You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'We evaluate the Green’s function of the D-dimensional relativistic Coulomb system via sum over perturbation series which is obtained by expanding the exponential containing the potential term $V({\bf x)}$ in the path integral into a power series. The energy spectra and wave functions are extracted from the resulting amplitude.'
address: |
    Department of Physics, National Tsing Hua University\
    Hsinchu 30043, Taiwan
author:
- 'De-Hone Lin [^1]'
title: 'Green’s function for the Relativistic Coulomb System via Sum Over Perturbation Series'
---

=10000

Introduction
============

      Most physical problems cannot be solved exactly. It is therefore necessary to develop approximation procedures which allow us to approach the exact result with appropriate accuracy. An important approximation method for solving problems in quantum mechanics (QM) is the Rayleigh-Schrödinger perturbation theory. It provides us an effective method to calculate the approximate solutions of many problems which can not be exactly solved by using the Schrödinger equation. Similar to the standard QM, the perturbation method can be developed in the path integral framework of QM [@1]. Historically of utmost importance was the application of the perturbation expansion of path integral to the quantum electrodynamics by Feynman [@2], from which he derived for the first time the “Feynman’s rules”, which provide an extremely effective method to calculate the perturbation series and a clear, neat interpretation of the interaction picture.

In the past 10 years, perturbation expansion of the path integral has been used to obtain the exact Green’s functions for $\delta $-function potential problems [@3; @3.5; @4; @4.5], non-relativistic Coulomb system [@5], and to yield the Dirichlet boundary conditions in Refs. [@6; @6.5] for the non-relativistic problems and in Ref. [@7] for the relativistic problems by summing the $\delta $-function perturbation series.

In this paper, we would like to add a further application of the perturbation method of the path integral. We calculate the Green’s function of a D-dimensional relativistic Coulomb system via summing over the perturbation series. The energy spectra and wave functions are extracted from the resulting amplitude.

Path integral for the relativistic Coulomb system via sum over the perturbation series
======================================================================================

      Let us first consider a point particle of mass $M$ moving at a relativistic velocity in a $(D+1)$-dimensional Minkowski space with a given electromagnetic field. By using $t=-i\tau =-ix^{4}/c$, the path integral representation of the Green’s function is conveniently formulated in a $(D+1)
$-Euclidean spacetime with the Euclidean metric, $$(g_{\mu \nu })={\rm diag}\;\,(1,\cdots ,1,c^{2}),  \label{1.1}$$ and it is given by [@8; @9] $$G({\bf {x}}_{b},{\bf {x}}_{a};E)=\frac{i\hbar }{2Mc}\int_{0}^{\infty }dS\int
D\rho \Phi \left[ \rho \right] \int D^{D}xe^{-A_{E}/\hbar }.\qquad \qquad 
\label{1.2}$$ The action integral $$A_{E}=\int_{\lambda _{a}}^{\lambda _{b}}d\lambda \left[ \frac{M}{2\rho
\left( \lambda \right) }{\bf {x}}^{\prime ^{2}}\left( \lambda \right) -i%
\frac{e}{c}{\bf A(x)\cdot x^{\prime }(}\lambda {\bf )}-\rho \left( \lambda
\right) \frac{(E-V({\bf x}))^{2}}{2Mc^{2}}+\rho \left( \lambda \right) \frac{%
Mc^{2}}{2}\right] ,  \label{1.3}$$ where $S$ is defined by $$S=\int_{\lambda _{a}}^{\lambda _{b}}d\lambda \rho (\lambda ),  \label{1.4}$$ in which $\rho (\lambda )$ is an arbitrary dimensionless fluctuating scale variable, and $\Phi \lbrack \rho ]$ is some convenient gauge-fixing functional, such as $\Phi \left[ \rho \right] =\delta \left[ \rho -1\right] $, to fix the value of $\rho (\lambda )$ to unity [@8; @9]. $\hbar /Mc$ is the well-known Compton wave length of a particle of mass $M$, ${\bf A(x)}$ is the vector potential, $V({\bf x})$ is the scalar potential, $E$ is the system energy, and ${\bf {x}}$ is the spatial part of the ($D+1$) vector $x=(%
{\bf {x}},\tau )$. This path integral forms the basis for studying relativistic potential problems.

