right)}^{\frac{1}{2}\gamma}{\left(\frac{r-r_{+}}{\delta_{r}}\right)}^{\frac{1}{2}\beta}&\emph{ingoing}\\\
\psi_{+out}=B{\left(\frac{r_{+}-r_{-}}{\delta_{r}}\right)}^{\frac{1}{2}\gamma}{\left(\frac{r-r_{+}}{\delta_{r}}\right)}^{-\frac{1}{2}\beta}&\emph{outgoing}\end{array}\right..$
(136)

Suppose there is an ingoing wave hitting the apparent horizon $r_{+}$. This
will induce a particle-antiparticle pair, with the particle being reflected,
while the antiparticle will be transmitted, going through the horizon. The
analytical continuation of the wave function can be calculated as follows,

$\begin{split}{\left(\frac{r-r_{+}}{\delta_{r}}\right)}^{\lambda}&={\left(\frac{r_{+}}{\delta_{r}}\right)}^{\lambda}{\left(\frac{r}{r_{+}}-1\right)}^{\lambda}\\\
&\to{{\left(\frac{r_{+}}{\delta_{r}}\right)}^{\lambda}\left[\left(\frac{r}{r_{+}}-1\right)+i\epsilon\right]}^{\lambda}\\\
&=\left\\{\begin{array}[]{cc}{\left(\frac{r_{+}}{\delta_{r}}\right)}^{\lambda}{\left(\frac{r-r_{+}}{\delta_{r}}\right)}^{\lambda}&,\
r>r_{+}\\\
{\left(\frac{r_{+}}{\delta_{r}}\right)}^{\lambda}{\left\lvert\frac{r-r_{+}}{\delta_{r}}\right\rvert}^{\lambda}e^{i\lambda\pi}&,\
r<r_{+}\end{array}\right.,\end{split}$ (137)

The analytical continuation enables us to obtain the expression
$\psi_{-out}=\psi_{+out}\left(\left(\frac{r-r_{+}}{\delta_{r}}\right)\to\left(\frac{r-r_{+}}{\delta_{r}}\right)e^{i\pi}\right)$
simply by replacing
$\left(\frac{r-r_{+}}{\delta_{r}}\right)\to-\left(\frac{r-r_{+}}{\delta_{r}}\right)=\left(\frac{r-r_{+}}{\delta_{r}}\right)e^{i\pi}$
as follows,

$\begin{split}\psi_{-out}&=B{\left(\frac{r_{+}-r_{-}}{\delta_{r}}\right)}^{\frac{1}{2}\gamma}{\left(\left(\frac{r-r_{+}}{\delta_{r}}\right)e^{i\pi}\right)}^{-\frac{1}{2}\beta},\\\
&=\psi_{+out}e^{-\frac{1}{2}i\pi\beta}.\end{split}$ (138)

One can also find the modulus square of the probability amplitude with respect
to the ingoing wave as follows,

$\displaystyle{\left\lvert\frac{\psi_{-out}}{\psi_{+in}}\right\rvert}^{2}$
$\displaystyle={\left\lvert\frac{\psi_{+out}}{\psi_{+in}}\right\rvert}^{2}e^{-i2\pi\beta}$
(139)
$\displaystyle={\left\lvert\frac{\psi_{+out}}{\psi_{+in}}\right\rvert}^{2}e^{\frac{4\pi}{\delta_{r}}\left[\frac{E}{\hslash
c}\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)-{m_{l}a}\right]}$ (140)
$\displaystyle={\left\lvert\frac{\psi_{+out}}{\psi_{+in}}\right\rvert}^{2}e^{\zeta}.$
(141)

The exponent $e^{-\zeta}$ represents the relative probability of radiation
emission. The amplitude of the pair productions that occur is described by
that function. Since the observer stays outside the black hole horizon, the
absolute probability of the processes occurring outside the horizon needs to
be found by summing all probabilities to create no pair, 1 pair, 2 pairs and
so on, as follows,

$C_{\omega}\left(1+e^{-\zeta}+{\left(e^{-\zeta}\right)}^{2}+\dots\right)=1\to
C_{\omega}=1-e^{-\zeta}.$ (142)

The probability to create $j$ pairs of particle-antiparticle is given by,

$C_{\omega}{\left(e^{-\zeta}\right)}^{j}=\left(1-e^{-\zeta}\right)e^{-j\zeta}.$
(143)

The normalized spectrum of all the possible pair productions is obtained by
calculating the mean number of the emitted particles,

