[table]capposition=bottom capbtabboxtable[][]

# Light-cone sum rules study on the purely non-factorizable
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ decay

Yu-Ji Shi<EMAIL_ADDRESS>School of Physics, East China University of
Science and Technology, Shanghai 200237, China  Zhen-Xing Zhao  School of
Physical Science and Technology, Inner Mongolia University, Hohhot 010021,
China

###### Abstract

We investigate the purely non-factorizable $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$
decay using light-cone sum rules. To extract the decay amplitudes, a three-
point correlation function is defined and calculated at hadron and quark-gluon
level, respectively. Both the W-exchange and the W-inward emission diagrams
are considered in the quark-gluon level calculation, where the two-particle
light-cone distribution amplitudes (LCDAs) of kaon are used as non-
perturbative input. We obtain the decay amplitudes contributed from the
twist-2 and twist-3 kaon LCDAs, respectively. The obtained P-wave amplitude is
consistent with those predicted in the literature, while the S-wave one is
much smaller. This significant difference between the S- and P -wave
amplitudes leads to small up-down spin asymmetry, closing to the experimental
measurement.

##  I Introduction

Studies on charmed baryon decays are significant for understanding the weak
and strong interactions in the Standard Model (SM) of particle physics. The
ground state of the singly-charmed baryons $\Lambda_{c}^{+}$ was firstly
discovered in 1979 Abrams:1979iu . In the past few decades, many experimental
and theoretical efforts have been made to investigate the properties of
charmed baryons, especially the weak decays. Among various of weak decay
modes, the non-leptonic two-body decays of singly-charmed baryon play a
crucial role in our understanding of non-perturbative QCD processes. Such
decay process is intriguing due to its two-body nature and the appearance of
non-factorizable contributions. Both experimental and theoretical
investigations into such decays not only deepen our comprehension of strong
interaction dynamics, but also provide insights into the underlying mechanisms
governing hadronic transitions.

Early in 2002 and 2007, the Belle and BaBar collaborations respectively
published the measurement for the decay process $\Lambda_{c}^{+}\to\Lambda
K^{+}$ Belle:2001hyr ; BaBar:2006eah , which was updated by BESIII in 2022
BESIII:2022tnm  with the newly measured branching fraction lying between the
former two results. Nowadays, further experimental measurements on the non-
leptonic two-body decays of singly-charmed baryon have been made by various of
experimental collaborations. In 2017 and 2021, BESIII and Belle separately
measured the decay processes $\Lambda_{c}^{+}\to p\eta$ and
$\Lambda_{c}^{+}\to p\pi^{0}$ BESIII:2017fim ; Belle:2021mvw . Both
collaborations provided consistent branching fraction for $\Lambda_{c}^{+}\to
p\eta$ and upper limit for $\Lambda_{c}^{+}\to p\pi^{0}$. Recently, BESIII has
successfully measured the branching fraction of $\Lambda_{c}^{+}\to p\pi^{0}$
BESIII:2023uvs  and for the first time measured the branching fraction of
$\Lambda_{c}^{+}\to n\pi^{+}$ BESIII:2022bkj . In 2022, Belle measured the
branching fractions of $\Lambda_{c}^{+}\to\Lambda K^{+}(\pi^{+})$ and
$\Lambda_{c}^{+}\to\Sigma^{0}K^{+}(\pi^{+})$, and for the first time provided
the CP violation in these decays Belle:2022uod . On the theoretical side, with
the use of flavor SU(3) analysis, pole model (PM) and quark model, there are
many works on the non-leptonic two-body decays of charmed baryon in literature
Sharma:1996sc ; Uppal:1994pt ; Lu:2016ogy ; Cheng:2018hwl ; Geng:2018plk ;
Chen:2002jr ; Geng:2017esc ; Zou:2019kzq ; Chen:2002jr . Recently, with the
use of all the existing experimental data, a latest and flavor SU(3) based
global analysis on these decays can be found in Refs.Xing:2023dni ;
Sun:2024mmk .

Figure 1: The W-exchange (WE) (left) and the W-inward emission (IE) (right)
diagrams for the $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ decay.

Generally, the non-leptonic two-body decay of singly-charmed baryons are
induced by factorizable and non-factorizable diagrams. The factorizable ones
are approximately treated by naive factorization, where the baryonic
transition matrix elements can be calculated by various of theoretical methods
such as perturbative QCD Rui:2023fiz ; Rui:2022sdc ; Zhang:2022iun ;
Han:2022srw , QCD/light cone sum rules Shi:2019hbf ; Xing:2021enr ;
Zhao:2021sje ; Huang:2022lfr ; Aliev:2021wat ; Aliev:2022maw ; Shi:2022zzh ;
Shi:2019fph ; Duan:2024lnw  and quark model Lu:2007sg ; Wang:2008xt ;
Wei:2009np ; Ke:2012wa ; Wang:2017mqp ; Ke:2019smy ; Zhao:2018mrg ;
Zhao:2018zcb ; Zhao:2023yuk . Furthermore, the factorization of baryonic
transitions in QCD has been proved in Ref.Wang:2011uv ; Li:2024htn , and the
corresponding light-cone distribution amplitudes (LCDAs) of baryon has been
studied in Refs.Ali:2012pn ; Han:2024ucv ; Deng:2023csv . Comparatively, the
non-factorizable diagrams cannot be divided into several simpler pieces so
that are much more difficult to be studied. Recently, the BESIII collaboration
has announced the first measurement of the decay asymmetry in the decay
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$BESIII:2023wrw , which is a purely non-
factorizable decay process. This measurement provides a great chance for us to
study the non-factorizable dynamics inside the non-leptonic charmed baryon
decays.

The $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ decay is induced by two diagrams as shown
in Fig.1 , the W-exchange (WE) and the W-inward emission (IE) diagrams. There
have been some theoretical studies on this decay in the literature, basing on
flavor SU(3) analysis Geng:2019xbo ; Zhong:2022exp  and current algebra (CA)
with Pole Model (PM) Zou:2019kzq . In this work, we attempt to study
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ by light-cone sum rules (LCSR), a model
independent method basing on QCD. The application of LCSR to non-leptonic
decays was firstly proposed in Refs. Khodjamirian:2000mi ; Khodjamirian:2003eq
; Khodjamirian:2017zdu , where the non-leptonic $B$ meson decays into two
light mesons are considered. This technique of LCSR for non-leptonic decays
has be successfully promoted to the case of heavy baryon in our previous study
on $\Xi_{cc}^{++}\to\Xi_{c}^{+(\prime)}\pi^{+}$ decays Shi:2022kfa . Now the
purely non-factorizable decay $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ can be studied
in a similar procedure.

This paper is organized as follows. In Sec. II, we introduce a three-point
correlation function to extract the decay amplitudes of
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$. In Sec. III, hadron level calculation for
the correlation function is performed. In Sec. IV, we calculate the
correlation function at quark-gluon level with the use of two-particle LCDAs
of the kaon. In Sec. V, we give the numerical results on the decay amplitudes,
the up-down spin asymmetry and the branching fractions of
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$. A comparison between our results and those
from the literature will also be presented. Sec. VI is a brief summary of this
work.

##  II Correlation function in LCSR

In this section, we construct a correlation function to calculate the decay
amplitudes of $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ in the framework of LCSR. The
corresponding effective Hamiltonian for this decay is

$\displaystyle{\cal H}_{\rm eff}=$
$\displaystyle\frac{G_{F}}{\sqrt{2}}V_{cs}V_{ud}^{*}\left(C_{1}{\cal
O}_{1}+C_{2}{\cal O}_{2}\right),$ $\displaystyle{\cal O}_{1}=$
$\displaystyle\bar{s}\gamma_{\mu}(1-\gamma_{5})c\bar{u}\gamma^{\mu}(1-\gamma_{5})d,$
$\displaystyle{\cal O}_{2}=$
$\displaystyle\bar{s}_{a}\gamma_{\mu}(1-\gamma_{5})c_{b}\
\bar{u}_{b}\gamma^{\mu}(1-\gamma_{5})d_{a},$ (1)

where $C_{1},C_{2}$ are Wilson coefficients, ${\cal O}_{1,2}$ are the four-
fermion operators, and the subscripts $a,b$ denote color indices. The
transition matrix element of $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ induced by
${\cal O}_{1,2}$ can be generally parameterized by two amplitudes $A^{1,2}$
and $B^{1,2}$ as

$\displaystyle\langle\Xi^{0}(p-q)K^{+}(q)|{\cal
O}_{i}(0)|\Lambda_{c}^{+}(p)\rangle$ $\displaystyle=$ $\displaystyle\ i\
\bar{u}_{\Xi}(p-q)(A^{i}-B^{i}\gamma_{5})u_{\Lambda_{c}}(p).$ (2)

Note that since all the external states are on shell, as functions of the
external invariant momentum squares, $A^{i}$ and $B^{i}$ are just constants
only depending on the external state masses: $m_{\Xi},m_{K},m_{\Lambda_{c}}$.
In this work, we use $A^{i}_{\rm WE},B^{i}_{\rm WE}$ and $A^{i}_{\rm
IE},B^{i}_{\rm IE}$ to denote the contributions from the WE and IE diagrams as
shown in Fig.1, respectively.

In the framework of LCSR, to calculate a transition matrix element one has to
construct an appropriate correlation function, which will be calculated both
at the hadron and quark-gluon level. In the case of
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$, the correlation function is chosen as

$\displaystyle\Pi^{{\cal O}_{i}}(p,q,k)=$ $\displaystyle\ i^{2}\int d^{4}x\
e^{-i(p-q)\cdot x}\int d^{4}y\ e^{i(p-k)\cdot y}$ $\displaystyle\times\langle
0|T\left\\{J_{\Xi}(y){\cal
O}_{i}(0)\bar{J}_{\Lambda_{c}}(x)\right\\}|K^{-}(q)\rangle,$ (3)

where the hadron currents are defined as Colangelo:2000dp ; Shi:2019hbf

$\displaystyle J_{\Xi}=$
$\displaystyle\varepsilon_{ijk}\left(s_{i}^{T}C\gamma^{\mu}s_{j}\right)\gamma_{\mu}\gamma_{5}Q_{k},$
$\displaystyle J_{\Lambda_{c}}=$
$\displaystyle\varepsilon_{ijk}\left(u_{i}^{T}C\gamma_{5}d_{j}\right)Q_{k}.$
(4)

Instead of setting the kaon as a final state, using the approach proposed in
Ref. Khodjamirian:2000mi , in Eq.(3) we set the kaon as an initial state when
constructing the correlation function. This arrangement dose not follow the
real physical process $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ where the Kaon is a
final state. However, its advantage is that it enables us to factorize out the
matrix element $\langle 0|J_{\Xi}|\Xi\rangle$ by inserting the $\Xi$ state in
the correlation function without any pollution from kaon. On the other hand,
we have introduced an auxiliary momentum $k$, leading to extra momentum
invariants relied by the correlation function. Conducting analytical
continuation for these extra invariants in the hadron level calculation, we
can transform the initial kaon state to the final state. This procedure will
be explained in the next section.

##  III Hadron level calculation in LCSR

In this section, we perform the hadron level calculation for the correlation
function defined in Eq. (3). It can be found that $q^{2}=m_{K}^{2}$ is a
constant, and $\Pi^{{\cal O}_{i}}(p,q,k)$ depends on 5 Lorentz invariants:

$\displaystyle
p^{2},~{}~{}k^{2},~{}~{}(p-q)^{2},~{}~{}(p-k)^{2},~{}~{}{\bar{p}}^{2}=(p-q-k)^{2}.$
(5)

Note that the correlation function should be calculated both at the hadron and
quark-gluon level. To realize the operator product expansion (OPE) at the
quark-gluon level, one has to choose following deep Euclidean regions:
$(p-q)^{2},(p-k)^{2},{\bar{p}}^{2}\leq 0$.

We insert a complete set of states with the same quantum numbers as $\Xi^{+}$
between $J_{\Xi}(y)$ and ${\cal O}_{i}(0)$ in the correlation function:

$\displaystyle\Pi_{H}^{{\cal O}_{i}}(p,q,k)$ $\displaystyle=$ $\displaystyle\
i^{2}\int d^{4}xd^{4}y\ e^{-i(p-q)\cdot x}e^{i(p-k)\cdot
y}\sum_{\pm^{\prime},\sigma^{\prime}}\int\frac{d^{3}\vec{l}}{(2\pi)^{3}}\frac{1}{2E_{l}}$
$\displaystyle\times\langle
0|J_{\Xi}(y)|l,\sigma^{\prime},\pm^{\prime}\rangle\langle
l,\sigma^{\prime},\pm^{\prime}|{\cal
O}_{i}(0)\bar{J}_{\Lambda_{c}}(x)|K^{-}(q)\rangle$
$\displaystyle+\int_{s_{\Xi}}^{\infty}ds^{\prime}\frac{\rho_{\Xi}(s^{\prime},\cdots)}{s^{\prime}-(p-k)^{2}},$
(6)

where the integration for $\rho_{\Xi}(s^{\prime},\cdots)$ represents the
contribution from the excited states and continuous spectrum, the $\cdots$
denotes the invariants in Eq.(5) except $(p-k)^{2}$. $s_{\Xi}$ is the
threshold parameter for the excited states. The index $\pm^{\prime}$ denotes
the positive or negative parity states: $\Xi(\frac{1}{2}^{\pm})$, both of
which can be created by the $J_{\Xi}$ current, and $\sigma^{\prime}$ denotes
their spins. As momentum of a one-particle state in the complete set, $l$ is
on-shell: $l^{2}=m_{\Xi}^{\pm 2}$. Thus the first matrix element in Eq. (6)
can be simply parameterized by the $\Xi(\frac{1}{2}^{\pm})$ pole residues
$\lambda_{\Xi^{\pm}}$:

$\displaystyle\langle 0|J_{\Xi}(y)|l,\sigma^{\prime},+\rangle$
$\displaystyle=\lambda_{\Xi}^{+}u_{\Xi}(l,\sigma^{\prime})e^{-il\cdot y},$
$\displaystyle\langle 0|J_{\Xi}(y)|l,\sigma^{\prime},-\rangle$
$\displaystyle=\lambda_{\Xi}^{-}i\gamma_{5}u_{\Xi}(l,\sigma^{\prime})e^{-il\cdot
y}.$ (7)

For simplicity we define $u^{+}\equiv u$ and $u^{-}\equiv i\gamma_{5}u$.