Expanding the potential term $V({\bf x})$ into a power series and interchanging the order of integration and summation, we obtain the result $$G({\bf {x}}_{b},{\bf {x}}_{a};E)=\frac{i\hbar }{2Mc}\int_{0}^{\infty }dS\int
D\rho \Phi \left[ \rho \right] e^{-\frac{1}{\hbar }\int_{\lambda
_{a}}^{\lambda _{b}}d\lambda \rho (\lambda ){\cal E}}K({\bf {x}}_{b},{\bf {x}%
}_{a};\lambda _{b}-\lambda _{a})  \label{1.5}$$ with the series expansion of the pseudotime propagator $$K({\bf {x}}_{b},{\bf {x}}_{a};\lambda _{b}-\lambda _{a})=\left\{ 
\vbox to
24pt{}K_{0}+\sum_{n=1}^{\infty }\frac{1}{n!}\left( -\frac{\beta }{\hbar }%
\right) ^{n}\right.$$ $$\times \int D^{D}xe^{-\frac{1}{\hbar }\int_{\lambda _{a}}^{\lambda
_{b}}d\lambda \left[ \frac{M}{2\rho \left( \lambda \right) }{\bf {x}}%
^{\prime ^{2}}\left( \lambda \right) -i\frac{e}{c}{\bf A(x)\cdot x^{\prime }(%
}\lambda {\bf )}-\rho \left( \lambda \right) \frac{V({\bf x})^{2}}{2Mc^{2}}%
\right] }$$ $$\times \left. \int_{\lambda _{a}}^{\lambda _{b}}d\lambda _{1}\rho {\bf (}%
\lambda _{1})V({\bf x(}\lambda _{1}))\int_{\lambda _{a}}^{\lambda
_{b}}d\lambda _{2}\rho {\bf (}\lambda _{2})V({\bf x(}\lambda _{2}))\cdots
\int_{\lambda _{a}}^{\lambda _{b}}d\lambda _{n}\rho {\bf (}\lambda _{n})V(%
{\bf x(}\lambda _{n}))\vbox to 24pt{}\right\} ,  \label{1.6}$$ where we have defined the quantities $\beta =E/Mc^{2},$ ${\cal E}=($ $%
M^{2}c^{4}-E^{2})/2Mc^{2}$, and $$K_{0}({\bf {x}}_{b},{\bf {x}}_{a};\lambda _{b}-\lambda _{a})=\int D^{D}xe^{-%
\frac{1}{\hbar }\int_{\lambda _{a}}^{\lambda _{b}}d\lambda \left[ \frac{M}{%
2\rho \left( \lambda \right) }{\bf {x}}^{\prime ^{2}}\left( \lambda \right)
-i\frac{e}{c}{\bf A(x)\cdot x^{\prime }(}\lambda {\bf )}-\rho \left( \lambda
\right) \frac{V({\bf x})^{2}}{2Mc^{2}}\right] }.  \label{1.7}$$ Ordering the $\lambda $ as $\lambda _{1}<\lambda _{2}<\cdots <\lambda
_{n}<\lambda _{b}$ and denoting ${\bf x(}\lambda _{k})={\bf x}_{k},$ the perturbative series in Eq. (\[1.6\]) turns into [@1] $$K({\bf {x}}_{b},{\bf {x}}_{a};\lambda _{b}-\lambda _{a})=K_{0}({\bf {x}}_{b},%
{\bf {x}}_{a};\lambda _{b}-\lambda _{a})+\sum_{n=1}^{\infty }\left( -\frac{%
\beta }{\hbar }\right) ^{n}\int_{\lambda _{a}}^{\lambda _{b}}d\lambda
_{n}\int_{\lambda _{a}}^{\lambda _{n}}d\lambda _{n-1}\cdots \int_{\lambda
_{a}}^{\lambda _{2}}d\lambda _{1}$$ $$\times \int \left[ \prod_{j=0}^{n}K_{0}({\bf {x}}_{j+1},{\bf {x}}%
_{j};\lambda _{j+1}-\lambda _{j})\right] \prod_{k=1}^{n}\rho _{k}V({\bf x}%
_{k})d{\bf x}_{k},  \label{1.8}$$ where $\lambda _{0}=\lambda _{a},\lambda _{n+1}=\lambda _{b},{\bf x}_{n+1}=%
{\bf x}_{b},$ and ${\bf x}_{0}={\bf x}_{a}.$ In the case of an attractive Coulomb potential, we have $${\bf A(x)=}0,\quad V(r)=-\frac{e^{2}}{r}.  \label{1.9}$$ The perturbative expansion in Eq. (\[1.8\]) becomes $$K({\bf {x}}_{b},{\bf {x}}_{a};\lambda _{b}-\lambda _{a})=K_{0}({\bf {x}}_{b},%
{\bf {x}}_{a};\lambda _{b}-\lambda _{a})+\sum_{n=1}^{\infty }\left( \frac{%
\beta e^{2}}{\hbar }\right) ^{n}\int_{\lambda _{a}}^{\lambda _{b}}d\lambda
_{n}\int_{\lambda _{a}}^{\lambda _{n}}d\lambda _{n-1}\cdots \int_{\lambda
_{a}}^{\lambda _{2}}d\lambda _{1}$$ $$\times \int \left[ \prod_{j=0}^{n}K_{0}({\bf {x}}_{j+1},{\bf {x}}%
_{j};\lambda _{j+1}-\lambda _{j})\right] \prod_{k=1}^{n}\rho _{k}\frac{d{\bf %
