无限深方势阱中分数阶量子力学核的变换Transformation of Fractional Quantum Mechanical Kernel in the Infinite Square Well

DOI: 10.12677/MP.2020.105009, PDF, HTML, XML, 下载: 112  浏览: 292  国家自然科学基金支持

Abstract: Fractional quantum mechanics kernel is a kind of wave function, which can describe the evolution process of fractional quantum system. In this paper, we study the Laplace transformation, ener-gy-time transformation and momentum representation of free particle quantum mechanics kernel in infinite square well. Firstly, we obtain the Laplace transformation of quantum mechanical kernel by using the Fox’s H function representation of free particle quantum mechanics kernel in the infinite square well, and then use Path integral form of quantum mechanics kernel to calculate its energy-time transformation and momentum representation. The transformation of quantum me-chanics kernel can simplify the calculation results in fractional quantum mechanics, so as to study its properties better.

1. 引言

2. 量子力学核的Laplace变换

$V\left(x\right)=\left\{\begin{array}{l}\text{0},\text{}|x|\le l,\\ \infty ,\text{}|x|>l,\end{array}$ (1)

${K}_{L}\left({x}_{b}{t}_{b}|{x}_{a}{t}_{a}\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left[{K}_{L}^{\left(0\right)}\left({x}_{2m}{t}_{b}|{x}_{a}{t}_{a}\right)-{K}_{L}^{\left(0\right)}\left({x}_{2m+1}{t}_{b}|{x}_{a}{t}_{a}\right)\right]\text{ }\text{ },$ (2)

${x}_{r}=\left\{\begin{array}{l}2lr+{x}_{b},\text{}r=2m,\\ 2lr-{x}_{b},\text{}r=2m+1,\end{array}$ (3)

${K}_{L}^{\left(0\right)}\left({x}_{2m}{t}_{b}|{x}_{a}{t}_{a}\right)$ 表示起点在 $\left({x}_{a},{t}_{a}\right)$，终点在 $\left({x}_{\text{2}m},{t}_{b}\right)$ 的自由粒子的量子力学核， ${K}_{L}^{\left(0\right)}\left({x}_{2m+1}{t}_{b}|{x}_{a}{t}_{a}\right)$ 表示起点在 $\left({x}_{a},{t}_{a}\right)$，终点在 $\left({x}_{\text{2}m+\text{1}},{t}_{b}\right)$ 的自由粒子的量子力学核。

${K}_{L}^{\left(0\right)}\left({x}_{b}{t}_{b}|{x}_{a}{t}_{a}\right)=\frac{1}{\alpha |{x}_{b}-{x}_{a}|}{H}_{2,2}^{1,1}\left[\frac{1}{\hslash }{\left(\frac{\hslash }{i{D}_{\alpha }\left({t}_{b}-{t}_{a}\right)}\right)}^{1/\alpha }|{x}_{b}-{x}_{a}||{}_{\left(1,1\right),\left(1,1/2\right)}^{\left(1,1/\alpha \right),\left(1,1/2\right)}\right],$ (4)

$\begin{array}{l}{K}_{L}\left({x}_{b}{t}_{b}|{x}_{a}{t}_{a}\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{1}{\alpha |{x}_{2m}-{x}_{a}|}{H}_{2,2}^{1,1}\left[\frac{1}{\hslash }{\left(\frac{\hslash }{i{D}_{\alpha }\left({t}_{b}-{t}_{a}\right)}\right)}^{1/\alpha }|{x}_{2m}-{x}_{a}||{}_{\left(1,1\right),\left(1,1/2\right)}^{\left(1,1/\alpha \right),\left(1,1/2\right)}\right]\\ \text{}-\frac{1}{\alpha |{x}_{2m+\text{1}}-{x}_{a}|}{H}_{2,2}^{1,1}\left[\frac{1}{\hslash }{\left(\frac{\hslash }{i{D}_{\alpha }\left({t}_{b}-{t}_{a}\right)}\right)}^{1/\alpha }|{x}_{2m+\text{1}}-{x}_{a}||{}_{\left(1,1\right),\left(1,1/2\right)}^{\left(1,1/\alpha \right),\left(1,1/2\right)}\right]\right\}.\end{array}$ (5)