Figure 2: A diagram for the hadronic transition sector
$M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}$ in Eq. (6). (a): In the region
${\bar{p}}^{2}\ll 0$ it can be considered as a t-channel scattering process
from $K^{-}(q)$ to $\Xi^{+}(l)$ with incoming momentum $l-q=\bar{p}$. (b): In
the region ${\bar{p}}^{2}>0$, the initial state $|K^{-}(q)\rangle$ of the
matrix element $M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}$ can be replaced by
a final state $\langle K^{+}(-q)|$.

On the other hand, in Eq. (6) the hadronic transition sector involving the
external states can be described by a following matrix element

$\displaystyle M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}(p,q,l)=$
$\displaystyle\int d^{4}x\ e^{-i(p-q)\cdot x}$ $\displaystyle\times\langle
l,\sigma^{\prime},\pm^{\prime}|{\cal
O}_{i}(0)\bar{J}_{\Lambda_{c}}(x)|K^{-}(q)\rangle,$ (8)

which depends on 4 invariants: $(p-q)^{2},p^{2},(l-q)^{2},(p-l)^{2}$, and
$l^{2}=m_{\Xi}^{\pm^{\prime}2}$ is a constant. This matrix element can be
considered as a flavor changing transition from $K^{-}(q)$ to $\Xi^{+}(l)$
with incoming momentum $l-q$, which is intuitively shown by the diagram (a) in
Fig.2. The momentum conservation in Eq.(6) leads to $l=p-k$ and the
correlation function can be expressed by
$M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}$ as

$\displaystyle\Pi_{H}^{{\cal
O}_{i}}(p,q,k)=i^{2}\sum_{\pm^{\prime},\sigma^{\prime}}\frac{i}{(p-k)^{2}-m_{\Xi}^{\pm^{\prime}2}}$
$\displaystyle\times\langle
0|J_{\Xi}(0)|p-k,\sigma^{\prime},\pm^{\prime}\rangle
M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}(p,q,p-k)$
$\displaystyle+\int_{s_{\Xi}}^{\infty}ds^{\prime}\frac{\rho_{\Xi}(s^{\prime},\cdots)}{s^{\prime}-(p-k)^{2}}.$
(9)

Now the incoming momentum of $M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}$
becomes $\bar{p}=p-k-q$, which is in the deep Euclidean region. Therefore, the
$M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}(p,q,p-k)$ in Eq.(9) is a function
of $(p-q)^{2},p^{2},k^{2}$ and ${\bar{p}}^{2}$, which describes a t-channel
scattering process, where a hard momentum $\bar{p}$ scatters on the
$K^{-}(q)$, and then produce a $\Xi^{+}(p-k)$.

At the quark-hadron level, the same correlation function will be calculated by
OPE, and the result at these two levels should be equivalent:

$\displaystyle\Pi_{H}^{{\cal O}_{i}}(p,q,k)=\Pi_{WE}^{{\cal O}_{i}}(p,q,k).$
(10)

Operating the Borel transformation for $(p-k)^{2}$ on the both sides above,
using quark-hadron duality to cancel the continuous spectrum contribution, one
arrives at

$\displaystyle-i^{3}\sum_{\pm^{\prime},\sigma^{\prime}}e^{-m_{\Xi}^{\pm^{\prime}2}/T^{\prime
2}}\lambda_{\Xi}^{\pm^{\prime}}u_{\Xi}^{\pm}(p-k)M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}(p,q,p-k)$
$\displaystyle=$ $\displaystyle\frac{1}{\pi}\int_{4m_{s}^{2}}^{s_{\rm
th}^{\Xi}}ds^{\prime}\ e^{-s^{\prime}/T^{\prime 2}}{\rm Im}\Pi_{WE}^{{\cal
O}_{i}}\left[s^{\prime},\cdots,{\bar{p}}^{2}\right],$ (11)

where $s_{\rm th}^{\Xi}$ is the threshold parameter for the continuous
spectrum in the $\Xi$ channel. Note that the right hand side of Eq. (11) is an
analytic function of ${\bar{p}}^{2}$, and the same as the left hand side.
Therefore one can use analytic continuation to extend ${\bar{p}}^{2}$ from
deep Euclidean region to the physical region: ${\bar{p}}^{2}>0$. Accordingly,
as shown by the diagram (b) in Fig.2, the initial state $|K^{-}(q)\rangle$ of
the matrix element $M_{K\to\Xi}^{\sigma^{\prime},\pm^{\prime}}$ can be
replaced by a final state $\langle K^{+}(-q)|$. From this diagram, it can also
be found that after the auxiliary momentum $k$ setting to zero in the end of
the calculation, ${\bar{p}}^{2}$ is actually the invariant mass square of the
two final states, equivalently, the mass square of the initial state:
${\bar{p}}^{2}=m_{\Lambda_{c}}^{+2}$. Now Eq. (11) becomes

$\displaystyle-i^{3}\sum_{\pm^{\prime},\sigma^{\prime}}e^{-m_{\Xi}^{\pm^{\prime}2}/T^{\prime
2}}\lambda_{\Xi}^{\pm^{\prime}}\int d^{4}x\ e^{-i(p-q)\cdot x}$
$\displaystyle\times u_{\Xi}^{\pm}(p-k,\sigma^{\prime})\langle
p-k,\sigma^{\prime},\pm^{\prime};K^{+}(-q)|{\cal
O}_{i}(0)\bar{J}_{\Lambda_{c}}(x)|0\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\pi}\int_{4m_{s}^{2}}^{\infty}ds^{\prime}\
e^{-s^{\prime}/T^{\prime 2}}{\rm Im}\Pi_{WE}^{{\cal
O}_{i}}\left[s^{\prime},\cdots,{\bar{p}}^{2}\right].$ (12)

On the left hand side of Eq. (12), we insert another complete set of states
with the same quantum number as $\Lambda_{c}^{+}$ between ${\cal O}_{i}(0)$
and ${\bar{J}}_{\Lambda_{c}}$, operate Borel transfromation for $(p-q)^{2}$,
and use quark-hadron duality to cancel the continuous spectrum contribution,
then arrive at

$\displaystyle\sum_{\pm^{\prime},\sigma^{\prime}}\sum_{\pm,\sigma}e^{-m_{\Xi}^{\pm^{\prime}2}/T^{\prime
2}}e^{-m_{\Lambda_{c}}^{\pm
2}/T^{2}}\lambda_{\Xi}^{\pm^{\prime}}\lambda_{\Lambda_{c}}^{\pm}u_{\Xi}^{\pm^{\prime}}(p-k,\sigma^{\prime})\langle
p-k,\sigma^{\prime},\pm^{\prime};K^{+}(-q)|{\cal
O}_{i}(0)|p-q,\sigma,\pm\rangle{\bar{u}}_{\Lambda_{c}}^{\pm}(p-q,\sigma)$
$\displaystyle=$ $\displaystyle\frac{1}{\pi^{2}}\int_{4m_{s}^{2}}^{s_{\rm
th}^{\Xi}}ds^{\prime}\int_{(m_{c}+m_{s})^{2}}^{s_{\rm th}^{\Lambda_{c}}}ds\
e^{-s^{\prime}/T^{\prime 2}}e^{-s/T^{2}}{\rm Im}^{2}\Pi_{WE}^{{\cal
O}_{i}}\left[s^{\prime},s,\cdots,{\bar{p}}^{2}\right],$ (13)

where $s_{\rm th}^{\Lambda_{c}}$ is the threshold parameter for the continuous
spectrum in the $\Lambda_{c}$ channel. Unlike Eq.(2), the matrix element above
depends on an extra momentum $k$, so that it should be parameterized by 4
amplitudes:

$\displaystyle\langle p-k,\sigma^{\prime},\pm^{\prime};K^{+}(-q)|{\cal
O}_{i}(0)|p-q,\sigma,\pm\rangle$ $\displaystyle=$ $\displaystyle\ i\
{\bar{u}}_{\Xi}^{\pm^{\prime}}(p-k,\sigma^{\prime})$
$\displaystyle\times\left[A_{1}^{i\pm^{\prime}\pm}-B_{1}^{i\pm^{\prime}\pm}\gamma_{5}-A_{2}^{i\pm^{\prime}\pm}\frac{\not{q}}{m_{\Lambda_{c}}^{\pm}}+B_{2}^{i\pm^{\prime}\pm}\frac{\not{q}\gamma_{5}}{m_{\Lambda_{c}}^{\pm}}\right]$
$\displaystyle\times u_{\Lambda_{c}}^{\pm}(p-q,\sigma).$ (14)

Now the amplitudes $A_{1,2}^{i\pm^{\prime}\pm},B_{1,2}^{i\pm^{\prime}\pm}$ are
functions of 2 Lorentz invariants: $p^{2}$ and $k^{2}$. Since $k$ is only an
auxiliary momentum, it can be set as $k^{2}=0$ during the calculation. For the
setting of $p^{2}$, the diagram (b) in Fig.2 shows that it corresponds to the
momentum of the $\Xi$ channel. In the same way as for setting ${\bar{p}}^{2}$,
one can set $p^{2}=m_{\Xi}^{\pm^{\prime}2}$. Inserting Eq.(14) into Eq.(13),
and summing up all the spin indices, one arrives at

$\displaystyle\sum_{\pm^{\prime},\sigma^{\prime}}\sum_{\pm,\sigma}e^{-m_{\Xi}^{\pm^{\prime}2}/T^{\prime
2}}e^{-m_{\Lambda_{c}}^{\pm
2}/T^{2}}\lambda_{\Xi}^{\pm^{\prime}}\lambda_{\Lambda_{c}}^{\pm}(\not{p}_{2}\pm^{\prime}m_{\Xi}^{\pm^{\prime}})\left[A_{1}^{\pm^{\prime}\pm}-B_{1}^{\pm^{\prime}\pm}\gamma_{5}-A_{2}^{\pm^{\prime}\pm}\frac{\not{q}}{m_{\Lambda_{c}}^{\pm}}+B_{2}^{\pm^{\prime}\pm}\frac{\not{q}\gamma_{5}}{m_{\Lambda_{c}}^{\pm}}\right](\not{p}_{1}\pm
m_{\Lambda_{c}}^{\pm})$ $\displaystyle=$
$\displaystyle\frac{1}{\pi^{2}}\int_{4m_{s}^{2}}^{s_{\rm
th}^{\Xi}}ds^{\prime}\int_{(m_{c}+m_{s})^{2}}^{s_{\rm th}^{\Lambda_{c}}}ds\
e^{-s^{\prime}/T^{\prime 2}}e^{-s/T^{2}}{\rm Im}^{2}\Pi_{WE}^{{\cal
O}_{i}}\left[s^{\prime},s,\cdots,{\bar{p}}^{2}\right].$ (15)

where $p_{1}=p-q,\ p_{2}=p-k$. The pole residues of $\Lambda_{c}$ is defined
in the same way as that in Eq.(7). At this step, the reason for inserting both
the positive and negative states becomes clear. Both the left and right hands
of Eq.(15) have 16 independent Dirac structures, and 16 independent amplitudes
$A_{1,2}^{i\pm^{\prime}\pm},B_{1,2}^{i\pm^{\prime}\pm}$. The correspondence
between the amplitudes and the Dirac structures is one by one, which enables
us to extract each amplitude unambiguously.

Finally, when $k$ is set to zero, through equation of motion the
parameterization in Eq.(14) will return to the form as Eq.(2), where

$\displaystyle A^{i}=$ $\displaystyle
A_{1}^{i++}+\Big{(}1-\frac{m_{\Xi}^{+}}{m_{\Lambda_{c}}^{+}}\Big{)}A_{2}^{i++},$
$\displaystyle B^{i}=$ $\displaystyle
B_{1}^{i++}-\Big{(}1+\frac{m_{\Xi}^{+}}{m_{\Lambda_{c}}^{+}}\Big{)}B_{2}^{i++}$
(16)

for the case of positive parity states. It should be mentioned that when using
Eq.(15), we will all adopt $p^{2}=m_{\Xi}^{+2},\
{\bar{p}}^{2}=m_{\Lambda_{c}}^{+2}$. This treatment is not consistent if one
wants to extract the amplitudes involving negative parity states, such as
$A_{1,2}^{i-+},A_{1,2}^{i--},B_{1,2}^{i+-}\cdots$. However, since we only care
about the case of positive parity, this treatment is feasible as long as
$A_{1,2}^{i++},B_{1,2}^{i++}$ can be extracted correctly. Other redundant
amplitude only serves to match redundant freedoms.

##  IV Quark-Gluon Level

This section serves to the OPE calculation on the correlation function in Eq.
(3). At the quark-gluon level, and in the deep Euclidean region:
$(p-k)^{2}\sim(p-q)^{2}\sim{\bar{p}}^{2}\ll 0$, the correlation function can
be expressed as a convolution of a perturbative kernel and a nonperturbative
matrix element of the kaon. The correlation function corresponding to the WE
diagram is .

