Energy levels of the improved Tietz oscillator in external magnetic and Aharonov-Bohm flux fields: the Pekeris approximation recipe

Abstract

In this work, the radial Schrödinger equation is solved with the improved Tietz oscillator in the presence of an external magnetic and Aharonov-Bohm (AB) flux fields. By employing the proper quantization rule, analytical equation for bound state energy levels was derived within the framework of Pekeris-type approximation scheme. The expression for the energy levels was used to generate numerical data for some diatomic substances including HF (X ¹Σ⁺), HCl (X ¹Σ⁺), HI (X ¹Σ⁺), CO (X ¹Σ⁺), MgH (X ²Σ⁺), ICl (A ³Π₁), K₂ (a ³Σ_u⁺), ⁷Li₂ (a ³Σ_u⁺), BrF (X ¹Σ⁺) and BCl (X ¹Σ⁺) molecules. In the absence of the external fields, the mean absolute deviation of the energy levels from experimental data of the molecules are 3.3494%, 2.9210%, 2.8613%, 0.3985%, 4.0886%, 0,9203%, 1.7691%, 0.4850%, 1.0628%, and 0.9010%. The study further reveal that in the absence of the external fields, the obtained energy levels are degenerate. However, if the fields are maintained at about 10 μT, the resulting energy of the molecules is nondegenerate. The results obtained are in good agreement with available literature on diatomic systems.

Graphical abstract

Appendices

Appendix A

Pekeris approximation models for the functions F/r and 1/r ²

In this section, approximate expressions are derived for the functions F/r and 1/r² in terms of component functions of the ITO. Firstly, we note that the ITO can be expressed as ${\text{U}}\left( r \right) = D_{{\text{e}}} - 2D_{{\text{e}}} b{\text{F}}\left( r \right) + D_{{\text{e}}} b^{2} {\text{G}}\left( r \right),$ where ${\text{F}}\left( r \right) = \left( {{\text{e}}^{\alpha r} + q} \right)^{ - 1}$ and G (r) = F² (r). The main target here is to express F/r, and 1/r² in similar identical functional form with the potential U(r). The functions F(r) and G(r) can be expanded in Taylor series about the point r ≈ r_e, or equivalently, about x = 0, where x = r/r_e – 1, leading to

$$\text{F}\left( x \right) = \frac{Q}{{\text{e}^{{ax}} + qQ}} \approx \text{F}_{0} + \text{F}^{\prime } _{0} x + \tfrac{1}{2}\text{F}^{{\prime \prime }} _{0} x^{2} + Ox^{3} ,$$

(38)

$$\text{G}\left( x \right) = \frac{{Q^{2} }}{{\left( {\text{e}^{{ax}} + qQ} \right)^{2} }} \approx \text{G}_{0} + \text{G}^{\prime } _{0} x + \tfrac{1}{2}\text{G}^{{\prime \prime }} _{0} x^{2} + Ox^{3} ,$$

(39)

where a = αr_e, Q = e^−a, prime denotes derivative with respect to x, $\text{F}_{0} = \text{F}\left( 0 \right),\;\text{F}^{\prime}_{0} = \text{F}^{\prime}\left( 0 \right),\;\text{F}^{\prime\prime}_{0} = \text{F}^{\prime\prime}\left( 0 \right)$. Similarly $\text{G}_{0} = \text{G}\left( 0 \right),\;\text{G}^{\prime}_{0} = \text{G}^{\prime}\left( 0 \right),\;\text{G}^{\prime\prime}_{0} = \text{G}^{\prime\prime}\left( 0 \right)$. Using the above equations, we obtained

$$\text{F}_{0} = \frac{Q}{{1 + qQ}}\;\text{F}^{\prime } _{0} = - \frac{{a\text{F}_{0} }}{{1 + qQ}}\;\text{F}^{{\prime \prime }} _{0} = \frac{{a^{2} \left( {1 - qQ} \right)\text{F}_{0} }}{{\left( {1 + qQ} \right)^{2} }}$$

(40)

$$\text{G}_{0} = \frac{{Q^{2} }}{{\left( {1 + qQ} \right)^{2} }}\;\text{G}^{\prime } _{0} = - \frac{{2aG_{0} }}{{1 + qQ}}\;\text{G}^{{\prime \prime }} _{0} = \frac{{2a^{2} \left( {2 - qQ} \right)\text{G}_{0} }}{{\left( {1 + qQ} \right)^{2} }}$$

(41)

Next, we consider an arbitrary function H(x) with the property that

$${\text{H}}\left( x \right) \, = {\text{ X }} + {\text{ YF}}\left( x \right) \, + {\text{ ZG}}\left( x \right),$$

(42)

where (X; Y; Z) ≡ (a₀, b₀, c₀, …; a₁, b₁, c₁, …; a₂, b₂, c₂, …). The function H(x) expanded in Taylor series about x = 0 gives

$$\text{H}\left( x \right) \approx \text{H}_{0} + \text{H}^{\prime } _{0} x + \tfrac{1}{2}\text{H}^{{\prime \prime }} _{0} x^{2} + Ox^{3}$$

(43)

