1 Introduction

Obtaining analytical and numerical solutions of the wave equations is a very important task in quantum mechanics to understand different physical systems at the atomic level [1,2,3,4]. Different analytical and numerical methods have been used in the past to obtain solutions to the associated eigenvalue problems such as the supersymmetry method (SUSY) [5, 6], the exact quantisation method (EQM) [7,8,9,10,11,12,13,14], the tridiagonal representation approach (TRA) [15,16,17], the asymptotic iteration method (AIM) [18,19,20,21,22,23] and many other methods [24,25,26,27].

Very recently, Assi et al [17] introduced the following one-dimensional (1D) and three-dimensional (3D) potentials:

$$\begin{aligned} V(x)= & {} \frac{B}{\cosh ^{2}(\alpha x)}+C\cosh (2\alpha x), \end{aligned}$$
(1.1)
$$\begin{aligned} V(r)= & {} \frac{A}{\sinh ^{2}(\alpha r)}+\frac{B}{\cosh ^{2}(\alpha r)}+C\cosh (2\alpha r), \end{aligned}$$
(1.2)

where \(A,\;C>0\), \(B<0\) for bound states and \(\alpha >0\) is just a scaling parameter. We shall solve the corresponding Schrödinger equation and obtain the energy eigenvalues for both potentials using the AIM.

The flow of this work is organised as follows. In §2 and §3, we apply the AIM to solve the wave equation for the 1D and the 3D potentials, respectively. In addition, §4 contains our results and discussions for both the problems. Finally, we conclude our work in §5.

2 The 1D potential well

The non-relativistic stationary wave equation (in units \(\hbar =m=1)\) for the potential in (1.1) reads [28, 29]

$$\begin{aligned}&\left[ {-\frac{1}{2}\frac{\hbox {d}^{2}}{\hbox {d}x^{2}}+\frac{B}{\cosh ^{2}(\alpha x)}+C\cosh (2\alpha x)} \right] \psi (x)\nonumber \\&\quad =E\psi (x), \end{aligned}$$
(2.1)

where \(x\in \left( {-\infty ,+\infty } \right) \). Using the change of variable \(y=2\tanh ^{2}\left( {\alpha x} \right) -1\), eq. (2.1) becomes

$$\begin{aligned}&\left\{ {-\alpha ^{2}(1+y)(1-y)^{2}\frac{\hbox {d}^{2}}{\hbox {d}y^{2}}+\frac{\alpha ^{2}}{2}(1-y)(1+3y)\frac{\hbox {d}}{\hbox {d}y}} \right. \nonumber \\&\quad \left. {+\,\frac{B}{2}(1-y)+\frac{4C}{1-y}} \right\} \psi (y)=(E+C)\psi (y),\nonumber \\ \end{aligned}$$
(2.2)

where \(y\in \left[ {-1,+1} \right] \). The above equation can be written in the following AIM form:

$$\begin{aligned} \frac{\hbox {d}^{2}\psi \left( y \right) }{\hbox {d}y^{2}}=\lambda _0 (y)\frac{\hbox {d}\psi \left( y \right) }{\hbox {d}y}+s_0 (y)\psi \left( y \right) , \end{aligned}$$
(2.3)

where

$$\begin{aligned} \lambda _0 (y)= & {} \frac{1}{2}\frac{1+3y}{1-y^{2}}, \end{aligned}$$
(2.4a)
$$\begin{aligned} s_0 (y)= & {} \frac{{\tilde{B}}}{1-y^{2}}+\frac{{\tilde{C}}}{\left( {1-y^{2}} \right) \left( {1-y} \right) ^{2}}\nonumber \\&-\,\frac{{\tilde{E}}}{\left( {1-y^{2}} \right) \left( {1-y} \right) } \end{aligned}$$
(2.4b)

and

$$\begin{aligned} {\tilde{B}}=\frac{B}{2\alpha ^{2}},\quad {\tilde{C}}=\frac{4C}{\alpha ^{2}},\quad {\tilde{E}}=\frac{E+C}{\alpha ^{2}}, \end{aligned}$$
(2.4c)