$\displaystyle\Pi_{WE}^{{\cal O}_{1}}(p,q,k)_{\kappa\tau}=-\Pi_{WE}^{{\cal
O}_{2}}(p,q,k)_{\kappa\tau}$ $\displaystyle=$ $\displaystyle\
-4\delta_{im}\int d^{4}x\ e^{-i(p-q)\cdot x}\int d^{4}y\ e^{i(p-k)\cdot y}$
$\displaystyle\times\left[\gamma_{\mu}\gamma_{5}D_{u}(y,0)\gamma^{\nu}(1-\gamma_{5})D_{d}(0,x)\gamma_{5}\right]_{\kappa\beta}$
$\displaystyle\times\left[C\gamma^{\mu}D_{s}(y,0)\gamma_{\nu}(1-\gamma_{5})D_{c}(0,x)\right]_{\alpha\tau}$
$\displaystyle\times\langle
0|\bar{u}_{m}^{\beta}(x)s_{i}^{\alpha}(y)|K^{-}(q)\rangle,$ (17)

where $\alpha,\beta,\kappa,\tau$ are spinor indices and $D_{c,u,s}$ are the
free propagators of the $c,u,s$ quarks. The last matrix element can be
expressed by the Light-Cone Distribution Amplitudes (LCDAs) of the kaon. Fig.3
shows the Feynman diagram for the WE correlation function, where the gray
bubble denotes the kaon LCDAs. The contribution from the two-particle LCDAs of
kaon up to twist-3 are defined as Ball:2004ye

$\displaystyle\langle
0|\bar{u}_{\alpha}^{i}(x)s_{\beta}^{j}(0)|K^{-}(q)\rangle=-\frac{i}{4Nc}\delta_{ij}\int_{0}^{1}du\
e^{-i\bar{u}q\cdot x}$
$\displaystyle\times\Big{[}(\not{p}\gamma_{5})_{\beta\alpha}f_{K}\phi_{K}(u)+(\gamma_{5})_{\beta\alpha}\mu_{K}^{2}\phi_{p}(u)$
$\displaystyle+\frac{1}{6}\mu_{K}^{2}(1-\rho_{K}^{2})(\gamma_{5}\sigma_{\mu\nu})_{\beta\alpha}q^{\mu}x^{\nu}\phi_{\sigma}^{\sigma}(u)\Big{]},$
(18)

where $\phi_{K}$ is the twist-2 LCDA, and $\phi_{p},\phi_{\sigma}$ are the
twist-3 LCDAs. $\mu_{K}^{2}=f_{K}m_{K}^{(0)},~{}~{}\rho_{K}=m_{K}/m_{K}^{(0)}$
with $m_{K}^{(0)}=1.6$ GeV being the chiral mass of kaon and $f_{K}=0.156$
GeV. The expressions of the LCDAs read as

$\displaystyle\phi_{K}=$ $\displaystyle
6u{\bar{u}}[1+a_{1}^{K}C_{1}^{3/2}(2u-1)+a_{2}^{K}C_{2}^{3/2}(2u-1)],$
$\displaystyle\phi_{p}=$ $\displaystyle
1+\left(30\eta_{3}-\frac{5}{2}\right)C_{2}^{1/2}(2u-1)$
$\displaystyle+\left(-3\eta_{3}w_{3}-\frac{27}{20}\rho_{K}^{2}-\frac{81}{10}\rho_{K}^{2}a_{2}^{K}\right)C_{4}^{1/2}(2u-1),$
$\displaystyle\phi_{\sigma}=$ $\displaystyle
6u{\bar{u}}\left[1+\left(5\eta_{3}-\frac{1}{2}\eta_{3}w_{3}-\frac{7}{20}\rho_{K}^{2}-\frac{3}{15}\rho_{K}^{2}a_{2}^{K}\right)\right],$
(19)

where ${\bar{u}}=1-u$ and $C^{\alpha}_{n}$ are the Gegenbauer polynomials. We
firstly take the twist-2 and twist-3p contributions as an example to
illustrate the calculation of Eq.(17). Their contribution reads as

$\displaystyle\Pi_{WE}^{{\cal
O}_{1}(2)}(p,q,k)=i\int_{0}^{1}du\int\frac{d^{4}k_{1}}{(2\pi)^{4}}\frac{d^{4}k_{2}}{(2\pi)^{4}}\frac{d^{4}k_{3}}{(2\pi)^{4}}\frac{d^{4}k_{4}}{(2\pi)^{4}}$
$\displaystyle\times(2\pi)^{8}\delta^{4}(p-uq-
k_{1}-k_{2})\delta^{4}({\bar{p}}+{\bar{u}}q-k_{3}-k_{4})$
$\displaystyle\times\frac{1}{k_{4}^{2}}\frac{1}{k_{2}^{2}}\frac{1}{k_{3}^{2}-m_{s}^{2}}\frac{1}{k_{1}^{2}-m_{c}^{2}}f_{K}\phi_{K}(u)$
$\displaystyle\times[\gamma_{\mu}\gamma_{5}\not{k}_{4}\gamma^{\nu}(1-\gamma_{5})\not{k}_{2}\gamma_{5}C](\not{q}\gamma_{5})^{T}$
$\displaystyle\times[C\gamma^{\mu}(\not{k}_{3}+m_{s})\gamma_{\nu}(1-\gamma_{5})(\not{k}_{1}+m_{c})].$
(20)

Figure 3: Feynman diagram for the WE correlation function. The gray bubble
denotes the kaon LCDAs. The black dots denote the baryon currents, and the
white crossed dot denote the four-fermion operator ${\cal O}_{i}$. The dashed
lines crossing the quark lines means the operation of cutting rules, where
each quark propagator is put on shell.

The double imaginary part of the correlation function in Eq.(15) is related to
its double discontinuity, which can be extracted by the cutting rules. Making
the replacement for each propagators in Eq.(17): $1/(k_{i}^{2}-m^{2})\to(-2\pi
i)\delta(k_{i}^{2}-m^{2})$, one obtains

$\displaystyle{\rm Im}^{2}\Pi_{WE}^{{\cal
O}_{1}(2)}(s^{\prime},s)_{(2)}=\frac{1}{(2i)^{2}}{\rm Disc^{2}}\Pi_{WE}^{{\cal
O}_{1}(2)}(s^{\prime},s)_{(2)}$ $\displaystyle=$
$\displaystyle-\frac{if_{K}}{64\pi^{4}}\int_{0}^{1}du\ \phi_{K}(u)\int
d\Phi_{2}[(p-uq))^{2},k_{1},k_{2}]\int
d\Phi_{2}[({\bar{p}}+{\bar{u}}q))^{2},k_{3},k_{4}]$
$\displaystyle\times\left[\gamma_{\mu}\gamma_{5}\not{k}_{4}\gamma^{\nu}(1-\gamma_{5})\not{k}_{2}\gamma^{\mu}(\not{k}_{3}+m_{s})\gamma_{\nu}(1-\gamma_{5})(\not{k}_{1}+m_{c})\right].$
(21)

We have expressed the double imaginary part of the correlation function as a
convolution of two two-body phase space integration, each one of which is
defined as

$\displaystyle\int d\Phi_{2}[P^{2},k_{i},k_{j}]$ $\displaystyle=$
$\displaystyle\ \int
d^{4}k_{i}d^{4}k_{j}\delta^{4}(P-k_{i}-k_{j})\delta(k_{i}^{2}-m_{i}^{2})\delta(k_{j}^{2}-m_{j}^{2}).$
(22)

It should be mentioned that in Fig.3, the momentum flowing into the left
bubble is $p-uq$, while the momentum flowing off the right bubble is
${\bar{p}}+{\bar{u}}q$. The on shell condition of internal quarks and
$k^{2}=0,\ k\neq 0$ demand that
$(p-uq)^{2}>m_{c}^{2}>({\bar{p}}+{\bar{u}}q)^{2}>m_{s}^{2}$.

The calculation for the $\phi_{p}$ contribution is almost the same as
$\phi_{K}$. The contribution of $\phi_{\sigma}$ to the correlation function is

$\displaystyle\Pi_{WE}^{{\cal
O}_{1}(\sigma)}(p,q,k)=\frac{i}{6}\mu_{K}^{2}(1-\rho_{K}^{2})\int\phi_{\sigma}(u)$
$\displaystyle\times\int d^{4}x\
d^{4}y\int\frac{d^{4}k_{1}}{(2\pi)^{4}}\frac{d^{4}k_{2}}{(2\pi)^{4}}\frac{d^{4}k_{3}}{(2\pi)^{4}}\frac{d^{4}k_{4}}{(2\pi)^{4}}$
$\displaystyle\times e^{-i(k+k_{3}+k_{4}-k_{1}-k_{2})\cdot
x}e^{-i({\bar{p}}+{\bar{u}}q-k_{3}-k_{4})\cdot y}$
$\displaystyle\times\frac{1}{k_{4}^{2}}\frac{1}{k_{2}^{2}}\frac{1}{k_{3}^{2}-m_{s}^{2}}\frac{1}{k_{1}^{2}-m_{c}^{2}}$
$\displaystyle\times\big{[}\gamma_{\mu}\not{k}_{4}\gamma^{\nu}(\gamma_{5}-1)\not{k}_{2}\sigma^{\alpha\beta}\gamma^{\mu}(\not{k}_{3}+m_{s})\gamma_{\nu}$
$\displaystyle\times(1-\gamma_{5})(\not{k}_{1}+m_{c})\big{]}q_{\alpha}y_{\beta}.$
(23)

The linear term of $y$ in the correlation function can be transformed to the
derivative on momentum: $y_{\beta}\to i(\partial/\partial p^{\beta})$. Thus
the calculation can still be done in momentum space.

Figure 4: Feynman diagram for the IE correlation function. The gray bubble
denotes the kaon LCDAs. The black dots denote the baryon currents, and the
white crossed dot denote the four-fermion operator ${\cal O}_{i}$. The dashed
lines crossing the quark lines means the operation of cutting rules, where
each quark propagator is put on shell.

The correlation function contributed by the IE diagram is shown in Fig.4,
which is expressed as

$\displaystyle\Pi_{IE}^{{\cal O}_{1}}(p,q,k)_{\kappa\tau}=-\Pi_{IE}^{{\cal
O}_{2}}(p,q,k)_{\kappa\tau}$ $\displaystyle=$ $\displaystyle\
-2\delta_{mi}\int d^{4}x\ e^{-i(p-q)\cdot x}\int d^{4}y\ e^{i(p-k)\cdot y}$
$\displaystyle\times\left[\gamma_{\mu}\gamma_{5}D_{u}(y,x)C^{T}\gamma_{5}D_{d}^{T}(0,x)(1-\gamma_{5})\gamma^{\nu
T}\right]_{\kappa\beta}$
$\displaystyle\times\left[C\gamma^{\mu}D_{s}(y,0)\gamma_{\nu}(1-\gamma_{5})D_{c}(0,x)\right]_{\alpha\tau}$
$\displaystyle\times\langle
0|\bar{u}_{m}^{\beta}(o)s_{i}^{\alpha}(y)|K^{-}(q)\rangle.$ (24)

Using cutting rules, one can extract the double imaginary part of the
correlation function in Eq.(24) as convolution of a two-body and a triangle
diagram integration. The triangle diagram integration is defined as

$\displaystyle\int d\Phi_{\Delta}(P_{1}^{2},P_{2}^{2},k,m_{1},m_{2},m)$
$\displaystyle=$ $\displaystyle\ \int
d^{4}kd^{4}k_{1}d^{4}k_{2}\delta(k_{1}^{2}-m_{1}^{2})\delta(k_{2}^{2}-m_{2}^{2})\delta(k^{2}-m^{2})$
$\displaystyle\times\delta^{4}(P_{1}-k_{1}-k)\delta^{4}(P_{2}-k_{2}-k).$ (25)

The contribution from $\phi_{2}$ and $\phi_{3p}$ to the imaginary part of the
IE correlation function reads as

$\displaystyle{\rm Im}^{2}\Pi_{IE}^{{\cal
O}_{1}(2+3p)}(s^{\prime},s)=\frac{1}{(2i)^{2}}{\rm Disc^{2}}\Pi_{IE}^{{\cal
O}_{1}(2+3p)}(s^{\prime},s)$ $\displaystyle=$
$\displaystyle-\frac{i}{64\pi^{4}}\int_{0}^{1}du\ \int dm_{12}^{2}\int
d\Phi_{\Delta}[(p-q)^{2},({\bar{p}}+{\bar{u}}q))^{2},k_{3},m_{12},m_{s},0]\int
d\Phi_{2}[m_{12}^{2},k_{1},k_{2}]$
$\displaystyle\times\left[\gamma_{\mu}\not{k}_{3}\not{k}_{2}(1-\gamma_{5})\gamma^{\nu}(f_{K}\phi_{K}(u)\gamma_{5}\not{q}-\mu_{K}^{2}\phi_{p}(u)\gamma_{5})\gamma^{\mu}(\not{k}_{4}+m_{s})\gamma_{\nu}(1-\gamma_{5})(\not{k}_{1}+m_{c})\right],$
(26)

The momentum flow in this correlation function can also be understood
intuitively by Fig.4. The $d,c$ quark bubble, combined with the $s$ and $u$
quark lines, compose a triangle sub-diagram, while the $d,c$ quark bubble is a
sub-diagram of this triangle. The momentum flowing into the left-upper corner
of the triangle diagram is $p_{1}$, while the momentum flowing off the right-
upper corner is not $p_{2}=p-k$ but ${\bar{p}}+{\bar{u}}q$ instead. The on
shell internal quarks require that the invariant mass of the $d,c$ quark
bubble must satisfy $m_{12}^{2}>m_{c}^{2}$.

The contribution from $\phi_{3\sigma}$ to the imaginary part of the IE
correlation function reads as

$\displaystyle\Pi_{IE}^{{\cal
O}_{1}(\sigma)}(p,q,k)=-\frac{i}{6}\mu_{K}^{2}(1-\rho_{K}^{2})\int\phi_{\sigma}(u)$
$\displaystyle\times\int d^{4}x\
d^{4}y\int\frac{d^{4}k_{1}}{(2\pi)^{4}}\frac{d^{4}k_{2}}{(2\pi)^{4}}\frac{d^{4}k_{3}}{(2\pi)^{4}}\frac{d^{4}k_{4}}{(2\pi)^{4}}$
$\displaystyle\times e^{-i(p-q-k_{1}-k_{2}-k_{3})\cdot
x}e^{-i({\bar{p}}+{\bar{u}}q-k_{3}-k_{4})\cdot y}$
$\displaystyle\times\frac{1}{k_{3}^{2}}\frac{1}{k_{2}^{2}}\frac{1}{k_{4}^{2}-m_{s}^{2}}\frac{1}{k_{1}^{2}-m_{c}^{2}}$
$\displaystyle\times\big{[}\gamma_{\mu}\not{k}_{3}\not{k}_{2}(1-\gamma_{5})\gamma^{\nu}\sigma^{\alpha\beta}\gamma^{\mu}(\not{k}_{4}+m_{s})\gamma_{\nu}$
$\displaystyle\times(1-\gamma_{5})(\not{k}_{1}+m_{c})\big{]}q_{\alpha}y_{\beta}.$
(27)

In this case, note that both the two exponential terms in Eq.(27) depend on
$p$. Therefore we cannot just replace $y_{\beta}$ by a derivative on
$p^{\beta}$, but

$\displaystyle y_{\beta}\to\frac{1}{i\bar{u}}\left(\frac{\partial}{\partial
p^{\beta}}+\frac{\partial}{\partial q^{\beta}}\right)$ (28)

instead. It should be mentioned that since this derivative operator places
$p,q$ on an equal footing, its operation on $p_{1}=p-q$ vanishes. Thus only
the terms depending on $p_{2}$ in Eq.(27) survive under this operation.