${\stackrel{˜}{K}}_{L}\left({x}_{b},{x}_{a};s\right)={\int }_{0}^{\infty }\text{d}\tau \text{ }\text{ }{\text{e}}^{-s\tau }{K}_{L}\left({x}_{b},{x}_{a};\tau \right),$ (6)

$\begin{array}{l}{\stackrel{˜}{K}}_{L}\left({x}_{b},{x}_{a};s\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{1}{\alpha |{x}_{2m}-{x}_{a}|}\underset{k=0}{\overset{\infty }{\sum }}\frac{\Gamma \left(\frac{1+k}{\alpha }\right)}{\Gamma \left(\frac{1+k}{2}\right)\Gamma \left(\frac{1-k}{2}\right)}\frac{{\left(-1\right)}^{k}}{k!}{\left(\frac{1}{\hslash }{\left(\frac{\hslash }{i{D}_{\alpha }}\right)}^{1/\alpha }|{x}_{2m}-{x}_{a}|\right)}^{1+k}\\ \text{}-\frac{1}{\alpha |{x}_{2m+\text{1}}-{x}_{a}|}\underset{k=0}{\overset{\infty }{\sum }}\frac{\Gamma \left(\frac{1+k}{\alpha }\right)}{\Gamma \left(\frac{1+k}{2}\right)\Gamma \left(\frac{1-k}{2}\right)}\frac{{\left(-1\right)}^{k}}{k!}{\left(\frac{1}{\hslash }{\left(\frac{\hslash }{i{D}_{\alpha }}\right)}^{1/\alpha }|{x}_{2m+\text{1}}-{x}_{a}|\right)}^{1+k}\right\}\text{}×{\int }_{\text{0}}^{\infty }\text{d}\tau \text{ }\text{ }{\text{e}}^{-s\tau }{\tau }^{-\frac{1+k}{\alpha }}\\ \text{}=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{1}{\alpha s|{x}_{2m}-{x}_{a}|}\underset{k=0}{\overset{\infty }{\sum }}\frac{\Gamma \left(\frac{1+k}{\alpha }\right)\Gamma \left(1-\frac{1+k}{\alpha }\right)}{\Gamma \left(\frac{1+k}{2}\right)\Gamma \left(\frac{1-k}{2}\right)}\frac{{\left(-1\right)}^{k}}{k!}{\left(\frac{1}{\hslash }{\left(\frac{\hslash s}{i{D}_{\alpha }}\right)}^{1/\alpha }|{x}_{2m}-{x}_{a}|\right)}^{1+k}\\ \text{}-\frac{1}{\alpha |{x}_{2m+\text{1}}-{x}_{a}|}\underset{k=0}{\overset{\infty }{\sum }}\frac{\Gamma \left(\frac{1+k}{\alpha }\right)\Gamma \left(1-\frac{1+k}{\alpha }\right)}{\Gamma \left(\frac{1+k}{2}\right)\Gamma \left(\frac{1-k}{2}\right)}\frac{{\left(-1\right)}^{k}}{k!}{\left(\frac{1}{\hslash }{\left(\frac{\hslash s}{i{D}_{\alpha }}\right)}^{1/\alpha }|{x}_{2m+\text{1}}-{x}_{a}|\right)}^{1+k}\right\},\end{array}$ (7)

${\int }_{\text{0}}^{\infty }\text{d}\tau \text{ }\text{ }{\text{e}}^{-s\tau }{\tau }^{-\frac{1+k}{\alpha }}={s}^{\frac{1+k}{\alpha }-1}\Gamma \left(1-\frac{1+k}{\alpha }\right).$ (8)

$\begin{array}{l}{\stackrel{˜}{K}}_{L}\left({x}_{b},{x}_{a};s\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{1}{\alpha s|{x}_{2m}-{x}_{a}|}{H}_{3,2}^{1,2}\left[\frac{1}{\hslash }{\left(\frac{\hslash s}{i{D}_{\alpha }}\right)}^{1/\alpha }|{x}_{2m}-{x}_{a}||{}_{\left(1,1\right),\left(1,1/2\right)}^{\left(1,1/\alpha \right),\left(0,-1/\alpha \right),\left(1,1/2\right)}\right]\\ \text{}-\frac{1}{\alpha s|{x}_{2m+\text{1}}-{x}_{a}|}{H}_{3,2}^{1,2}\left[\frac{1}{\hslash }{\left(\frac{\hslash s}{i{D}_{\alpha }}\right)}^{1/\alpha }|{x}_{2m+\text{1}}-{x}_{a}||{}_{\left(1,1\right),\left(1,1/2\right)}^{\left(1,1/\alpha \right),\left(0,-1/\alpha \right),\left(1,1/2\right)}\right]\right\}.\end{array}$ (9)