x}_{k}}{r_{k}}.  \label{1.10}$$ The corresponding amplitude $K_{0}$ takes the form $$K_{0}({\bf {x}}_{b},{\bf {x}}_{a};\lambda _{b}-\lambda _{a})=\int D^{D}xe^{-%
\frac{1}{\hbar }\int_{\lambda _{a}}^{\lambda _{b}}d\lambda \left[ \frac{M}{%
2\rho \left( \lambda \right) }{\bf {x}}^{\prime ^{2}}\left( \lambda \right)
-\rho \left( \lambda \right) \frac{\hbar ^{2}}{2M}\frac{\alpha ^{2}}{r^{2}}%
\right] },  \label{1.11}$$ where $\alpha =e^{2}/\hbar c$ is the fine structure constant. We now choose $%
\Phi \left[ \rho \right] =\delta \left[ \rho -1\right] $ to fix the value of $\rho (\lambda )$ to unity. The Green’s function in Eq. (\[1.5\]) becomes $$G({\bf {x}}_{b},{\bf {x}}_{a};E)=\frac{i\hbar }{2Mc}\int_{0}^{\infty }dSe^{-%
\frac{{\cal E}}{\hbar }S}\left\{ \vbox to 24pt{}K_{0}({\bf {x}}_{b},{\bf {x}}%
_{a};S)\right.$$ $$+\left. \sum_{n=1}^{\infty }\left( \frac{\beta e^{2}}{\hbar }\right)
^{n}\int_{\lambda _{a}}^{\lambda _{b}}d\lambda _{n}\int_{\lambda
_{a}}^{\lambda _{n}}d\lambda _{n-1}\cdots \int_{\lambda _{a}}^{\lambda
_{2}}d\lambda _{1}\int \left[ \prod_{j=0}^{n}K_{0}({\bf {x}}_{j+1},{\bf {x}}%
_{j};\lambda _{j+1}-\lambda _{j})\right] \prod_{k=1}^{n}\frac{d{\bf x}_{k}}{%
r_{k}}\vbox to 24pt{}\right\} .  \label{1.12}$$ We observe that the integration over $S$ is a Laplace transformation. Because of the convolution property of the Laplace transformation, we obtain $$G({\bf {x}}_{b},{\bf {x}}_{a};E)=\frac{i\hbar }{2Mc}$$ $$\times \left\{ G_{0}({\bf {x}}_{b},{\bf {x}}_{a};{\cal E})+\sum_{n=1}^{%
\infty }\left( \frac{\beta e^{2}}{\hbar }\right) ^{n}\int \left[
\prod_{j=0}^{n}G_{0}({\bf {x}}_{j+1},{\bf {x}}_{j};{\cal E})\right]
\prod_{k=1}^{n}\frac{d{\bf x}_{k}}{r_{k}}\right\} .  \label{1.13}$$ We now perform the angular decomposition of Eq. (\[1.13\]) [@9; @10; @11]. This can be reached by inserting in Eq. (\[1.13\]) the expansion of $G_{0}$ in term of the D-dimensional hyperspherical harmonics $Y_{l{\bf m}}({\bf 
\hat{x}})$ [@12]: $$G_{0}({\bf {x}}_{j+1},{\bf {x}}_{j};{\cal E})=\frac{M}{\hbar
(r_{j+1}r_{j})^{D/2-1}}\sum_{l=0}^{\infty }g_{l}^{0}(r_{j+1},r_{j};{\cal E}%
)\sum_{{\bf m}}Y_{l{\bf m}}({\bf \hat{x}}_{j+1})Y_{l{\bf m}}^{\ast }({\bf 
\hat{x}}_{j}),  \label{1.14}$$ where the $g_{l}^{0}$ is given by [@11] $$\int_{0}^{\infty }\frac{dS}{S}e^{-\frac{{\cal E}}{\hbar }%
S}e^{-M(r_{j+1}^{2}+r_{j}^{2})/2\hbar S}I_{\sqrt{(l+D/2-1)^{2}-\alpha ^{2}}%
}\left( \frac{M}{\hbar }\frac{r_{j+1}r_{j}}{S}\right) .  \label{1.15}$$ The notation $I$ denotes the modified Bessel function. Integrating over the intermediate angular part of Eq. (\[1.13\]), we arrive at $$G({\bf {x}}_{b},{\bf {x}}_{a};E)=\frac{i\hbar }{2Mc}\sum_{l=0}^{\infty
}G_{l}(r_{b},r_{a};{\cal E})\sum_{{\bf m}}Y_{l{\bf m}}({\bf \hat{x}}_{b})Y_{l%
{\bf m}}^{\ast }({\bf \hat{x}}_{a}).  \label{1.16}$$ The pure radial amplitude $G_{l}(r_{b},r_{a};{\cal E})$ has the form $$G_{l}(r_{b},r_{a};{\cal E})=\frac{M}{\hbar }\frac{1}{(r_{b}r_{a})^{D/2-1}}%
\sum_{n=0}^{\infty }\left( \frac{M\beta e^{2}}{\hbar ^{2}}\right)
^{n}g_{l}^{(n)}(r_{b},r_{a};{\cal E})  \label{1.17}$$ with $g_{l}^{(n)}$ is given by $$g_{l}^{(n)}(r_{b},r_{a};{\cal E})=\int_{0}^{\infty }\cdots \int_{0}^{\infty }%
\left[ \prod_{j=0}^{n}g_{l}^{(0)}(r_{j+1},r_{j};{\cal E})\right]
\prod_{k=1}^{n}dr_{k}.  \label{17.5}$$ To obtain the explicit result of $g_{l}^{(n)}$, we note that $$\int_{0}^{\infty }\frac{dS}{S}e^{-\frac{{\cal E}}{\hbar }%