##  V Numerical Results

Table 1: The masses and decay constants of the charmed baryons with positive or negative parity. Baryon | $\Lambda_{c}^{+}(\frac{1}{2}^{+})$ | $\Lambda_{c}^{+}(\frac{1}{2}^{-})$ | $\Xi^{0}(\frac{1}{2}^{+})$ | $\Xi^{0}(\frac{1}{2}^{-})$
---|---|---|---|---
Mass ${\rm[GeV]}$ |  $2.286$ ParticleDataGroup:2020ssz  |  $2.592$ ParticleDataGroup:2020ssz  |  $1.315$ ParticleDataGroup:2020ssz  |  $1.62$ ParticleDataGroup:2020ssz
$\lambda$ ${\rm[GeV^{3}]}$ |  $0.022\pm 0.003$ Zhao:2020mod  | - |  $0.0425$ Zhao:2021sje  | -

In this section, we give the numerical results for the amplitudes of the
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ decay: $A_{\rm WE},B_{\rm WE}$ and $A_{\rm
IE},B_{\rm IE}$, corresponding to the WE and IE diagrams, respectively. In
this work, we use the $\overline{\rm MS}$ masses for the quarks,
$m_{c}(\mu)=1.27$ GeV and $m_{s}(\mu)=0.103$ GeV with $\mu=1.27$ GeV
ParticleDataGroup:2020ssz . The masses of the $u,d$ quarks are neglected. The
masses and pole residues of $\Lambda_{c}^{+}$ and $\Xi^{0}$ with positive or
negative parity are listed in Table 1, where the pole residues of negative
parity baryons are not presented since they are irrelevant to the amplitudes
for the positive parity baryons. The uncertainty of the $\Lambda_{c}^{+}$ pole
residue will be considered for error analysis, while the uncertainty of the
$\Xi^{0}$ pole residue given in Ref.Zhao:2021sje  is tiny so that can be
neglected. Nowadays, the leading two moments of twist-2 kaon LCDA have been
calculated by lattice QCD as $a_{1}^{K}=0.053$ and $a_{2}^{K}=0.106$
RQCD:2019osh . Other parameters for the kaon LCDA are obtained by QCD sum
rules as $\eta_{3}=0.015$ and $w_{3}=-3$ Ball:2004ye .

On the other hand, there are two parameters in the LCSR, the thresholds
$s_{\rm th}^{\Xi},s_{\rm th}^{\Lambda_{c}}$ and the Borel parameters
$T^{2},T^{\prime 2}$. The threshold parameters are the beginning of the
continuous spectrum above the poles we have inserted into the correlation
function, namely: $\Lambda_{c}^{+}(\frac{1}{2}^{\pm})$ and
$\Xi^{0}(\frac{1}{2}^{\pm})$. An empirical approach to determine the threshold
parameters is setting them nearly $0.5$ GeV above the lowest pole mass. Thus
in our case we set:

$\displaystyle(m_{\Xi}^{+}+0.4\ {\rm GeV})^{2}<s_{\rm
th}^{\Xi}<(m_{\Xi}^{+}+0.6\ {\rm GeV})^{2},$
$\displaystyle(m_{\Lambda_{c}}^{+}+0.4\ {\rm GeV})^{2}<s_{\rm
th}^{\Lambda_{c}}<(m_{\Lambda_{c}}^{+}+0.6\ {\rm GeV})^{2},$ (29)

where uncertainty of $\pm\ 0.1$ GeV is considered. In the following numberical
calculation, we will consider the uncertainty of the thresholds when
evaluating the error of numerical results.

In principle, the physical results should be independent of the Borel
parameters. Thus we have to study the Borel parameter dependence of the
obtained amplitudes, so that a suitable Borel parameter region. To simplify
the problem, we use the following relation to constrain the two Borel
parameters corresponding to the $s$ and $s^{\prime}$ channels Ball:1991bs :

$\displaystyle\frac{T^{2}}{T^{\prime
2}}=\frac{m_{\Lambda_{c}}^{2}-m_{c}^{2}}{m_{\Xi}^{2}-m_{s}^{2}}.$ (30)

In Fig. 5 and Fig. 6 we present the Borel parameter dependence of the
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ decay amplitudes $A_{\rm WE},B_{\rm WE}$ and
$A_{\rm IE},B_{\rm IE}$, respectively. The contribution from the twist-2,
twist-3p and twist-3$\sigma$ LCDAs are separated presented. In each diagram,
the blue band denotes the uncertainty from the error of the threshold given in
Eq.(29). The upper and lower red bands denote the uncertainty from the pole
residue of $\Lambda_{c}^{+}$.

Figure 5: The Borel parameter dependence of the WE amplitudes for
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ decay : $A_{\rm WE},B_{\rm WE}$. The
contribution from the twist-2, twist-3p and twist-3$\sigma$ LCDAs are
separated presented. In each diagram, the blue band denotes the numerical
error from the uncertainty of threshold parameters given in Eq.(29). The upper
and lower red bands denote the uncertainty from the pole residue of
$\Lambda_{c}^{+}$. Figure 6: The Borel parameter dependence of the IE
amplitudes for $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ decay : $A_{\rm IE},B_{\rm
IE}$. The contribution from the twist-2, twist-3p and twist-3$\sigma$ LCDAs
are separated presented. In each diagram, the blue band denotes the numerical
error from the uncertainty of threshold parameters given in Eq.(29). The upper
and lower red bands denote the uncertainty from the pole residue of
$\Lambda_{c}^{+}$.

From Fig. 5 and Fig. 6, it can be found that the behavior of the amplitudes as
functions of $T^{2}$ become stable in the large $T^{2}$ region, which is
arround $T^{2}>12\ {\rm GeV}^{2}$. In this work, we choose the Borel parameter
in the stable region $12{\rm GeV}^{2}<T^{2}<20{\rm GeV}^{2}$. The amplitudes
of $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ contributed by the WE and IE diagrams are
listed in Table 2, where the central values and uncertainties are obtained
from the results shown in Fig. 5 and Fig. 6. It can be found that the
contribution of WE diagram to the amplitude $A$ is comparable. However, in
terms of the amplitude $B$, the contribution from IE diagram dominates. The
twist-2 contribution to $A_{WE},B_{WE}$ is small because they are proportional
to $m_{s}$, which receives large suppression. Table 3 gives a comparison of
our results with those from the literature, including the studies by flavor
SU(3) analysis Geng:2019xbo ; Zhong:2022exp ; Xing:2023dni ; Sun:2024mmk  and
current algebra (CA) Zou:2019kzq . The comparison of the branching fraction
with the BESIII measurement BESIII:2018cvs  and the PDG fitting
ParticleDataGroup:2020ssz  is also presented. Instead of ${\cal O}_{1,2}$, the
amplitudes presented in this table are defined through the matrix element of
the effective Hamiltonian in Eq. (1):

$\displaystyle\langle\Xi^{0}(p-q)K^{+}(q)|{\cal H}_{\rm
eff}(0)|\Lambda_{c}^{+}(p)\rangle$ $\displaystyle=$ $\displaystyle i\
\bar{u}_{\Xi}(p-q)\big{[}A_{\rm tot}-B_{\rm
tot}\gamma_{5}\big{]}u_{\Lambda_{c}}(p).$ (31)

$A_{\rm tot}$ and $B_{\rm tot}$ are the S-wave and P-wave amplitudes of the
decay. The Wilson coefficients are taken as $C_{1}=1.346$ and $C_{2}=-0.636$
at $\mu=1.27$ GeV Buchalla:1995vs . The up-down spin asymmetry
$\alpha_{\Xi^{0}K^{+}}$ is defined as

$\displaystyle\alpha_{\Xi^{0}K^{+}}=\frac{2\kappa{\rm Re}(A_{\rm
tot}^{*}B_{\rm tot})}{|A_{\rm tot}|^{2}+\kappa^{2}|B_{\rm tot}|^{2}},$ (32)

where $\kappa=p_{c}/(E_{\Xi}+m_{\Xi})$ and $p_{c}$ is the three-momentum in
the rest frame of the mother particle $\Lambda_{c}$. It can be found that the
$A_{\rm tot}$ obtained in this work is much smaller than $B_{\rm tot}$, and
also one order smaller than those from SU(3) and CA. On the other hand, the
obtained $B_{\rm tot}$ is consistent with those from SU(3) and CA within the
error. Due to the significant difference between the $A_{\rm tot}$ and $B_{\rm
tot}$ in this work, the resulting up-down spin asymmetry
$\alpha_{\Xi^{0}K^{+}}$ is extremely small compared with CA and most of the
SU(3) studies. However, it agrees with the BESIII measurement within the
error.

There are two issues deserve further discussion:

  * •

The absolute value of the up-down spin asymmetry measured by BESIII is very
small. This implies that one of the amplitudes of $A_{\rm tot},B_{\rm tot}$
must be much smaller than another one. Although our result on $A_{\rm
tot},B_{\rm tot}$ obeys this rule, their sign difference produce a negative
$\alpha_{\Xi^{0}K^{+}}$, which is inconsistent with other theoretical
predictions. However, since the uncertainty of our $A_{\rm tot}$ is large, the
actual sign of $\alpha_{\Xi^{0}K^{+}}$ is still undetermined.

  * •

In this work, we have only used the two-particle LCDAs of kaon. Although the
contribution from the 3-particle LCDAs is expected to be suppressed, in
principle they are necessary for improving the accuracy of calculation, which
will be included in a future study. On the other hand, until now only the
twist-2 LCDA of kaon has been studied by lattice QCD. It can be expected that
the calculation precision will be improved if the twist-3 LCDA of kaon is
updated by future lattice calculation.

Table 2: Decay amplitudes of $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ from the WE and IE contribution (in unit $10^{-2}$). The central values and uncertainties are obtained from the results shown in Fig. 5 and Fig. 6 with $12\ {\rm GeV}^{2}<T^{2}<20\ {\rm GeV}^{2}$. WE diagram | Twist-$2$ | Twist-$3p$ | Twist-$3\sigma$ | Total
---|---|---|---|---
$A_{{\rm WE}}$ | $-0.0067\pm 0.0011$ | $0.16\pm 0.03$ | $0.056\pm 0.004$ |  $0.21\pm$ 0.04
$B_{{\rm WE}}$ | $-0.025\pm 0.004$ | $-0.066\pm 0.012$ | $-0.94\pm 0.06$ | $-1.04\pm 0.08$
IE diagram | Twist-$2$ | Twist-$3p$ | Twist-$3\sigma$ | Total
$A_{{\rm IE}}$ | $0.02\pm 0.003$ | $0.36\pm 0.12$ | $-0.47\pm 0.05$ | $-0.08\pm 0.14$
$B_{{\rm IE}}$ | $-0.075\pm 0.012$ | $0.46\pm 0.15$ | $-11.62\pm 1.16$ | $-11.23\pm 1.32$
Table 3: Comparison of the decay amplitudes of $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ from this work with those from the literature. All the amplitudes are in the unit of $10^{-2}G_{F}$ GeV2. Here we list the amplitudes from the various of studies basing on flavor SU(3) analysis Geng:2019xbo ; Zhong:2022exp  and current algebra (CA) Zou:2019kzq . The superscript $a$ denotes a model with SU(3) symmetry, while model $b$ includes SU(3) symmetry- breaking effects. The superscripts $c$ and $d$ mean the fitting using real and complex form factors, respectively. $\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ | $A_{{\rm tot}}$ | $B_{{\rm tot}}$ | $\alpha_{\Xi^{0}K^{+}}$ | ${\cal B}(\Lambda_{c}^{+}\to\Xi^{0}K^{+})$
---|---|---|---|---
This Work | $0.17\pm 0.23$ | $-15.95\pm 1.83$ | $-0.09\pm 0.13$ | $3.56\pm 0.9$
Geng (2019), SU(3) Geng:2019xbo  | $2.7\pm 0.6$ | $16.1\pm 2.6$ | $0.94^{+0.06}_{-0.11}$ | $5.7\pm 0.9$
Zou (2020), CA Zou:2019kzq  | $-4.48$ | $-12.1$ | $0.9$ | $7.1$
Zhong (2022), SU(3)a Zhong:2022exp  | $3.2\pm 0.2$ | $8.7^{+0.6}_{-0.8}$ | $0.91^{+0.03}_{-0.04}$ | $3.8^{+0.4}_{-0.5}$
Zhong (2022), SU(3)b Zhong:2022exp  | $3.3^{+0.5}_{-0.7}$ | $12.3^{+1.2}_{-1.8}$ | $0.99\pm 0.01$ | $5.0^{+0.6}_{-0.9}$
Xing (2023), SU(3) Xing:2023dni  | – | – | $0.96\pm 0.02$ | $4.3\pm 0.3$
Sun (2024), SU(3)c Sun:2024mmk  | – | – | $0.96\pm 0.17$ | $4.17\pm 0.29$
Sun (2024), SU(3)d Sun:2024mmk  | – | – | $0.02\pm 0.15$ | $5.49\pm 0.69$
BESIII BESIII:2018cvs ; BESIII:2023wrw  | – | – | $0.01\pm 0.16\pm 0.03$ | $5.9\pm 0.86\pm 0.39$
PDG Fit (2022) ParticleDataGroup:2020ssz  | – | – | – | $5.5\pm 0.7$

##  VI Conclusion

In this work, we have investigated the purely non-factorizable
$\Lambda_{c}^{+}\to\Xi^{0}K^{+}$ decay using light-cone sum rules. A three-
point correlation is defined and calculated for extracting the S-wave and
P-wave decay amplitudes. This decay is contributed by W-exchange and the
W-inward emission diagrams, and both of them considered in the quark-gluon
level calculation. The two-particle light-cone distribution amplitudes (LCDAs)
of kaon are used as non-perturbative input. We seperately presented the decay
amplitudes contributed from the twist-2, twist-3p and twist-$3\sigma$ kaon
LCDAs, respectively. The obtained P-wave amplitude is consistent with those
predicted in the literature, while the S-wave one is one order smaller than
those in the literature, and also has a different sign. Such significant
difference between the S- and P -wave amplitudes leads to small but negative
up-down spin asymmetry, and the magnitude is close to the latest BESIII
measurement.