$\alpha =\text{2}$ 时，利用H函数的级数展开，我们得到了标准量子力学中无限深方势阱自由粒子核的Laplace变换

${\stackrel{˜}{K}}_{L}\left({x}_{b},{x}_{a};s\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\sqrt{\frac{M}{2si\hslash }}\mathrm{exp}\left(-\sqrt{\frac{2Ms}{i\hslash }}|{x}_{2m}-{x}_{a}|\right)-\sqrt{\frac{M}{2si\hslash }}\mathrm{exp}\left(-\sqrt{\frac{2Ms}{i\hslash }}|{x}_{2m+1}-{x}_{a}|\right)\right\},$ (10)

3. 量子力学核的能量–时间变换

${K}_{L}\left({x}_{b},{x}_{a};E\right)={\int }_{{t}_{a}}^{\infty }\text{d}{t}_{b}\text{ }{\text{e}}^{iE\left({t}_{b}-{t}_{a}\right)/\hslash }{K}_{L}\left({x}_{b}{t}_{b}|{x}_{a}{t}_{a}\right),$ (11)

${K}_{L}\left({x}_{b}{t}_{b}|{x}_{a}{t}_{a}\right)=\frac{1}{2\pi \hslash }{\int }_{-\infty }^{\infty }\text{d}E\text{ }\text{ }{\text{e}}^{-iE\left({t}_{b}-{t}_{a}\right)/\hslash }{K}_{L}\left({x}_{b},{x}_{a};E\right).$ (12)

${K}_{L}\left({x}_{b},{x}_{a};E\right)=\underset{n}{\sum }{\varphi }_{n}\left({x}_{b}\right){\varphi }_{n}^{*}\left({x}_{a}\right)\frac{i\hslash }{E-{E}_{n}+i\eta },$ (13)

${K}_{L}^{\left(\text{0}\right)}\left({x}_{b}{t}_{b}|{x}_{a}{t}_{a}\right)=\frac{1}{2\pi \hslash }{\int }_{-\infty }^{\infty }\text{d}p\mathrm{exp}\left\{i\frac{p\left({x}_{b}-{x}_{a}\right)}{\hslash }-i\frac{{D}_{\alpha }\left({t}_{b}-{t}_{a}\right){|p|}^{\alpha }}{\hslash }\right\},$ (14)

$\begin{array}{l}{K}_{L}\left({x}_{b}{t}_{b}|{x}_{a}{t}_{a}\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{1}{2\pi \hslash }{\int }_{-\infty }^{\infty }\text{d}p\mathrm{exp}\left\{i\frac{p\left({x}_{2m}-{x}_{a}\right)}{\hslash }-i\frac{{D}_{\alpha }\left({t}_{b}-{t}_{a}\right){|p|}^{\alpha }}{\hslash }\right\}\\ \text{}-\frac{1}{2\pi \hslash }{\int }_{-\infty }^{\infty }\text{d}p\mathrm{exp}\left\{i\frac{p\left({x}_{2m+\text{1}}-{x}_{a}\right)}{\hslash }-i\frac{{D}_{\alpha }\left({t}_{b}-{t}_{a}\right){|p|}^{\alpha }}{\hslash }\right\}\right\}.\end{array}$ (15)

$\begin{array}{l}{K}_{L}\left({x}_{b},{x}_{a};E\right)={\int }_{0}^{\infty }\text{d}\tau \text{ }{\text{e}}^{iE\tau /\hslash }\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{1}{2\pi \hslash }{\int }_{-\infty }^{\infty }\text{d}p\mathrm{exp}\left[i\frac{p\left({x}_{2m}-{x}_{a}\right)}{\hslash }-i\frac{{D}_{\alpha }\tau {|p|}^{\alpha }}{\hslash }\right]\\ \text{}-\frac{1}{2\pi \hslash }{\int }_{-\infty }^{\infty }\text{d}p\mathrm{exp}\left[i\frac{p\left({x}_{2m+1}-{x}_{a}\right)}{\hslash }-i\frac{{D}_{\alpha }\tau {|p|}^{\alpha }}{\hslash }\right]\right\},\end{array}$ (16)