S}e^{-M(r_{b}^{2}+r_{a}^{2})/2\hbar S}I_{\sqrt{(l+D/2-1)^{2}-\alpha ^{2}}%
}\left( \frac{M}{\hbar }\frac{r_{b}r_{a}}{S}\right)$$ $$=2\int_{0}^{\infty }dz\frac{1}{\sinh z}e^{-\kappa (r_{b}+r_{a})\coth z}I_{2%
\sqrt{(l+D/2-1)^{2}-\alpha ^{2}}}\left( \frac{2\kappa \sqrt{r_{b}r_{a}}}{%
\sinh z}\right)  \label{1.18}$$ with $\kappa =\sqrt{M^{2}c^{4}-E^{2}}/\hbar c$. The equality in Eq. (\[1.18\]) can be easily proved by the formulas $$\int_{0}^{\infty }dy\frac{e^{2\nu y}}{\sinh y}\exp \left[ -\frac{t}{2}\left(
\zeta _{a}+\zeta _{b}\right) \coth y\right] I_{\mu }\left( \frac{t\sqrt{%
\zeta _{b}\zeta _{a}}}{\sinh y}\right)$$ $$=\frac{\Gamma \left( \left( 1+\mu \right) /2-\nu \right) }{t\sqrt{\zeta
_{b}\zeta _{a}}\Gamma \left( \mu +1\right) }W_{\nu ,\mu /2}\left( t\zeta
_{b}\right) M_{\nu ,\mu /2}\left( t\zeta _{a}\right) ,  \label{1.20}$$ with the range of validity $$\begin{array}{l}
\zeta _{b}>\zeta _{a}>0, \\ 
{Re}[(1+\mu )/2-\nu ]>0, \\ 
{Re}(t)>0,\mid \arg t\mid <\pi ,
\end{array}$$ where $M_{\mu ,\nu }$ and $W_{\mu ,\nu }$ are the Whittaker functions, and $$\int_{0}^{\infty }\frac{dy}{y}e^{-zy}e^{-(a^{2}+b^{2})/y}I_{\nu }(\frac{2ab}{%
y})=2I_{\nu }(2a\sqrt{z})K_{\nu }(2b\sqrt{z}),  \label{1.21}$$ with the range of validity $$a<b,\quad {Re}z>0.$$ From Eq. (\[1.18\]), we get, by using the formula $$\int_{0}^{\infty }drre^{-r^{2}/a}I_{\nu }(\varsigma r)I_{\nu }(\xi r)=\frac{a%
}{2}e^{a(\xi ^{2}+\varsigma ^{2})/4}I_{\nu }\left( \frac{a\xi \varsigma }{2}%
\right) ,  \label{21.5}$$ the result $$g_{l}^{(1)}(r_{b},r_{a};{\cal E})=\int_{0}^{\infty }g_{l}^{(0)}(r_{b},r;%
{\cal E})g_{l}^{(0)}(r,r_{a};{\cal E})dr$$ $$=\frac{2^{2}}{\kappa }\int_{0}^{\infty }zh(z)dz,  \label{1.22}$$ where the function $h(z)$ is defined as $$h(z)=\frac{1}{\sinh z}e^{-\kappa (r_{b}+r_{a})\coth z}I_{2\sqrt{%
(l+D/2-1)^{2}-\alpha ^{2}}}\left( \frac{2\kappa \sqrt{r_{b}r_{a}}}{\sinh z}%
\right) .  \label{1.23}$$ The expression for $g_{l}^{(n)}(r_{b},r_{a};{\cal E})$ can be obtained by induction with respect to $n$, and is given by $$g_{l}^{(n)}(r_{b},r_{a};{\cal E})=\frac{2^{n+1}}{n!}\frac{1}{\kappa ^{n}}%
\int_{0}^{\infty }z^{n}h(z)dz.  \label{1.25}$$ Inserting the expression in Eq. (\[1.17\]), we obtain $$G_{l}(r_{b},r_{a};{\cal E})=\frac{M}{\hbar }\frac{2}{(r_{b}r_{a})^{D/2-1}}$$ $$\times \int_{0}^{\infty }dze^{\left( \frac{2M\beta e^{2}}{\hbar ^{2}\kappa }%
\right) z}\frac{1}{\sinh z}e^{-\kappa (r_{b}+r_{a})\coth z}I_{2\sqrt{%
(l+D/2-1)^{2}-\alpha ^{2}}}\left( \frac{2\kappa \sqrt{r_{b}r_{a}}}{\sinh z}%
\right) .  \label{1.26}$$ With help of the formula in Eq. (\[1.20\]), we complete the integration of Eq. (\[1.26\]), and find the radial Green’s function for $r_{b}>r_{a}$ in the closed form, $$G_{l}(r_{b},r_{a};E)=\frac{1}{(r_{b}r_{a})^{(D-1)/2}}\frac{Mc}{\sqrt{%
M^{2}c^{4}-E^{2}}}\qquad \qquad \qquad$$ $$\times \frac{\Gamma \left( 1/2+\sqrt{(l+D/2-1)^{2}-\alpha ^{2}}-\frac{%
E\alpha }{\sqrt{M^{2}c^{4}-E^{2}}}\right) }{\Gamma \left( 1+2\sqrt{%
(l+D/2-1)^{2}-\alpha ^{2}}\right) }$$ $$\times W_{\frac{E\alpha }{\sqrt{M^{2}c^{4}-E^{2}}},\sqrt{(l+D/2-1)^{2}-%
\alpha ^{2}}}\left( \frac{2}{\hbar c}\sqrt{M^{2}c^{4}-E^{2}}r_{b}\right)$$ $$\times M_{\frac{E\alpha }{\sqrt{M^{2}c^{4}-E^{2}}},\sqrt{(l+D/2-1)^{2}-%
\alpha ^{2}}}\left( \frac{2}{\hbar c}\sqrt{M^{2}c^{4}-E^{2}}r_{a}\right)
\label{1.27}$$