## Acknowledgement

We thank Zhi-Peng Xing for valuable discussions. This work is supported in
part by Natural Science Foundation of China under Grants No.12305103,
No.12065020. The work of Y.J. Shi is also supported by Opening Foundation of
Shanghai Key Laboratory of Particle Physics and Cosmology under Grant
No.22DZ2229013-2.

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# A note on entire functions sharing a finite set with applications to
difference equations

Molla Basir Ahamed  Department of Mathematics, Kalipada Ghosh Tarai
Mahavidyalya, West Bengal, 734014, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Value distribution and uniqueness problems of difference operator of an entire
function have been investigated in this article. This research shows that a
finite ordered entire function $f$ when sharing a set
$\mathcal{S}=\\{\alpha(z),\beta(z)\\}$ of two entire functions $\alpha$ and
$\beta$ with $\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$ with its difference
$\mathcal{L}^{n}_{c}(f)=\sum_{j=0}^{n}a_{j}f(z+jc)$, then
$\mathcal{L}^{n}_{c}(f)\equiv f$, and more importantly certain form of the
function $f$ has been found. The results in this paper improve those given by
_k. Liu_ , _X. M. Li_ , _J. Qi, Y. Wang and Y. Gu_ etc. Some examples have
been exhibited to show the condition
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$ is sharp in our main result.
Examples have been also exhibited to show that if $CM$ sharing is replaced by
$IM$ sharing, then conclusion of the main results ceases to hold.

###### Key words and phrases:

Uniqueness, difference operator, entire function, shared set.

###### 2010 Mathematics Subject Classification:

Primary 30D35 Secondary 30D30

##  1\. Introduction

In this paper, a meromorphic function will always be non-constant and
meromorphic in the complex plane C, unless specifically stated otherwise. In
what follows, we assume that the reader is familiar with the elementary
Nevanlinna theory,(see [16, 19, 30]). In particular, for a meromorphic
function $f$, $\mathcal{S}(f)$ denotes the family of all meromorphic function
$\zeta$ for which $T(r,\zeta)=S(r,f)=o(T(r,f))$, where $r\rightarrow\infty$
outside of a possible set of finite logarithmic measure.For convenience, we
agree that $\mathcal{S}(f)$ includes all constant functions and
$\overline{\mathcal{S}}(f):=\mathcal{S}(f)\cup\\{\infty\\}.$

A set $\mathcal{S}$ is called a unique range set (URSE) for a certain class of
entire functions if each inverse image of the set uniquely determines a
function from the given class. Let $\mathcal{S}$ be a finite set of some
entire functions and $f$ an entire function. Then, a set $E_{f}(\mathcal{S})$
is defined as

$\displaystyle
E_{f}(\mathcal{S})=\\{(z,m)\in\mathbb{C}\times\mathbb{Z}:f(z)-a(z)=0\text{\;with
multiplicity}\;m,a\in\mathcal{S}\\}.$

If we do not the count multiplicities, then we denote the set as
$\overline{E}_{f}(\mathcal{S})$.Assume that $g$ is another entire funct ion.
We say that $f$ and $g$ share $\mathcal{S}$ $CM$ if
$E_{f}(\mathcal{S})=E_{g}(\mathcal{S})$, and we say $f$ and $g$ share
$\mathcal{S}$ $IM$ if
$\overline{E}_{f}(\mathcal{S})=\overline{E}_{g}(\mathcal{S})$. Thus, a set
$\mathcal{S}$ is called $URSE$ if $E_{f}(\mathcal{S})=E_{g}(\mathcal{S}),$
where $f$ and $g$ are two entire functions, then $f\equiv g.$

_Gross_ and _Yang_ [13], first found an example of a $URSE$ which is
$\mathcal{S}=\\{z:e^{z}+z=0\\}$, and as this an infinite set, so it is very
natural to investigate whether there exists a finite $URSE$ or not. In $1995,$
_Yi_ [31] who found a $URSE$ $\mathcal{S}=\\{z:z^{n}+az^{m}+b=0\\}$, where
$n>2m+4$ and $a,\;b$ chosen in a such a way that the equations
$z^{n}+az^{m}+b=0$ has no repeated roots. Since then there have been many
efforts to study the problem of constructing $URSE$ time to time (see [11, 12,
23]). There is another study on the $URSE$ of entire functions, which is to
seek a set $\mathcal{S}$ for which if
$E_{f}(\mathcal{S})=E_{f^{\prime}}(\mathcal{S})$, then $f\equiv f^{\prime}$.
One can verify that the form of the function will be $f(z)=ce^{z}.$ where $c$
is a non-zero complex number. _Li_ and _Yang_ [21] also deduced that if
$E_{f}(\mathcal{S})=E_{f^{\prime}}(\mathcal{S})$, where
$\mathcal{S}=\\{a,b\\}$ with $a+b\neq 0$, then either $f(z)=\mathcal{A}e^{z}$
or $f(z)=\mathcal{A}e^{z}+a+b$, where $\mathcal{A}$ is a non-zero complex
number. Later, _Fang_ and _Zalcman_ [10], using the theory of normal families,
proved that there exists a set $\mathcal{S}=\\{a,b,c\\}$ such that
$E_{f}(\mathcal{S})=E_{f^{\prime}}(\mathcal{S})$, then $f\equiv f^{\prime}.$ A
special topic widely studied in the uniqueness theory is the case when $f(z)$
shares value(s) or set(s) with its derivatives or differential polynomials. We
recall a result of this type from the preceding literature:

######  Theorem A.

[21] Let $f$ be a non-constant entire function and $a_{1},a_{2}$ be two
distinct complex numbers. If
$E_{f}(\\{a_{1},a_{2}\\})=E_{f^{\prime}}(\\{a_{1},a_{2}\\})$, then $f$ takes
one of the following conclusions:

  1. (1)

$f=f^{\prime}$;

  2. (2)

$f+f^{\prime}=a_{1}+a_{2}$;

  3. (3)

$f(z)=c_{1}e^{cz}+c_{2}e^{-cz},$ with $a_{1}+a_{2}=0$, where $c_{1},c_{2},c$
are all non-zero complex numbers satisfying $c^{2}\neq 1$ and
$c_{1}c_{2}=\frac{1}{4}a_{1}^{2}\left(1-\frac{1}{c^{2}}\right).$

Recently, with a more general setting, namely $k$-th derivative $f^{(k)}$ or
differential monomial $M[f]$ or a more general setting namely differential
polynomial $P[f]$, the present author, deduced that, when some power of a
meromorphic function $f$ and its differential monomial (or polynomial) sharing
a set [1] or a small function [4], then even in this case, the function $f$
also assumes a certain form. From the literature of meromorphic functions on
sharing value problems, finding the class of the functions satisfying some
differential or difference equations gained a valuable space.

Throughout this paper, we denote by $\rho(f)$ and $\lambda(f)$, the order of
$f$ and the exponent of convergence of zeros of $f$ respectively (see [29]):
We also need the following notation: Let $f$ be a non-constant meromorphic
function, and we define difference operators as $\Delta_{c}(f)=f(z+c)-f(z)$,
and for $n\geqslant 2$, $\Delta^{n}_{c}(f)=\Delta^{n-1}_{c}(\Delta_{c}(f))$.

In the past recent years, the Navanlinna characteristic of $f(z+c)$, the value
distribution theory for difference polynomials, the Nevanlinna theory for the
difference operator and most importantly, the difference analogue of the
lemmas on the logarithmic derivative had been established (see [5, 9, 14, 15,
17, 18, 27, 28, 32]). For these theories, since the derivative is a difference
counterpart of a function, hence there has been recent study of whether the
derivative $f^{\prime}$ of a entire (meromorphic) function $f$ can be replaced
by the difference $\Delta_{c}(f)=f(z+c)-f(z)$ in the above mentioned results.
Number of researches have been done with difference operator (see[2, 6, 7, 20,
24, 25, 26, 27, 29])

In this direction, in $2009$, _Liu_ [27] considered the problem of sharing set
by an entire function and its difference, and obtained the following result.

######  Theorem B.

[27] Suppose that $\alpha$ is is a non-zero complex number, and $f$ is a
transcendental entire function with finite order. If
$E_{f}(\\{-\alpha,\alpha\\})=E_{\Delta_{c}(f)}(\\{-\alpha,\alpha\\})$, then
$\Delta_{c}(f(z))=f(z)$ for all $z\in\mathbb{C}$.

Since $\Delta_{c}(f)$ is a very special form of the setting
$\mathcal{L}_{c}(f):=a_{1}f(z+c)+a_{0}f(z)$, where $a_{1}(\neq
0),a_{0}\in\mathbb{C}$ (see [3]). Therefore a natural quarry would be as the
following:

######  Question 1.1.

Does _Theorem B_ still hold if we replace $\Delta_{c}(f)$ by
$\mathcal{L}_{c}(f)$ ?

From the following two examples, one can ensure that answer of _Question 1.1 _
is not affirmative.

######  Example 1.1.

Let $f(z)=e^{z}$, and for $c=\log(\log 2),$ suppose that

$\displaystyle\mathcal{L}_{c}(f)=\frac{\sqrt{5}+\i\sqrt{7}}{\log
2}f(z+c)+(1-\sqrt{5}-i\sqrt{7})f(z).$

It is clearly $f$ is a finite order entire function and
$E_{f}(\mathcal{S})=E_{\mathcal{L}_{c}(f)}(\mathcal{S})$, where
$\mathcal{S}=\\{-\alpha,\alpha\\}$, $\alpha\in\mathbb{C}\smallsetminus\\{0\\}$
but $\mathcal{L}_{c}(f)\neq f.$

######  Example 1.2.

Let $f(z)=sin(z)$ or $cos(z)$ and

$\displaystyle\mathcal{L}_{\pi}(f)=(1+i\sqrt{2})f(z+\pi)+i\sqrt{2}f(z).$

Clearly $f$ is of finite order and
$E_{f}(\mathcal{S})=E_{\mathcal{L}_{\pi}(f)}(\mathcal{S})$, where
$\mathcal{S}=\\{-\alpha,\alpha\\}$, $\alpha\in\mathbb{C}\smallsetminus\\{0\\}$
but $\mathcal{L}_{\pi}(f)\neq f.$

To investigate with $\mathcal{L}_{c}(f)$ to get the similar conclusion as of
_Theorem B_ , we require some extra conditions. Instead of looking for some
general setting of the difference, it is therefore reasonable to concentrate
for the generalization of the shared set $\\{-\alpha,\;\alpha\\}$.

In this direction, we recall here a question proposed by _Liu_ in [27] as
follows.

######  Question 1.2.

Let $\alpha$ and $\beta$ be two small functions of $f$ with period $c$. When a
transcendental entire function $f$ of finite order and its difference
$\Delta_{c}(f)$ share the set $\\{\alpha,\beta\\}$ $CM$, what can we say about
the relationship between $f$ and $\Delta_{c}(f)$ ?

In connection with the _Question 1.2_, in $2012$, _Li_ [24] established the
following result.

######  Theorem C.

[24] Suppose that $\alpha$, $\beta$ are two distinct entire functions, and $f$
is a non-constant entire function with $\rho(f)\neq 1$ and
$\lambda(f)<\rho(f)<\infty$ such that
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$. If
$E_{f}(\\{\alpha,\beta\\})=E_{\Delta_{c}(f)}(\\{\alpha,\beta\\})$, then
$\Delta_{c}(f(z))\equiv f(z)$ for all $z\in\mathbb{C}$.

######  Remark 1.1.

Next example confirms that conclusion of _Theorem C_ still holds if we remove
the condition $\rho(f)\neq 1$.

######  Example 1.3.

Let $f(z)=2^{z/{2\pi}}\cos(z)$, and $\mathcal{S}=\\{\alpha(z),\beta(z)\\}$,
where $\alpha(z)$ and $\beta(z)$ are two non-constant polynomials in $z$. It
is clear that $\rho(f)=1$, and
$\max\\{\rho(\alpha),\rho(\beta)\\}=0<1=\rho(f)$, and
$E_{f}(\\{\alpha,\beta\\})=E_{\Delta_{2\pi}(f)}(\\{\alpha,\beta\\})$, and
$\Delta_{c}(f(z))=f(z)$ for all $z\in\mathbb{C}$.

Thus, one natural question is : Can we prove Theorem C by omitting the
restriction $\rho(f)\neq 1$. Recently, _Qi, Wang_ and _Gu_ [29] answered this
question by proving the following result.

######  Theorem D.

[29] Suppose that $\alpha$, $\beta$ are two distinct entire functions, and $f$
is a non-constant entire function with $\lambda(f)<\rho(f)<\infty$ such that
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$. If
$E_{f}(\\{\alpha,\beta\\})=E_{\Delta_{c}(f)}(\\{\alpha,\beta\\})$, then
$f(z)=\mathcal{A}e^{\lambda z}$, where $\mathcal{A}$, $\lambda$ are two non-
zero complex numbers satisfying $e^{\lambda c}=2$. Furthermore
$\Delta_{c}(f(z))\equiv f(z)$ for all $z\in\mathbb{C}$.