By inserting equations (38), (39) and (43) into (A5) and equating corresponding coefficients of X, Y, and Z, one obtains the following set of relations

$$\text{X} + \text{F}_{0} \text{Y} + \text{G}_{0} \text{Z} = \text{H}_{0} ,$$

(44)

$$\text F^{\prime } _{0} \text Y + \text G^{\prime } _{0} \text Z = \text H^{\prime } _{0} ,$$

(45)

$$\text{F}^{{\prime \prime }} _{0} \text{Y} + \text{G}^{{\prime \prime }} _{0} \text{Z} = \text{H}^{{\prime \prime }} _{0} .$$

(46)

Solving equations (44)–(A9) yields

$$\text{X} = \text{H}_{0} + \frac{{\text{H}^{\prime } _{0} }}{{2a}}\left( {3 - q} \right)\left( {1 + qQ} \right) + \frac{{\text{H}^{{\prime \prime }} _{0} }}{{2a^{2} }}\left( {1 + qQ} \right)^{2}$$

(47)

$$\text{Y} = - \frac{{\text{H}^{\prime } _{0} }}{{aQ}}\left( {2 - q} \right)\left( {1 + qQ} \right)^{2} - \frac{{\text{H}^{{\prime \prime }} _{0} }}{{a^{2} Q}}\left( {1 + qQ} \right)^{3} ,$$

(48)

$$\text{Z} = \frac{{\text{H}^{\prime } _{0} }}{{aQ^{2} }}\left( {1 - q} \right)\left( {1 + qQ} \right)^{2} + \frac{{\text{H}^{{\prime \prime }} _{0} }}{{a^{2} Q^{2} }}\left( {1 + qQ} \right)^{4}$$

(49)

The coefficients (X, Y, Z) ≡ (a₀, a₁, a₂) in Eq. (17) can be obtained by letting H(r) = F/r, thus, in terms of parameter x, we have

$$\text{H}\left( x \right) = \frac{{\text{F}\left( r \right)}}{r} \equiv \frac{Q}{{r_{\text{e}} \left( {1 + x} \right)\left( {\text{e}^{{ax}} + qQ} \right)}}$$

(50)

Equation (50), and its first-two derivatives at x = 0 gives

$$\text{H}_{0} = \frac{Q}{{r_{\text{e}} \left( {1 + qQ} \right)}}\;\text{H}^{\prime } _{0} = - \left( {1 + \frac{a}{{1 + qQ}}} \right)\text{H}_{0} \;\text{H}^{{\prime \prime }} _{0} = - \left\{ {2 + \frac{{2a}}{{1 + qQ}} + \frac{{a^{2} \left( {1 - qQ} \right)}}{{\left( {1 + qQ} \right)^{2} }}} \right\}\text{H}_{0}$$

(51)

Substituting (A14) into (A10)–(A12) gives

$$a_{0} = - \frac{Q}{{2r_{{\text{e}}} }}\left\{ {\frac{{2 - q\left( {1 + Q} \right)}}{1 + qQ} + \frac{5 - q}{a} + \frac{{2\left( {1 + qQ} \right)}}{{a^{2} }}} \right\},$$

(52)

$$a_{1} = - \frac{1}{{r_{{\text{e}}} }}\left\{ {3 - q\left( {1 + Q} \right) + \frac{1}{a}\left( {4 - q} \right)\left( {1 + qQ} \right) + \frac{2}{{a^{2} }}\left( {1 + qQ} \right)^{2} } \right\},$$

(53)

$$a_{2} = \frac{1}{{Qr_{{\text{e}}} }}\left\{ {2 - q\left( {1 + qQ^{2} } \right) + \frac{1}{a}\left( {3 - q + 2qQ} \right)\left( {1 + qQ} \right) + \frac{2}{{a^{2} }}\left( {1 + qQ} \right)^{3} } \right\}.$$

(54)

To obtain the coefficients (X, Y, Z) ≡ (b₀, b₁, b₂) in (18), we let H(r) = 1/r², r = r_e (1 + x). This gives $\text{H}_{0} = r_{\text{e}}^{ - 2} \;\text{H}^{\prime}_{0} = - 2r_{\text{e}}^{ - 2} \;{\text{and}}\;\text{H}^{\prime\prime}_{0} = 6r_{\text{e}}^{ - 2}$,. With the aid of equations (47)–(A12), we have

$$b_{0} = \frac{1}{{r_{{\text{e}}}^{2} }} - \frac{1}{{ar_{{\text{e}}}^{2} }}\left( {3{-}q} \right)\left( {1 + qQ} \right) + \frac{3}{{a^{2} r_{{\text{e}}}^{2} }}\left( {1 + qQ} \right)^{2} ,$$

(55)

$$b_{1} = \frac{2}{{aQr_{{\text{e}}}^{2} }}\left( {2 - q} \right)\left( {1 + qQ} \right)^{2} - \frac{6}{{a^{2} Qr_{{\text{e}}}^{2} }}\left( {1 + qQ} \right)^{3} ,$$

(56)

$$b_{2} = - \frac{2}{{aQ^{2} r_{{\text{e}}}^{2} }}\left( {1 - q} \right)\left( {1 + qQ} \right)^{2} + \frac{6}{{a^{2} Q^{2} r_{{\text{e}}}^{2} }}\left( {1 + qQ} \right)^{4} .$$

(57)