To improve the accuracy and fast convergence of the AIM, in particular, we would like to reduce the order of the regular singularity at \(y= 1\), and we use the following transformation in (2.3):

$$\begin{aligned} \psi (y)=(1+y)^{a}(1-y)^{b}g(y), \end{aligned}$$
(2.5)

where a and b are some non-negative parameters selected so as to improve the convergence of our AIM algorithm, and g(y) is some function of y. Using (2.5) back in eq. (2.3), we get

$$\begin{aligned} \frac{\hbox {d}^{2}g\left( y \right) }{\hbox {d}y^{2}}=\tilde{\lambda }_0 (y)\frac{\hbox {d}g\left( y \right) }{\hbox {d}y}+{\tilde{s}}_0 (y)g\left( y \right) , \end{aligned}$$
(2.6)

where

$$\begin{aligned} {\tilde{\lambda }}_0 (y)=\lambda _0 (y)-2\left( {\frac{a}{1+y}-\frac{b}{1-y}} \right) \end{aligned}$$
(2.7a)

and

$$\begin{aligned} {\tilde{s}}_0 (y)= & {} s_0 (y)+\lambda _0 (y)\left\{ {\frac{a}{1+y}-\frac{b}{1-y}} \right\} -\frac{a(a-1)}{(1+y)^{2}} \nonumber \\&+\,\frac{b(b-1)}{(1-y)^{2}}-\frac{2ab}{1-y^{2}}. \end{aligned}$$
(2.7b)

Next, we should find, iteratively, a set of functions \(\big \{{{\tilde{\lambda }}_n (y), {\tilde{s}}_n (y)} \big \}_{n=1} \) using the following recursion relations [18]:

$$\begin{aligned} {\tilde{\lambda }}_n ={\tilde{\lambda }^{\prime }}_{n-1} +{\tilde{s}}_{n-1} +{\tilde{\lambda }}_0 {\tilde{\lambda }}_{n-1} ,\quad {\tilde{s}}_n ={\tilde{{s}}^{\prime }}_{n-1} +{\tilde{s}}_0 {\tilde{\lambda }}_{n-1} , \end{aligned}$$
(2.8)

where the primes stand for the derivatives with respect to y. This process is done using a suitable programming language. The next step in the computation will be finding the eigenvalues of the original problem by solving the following quantisation condition [18]:

$$\begin{aligned} \left| {{\begin{array}{ll} {{\tilde{\lambda }}_n }&{} \quad {{\tilde{s}}_n } \\ {{\tilde{\lambda }}_{n-1} }&{}\quad {{\tilde{s}}_{n-1} } \\ \end{array} }} \right| =0. \end{aligned}$$
(2.9)

The evaluation of the eigenenergies using eq. (2.9) will force us to use a suitable seed value for y to obtain the actual energy eigenvalues, and we use \(y=0\) for our case. The wave function, however, is obtained using the following recursive relation [18]:

$$\begin{aligned} \psi (y)=A(1+y)^{a}(1-y)^{b}\exp \left( {-\int _0^y {\frac{\tilde{s}_n (t)}{{\tilde{\lambda }}_n (t)}\hbox {d}t} } \right) , \end{aligned}$$
(2.10)

where A is the normalisation constant. The above solution must match with that which was obtained earlier [17]. Note that because our potential is symmetric around the origin, the physical solutions of the wave equation must have definite parity [28, 29]. Thus, the physical solutions of eq. (2.1) are written in a compact form as follows:

$$\begin{aligned} \varphi _\pm (y)=\frac{1}{\sqrt{2}}\left[ {\psi (y)\pm \psi (-y)} \right] , \end{aligned}$$
(2.11)

where the plus sign corresponds to the even states and the minus sign corresponds to the odd states. In the next section, we apply the AIM to the 3D problem.

Table 1 Lowest energy eigenvalues of the 1D potential (1.1) for \(B=-100, C=5\) and \(\alpha =1/\sqrt{2}\) obtained using AIM in this work against the results obtained in [17].
Full size table

3 Three-dimensional spherically symmetric confined potential

The radial stationary wave equation for the second potential in (1.2) reads as

$$\begin{aligned}&\left[ -\frac{1}{2}\frac{\hbox {d}^{2}}{\hbox {d}r^{2}}+\frac{A}{\sinh ^{2}\left( {\alpha r} \right) }+\frac{B}{\cosh ^{2}\left( {\alpha r} \right) }\right. \nonumber \\&\quad \left. +\,C\cosh \left( {2\alpha r} \right) +\frac{\ell \left( {\ell +1} \right) }{2r^{2}} \right] \psi \left( r \right) =E\psi \left( r \right) , \end{aligned}$$
(3.1)

where \(\ell \) is the angular momentum quantum number. To simplify the calculations, we use the following approximation for the centrifugal term suggested recently by Assi et al [17]:

$$\begin{aligned} \frac{1}{r^{2}}\approx & {} \alpha ^{2}\left[ \frac{1}{\sinh ^{2}\left( {\alpha r} \right) }+\frac{31}{945}\frac{1}{\cosh ^{2}\left( {\alpha r} \right) }\right. \nonumber \\&\left. -\,\frac{16}{945}\cosh \left( {2\alpha r} \right) +\frac{20}{63} \right] . \end{aligned}$$
(3.2)