######  Note 1.1.

In view of _Examples 1.1_ and _1.2_ , we have seen that in _Theorem B_ , it is
not possible to replace $\Delta_{c}(f)$ by $\mathcal{L}_{c}(f)$ in general,
but, the next two examples show that in case of _Theorem D_ , one can do it.

######  Example 1.4.

Let $f(z)=\displaystyle\left(\frac{\pi}{2}\right)^{z/c}$ and
$\mathcal{L}_{c}(f)=2f(z+c)+(1-\pi)f(z).$ Let
$\mathcal{S}=\\{\alpha,\beta\\}$, where $\alpha$ and $\beta$ are two entire
functions with $\rho(\alpha)<1$ and $\rho(\beta)<1$. Evidently,
$E_{f}(\mathcal{S})=E_{\mathcal{L}_{c}(f)}(\mathcal{S})$ with
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$, and also
$\mathcal{L}_{c}(f(z))=f(z)$.

######  Example 1.5.

Let $f(z)=(1+i)^{z/c}$ and

$\displaystyle\mathcal{L}_{c}(f)=\left(1+i\sqrt{3}\right)f(z+c)+\left(\sqrt{3}-i\left(\sqrt{3}+1\right)\right)f(z).$

Let $\mathcal{S}=\\{\alpha,\beta\\}$, where $\alpha$ and $\beta$ are two
entire functions with $\rho(\alpha)<1$ and $\rho(\beta)<1$. Then, clearly
$E_{f}(\mathcal{S})=E_{\mathcal{L}_{c}}(\mathcal{S})$ with
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$, and also
$\mathcal{L}_{c}(f(z))=f(z)$.

######  Note 1.2.

It is not hard to check that the functional form is
$f(z)=\displaystyle\left(\frac{1-a_{0}}{a_{2}}\right)^{z/c}g(z)$, where $g$ is
a $c$-periodic function, when $f$ satisfies the relation
$\mathcal{L}_{c}(f)=f$.

In this paper, we are mainly concerned for a more generalization of
$\Delta_{c}(f)$, and $\mathcal{L}_{c}(f)$, hence we define

$\displaystyle\mathcal{L}^{n}_{c}(f)=a_{n}f(z+nc)+a_{n-1}f(z+(n-1)c)+\ldots+a_{1}f(z+c)+a_{0}f(z),$

where $a_{n}(\neq 0),a_{i}\in\mathbb{C}$ for $(i=0,1,2,\ldots,n-1)$. If we
choose the coefficients as $a_{j}=\displaystyle(-)^{j}\binom{n}{j}$,
$j=0,1,2,\ldots,n$, then $\mathcal{L}^{n}_{c}(f)=\Delta^{n}_{c}(f)$. With this
generalization, our aim is to study _Theorems C_ and _D_ further. So it is
interesting to ask the following questions regarding _Theorems C_ and _D_.

######  Question 1.3.

In _Theorems C_ and _D_ , what happen if we replace $\Delta_{c}(f)$ by
$\mathcal{L}^{n}_{c}(f)$ ?

######  Question 1.4.

Can we get a corresponding result like _Theorems C_ and _D_ , in which the
condition $\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$ is sharp ?

######  Question 1.5.

What can be say about the specific form of the function $f$ when
$\Delta_{c}(f)$ is replaced by $\mathcal{L}_{c}(f)$ ?

##  2\. Main result

In this article, we dealt with the above questions and answered them all
affirmatively. Following is the main result of this paper.

######  Theorem 2.1.

Suppose that $\alpha$, $\beta$ are two distinct entire functions, and $f$ is a
non-constant entire function with $\lambda(f)<\rho(f)<\infty$ such that
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$ and
$\mathcal{L}^{n}_{c}(f)(\not\equiv 0)$. If
$E_{f}(\\{\alpha,\beta\\})=E_{\mathcal{L}^{n}_{c}(f)}(\\{\alpha,\beta\\})$,
then

$\displaystyle
f(z)=\mathcal{A}_{1}\lambda_{1}^{z/c}+\mathcal{A}_{2}\lambda_{2}^{z/c}+\ldots+\mathcal{A}_{n}\lambda_{n}^{z/c},$

where $\mathcal{A}_{i},\lambda_{i}\in\mathbb{C}\smallsetminus\\{0\\}$ for
$i=1,2,\ldots,n$, and $\lambda_{i}$ are the roots of the equation

(2.1) $\displaystyle a_{n}w^{n}+a_{n-1}w^{(n-1)c}+\ldots+a_{1}w+a_{0}-1=0.$

Furthermore $\mathcal{L}^{n}_{c}(f(z))\equiv f(z)$ for all $z\in\mathbb{C}$.

######  Remark 2.1.

Entire functions satisfying _Theorem 2.1_ do exist, and it is shown here for
the case $n=1$ and $n=3$ only, and we discussed the case $n=2$ later in a
corollary.

######  Example 2.1.

Let $f(z)=\left(1+i\right)^{z/3}$ and $\mathcal{L}_{3}(f)=2f(z+3)-(1+2i)f(z)$.
Let $\mathcal{S}=\\{z^{2}-2,\;2z^{3}-z+1\\}.$ Evidently,
$\max\\{\rho(\alpha),\rho(\beta)\\}=0<1=\rho(f)$ and
$E_{f}(\mathcal{S})=E_{\mathcal{L}_{3}(f)}(\mathcal{S})$, and $f$ has the
specific form and also satisfying the relation $\mathcal{L}_{3}(f)\equiv f.$

######  Example 2.2.

Let $f(z)=\displaystyle\left(\frac{i\sqrt{3}}{2}\right)^{z/\sqrt{2}}$ and

$\displaystyle\mathcal{L}_{\sqrt{2}}(f)=-2\left(\sqrt{2}+i\sqrt{3}\right)f(z+\sqrt{2})+\left(1+i\sqrt{3}\left(\sqrt{2}+i\sqrt{3}\right)\right)f(z).$

Let
$\mathcal{S}=\\{2z^{2}-3z+\sqrt{2},\sqrt{3}z^{3}-\sqrt{5}z^{2}-i\sqrt{2}\beta\\}$.
Clearly, $\max\\{\rho(\alpha),\rho(\beta)\\}=0<1=\rho(f)$ and
$E_{f}(\mathcal{S})=E_{\mathcal{L}_{\sqrt{2}}(f)}(\mathcal{S})$, and $f$ has
the specific form and also satisfying the relation
$\mathcal{L}_{\sqrt{2}}(f)\equiv f.$

######  Example 2.3.

Let
$f(z)=6^{z/c}+\left(1+2\omega+3\omega^{2}\right)^{z/c}+\left(1+2\omega^{2}+3\omega\right)^{z/c}$
and $\mathcal{S}=\\{\alpha,\beta\\}$, where $\alpha$ and $\beta$ are any two
polynomials in $z$. Then, we see that
$\max\\{\rho(\alpha),\rho(\beta)\\}=0<1=\rho(f)$ and
$E_{f}(\mathcal{S})=E_{\mathcal{L}^{3}_{c}(f)}(\mathcal{S})$, where

$\displaystyle\mathcal{L}^{3}_{c}(f)=f(z+3c)-3f(z+2c)-15f(z+c)-17f(z).$

Clearly $f$ has the specific form and also satisfying the relation
$\mathcal{L}^{3}_{c}(f)\equiv f$.

######  Example 2.4.

Let $f(z)=\mathcal{A}^{z/c}+\mathcal{B}^{z/c}+\mathcal{C}^{z/c}$ and
$\mathcal{S}=\\{\alpha,\beta\\}$, where $\alpha$ and $\beta$ are any two
polynomials in $z$. Then, we see that
$\max\\{\rho(\alpha),\rho(\beta)\\}=0<1=\rho(f)$ and
$E_{f}(\mathcal{S})=E_{\mathcal{L}^{3}_{c}(f)}(\mathcal{S})$, where

$\displaystyle\mathcal{L}^{3}_{c}(f)=f(z+3c)-\left(\mathcal{A}+\mathcal{B}+\mathcal{C}\right)f(z+2c)+\left(\mathcal{AB}+\mathcal{BC}+\mathcal{CA}\right)f(z+c)-\left(\mathcal{ABC}-1\right)f(z).$

Clearly $f$ has the specific form and also satisfying the relation
$\mathcal{L}^{3}_{c}(f)\equiv f$

######  Remark 2.2.

In _Theorem 2.1_, the condition $\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$
can not be replaced by $\max\\{\rho(\alpha),\rho(\beta)\\}=\rho(f)$ i.e., the
condition is sharp, which can be seen in the next examples for the case
$n=1,2$ and $3$ only.

######  Example 2.5.

Let $f(z)=e^{2z}+e^{z}$, and

$\displaystyle\mathcal{L}_{c}(f)=\sqrt{3}f(z+c)+\frac{-\sqrt{3}-i\sqrt{4\sqrt{3}-3}}{2}f(z),$

where
$c=\log\left(\displaystyle\frac{\sqrt{3}+i\sqrt{4\sqrt{3}-3}}{2\sqrt{3}}\right)$.
Let $\mathcal{S}=\\{\alpha(z),\beta(z)\\}$, where $\alpha(z)=\mathcal{A}e^{z}$
and $\beta(z)=(1-\mathcal{A})e^{z}$,
$\mathcal{A}\in\mathbb{C}\smallsetminus\bigg{\\{}0,\displaystyle\frac{1}{2},1\bigg{\\}}$.
We see that $\max\\{\rho(\alpha),\rho(\beta)\\}=1=\rho(f)$, and
$E_{f}(\mathcal{S})=E_{\mathcal{L}_{c}(f)}(\mathcal{S})$ but neither $f$ has
the specific form nor satisfying the relation $\mathcal{L}_{c}(f)=f.$

######  Example 2.6.

Let $f(z)=e^{2z}+e^{z}+e^{-z}$, and $\mathcal{L}_{\pi
i}(f)=-\frac{1}{2}f(z+\pi i)-\frac{1}{2}f(z).$ Let
$\mathcal{S}=\\{\alpha(z),\beta(z)\\}$, where $\alpha(z)=e^{z}$ and
$\beta(z)=e^{-z}$. Clearly $\max\\{\rho(\alpha),\rho(\beta)\\}=1=\rho(f)$, and
$E_{f}(\mathcal{S})=E_{\mathcal{L}_{\pi i}(f)}(\mathcal{S})$ but note that $f$
is neither in the form nor satisfying the relation $\mathcal{L}_{\pi i}(f)=f.$

######  Example 2.7.

Let $f(z)=\sin z+e^{z}$ and

$\displaystyle\mathcal{L}^{2}_{\pi}(f)=\left(\frac{1+2e^{\pi}}{1-e^{2\pi}}\right)f(z+2\pi)+2f(z+\pi)+\left(\frac{\left(e^{\pi}+2\right)e^{\pi}}{e^{2\pi}-1}\right)f(z).$

Let $\mathcal{S}=\\{\alpha,\beta\\}$, where $\alpha(z)=\mathcal{B}e^{z}$ and
$\beta(z)=\left(1-\mathcal{B}\right)e^{z}$,
$\mathcal{B}\in\mathbb{C}\smallsetminus\bigg{\\{}0,\displaystyle\frac{1}{2},1\bigg{\\}}$.
We check that $E_{f}(\mathcal{S})=E_{\mathcal{L}^{2}_{\pi}(f)}(\mathcal{S})$
and $\max\\{\rho(\alpha),\rho(\beta)\\}=1=\rho(f)$ but $f$ is neither in the
specific form nor satisfying $\mathcal{L}^{2}_{\pi}(f)=f$.

######  Example 2.8.

Let $f(z)=\cos z+e^{-z}$ and

$\displaystyle\mathcal{L}^{3}_{\pi}(f)$ $\displaystyle=$
$\displaystyle\left(\frac{e^{2\pi}+e^{\pi}-1}{e^{-3\pi}+1}\right)f(z+3\pi)+e^{2\pi}f(z+\pi)-e^{\pi}f(z+\pi)$
$\displaystyle+\left(\frac{e^{2\pi}+e^{\pi}-1}{e^{-3\pi}+1}-e^{2\pi}-e^{\pi}\right)f(z).$

Let $\mathcal{S}=\\{\alpha,\beta\\}$, where $\alpha(z)=\mathcal{C}\cos z$ and
$\beta(z)=\left(1-\mathcal{C}\right)\cos z$,
$\mathcal{C}\in\mathbb{C}\smallsetminus\bigg{\\{}0,\displaystyle\frac{1}{2},1\bigg{\\}}$.
We check that $E_{f}(\mathcal{S})=E_{\mathcal{L}^{3}_{\pi}(f)}(\mathcal{S})$
and $\max\\{\rho(\alpha),\rho(\beta)\\}=1=\rho(f)$ but $f$ is not in the
specific form and also not satisfying $\mathcal{L}^{3}_{\pi}(f)=f$.

######  Remark 2.3.

The next examples show that, conclusion of _Theorem 2.1_ ceases to hold if one
replace the $CM$ sharing by $IM$ sharing. The examples have been exhibited for
$n=1$ and $n=2$ only.

######  Example 2.9.

Let $f(z)=-\displaystyle\frac{1}{2}\left(e^{z}+a^{2}e^{-z}\right)$ for
$a\in\mathbb{C}\smallsetminus\\{0\\}$. We choose
$c\in\mathbb{C}\smallsetminus\\{k\pi i\\},$ for $k\in\mathbb{Z}$ be such that

$\displaystyle\mathcal{L}_{c}(f)=\frac{2e^{c}}{1-e^{2c}}f(z+c)-\frac{2}{1-e^{2c}}f(z).$

Evidently
$\overline{E}_{f}(\mathcal{S})=\overline{E}_{\mathcal{L}_{c}(f)}(\mathcal{S})$
where $\mathcal{S}=\\{\alpha,\beta\\}=\\{-a,a\\}$ and
$\max\\{\rho(\alpha),\rho(\beta)\\}=0<1=\rho(f)$ but $f$ has neither the
specific form as in _Theorem 1.1_ nor satisfies the relation
$\mathcal{L}_{c}(f)\equiv f$.

