Effect of the Screening Parameter on Shannon Entropy and Thermal Properties for Exponential Kratzer–Feus Potential

Abstract

In this study, we propose the exponential Kratzer–Feus potential and study the effect of the screening parameter on the diatomic molecules of CH, $H_2$, NO, HCL, and LiH. We first solve the Schrödinger equation using the Nikiforov–Uvarov functional analysis method to obtain the energy eigenvalue. Interestingly, the proposed exponential Kratzer–Feus potential exhibits a repulsive interaction for diatomic molecules. We also compute the energy spectra for diatomic molecules for different values of the screening parameter $\alpha = 0$, 0.2, 0.4, 0.8, and 1.0. For a more complete study, we analyse the thermodynamic properties of the model. Furthermore, the quantum information measurements are calculated and used to study particle locations.

Appendix

Ikot et al. [42] proposed a simple and elegant method for solving a second-order differential equation of the hypergeometric type called the Nikiforov–Uvarov functional analysis method (NUFA) method. This method is easy and simple, just like the parametric NU method. As well known, the NU is used to solve a second-order differential equation of the form

$$\begin{aligned} \psi ''_n(s) + \frac{\tilde{\tau }(s)}{\sigma (s)}\psi '_n(s)+\frac{\tilde{\sigma }(s)}{\sigma ^2(s)}\psi _n(s) = 0, \end{aligned}$$

(24)

where $\sigma (s)$ and $\tilde{\sigma }(s)$ are polynomial, at most of second degree, and $\tilde{\tau }(s)$ is first-degree polynomial. The parametric form of the NU method is the form

$$\begin{aligned} \psi ''_n(s) + \frac{\alpha _1-\alpha _2s}{s(1-\alpha _3s)}\psi '+\frac{1}{s^2(1-\alpha _3s)^2} [-\xi _1s^2+\xi _2s-\xi _3] \psi _n(s) = 0, \end{aligned}$$

(25)

where $\alpha _i$ and $_i(i=1,2,3)$ are all parameters. It can be observed in equation (25) two singularities at $s\rightarrow 0$ and $s\rightarrow 1$, and the wave function takes the form,

$$\begin{aligned} \psi (s) = s^{\lambda }(1-s)^v f(s). \end{aligned}$$

(26)

Substituting equation (26) into (25), we have

$$\begin{aligned} s(1-\alpha _1 s)f''(s) +[\alpha _1 + 2\lambda (2\lambda \alpha _3+2v\alpha _3+\alpha _2)s]f'(s){} & {} \nonumber \\ - \alpha _3 \left( \lambda +v+\frac{\alpha _2}{\alpha _3}-1 + \sqrt{\left( \frac{\alpha _2}{\alpha _3}-1\right) ^2+\frac{\xi _1}{\alpha _3}}\right){} & {} \nonumber \\ \times \left( \lambda +v+\frac{\alpha _2}{\alpha _3^2}-1 + \sqrt{\left( \frac{\alpha _2}{\alpha _3}-1\right) ^2+\frac{\xi _1}{\alpha _3^2}}\right){} & {} \nonumber \\ +\Bigg [\frac{\alpha _2 v - \alpha _1\alpha _3v +v(v+1)\alpha _3 -\frac{\xi _1}{\alpha _3}+\xi _2-\xi _3\alpha _3}{1-\alpha _3s}{} & {} \nonumber \\ + \frac{\lambda (\lambda -1)+\alpha _1\lambda -\xi _3}{s}\Bigg ] f(s)= & {} 0. \end{aligned}$$

(27)

Equation (27) can be reduced to a Gauss hypergeometric equation with the following condition,

$$\begin{aligned} \lambda (\lambda -1)+\alpha _1\lambda -\xi _3 = 0 \end{aligned}$$

(28)

$$\begin{aligned} \alpha _2v - \alpha _1\alpha _2v + v(v-1)\alpha - \frac{\xi _1}{\alpha _3} + \xi _2 - \xi _3\alpha = 0. \end{aligned}$$

(29)

Equation (27) becomes

$$\begin{aligned} s(1-\alpha _1 s)f''(s) +[\alpha _1 + 2\lambda (2\lambda \alpha _3+2v\alpha _3+\alpha _2)s]f'(s){} & {} \nonumber \\ - \alpha _3 \left( \lambda +v+\frac{\alpha _2}{\alpha _3}-1 + \sqrt{\left( \frac{\alpha _2}{\alpha _3}-1\right) ^2+\frac{\xi _1}{\alpha _3}}\right){} & {} \nonumber \\ \quad \times ~ \left( \lambda +v+\frac{\alpha _2}{\alpha _3^2}-1 + \sqrt{\left( \frac{\alpha _2}{\alpha _3}-1\right) ^2+\frac{\xi _1}{\alpha _3^2}}\right) f(s)= & {} 0. \end{aligned}$$

(30)

Solving equations (28) and (29) completely gives

$$\begin{aligned} \lambda= & {} \frac{(1-\alpha _1)\pm \sqrt{(1-\alpha _1)^2+4\xi }}{2} \nonumber \\ v= & {} \frac{\alpha _3(1+\alpha _1)-\alpha _2 \pm \sqrt{(\alpha _3(1+\alpha _1)\alpha _2)^2 +4\left( \frac{\xi _1}{\alpha _3}+\alpha _3\xi _3-\xi _2\right) }}{2}. \end{aligned}$$

(31)

The hypergeometric equation type takes the form,

$$\begin{aligned} x(1-x)f''(x)+[c+(a+b+1)x]f'(x)-abf(x) = 0. \end{aligned}$$

(32)

The energy equation and its wave equation are obtained as follows:

$$\begin{aligned} \lambda ^2 + 2\lambda \left[ v+\frac{\alpha _2}{\alpha _3} -1+\frac{n}{\sqrt{\alpha _3}}\right] + \left[ v+\frac{\alpha _2}{\alpha _3} -1+\frac{n}{\sqrt{\alpha _3}}\right] ^2 - \left[ \frac{\alpha _2}{\alpha _3} -1\right] - \frac{\xi ^3}{\alpha _3^2} = 0, \end{aligned}$$

(33)

and

$$\begin{aligned} \psi (s) = Ns^{\lambda }(1-\alpha _3s)^{v} {}_2F_1(a,b,c;s), \end{aligned}$$

(34)

where a, b, c are given by

$$\begin{aligned} a= & {} \sqrt{\alpha _3} \left[ \lambda +v+\frac{\alpha _2}{\alpha _3}-1+\sqrt{\left( \frac{\alpha _2}{\alpha _3}-1\right) ^2 +\frac{\xi _1}{\alpha _3}}\right] , \end{aligned}$$

(35)

$$\begin{aligned} b= & {} \sqrt{\alpha _3}\left[ \lambda +v+\frac{\alpha _2}{\alpha _3}-1-\sqrt{\left( \frac{\alpha _2}{\alpha _3}-1\right) ^2 +\frac{\xi _1}{\alpha _3}}\right] , \end{aligned}$$

(36)

$$\begin{aligned} c= & {} \alpha _1 + 2\lambda . \end{aligned}$$

(37)