The energy spectra and wave functions can be extracted from the poles of Eq. (\[1.27\]). For convenience, we define the following variables $$\left\{ 
\begin{array}{l}
\begin{array}{rcl}
\kappa & = & \frac{1}{\hbar c}\sqrt{M^{2}c^{4}-E^{2}},
\end{array}
\\ 
\begin{array}{rcl}
\nu & = & \frac{\alpha E}{\sqrt{M^{2}c^{4}-E^{2}}},
\end{array}
\\ 
\begin{array}{rcl}
\tilde{l} & = & \sqrt{(l+D/2-1)^{2}-\alpha ^{2}}-1/2,
\end{array}
\end{array}
\right.  \label{1.28}$$ From the poles of $G_{l}(r_{b},r_{a};E)$, we find that the energy levels must satisfy the equality $$-\nu +\tilde{l}+1=-n_{r},\quad n_{r}=0,1,2,3,\cdots .  \label{1.29}$$ Expanding this equation into powers of $\alpha $, we get $$E_{nl}\approx \pm Mc^{2}\left\{ 1-\frac{1}{2}\left[ \frac{\alpha }{n+1/2(D-3)%
}\right] ^{2}-\frac{\alpha ^{4}}{\left[ n+1/2(D-3)\right] ^{3}}\right.$$ $$\left. \times \left[ \frac{1}{2\left[ l+1/2(D-2)\right] }-\frac{3}{8}\frac{1%
}{\left[ n+1/2(D-3)\right] }\right] +O(\alpha ^{6})\right\} .  \label{1.30}$$ Here $n$ is defined by $n_{r}=n-l-1$. We point out that by setting D=3, the energy levels reduce to the well-known form $$E_{nl}\approx \pm Mc^{2}\left\{ 1-\frac{1}{2}\left( \frac{\alpha }{n}\right)
^{2}-\frac{\alpha ^{4}}{n^{3}}\left[ \frac{1}{2l+1}-\frac{3}{8n}\right]
+O(\alpha ^{6})\right\} .  \label{1.31}$$