######  Example 2.10.

Let $f(z)=\displaystyle\frac{1}{2}\left(e^{z}+e^{-z}\right)$ and for
$c\in\mathbb{C}\smallsetminus\bigg{\\{}\displaystyle\frac{k\pi
i}{2}:k\in\mathbb{Z}\bigg{\\}},$

$\displaystyle\mathcal{L}^{2}_{c}(f)=\frac{2}{1-e^{4c}}f(z+2c)-2e^{-c}f(z+c)+\frac{2e^{2c}}{e^{4c}-1}f(z).$

Evidently $E_{f}(\mathcal{S})=E_{\mathcal{L}^{2}_{c}(f)}(\mathcal{S})$, where
$\mathcal{S}=\\{-1,1\\}$ with $\max{\rho(\alpha),\rho(\beta)}=0<1=\rho(f)$ but
$f$ has neither the specific form nor satisfying $\mathcal{L}^{2}_{c}(f)=f.$

For $a_{2}(\neq 0),a_{1},a_{0}\in\mathbb{C}$, we define two constants
$\mathcal{D}_{1}$ and $\mathcal{D}_{2}$ as follows

$\displaystyle\mathcal{D}_{1}=\frac{-a_{1}+\sqrt{a_{1}^{2}-4a_{2}(a_{0}-1)}}{2a_{2}}\;\;\text{and}\;\;\mathcal{D}_{2}=\frac{-a_{1}-\sqrt{a_{1}^{2}-4a_{2}(a_{0}-1)}}{2a_{2}}.$

We have a corollary of the main result.

######  Corollary 2.1.

Suppose that $\alpha$, $\beta$ are two distinct entire functions, and $f$ is a
non-constant entire function with $\lambda(f)<\rho(f)<\infty$ such that
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$ and
$\mathcal{L}^{2}_{c}(f)(\not\equiv 0)$. If
$E_{f}(\\{\alpha,\beta\\})=E_{\mathcal{L}^{2}_{c}(f)}(\\{\alpha,\beta\\})$,
then

$f(z)=\begin{cases}\displaystyle\mathcal{A}\left(-\frac{a_{1}}{2a_{2}}\right)^{z/c}\;\text{when}\;\;a_{1}^{2}+4a_{2}=4a_{0}a_{2}\\\
\mathcal{A}_{1}\mathcal{D}_{1}^{z/c}+\mathcal{A}_{2}\mathcal{D}_{2}^{z/c},\;\text{otherwise}\\\
\end{cases}$

where
$\mathcal{A},\mathcal{A}_{1},\mathcal{A}_{2}\in\mathbb{C}\smallsetminus\\{0\\}.$
Furthermore $\mathcal{L}^{2}_{c}(f(z))\equiv f(z)$ for all $z\in\mathbb{C}$.

######  Remark 2.4.

From the next examples, one can observe that entire functions in support of
_Corollary 2.1_ when (i) $a_{1}^{2}+4a_{2}=4a_{0}a_{2}$ and (ii)
$a_{1}^{2}+4a_{2}\neq 4a_{0}a_{2}.$ Here $\mathcal{S}=\\{\alpha,\beta\\}$,
where $\alpha$ and $\beta$ are two polynomials in $z$.

######  Example 2.11.

Let
$f(z)=\displaystyle\frac{2\sqrt{2}}{\sqrt{3}}\left(\frac{\sqrt{3}}{2\sqrt{2}}\right)^{z/c}$,
and

$\displaystyle\mathcal{L}^{2}_{c}(f)=\sqrt{2}f(z+2c)-\sqrt{3}f(z+c)+\frac{3+4\sqrt{2}}{4\sqrt{2}}f(z).$

One can check that $a_{1}^{2}+4a_{2}=4a_{0}a_{2}$, and all the conditions of
_Corollary 2.1_ are satisfied and $f$ has the specific form and also
$\mathcal{L}^{2}_{c}(f)=f$.

######  Example 2.12.

Let

$\displaystyle
f(z)=\sqrt{3}\displaystyle\left(\frac{\sqrt{16+\sqrt{17}}+\sqrt{\sqrt{17}}}{4\sqrt{2}}\right)^{z/c}+\sqrt{2}\displaystyle\left(\frac{\sqrt{16+\sqrt{17}}-\sqrt{\sqrt{17}}}{4\sqrt{2}}\right)^{z/c}$

and

$\displaystyle\mathcal{L}^{2}_{c}(f)=2\sqrt{2}f(z+2c)-\sqrt{16+\sqrt{17}}f(z+c)+(1+\sqrt{2})f(z).$

Verify that $a_{1}^{2}+4a_{2}\neq 4a_{0}a_{2}$, and all the conditions of
_Corollary 2.1_ are satisfied and $f$ is in the form satisfying
$\mathcal{L}^{2}_{c}(f)=f$.

##  3\. Some useful lemmas

As discussed in section 1, _Halburd - Korhonen_ [14] and _Chiang - Feng_ [9]
inspected the value distribution theory of difference expressions, including
the difference analogue of the logarithmic derivative lemma, independently, we
recall here their results.

######  Lemma 3.1.

[14] Let $f(z)$ be a non-constant meromorphic function of finite order, and
$c\in\mathbb{C}\smallsetminus\\{0\\}$, $\delta<1.$ Then

$\displaystyle
m\left(r,\frac{f(z+c)}{f(z)}\right)=o\left(\frac{T(r+|z|,f)}{r^{\delta}}\right),$

for all $r$ outside of a possible exceptional set with finite logarithmic
measure.

By [15, Lemma 2.1], we have $T(r+|c|,f(z))=(1+o(1))T(r,f)$ for all r outside
of a set with finite logarithmic measure, when $f$ is of finite order.

######  Lemma 3.2.

[9] Let $f$ be a meromorphic function of finite order, and $\eta_{1}$,
$\eta_{2}$ be two distinct arbitrary complex numbers. Assume that $\sigma$ is
the order of $f$, then for each $\epsilon>0$, we have

$\displaystyle
m\left(r,\frac{f(z+\eta_{1})}{f(z+\eta_{2})}\right)=O\left(r^{\sigma-1+\epsilon}\right).$

######  Lemma 3.3.

[8] Let $g$ be a function transcendental and meromorphic in the plane of order
less than $1$. Set $h>0$. Then there exists an $\epsilon$-set $E$ such that

$\displaystyle\frac{g(z+\eta)}{g(z)}\rightarrow 1,\text{when
$z\rightarrow\infty$ in $\mathbb{C}\smallsetminus E$}$

uniformly in $\eta$ for $|\eta|\leqslant h$.

######  Lemma 3.4.

Let $\alpha$, $\beta$ are two distinct entire functions, and $f$ is a non-
constant entire function such that
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)$ and
$E_{f}(\\{\alpha,\beta\\})=E_{\mathcal{L}_{c}(f)}(\\{\alpha,\beta\\})$. Let
$f(z)=\mathcal{G}(z)e^{P(z)}$, where $\mathcal{G}(\not\equiv 0)$ is an entire
function and $P$ be a polynomial in $z$. If

$\displaystyle\mathcal{W}(z)=\sum_{j=0}^{n}a_{j}\mathcal{G}(z+jc)e^{P(z+jc)-P(z)},$

then $\mathcal{W}(z)\not\equiv 0$.

###### Proof.

By the _Hadamard factorization theorem_ , since $f$ is an entire function, we
can write $f(z)=\mathcal{G}(z)e^{P(z)}$, where $\mathcal{G}(\not\equiv 0)$ is
an entire function and $P$ be a polynomial in $z$. Since
$E_{f}(\\{\alpha,\beta\\})=E_{\mathcal{L}^{n}_{c}(f)}(\\{\alpha,\beta\\})$, so
we can write

(3.1)
$\displaystyle\frac{(\mathcal{L}^{n}_{c}(f)-\alpha)((\mathcal{L}^{n}_{c}(f)-\beta))}{(f-\alpha)(f-\beta)}=e^{\mathcal{Q}},$

where $\mathcal{Q}$ is an entire function. Furthermore, it follows from (3.1)
and the condition $\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)<\infty,$ that
$\mathcal{Q}$ is a polynomial in $z$. Since $f(z)=\mathcal{G}(z)e^{P(z)}$,
then we can write $\mathcal{L}^{n}_{c}(f)$ as

(3.2) $\displaystyle\mathcal{L}^{n}_{c}(f)$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{n}a_{j}f(z+jc)=\bigg{[}\sum_{j=0}^{n}a_{j}\mathcal{G}(z+jc)e^{P(z+jc)-P(z)}\bigg{]}e^{P(z)}.$

Substituting the forms of the functions $f$ and $\mathcal{L}^{n}_{c}(f)$ in
(3.1), we have

$\displaystyle\bigg{\\{}\bigg{[}\sum_{j=0}^{n}a_{j}\mathcal{G}(z+jc)e^{P(z+jc)-P(z)}\bigg{]}e^{P(z)}-\alpha(z)\bigg{\\}}$
$\displaystyle\times$
$\displaystyle\bigg{\\{}\bigg{[}\sum_{j=0}^{n}a_{j}\mathcal{G}(z+jc)e^{P(z+jc)-P(z)}\bigg{]}e^{P(z)}-\beta(z)\bigg{\\}}$
$\displaystyle=$
$\displaystyle\bigg{\\{}\mathcal{H}(z)e^{P(z)}-\alpha(z)\bigg{\\}}\bigg{\\{}\mathcal{H}(z)e^{P(z)}-\beta(z)\bigg{\\}}e^{\mathcal{Q}(z)}.$

On contrary, let if possible $\mathcal{W}\equiv 0$. This shows that

(3.4) $\displaystyle\sum_{j=0}^{n}a_{j}\mathcal{G}(z+jc)e^{P(z+jc)-P(z)}=0.$

In view of (3.2) and (3.4), we see that $\mathcal{L}^{n}_{c}(f)\equiv 0,$
which contradicts $\mathcal{L}^{n}_{c}(f)\not\equiv 0$. Therefore, we must
have $\mathcal{W}(z)\not\equiv 0.$ ∎

######  Lemma 3.5.

Let $f$ and $\mathcal{G}$ be two entire functions and $P(z)$ is a polynomial
in $zj$, and $a_{j}$, $j=0,1,\ldots,n$ be all complex constants with
$a_{n}\neq 0$, such that

$\displaystyle\sum_{j=0}^{n}a_{j}\mathcal{G}(z+jc)e^{P(z+jc)-P(z)}=\mathcal{G}(z),$

then for each $j=1,2,\ldots,n$ and $\epsilon>0$,

$\displaystyle
m\left(r,e^{P(z+jc)-P(z)}\right)=[\mathcal{A}+o(1)]r^{\rho(f)-1+\epsilon},$

for a complex number $\mathcal{A}.$

###### Proof.

We prove this lemma by inductive way. Let $n=1$.

Then, we have

$\displaystyle
a_{0}\mathcal{G}(z)+a_{1}\mathcal{G}(z+c)e^{P(z+c)-P(z)}=\mathcal{G}(z)$

which we can rewrite as

(3.5) $\displaystyle
e^{P(z+c)-P(z)}=\frac{1-a_{0}}{a_{1}}\frac{\mathcal{G}(z)}{\mathcal{G}(z+c)}.$

In view of _Lemma 3.3_, we get from (3.5) that

$\displaystyle m\left(r,e^{P(z+c)-P(z)}\right)$ $\displaystyle=$
$\displaystyle
m\left(r,\frac{1-a_{0}}{a_{1}}\frac{\mathcal{G}(z)}{\mathcal{G}(z+c)}\right)$
$\displaystyle=$ $\displaystyle
m\left(r,\frac{\mathcal{G}(z)}{\mathcal{G}(z+c)}\right)+O(1)$ $\displaystyle=$
$\displaystyle O\left(r^{\rho(\mathcal{G})-1+\epsilon}\right)+O(1).$

On the other hand we have
$m\left(r,e^{P(z+c)-P(z)}\right)=[\mathcal{A}+o(1)]r^{\rho(f)-1+\epsilon},$
where $\mathcal{A}$ is a complex number.

Let $n=2$, then we have

(3.6) $\displaystyle
e^{P(z+2c)-P(z)}=\frac{1-a_{0}}{a_{2}}\frac{\mathcal{G}(z)}{\mathcal{G}(z+2c)}-\frac{a_{1}}{a_{2}}\frac{\mathcal{G}(z+c)}{\mathcal{G}(z+2c)}.$

By _Lemma 3.3_, we get from (3.6)

$\displaystyle m\left(r,e^{P(z+2c)-P(z)}\right)$ $\displaystyle=$
$\displaystyle
m\left(r,\frac{\mathcal{G}(z)}{\mathcal{G}(z+2c)}\right)+m\left(r,\frac{\mathcal{G}(z+c)}{\mathcal{G}(z+2c)}\right)+m\left(r,e^{P(z+c)-P(z)}\right)$
$\displaystyle+O(1)$ $\displaystyle=$ $\displaystyle
O\left(r^{\rho(\mathcal{G})-1+\epsilon}\right)+O(1).$

On the other hand we have
$m\left(r,e^{P(z+2c)-P(z)}\right)=[\mathcal{A}+o(1)]r^{\rho(f)-1+\epsilon}.$

So, continuing in this way, one can prove that

$\displaystyle
m\left(r,e^{P(z+jc)-P(z)}\right)=[\mathcal{A}+o(1)]r^{\rho(f)-1+\epsilon},\;\text{for}\;\;j=3,4,\ldots,n.$

This completes the proof.

∎

##  4\. Proof of Theorem

In this section, we give the proof of the main result of this article.