$\begin{array}{l}{K}_{L}\left({x}_{b},{x}_{a};E\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{1}{2\pi \hslash }{\int }_{0}^{\infty }\frac{i\hslash }{E-{D}_{\alpha }{|p|}^{\alpha }}\left[\mathrm{exp}\left(i\frac{p\left({x}_{2m}-{x}_{a}\right)}{\hslash }\right)+\mathrm{exp}\left(-i\frac{p\left({x}_{2m}-{x}_{a}\right)}{\hslash }\right)\right]\text{d}p\\ \text{}-\frac{1}{2\pi \hslash }{\int }_{0}^{\infty }\frac{i\hslash }{E-{D}_{\alpha }{|p|}^{\alpha }}\left[\mathrm{exp}\left(i\frac{p\left({x}_{2m+1}-{x}_{a}\right)}{\hslash }\right)+\mathrm{exp}\left(-i\frac{p\left({x}_{2m+1}-{x}_{a}\right)}{\hslash }\right)\right]\text{d}p\right\}.\end{array}$ (17)

$\frac{{z}^{\beta }}{1+a{z}^{\alpha }}={a}^{-\beta /\alpha }{H}_{1,1}^{1,1}\left[{a{z}^{\alpha }|}_{\left(\beta /\alpha ,1\right)}^{\left(\beta /\alpha ,1\right)}\right],$ (18)

$\begin{array}{l}{K}_{L}\left({x}_{b},{x}_{a};E\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{1}{\pi \hslash }{\int }_{0}^{\infty }\frac{i\hslash }{E-{D}_{\alpha }{|p|}^{\alpha }}\mathrm{cos}\left[\frac{p\left({x}_{2m}-{x}_{a}\right)}{\hslash }\right]\text{d}p\\ \text{}-\frac{1}{\pi \hslash }{\int }_{0}^{\infty }\frac{i\hslash }{E-{D}_{\alpha }{|p|}^{\alpha }}\mathrm{cos}\left[\frac{p\left({x}_{2m+1}-{x}_{a}\right)}{\hslash }\right]\text{d}p\right\}\\ \text{}=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{i}{\pi E}{\int }_{0}^{\infty }{H}_{1,\text{1}}^{1,1}\left[{\left(-\frac{{D}_{\alpha }}{E}\right){p}^{\alpha }|}_{\left(0,1\right)}^{\left(0,1\right)}\right]\mathrm{cos}\left[\frac{p\left({x}_{2m}-{x}_{a}\right)}{\hslash }\right]\text{d}p\\ \text{}-\frac{i}{\pi E}{\int }_{0}^{\infty }{H}_{1,\text{1}}^{1,1}\left[{\left(-\frac{{D}_{\alpha }}{E}\right){p}^{\alpha }|}_{\left(0,1\right)}^{\left(0,1\right)}\right]\mathrm{cos}\left[\frac{p\left({x}_{2m+1}-{x}_{a}\right)}{\hslash }\right]\text{d}p\right\},\end{array}$ (19)

$\begin{array}{l}{K}_{L}\left({x}_{b},{x}_{a};E\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{\frac{i\hslash }{\alpha E|{x}_{2m}-{x}_{a}|}{H}_{2,3}^{2,1}\left[{\left(-\frac{E}{{D}_{\alpha }{\hslash }^{2}}\right)}^{1/\alpha }|{x}_{2m}-{x}_{a}||{}_{\left(1,1\right),\left(1,1/\alpha \right),\left(1,1/2\right)}^{\left(1,1/\alpha \right),\left(1,1/2\right)}\right]\\ \text{}-\frac{i\hslash }{\alpha E|{x}_{2m+1}-{x}_{a}|}{H}_{2,3}^{2,1}\left[{\left(-\frac{E}{{D}_{\alpha }{\hslash }^{2}}\right)}^{1/\alpha }|{x}_{2m+1}-{x}_{a}||{}_{\left(1,1\right),\left(1,1/\alpha \right),\left(1,1/2\right)}^{\left(1,1/\alpha \right),\left(1,1/2\right)}\right]\right\}.\end{array}$ (20)