The pole positions, which satisfy $\nu =\tilde{n}_{l}\equiv n+\tilde{l}-l$ ($%
n=l+1,l+2,l+3,\cdots $), correspond to the bound states of the D-dimensional relativistic Coulomb system. Near the positive-energy poles, we use the behavior for $\nu \approx \tilde{n}_{l}$, $$-\Gamma (-\nu +\tilde{l}+1)\frac{M}{\hbar \kappa }\approx \frac{(-)^{n_{r}}}{%
\tilde{n}_{l}^{2}n_{r}!}\frac{1}{\tilde{a}_{H}}\left( \frac{E}{Mc^{2}}%
\right) ^{2}\frac{2\hbar Mc^{2}}{E^{2}-E_{nl}^{2}}  \label{1.32}$$ with $\tilde{a}_{H}\equiv a_{H}\frac{Mc^{2}}{E}$ being the modified energy-dependent Bohr radius and $n_{r}=n-l-1$ the radial quantum number, to extract the wave functions of the D-dimensional Coulomb system $$G_{l}(r_{b},r_{a};E)=-\frac{i}{(r_{b}r_{a})^{(D-1)/2}}\sum_{n=l+1}^{\infty
}\left( \frac{E}{Mc^{2}}\right) ^{2}\frac{2\hbar Mc^{2}}{E^{2}-E_{nl}^{2}}%
\qquad \qquad$$ $$\times \frac{1}{\left[ (2\tilde{l}+1)!\right] ^{2}}\frac{1}{\tilde{n}_{l}^{2}%
\tilde{a}_{H}}\frac{(\tilde{n}_{l}+\tilde{l})!}{(n-l-1)!}e^{-(r_{b}+r_{a})/%
\tilde{a}_{H}\tilde{n}_{l}}\left( \frac{2r_{b}}{\tilde{a}_{H}\tilde{n}_{l}}%
\frac{2r_{a}}{\tilde{a}_{H}\tilde{n}_{l}}\right) ^{\tilde{l}+1}$$ $$\times M(-n+l+1,2\tilde{l}+2;\frac{2r_{b}}{\tilde{a}_{H}\tilde{n}_{l}}%
)M(-n+l+1,2\tilde{l}+2;\frac{2r_{a}}{\tilde{a}_{H}\tilde{n}_{l}})\label{1.33}$$ $$=-\frac{i}{(r_{b}r_{a})^{(D-1)/2}}\sum_{n=l+1}^{\infty }(\frac{E}{Mc^{2}}%
)^{2}\frac{2\hbar Mc^{2}}{E^{2}-E_{nl}^{2}}R_{nl}(r_{b})R_{nl}^{*}(r_{a})+%
\cdots ,  \label{1.34}$$ where we have expressed the Whittaker function $M_{\lambda ,\mu }(z)$ in terms of the Kummer functions $M(a,b;z)$, $$M_{\lambda ,\mu }(z)=z^{\mu +1/2}e^{-z/2}M(\mu -\lambda +1/2,2\mu +1;z).
\label{1.35}$$ From this we obtain the radial wave functions $$R_{nl}(r)=\frac{1}{\tilde{n}_{l}\tilde{a}_{H}^{1/2}}\frac{1}{(2\tilde{l}+1)!}%
\sqrt{\frac{(\tilde{n}_{l}+\tilde{l})!}{(n-l-1)!}}\qquad \qquad \qquad
\qquad$$ $$\times \left( \frac{2r}{\tilde{a}_{H}\tilde{n}_{l}}\right) ^{^{\tilde{l}%
+1}}e^{-r/\tilde{a}_{H}\tilde{n}_{l}}M(-n+l+1,2\tilde{l}+2;\frac{2r}{\tilde{a%
}_{H}\tilde{n}_{l}}).\qquad \qquad  \label{1.36}$$ The normalized wave functions are given by $$\Psi _{nl{\bf m}}({\bf x})=\frac{1}{r^{(D-1)/2}}R_{nl}(r)Y_{l{\bf m}}({\bf 
\hat{x}}).  \label{1.37}$$