###  4.1. Proof of Theorem 2.1

From the conditions of _Theorem 2.1_, since we have
$E_{f}(\\{\alpha,\beta\\})=E_{\mathcal{L}^{n}_{c}(f)}(\\{\alpha,\beta\\})$,
there must exists an entire function $\mathcal{Q}$ such that

(4.1)
$\displaystyle\frac{(\mathcal{L}^{n}_{c}(f)-\alpha)((\mathcal{L}^{n}_{c}(f)-\beta))}{(f-\alpha)(f-\beta)}=e^{\mathcal{Q}}.$

Again, it follows from (4.1) and the condition
$\max\\{\rho(\alpha),\rho(\beta)\\}<\rho(f)<\infty,$ that $\mathcal{Q}$ is a
polynomial in $z$. By the _Hadamard factorization theorem_ , we may suppose
that $f(z)=\mathcal{G}(z)e^{P(z)}$, where $\mathcal{G}(\not\equiv 0)$ is an
entire function, and $P$ is a polynomial satisfying
$\lambda(f)=\rho(\mathcal{G})<\rho(f)=\deg(P).$ We set

$\displaystyle\mathcal{W}(z)=\sum_{j=0}^{n}a_{j}\mathcal{G}(z+jc)e^{P(z+jc)-P(z)}.$

By _Lemma 3.4_, we must have $\mathcal{W}(z)\not\equiv 0$. It is not hard to
check that $\mathcal{W}$ is a small function of $e^{P(z)}$. Rewriting (3), we
have

(4.2) $\displaystyle
e^{\mathcal{Q}(z)}=\frac{\mathcal{W}^{2}(z)\bigg{[}e^{P(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{W}(z)}\bigg{]}\bigg{[}e^{P(z)}-\displaystyle\frac{\beta(z)}{\mathcal{W}(z)}\bigg{]}}{\mathcal{G}^{2}(z)\bigg{[}e^{P(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}\bigg{]}\bigg{[}e^{P(z)}-\displaystyle\frac{\beta(z)}{\mathcal{G}(z)}\bigg{]}}.$

By our assumption, since $\alpha(z)\not\equiv\beta(z)$, without any loss of
generality, we may suppose that $\alpha(z)\not\equiv 0.$

Let $z_{0}$ be a zero of
$e^{P(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{H}(z)}$ such that
$\mathcal{W}(z_{0})\neq 0$, then it follows from (4.2) that $z_{0}$ must be a
zero of $e^{P(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{W}(z)}$ or
$e^{P(z)}-\displaystyle\frac{\beta(z)}{\mathcal{W}(z)}$. We denote now two
counting functions here, one is $N_{1}\left(r,e^{P}\right)$ of the common
zeros of $e^{P(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}$ and
$e^{P(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{W}(z)},$ and the second one
is $N_{2}\left(r,e^{P}\right)$ be the reduced counting of those zeros of
$e^{P(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}$ and
$e^{P(z)}-\displaystyle\frac{\beta(z)}{\mathcal{W}(z)}.$ Since
$\mathcal{G}(z)$ is a small function of $e^{P(z)}$, then by applying _First_
and _Second Fundamental Theorem_ , we deduce that

$\displaystyle T\left(r,e^{P}\right)$ $\displaystyle=$
$\displaystyle\overline{N}\left(r,\frac{1}{e^{P(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{H}(z)}}\right)+S\left(r,e^{P(z)}\right)$
$\displaystyle=$ $\displaystyle
N_{1}\left(r,e^{P(z)}\right)+N_{2}\left(r,e^{P(z)}\right)+S\left(r,e^{P(z)}\right).$

It is clear from (4.1) that

$\displaystyle\text{either}\;\;N_{1}\left(r,e^{P(z)}\right)\neq
S\left(r,e^{P(z)}\right)\;\;\text{or}\;\;N_{2}\left(r,e^{P(z)}\right)\neq
S\left(r,e^{P(z)}\right).$

In our next discussions, we are going to appraise the following two cases.
Case 1. Suppose $N_{1}\left(r,e^{P(z)}\right)\neq S\left(r,e^{P(z)}\right).$

Let $z_{1}$ be a common zero of
$e^{P}-\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}$ and
$e^{P}-\displaystyle\frac{\alpha(z)}{\mathcal{W}(z)}.$ Therefore, $z_{1}$ must
be a zero of

$\displaystyle\left(e^{P}-\displaystyle\frac{\alpha(z)}{\mathcal{W}(z)}\right)-\left(e^{P}-\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}\right)=\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}-\displaystyle\frac{\alpha(z)}{\mathcal{W}(z)}.$

Subcase 1.1. Let is possible, $\mathcal{G}(z)\not\equiv\mathcal{W}(z)$. Then
it follows that
$\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}-\frac{\alpha(z)}{\mathcal{W}(z)}\not\equiv
0.$ We next deduce that

$\displaystyle\;\;\;\;\;\;\;\;\;S\left(r,e^{P}\right)\neq
N_{1}\left(r,e^{P}\right)\leqslant
N\left(r,\frac{1}{\displaystyle\frac{\alpha(z)}{\mathcal{H}(z)}-\frac{\alpha(z)}{\mathcal{W}(z)}}\right)\leqslant
T\left(r,\frac{\alpha(z)}{\mathcal{H}(z)}-\frac{\alpha(z)}{\mathcal{W}(z)}\right)=S(r,e^{P}),$

which is a contradiction.
Subcase 1.2. Suppose $\mathcal{G}(z)\equiv\mathcal{W}(z)$. Which in turn shows
that

(4.4)
$\displaystyle\sum_{j=0}^{n}a_{j}\mathcal{G}(z+jc)e^{P(z+jc)-P(z)}=\mathcal{G}(z),$

In view of _Lemma 3.5_ and (4.4), for each $j=1,2,\ldots,n$ and $\epsilon>0$,
we have

$\displaystyle
m\left(r,e^{P(z+jc)-P(z)}\right)=[\mathcal{A}+o(1)]r^{\rho(f)-1+\epsilon},$

for a complex number $\mathcal{A}.$

Let if possible $\rho(f)>1$.
Then, by $\rho(f)>\rho(\mathcal{G})$ and the estimates of
$m\left(r,e^{P(z+jc)-P(z)}\right)$, we can easily get a contradiction.

Thus, we see that $\rho(f)\leqslant 1$, and which in turn shows that
$e^{P(z+jc)-P(z)}$ is a non-zero constant, say, $\eta_{j}$ for
$j=0,1,2,\ldots,n$. We have $e^{P(z+c)}=\eta_{j}e^{P(z)}$. Therefore, it
follows from (4.4) that,

(4.5)
$\displaystyle\sum_{j=0}^{n}a_{j}\eta_{j}\frac{\mathcal{G}(z+jc)}{\mathcal{G}(z)}=1.$

Since $\rho(\mathcal{G})<\rho(f)\leqslant 1$, then in view of _Lemma 3.3_,
there exists an $\epsilon$-set $E$, as $z\not\in E$ and
$|z|\rightarrow\infty$, such that for each $j=1,2,\ldots,n$ we can obtained

$\displaystyle\frac{\mathcal{G}(z+jc)}{\mathcal{G}(z)}\rightarrow 1$

which shows that $\displaystyle\sum_{j=0}^{n}a_{j}\eta_{j}=1.$ If we approach
inductively, it follows from (4.5) that $\mathcal{G}(z+c)=\mathcal{G}(z)$ for
all $z\in\mathbb{C}$ i.e., $\mathcal{G}$ is a periodic entire function of
period $c$. Let if possible $\mathcal{G}$ is non-constant, then $\mathcal{G}$
must be transcendental, and hence $\rho(\mathcal{G})\geqslant 1$, which
contradicts $\rho(\mathcal{G})\leqslant\rho(f)<1$.

Therefore, $\mathcal{G}$ is constant. Since, $f$ is a finite order non-
constant entire function with $\deg(P)=\rho(f)\leqslant 1$, so we must have
$\deg(P)=1$, and hence $P(z)$ will be of the form $P(z)=d_{1}z+d_{0}$, for
some $d_{1}(\neq 0),d_{0}\in\mathbb{C}$. Therefore, we can write the function
$f$ as $f(z)=\mathcal{C}e^{\mu z},$ where $\mathcal{C}$ and $\mu$ are two non-
zero complex constants.

By our assumption in Case 1, we see that $f-\alpha$ and
$\mathcal{L}^{n}_{c}(f)-\alpha$ have common zeros, which are not the zeros of
$\alpha(z)$. Let $z_{3}$ be a common zero of $f-a$ and
$\mathcal{L}^{n}_{c}(f)$ so that $\alpha(z_{3})\neq 0.$ Then, we see that
$z_{3}$ must be a zero

$\displaystyle
f(z)-\alpha(z)+\mathcal{L}^{n}_{c}(f)-\alpha(z)=\sum_{j=1}^{n}a_{j}f(z+jc)+(a_{0}+1)f(z)-2\alpha(z).$

Thus we have

$\begin{cases}\mathcal{C}e^{\mu z_{3}}-\alpha(z_{3})=0\\\
\displaystyle\sum_{j=1}^{n}a_{j}\mathcal{C}e^{\mu z_{3}}e^{\mu
jc}+(a_{0}+1)\mathcal{C}e^{\mu z_{3}}-2\alpha(z_{3})=0\\\ \end{cases}$

Thus we must have $\displaystyle\sum_{j=1}^{n}a_{j}e^{\mu jc}+a_{0}=1$ which
can written as

(4.6) $\displaystyle a_{n}\left(e^{\mu c}-\lambda_{1}\right)\left(e^{\mu
c}-\lambda_{2}\right)\ldots\left(e^{\mu c}-\lambda_{n}\right)=0,$

where $\lambda_{1},\ldots,\lambda_{n}$ are distinct roots of the equation

$\displaystyle a_{n}z^{n}+a_{n-1}z^{n-1}+\ldots+a_{0}=1.$

From (4.6) , we obtained that

$\displaystyle e^{\mu c}=\lambda_{i},\;\text{for some}\;i=1,2,\ldots,n,$

and thus, the general form of the function is

$\displaystyle
f(z)=\mathcal{C}_{1}\lambda_{1}^{z/c}+\mathcal{C}_{2}\lambda_{2}^{z/c}+\ldots+\mathcal{C}_{n}\lambda_{n}^{z/c},$

where $\mathcal{C}_{1},\mathcal{C}_{2},\ldots,\mathcal{C}_{n}$ are all complex
constants.
Case 2. Let $N_{2}\left(r,e^{P}\right)\neq S\left(r,e^{P}\right).$

Let $z_{4}$ be a common zero of $\displaystyle
e^{P(z)}-\frac{\alpha(z)}{\mathcal{G}(z)}$ and $\displaystyle
e^{P(z)}-\frac{\beta(z)}{\mathcal{W}(z)}.$ One can check that $z_{4}$ must be
a zero of
$\displaystyle\frac{\alpha(z)}{\mathcal{H}(z)}-\frac{\beta(z)}{\mathcal{W}(z)}.$

Let
$\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}-\frac{\beta(z)}{\mathcal{W}(z)}\not\equiv
0$. Then we see that

$\displaystyle\;\;\;\;\;\;\;S\left(r,e^{P}\right)\neq
N_{2}\left(r,e^{P}\right)\leqslant
N\left(r,\frac{1}{\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}-\frac{\beta(z)}{\mathcal{W}(z)}}\right)\leqslant
T\left(r,\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}-\frac{\beta(z)}{\mathcal{W}(z)}\right)=S\left(r,e^{P}\right),$

which is clearly a contradiction.

Thus we have

(4.7)
$\displaystyle\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}\equiv\frac{\beta(z)}{\mathcal{W}(z)}.$

Suppose $\beta(z)\equiv 0$, then
$\displaystyle\frac{\alpha(z)}{\mathcal{G}(z)}\equiv 0$ i.e., $\alpha(z)\equiv
0$, which is absurd.
Let $z_{5}$ is a zero of $\displaystyle e^{P}-\frac{\beta}{\mathcal{G}}$ but
not a zero of $\mathcal{W}$. Then it follows from (4.2) that $z_{5}$ is a zero
of $\displaystyle e^{P}-\frac{\alpha}{\mathcal{W}}$ or $\displaystyle
e^{P}-\frac{\beta}{\mathcal{G}}.$ Here, we denote by
$N_{3}\left(r,e^{P}\right)$ the reduced counting function of those common
zeros of $e^{P}-\frac{\beta}{\mathcal{G}}$ and
$e^{P}-\frac{\alpha}{\mathcal{W}}$. Similarly, we denote by
$N_{4}\left(r,e^{P}\right)$ the reduced counting function of those common
zeros of $e^{P}-\frac{\beta}{\mathcal{G}}$ and
$e^{P}-\frac{\beta}{\mathcal{W}}.$

Applying _First_ and _Second Fundamental Theorem_ , we have

$\displaystyle T\left(r,e^{P}\right)$ $\displaystyle=$
$\displaystyle\overline{N}\left(r,\frac{1}{e^{P}-\frac{\beta}{\mathcal{H}}}\right)+S\left(r,e^{P}\right)$
$\displaystyle=$ $\displaystyle
N_{3}\left(r,e^{P}\right)+N_{4}\left(r,e^{P}\right)+S\left(r,e^{P}\right),$

which implies that either $N_{3}\left(r,e^{P}\right)\neq
S\left(r,e^{P}\right)$ or $N_{4}\left(r,e^{P}\right)\neq
S\left(r,e^{P}\right)$.
Subcase 2.1. If $N_{4}\left(r,e^{P}\right)\neq S\left(r,e^{P}\right)$, then in
the same manner as in Case 1, we get our desired result.
Subcase 2.2. Next we assume that $N_{3}\left(r,e^{P}\right)\neq
S\left(r,e^{P}\right)$.
Then similar to the Case 2, one can simply deduce that

(4.8)
$\displaystyle\displaystyle\frac{\beta}{\mathcal{G}}-\frac{\alpha}{\mathcal{W}}\equiv
0.$

Combining (4.7) and (4.8) yields that $\alpha^{2}=\beta^{2}$ which in turn
implies that $\alpha=-\beta$ as $\alpha\neq\beta$. Thus, we have from (4.8)
that $\mathcal{W}=\mathcal{G}$. Rest of the proof of the result next follows
from the Subcase 1.2.

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