$\alpha =\text{2}$ 时，根据H函数的性质，并且运用公式

${H}_{0,\text{1}}^{1,0}\left[{z|}_{\left(b,\beta \right)}^{-}\right]\text{}=\frac{1}{\beta }{z}^{b/\beta }\mathrm{exp}\left\{-{z}^{1/\beta }\right\},$ (21)

$\begin{array}{l}{K}_{L}\left({x}_{b},{x}_{a};E\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\left\{-i\frac{M}{\hslash }{\left(-\frac{E}{{D}_{\alpha }{\hslash }^{2}}\right)}^{-1/\alpha }\mathrm{exp}\left[-{\left(-\frac{E}{{D}_{\alpha }{\hslash }^{2}}\right)}^{1/\alpha }|{x}_{2m}-{x}_{a}|\right]\\ \text{}+i\frac{M}{\hslash }{\left(-\frac{E}{{D}_{\alpha }{\hslash }^{2}}\right)}^{-1/\alpha }\mathrm{exp}\left[-{\left(-\frac{E}{{D}_{\alpha }{\hslash }^{2}}\right)}^{1/\alpha }|{x}_{2m+1}-{x}_{a}|\right]\right\}.\end{array}$ (22)

4. 量子力学核的动量表示

${K}_{L}\left({p}_{b}{t}_{b}|{p}_{a}{t}_{a}\right)=\int \text{d}{x}_{b}\text{d}{x}_{a}{\text{e}}^{-i{p}_{b}{x}_{b}/\hslash +i{p}_{a}{x}_{a}/\hslash }{K}_{L}\left({x}_{b}{t}_{b}|{x}_{a}{t}_{a}\right),$ (23)

$\begin{array}{l}{K}_{L}\left({p}_{b}{t}_{b}|{p}_{a}{t}_{a}\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}\frac{1}{2\pi \hslash }\left\{\int \text{d}{x}_{\text{2}m}\text{d}{x}_{a}{\text{e}}^{-i{p}_{b}{x}_{\text{2}m}/\hslash +i{p}_{a}{x}_{a}/\hslash }{\int }_{-\infty }^{\infty }\text{d}p\mathrm{exp}\left\{i\frac{p\left({x}_{2m}-{x}_{a}\right)}{\hslash }-i\frac{{D}_{\alpha }\left({t}_{b}-{t}_{a}\right){|p|}^{\alpha }}{\hslash }\right\}\\ \text{}-\int \text{d}{x}_{\text{2}m+1}\text{d}{x}_{a}{\text{e}}^{-i{p}_{b}{x}_{\text{2}m+1}/\hslash +i{p}_{a}{x}_{a}/\hslash }{\int }_{-\infty }^{\infty }\text{d}p\mathrm{exp}\left\{i\frac{p\left({x}_{2m+\text{1}}-{x}_{a}\right)}{\hslash }-i\frac{{D}_{\alpha }\left({t}_{b}-{t}_{a}\right){|p|}^{\alpha }}{\hslash }\right\}\right\}\\ \text{}=\underset{m=-\infty }{\overset{\infty }{\sum }}2\pi \hslash \mathrm{exp}\left\{-i\frac{{D}_{\alpha }\left({t}_{b}-{t}_{a}\right){|{p}_{a}|}^{\alpha }}{\hslash }\right\}\left[\delta \left({p}_{a}-{p}_{2m}\right)-\delta \left({p}_{a}-{p}_{2m+1}\right)\right]\text{ }\text{ }.\end{array}$ (24)

$\alpha =\text{2}$ 时，等式(24)可以转化成

${K}_{L}\left({p}_{b}{t}_{b}|{p}_{a}{t}_{a}\right)=\underset{m=-\infty }{\overset{\infty }{\sum }}2\pi \hslash \mathrm{exp}\left\{-i\frac{\left({t}_{b}-{t}_{a}\right){|{p}_{a}|}^{2}}{\text{2}M\hslash }\right\}\left[\delta \left({p}_{a}-{p}_{2m}\right)-\delta \left({p}_{a}-{p}_{2m+1}\right)\right]\text{ }\text{ }.$ (25)

5. 结论

NOTES

*通讯作者。

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