$n\left(\omega\right)=\sum^{\infty}_{n=0}n\left(1-e^{-\zeta}\right)e^{-n\zeta}=\frac{1}{e^{\zeta}-1}.$
(144)

The same distribution function can also be obtained via the normalization
condition of the wave function, following the method developed by Zhao . Now,
let us write the total outgoing wave that consists of the particle wave
outside the black hole’s horizon and the antiparticle wave inside the black
hole’s horizon. With the help of the Heaviside step function, the total
outgoing wave can be written in a unique form as follows,

$\displaystyle\psi_{out}=\psi_{+out}\Theta\left(r-r_{+}\right)+\psi_{-out}\Theta\left(r_{+}-r\right),$
(145)

or,

$\displaystyle\frac{\psi_{out}}{\psi_{+in}}=\frac{\psi_{+out}}{\psi_{+in}}\Theta\left(r-r_{+}\right)+\frac{\psi_{-out}}{\psi_{+in}}\Theta\left(r_{+}-r\right),$
(146)

The total probability of the emission of the particles and of the
antiparticles (remember that the antiparticle has a negative probability) must
be normalized to be one. Hence, we can write,

$\left\langle{\left\lvert\frac{\psi_{out}}{\psi_{+in}}\right\rvert}^{2}\right\rangle=\left\langle{\left\lvert\frac{\psi_{+out}}{\psi_{+in}}\right\rvert}^{2}\right\rangle-\left\langle{\left\lvert\frac{\psi_{-out}}{\psi_{+in}}\right\rvert}^{2}\right\rangle=1,$
(147)

and by using (141) to substitute
${\left\lvert\frac{\psi_{-out}}{\psi_{+in}}\right\rvert}^{2}$, we obtain the
following expression,

$\displaystyle\left\langle{\left\lvert\frac{\psi_{+out}}{\psi_{+in}}\right\rvert}^{2}\right\rangle\left\lvert
1-e^{\zeta}\right\rvert=1,$ (148)
$\displaystyle\left\langle{\left\lvert\frac{\psi_{+out}}{\psi_{+in}}\right\rvert}^{2}\right\rangle=\frac{1}{e^{\zeta}-1}.$
(149)

The Hawking temperature, $T_{H}$, is to be found by first rewriting $\zeta$ as
follows,

$\displaystyle\zeta$
$\displaystyle=\frac{4\pi}{\delta_{r}}\left[\frac{E}{\hslash
c}\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)-{m_{l}a}\right]$
$\displaystyle=\frac{4\pi}{\delta_{r}}\left[\frac{\omega}{c}\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)-{m_{l}a}\right]$
$\displaystyle=\frac{\hslash\left(\omega-{\omega}_{J}\right)}{\left[\frac{\delta_{r}c\hslash}{4\pi\left(r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right)}\right]},$
(150)

where we have defined,

${\omega}_{J}=\frac{m_{l}ac}{r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}}.$ (151)

Comparing the black hole’s radiation distribution function (144) with the
Bose-Einstein distribution function,

$n\left(\omega\right)=\frac{1}{e^{\frac{\hbar\omega-\mu}{k_{B}T}}-1},$ (152)

the apparent horizon’s Hawking temperature is found as follows,

$\displaystyle T_{H}=\frac{\delta_{r}c\hslash}{4\pi
k_{B}\left[r_{+}\left(r_{+}-2d\right)-k^{2}+a^{2}\right]}.$ (153)

Interestingly enough, similar to the case of the Kerr black hole, the angular
momentum plays the role of the chemical potential in the thermodynamics of
Kerr-Sen black hole, with $\mu=\hbar\omega_{J}$. For the Kerr-Sen black hole,
$\mu$ also depends on the axion and dilaton (or electric and magnetic) charges
of the black hole.

One can check that by setting the black hole’s angular momentum and charges to
be zero, we simultaneously set $r_{+}\to r_{s}$, $r_{-}\to 0$ and
$\delta_{r}\to r_{s}$. In this case, we recover the Hawking temperature of the
Schwarzschild black hole Harris ,

$T_{H}=\frac{c\hslash}{4\pi k_{B}r_{s}}.$ (154)

##  V Conclusions

In this work, we have presented the novel exact analytical general solutions
of the covariant massive and massless Klein-Gordon equations in the Dyonic
Kerr-Sen black hole spacetime. By exactly solving the Klein-Gordon equation,
we have also obtained the solutions for the canonical relativistic quantum
mechanical problem of the energy levels of the scalar fields, gravitationally
bounded by the Dyonic Kerr-Sen black hole. We have showed in detail the
derivations of our results, and we have presented the energy expressions of
the all possible sixteen quantize modes for both the massive and massless
cases III.4.1, together with the corresponding exact wave functions (104).
Since we did not use any approximations to derive the analytical solutions,
the obtained wave functions are valid for all regions of interest, i.e.
$r_{+}\leq r<\infty$. This is a remarkable improvement as compared to the
previous approaches, and, in contrast with the asymptotical methods, whose
solutions are correct only for regions closed to the horizon, or
asymptotically far away from the horizon.