Before extracting the continuous wave function we note that the parameter $%
\kappa $ is real for $\mid E\mid <Mc^{2}$. For $\mid E\mid >Mc^{2}$, the square root in Eq. (\[1.28\]) has two imaginary solutions $$\kappa =\mp i\tilde{k},\quad \tilde{k}=\frac{1}{\hbar c}\sqrt{%
E^{2}-M^{2}c^{4}},  \label{1.38}$$ corresponding to $$\nu =\pm i\tilde{\nu},\quad \tilde{\nu}=\frac{E\alpha }{\hbar c\tilde{k}}%
.\quad \qquad \qquad  \label{1.39}$$ Therefore the amplitude has a right-handed cut for $E>Mc^{2}$ and $E<-Mc^{2}$. For simplicity, we will only consider the positive energy cut.

The continuous wave function are recovered from the discontinuity of the amplitudes $G_{l}(r_{b},r_{a};E)$ across the cut in the complex $E$ plane. Hence we have $${\rm disc}G_{l}(r_{b},r_{a};E>Mc^{2})=G_{l}(r_{b},r_{a};E+i\eta
)-G_{l}(r_{b},r_{a};E-i\eta )=-\frac{i}{(r_{b}r_{a})^{(D-1)/2}}$$ $$\times \frac{M}{\hbar \tilde{k}}\left[ \frac{\Gamma (-i\tilde{\nu}+\tilde{l}%
+1)}{(2\tilde{l}+1)!}W_{i\tilde{\nu},\tilde{l}+1/2}(-2i\tilde{k}r_{b})M_{i%
\tilde{\nu},\tilde{l}+1/2}(-2i\tilde{k}r_{a})+(\tilde{\nu}\rightarrow -%
\tilde{\nu})\right] .  \label{1.40}$$ Using the relations $$M_{\kappa ,\mu }(z)=e^{\pm i\pi (2\mu +1)/2}M_{-\kappa ,\mu }(-z),
\label{1.41}$$ where the sign is positive or negative depending on whether Im$z>0$ or Im$%
z<0 $, and $$W_{\lambda ,\mu }(z)=e^{i\pi \lambda }e^{-i\pi (\mu +1/2)}\frac{\Gamma (\mu
+\lambda +1/2)}{\Gamma (2\mu +1)}$$ $$\times \left[ M_{\lambda ,\mu }(z)-\frac{\Gamma (2\mu +1)}{\Gamma (\mu
-\lambda +1/2)}e^{-i\pi \lambda }W_{-\lambda ,\mu }(e^{-i\pi }z)\right] ,
\label{1.42}$$ which is valid only for arg $(z)\in (-\pi /2,3\pi /2)$ and $\quad 2\mu \neq
-1,-2,-3,\cdots $. The discontinuity of the amplitude is found to be $${\rm disc}G_{l}(r_{b},r_{a};E>Mc^{2})=-\frac{i}{(r_{b}r_{a})^{(D-1)/2}}\frac{%
M}{\hbar \tilde{k}}\frac{\mid \Gamma (-i\tilde{\nu}+\tilde{l}+1)\mid ^{2}}{%
\mid \Gamma (2\tilde{l}+2)\mid ^{2}}$$ $$\times e^{\pi \tilde{\nu}}M_{-i\tilde{\nu},\tilde{l}+1/2}(2i\tilde{k}%
r_{b})M_{i\tilde{\nu},\tilde{l}+1/2}(-2i\tilde{k}r_{a}).  \label{1.43}$$ Thus we have $$\int_{Mc^{2}}^{\infty }\frac{dE}{2\pi \hbar }{\rm disc}%
G_{l}(r_{b},r_{a};E>Mc^{2})\qquad \qquad \qquad \qquad \qquad$$ $$=\frac{1}{2\pi \hbar }\int_{-\infty }^{\infty }\frac{(\hbar c)^{2}\tilde{k}d%
\tilde{k}}{\sqrt{M^{2}c^{4}+(\hbar c\tilde{k})^{2}}}{\rm disc}%
G_{l}(r_{b},r_{a};E>Mc^{2})$$ $$=-\frac{i}{(r_{b}r_{a})^{(D-1)/2}}\int_{-\infty }^{\infty }d\tilde{k}(\frac{E%
}{Mc^{2}})R_{\tilde{k}l}(r_{b})R_{\tilde{k}l}^{*}(r_{a}).\qquad \qquad
\label{1.44}$$ From this, we obtain the continuous radial wave function of the D-dimensional relativistic Coulomb system $$R_{\tilde{k}l}(r)=\sqrt{\frac{1}{2\pi }}\frac{1}{\left[ 1+(\frac{c\hbar 
\tilde{k}}{Mc^{2}})^{2}\right] ^{1/2}}\frac{\mid \Gamma (-i\tilde{\nu}+%
\tilde{l}+1)\mid }{(2\tilde{l}+1)!}e^{\pi \tilde{\nu}/2}M_{i\tilde{\nu},%
\tilde{l}+1/2}(-2i\tilde{k}r)$$ $$=\sqrt{\frac{1}{2\pi }}\frac{1}{\left[ 1+(\frac{c\hbar \tilde{k}}{Mc^{2}}%
)^{2}\right] ^{1/2}}\frac{\mid \Gamma (-i\tilde{\nu}+\tilde{l}+1)\mid }{(2%
\tilde{l}+1)!}\qquad \qquad \qquad \qquad$$ $$\times e^{\pi \tilde{\nu}/2}e^{i\tilde{k}r}(-2i\tilde{k}r)^{\tilde{l}%
+1}\times M(-i\tilde{\nu}+\tilde{l}+1,2\tilde{l}+2;-2i\tilde{k}r).\qquad
\qquad \qquad  \label{1.45}$$ It is easy to check the result is in accordance with the non-relativistic wave function when we take the non-relativistic limit.