We can recover the Schwarzschild massive quasibound state’s real valued energy
levels expression in the small black hole limit, given by,

$\displaystyle\frac{E_{n}}{E_{0}}\approx 1-\frac{\kappa^{2}}{8n^{2}},$ (155)
$\displaystyle\kappa={\left(\frac{E_{0}r_{s}}{\hbar c}\right)}^{2},$ (156)

by applying the small black hole limit $Er_{s}\to 0$ and taking as zero the
spin $a$, and the charges of the Dyonic Kerr-Sen black holes, $Q,P,k,d$, in
the equations (116) and (118). The $\frac{1}{n^{2}}$ Hydrogenic-atom-like
energy expression was also obtained in the previous works Daniel ; Anal1 ;
Anal2 ; Anal3 ; Anal10 ; David-Senjaya .

In the last section, by making use of the exact radial solutions, we have
investigated the Hawking radiation of the Dyonic Kerr-Sen black hole’s outer
horizon $r_{+}$. We have followed the Sigurd-Sannan method Sannan:1988eh ;
Damour  to treat the Klein pair production scenario as describing an incoming
particle hitting the black hole’s horizon. The radiation distribution function
is derived by summing all of the possible pair productions rates, and is
presented in (144). Comparing it with the bosonic distribution function, the
Hawking temperature of the Dyonic Kerr-Sen black hole’s apparent horizon was
obtained in (153).

## Acknowledgments

DS acknowledges this work is supported by the Second Century Fund (C2F),
Chulalongkorn University, Thailand. P.B. is supported in part by National
Research Council of Thailand (NRCT) and Chulalongkorn University under Grant
N42A660500. This research has received funding support from the NSRF via the
Program Management Unit for Human Resources & Institutional Development,
Research and Innovation [grant number B41G670027].

##  Appendix A The Confluent Heun Functions and S-Homotopic Transformation

The Confluent Heun equation is a second-order linear differential equation
that has three regular singularities given in the following canonical form
Heun ,

$\frac{d^{2}y_{H}}{dx^{2}}+\left(\alpha+\frac{\beta+1}{x}+\frac{\gamma+1}{x-1}\right)\frac{dy_{H}}{dx}\\\
+\left(\frac{\mu}{x}+\frac{\nu}{x-1}\right)y_{H}=0,$ (157)

where

$\displaystyle\mu=\frac{1}{2}\left(\alpha-\beta-\gamma+\alpha\beta-\beta\gamma\right)-\eta,$
(158)
$\displaystyle\nu=\frac{1}{2}\left(\alpha+\beta+\gamma+\alpha\gamma+\beta\gamma\right)+\delta+\eta.$
(159)

The solutions of the differential equation are given in terms of two
independent Confluent Heun functions as follows,

$y_{H}=A\operatorname{HeunC}\left(\alpha,\beta,\gamma,\delta,\eta,x\right)\\\
+Bx^{-\beta}\operatorname{HeunC}\left(\alpha,-\beta,\gamma,\delta,\eta,x\right).$
(160)

The Confluent Heun function can be reduced to an $n^{th}$ order polynomial
function if the following series termination condition is fulfilled,

$\frac{\delta}{\alpha}+\frac{\beta+\gamma}{2}+1=-n_{r},\quad
n_{r}\in\mathbb{Z}.$ (161)

Given a natural general form of the asymmetrical Confluent Heun equation Heun
; Vier22 ,

$\frac{d^{2}y}{dx^{2}}+\left(\frac{1}{x}+\frac{1}{x-1}\right)\frac{dy}{dx}\\\
+\left(\frac{A_{1}}{x}+\frac{A_{2}}{x-1}+\frac{A_{3}}{x^{2}}+\frac{A_{4}}{(x-1)^{2}}+A_{5}\right)y=0,$
(162)

in order to find the solution of (162), we have to apply the s-homotopic
transformation Alb ; Heun ; Vier22 ; Sylv ; chen  by transforming the
dependent variable $y(x)\to u(x)$ as follows,

$y(x)=e^{B_{0}}x^{B_{1}}(x-1)^{B_{2}}u(x).$ (163)

Substituting the transformation into the equation (162), we find the values of
the exponents $B_{0},B_{1},B_{2}$ from the initial equation as follows,

$\displaystyle B_{0}(B_{0}-1)+B_{0}+A_{5}=0\to B_{0}=\pm i\sqrt{A_{5}},$ (164)
$\displaystyle B_{1}(B_{1}-1)+B_{1}+A_{3}=0\to B_{1}=\pm i\sqrt{A_{3}},$ (165)
$\displaystyle B_{2}(B_{2}-1)+B_{2}+A_{4}=0\to B_{2}=\pm i\sqrt{A_{4}}.$ (166)

Thus, after obtaining the three exponents, substitution of the s-homotopic
transformation (163) into (162) leads to a differential equation for $u(x)$ as
follows,

$\frac{d^{2}u}{dx^{2}}+\left(2B_{0}+\frac{2B_{1}+1}{x}+\frac{2B_{2}+1}{x-1}\right)\frac{du}{dx}\\\
+\left(\frac{\sigma}{x}+\frac{\chi}{x-1}\right)u=0,$ (167)

where

$\displaystyle\sigma=-B_{1}-B_{2}-2B_{1}B_{2}+B_{0}+2B_{0}B_{1}+A_{1},$ (168)
$\displaystyle\chi=B_{1}+B_{2}+2B_{1}B_{2}+B_{0}+2B_{0}B_{2}+A_{2}.$ (169)

By comparing (167) with (157), we can write the solution for $u(x)$ in terms
of the Confluent Heun functions as follows,

$u=A\operatorname{HeunC}\left(\alpha,\beta,\gamma,\delta,\eta,x\right)\\\
+Bx^{-\beta}\operatorname{HeunC}\left(\alpha,-\beta,\gamma,\delta,\eta,x\right),$
(170)

where,

$\displaystyle\alpha=2B_{0}=\pm 2i\sqrt{A_{5}},$ (171)
$\displaystyle\beta=2B_{1}=\pm 2i\sqrt{A_{3}},$ (172)
$\displaystyle\gamma=2B_{2}=\pm 2i\sqrt{A_{4}},$ (173)
$\displaystyle\delta=A_{1}+A_{2},$ (174) $\displaystyle\eta=-A_{1}.$ (175)

Hence, the complete solutions for the natural general form of the asymmetrical
Confluent Heun equation (162) are obtained as follows,

$y=e^{\pm i\sqrt{A_{5}}}x^{\pm i\sqrt{A_{3}}}(x-1)^{\pm
i\sqrt{A_{4}}}\left[A\operatorname{HeunC}\left(\alpha,\beta,\gamma,\delta,\eta,x\right)\right.\\\
\left.+Bx^{-\beta}\operatorname{HeunC}\left(\alpha,-\beta,\gamma,\delta,\eta,x\right)\right],$
(176)

with $\alpha,\beta,\gamma,\delta,\eta$ are given by (A.15)-(A.19).

For $x\to\infty$, the approximate solution of (157) is given as follows,

$\displaystyle y_{H}$
$\displaystyle=Ax^{-\left(\frac{\delta}{\alpha}+\frac{\beta+\gamma+2}{2}\right)}+Be^{-\alpha
x}x^{\left(\frac{\delta}{\alpha}-\frac{\beta+\gamma+2}{2}\right)}$ (177)
$\displaystyle=e^{-\frac{1}{2}\alpha
x}x^{-\frac{\beta+\gamma+2}{2}}\left[Ae^{\frac{1}{2}\alpha
x}x^{-\frac{\delta}{\alpha}}+Be^{-\frac{1}{2}\alpha
x}x^{\frac{\delta}{\alpha}}\right]$ (178)
$\displaystyle=y_{0}e^{-\frac{1}{2}\alpha
x}x^{-\frac{\beta+\gamma+2}{2}}\sin\left[-i\frac{\alpha}{2}x+i\frac{\delta}{\alpha}\ln{x}+\phi_{0}\right]$
(179) $\displaystyle=y_{0}e^{-\frac{1}{2}\alpha
x}x^{-\frac{\delta}{\alpha}}x^{n_{r}}\sin\left[-i\frac{\alpha}{2}x+i\frac{\delta}{\alpha}\ln{x}+\phi_{0}\right].$
(180)

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