Concluding remarks
==================

       In this paper we have calculated the Green’s function of the relativistic Coulomb system via sum over perturbation series. From the resulting amplitude, the energy levels and wave functions are given. Different from the conventional treatment in path integral using the space-time and Kustaanheimo-Stiefel transformation techniques (e.g. [@9; @11]), the method presented here just involves the computation of the expectation value of the moments $Q^{n}$ $(Q=\int_{\lambda _{a}}^{\lambda
_{b}}d\lambda \rho {\bf (}\lambda )V({\bf x(}\lambda )))$ over the measure $$K_{0}({\bf {x}}_{b},{\bf {x}}_{a};\lambda _{b}-\lambda _{a})=\int D^{D}xe^{-%
\frac{1}{\hbar }\int_{\lambda _{a}}^{\lambda _{b}}d\lambda \left[ \frac{M}{%
2\rho \left( \lambda \right) }{\bf {x}}^{\prime ^{2}}\left( \lambda \right)
-\rho \left( \lambda \right) \frac{V({\bf x})^{2}}{2Mc^{2}}\right] }$$ and summing them in accordance with the Feynman-Kac formula [@14] $$G({\bf {x}}_{b},{\bf {x}}_{a};E)=\frac{i\hbar }{2Mc}\int_{0}^{\infty }dS\int
D\rho \Phi \left[ \rho \right] e^{-\frac{1}{\hbar }\int_{\lambda
_{a}}^{\lambda _{b}}d\lambda \rho (\lambda ){\cal E}}$$ $$\times E\left[ \exp \left\{ -\frac{1}{\hbar }\int_{\lambda _{a}}^{\lambda
_{b}}d\lambda \rho {\bf (}\lambda )V({\bf x(}\lambda ))\right\} \right]$$ $$=\frac{i\hbar }{2Mc}\int_{0}^{\infty }dS\int D\rho \Phi \left[ \rho \right]
e^{-\frac{1}{\hbar }\int_{\lambda _{a}}^{\lambda _{b}}d\lambda \rho (\lambda
){\cal E}}$$ $$\times \sum_{n=1}^{\infty }\frac{(-\beta /\hbar )}{n!}^{n}E\left[ \left(
\int_{\lambda _{a}}^{\lambda _{b}}d\lambda \rho {\bf (}\lambda )V({\bf x(}%
\lambda ))\right) ^{n}\right] ,  \label{1.46}$$ where the notation $E\left[ \star \right] $ stands for the expectation value of the moment $\star $.

We hope that the procedure presented in this article may help us to obtain the results of other interesting relativistic systems.

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