The reader should refer to other approximations used before [30, 31]. Substituting eq. (3.2) back in (3.1), we get [17]

$$\begin{aligned}&\left[ -\frac{1}{2}\frac{\hbox {d}^{2}}{\hbox {d}r^{2}}+\frac{\tilde{A}}{\sinh ^{2}\left( {\alpha r} \right) }+\frac{{\tilde{B}}}{\cosh ^{2}\left( {\alpha r} \right) }\right. \nonumber \\&\left. \qquad +\,{\tilde{C}}\cosh \left( {2\alpha r} \right) \right] \psi \left( r \right) ={\tilde{E}}\psi \left( r \right) , \end{aligned}$$
(3.3)

where

$$\begin{aligned} {\tilde{A}}= & {} A+\frac{\alpha ^{2}\ell (\ell +1)}{2}, \end{aligned}$$
(3.4a)
$$\begin{aligned} {\tilde{B}}= & {} B+\frac{31}{945}\frac{\alpha ^{2}\ell (\ell +1)}{2}, \end{aligned}$$
(3.4b)
$$\begin{aligned} {\tilde{C}}= & {} C-\frac{16}{945}\frac{\alpha ^{2}\ell (\ell +1)}{2} \end{aligned}$$
(3.4c)

and

$$\begin{aligned} {\tilde{E}}=E-\frac{20}{63}\frac{\alpha ^{2}\ell (\ell +1)}{2}. \end{aligned}$$
(3.4d)

Now, by making the change in the variable \(y=2\tanh ^{2}(\alpha r)-1\) back in eq. (3.3), we get

$$\begin{aligned} \frac{\hbox {d}^{2}\psi \left( y \right) }{\hbox {d}y^{2}}=\lambda _0 (y)\frac{\hbox {d}\psi \left( y \right) }{\hbox {d}y}+s_0 (y)\psi \left( y \right) , \end{aligned}$$
(3.5)

where

$$\begin{aligned} \lambda _0 (y)= & {} \frac{1}{2}\frac{1+3y}{1-y^{2}}, \end{aligned}$$
(3.6a)
$$\begin{aligned} s_0 (y)= & {} \frac{\bar{{A}}}{(1-y^{2})^{2}}+\frac{\bar{{B}}}{1-y^{2}} +\frac{\bar{{C}}}{(1-y^{2})(1-y)^{2}}\nonumber \\&-\,\frac{\varepsilon }{(1-y^{2})(1-y)} \end{aligned}$$
(3.6b)

and

$$\begin{aligned} \bar{{A}}&=\frac{2{\tilde{A}}}{\alpha ^{2}},\quad \bar{{B}}=\frac{\tilde{B}}{2\alpha ^{2}}, \nonumber \\ \bar{{C}}&=\frac{4{\tilde{C}}}{\alpha ^{2}},\quad \varepsilon =\frac{{\tilde{E}}+{\tilde{C}}+{\tilde{A}}}{\alpha ^{2}}. \end{aligned}$$
(3.7)

The next steps are mathematically similar to those followed in the previous section from eq. (2.5) up to eq. (2.10) but with the new form of \(s_0 (y)\) given in eq. (3.6b). So, we are not going to rewrite them again here. In the next section, we shall present our numerical results for the eigenenergies of both potentials which agree with what have been obtained in [17].

Table 2 Lowest 10 eigenvalues of the 1D eq. (2.1) for \(B = -10000, C=5\) and \(\alpha =1/\sqrt{2}\) corresponding to the even states obtained in this work vs. those found using different approaches in [17].
Full size table

4 Results and discussion

4.1 Lowest energy eigenvalues of the 1D potential

Here, we have numerically calculated the lowest energy eigenvalues for two different cases with the same values of the potential parameters taken in ref. [17]. In the first case, we took \(B=-100, C=5\) and \(\alpha =1/\sqrt{2}\). Our results are tabulated in table 1 against the results obtained using different methods [17], where the stability of the results obtained using \(a=b=0\) for the even state’s eigenvalues and the odd state’s energies obtained using \(a=b=1/2\). Comparing columns two with three, and four with five, we notice that the results are very close to a few decimal points which verify the results obtained in [17] using the AIM. Similarly, we have also considered another case with \(B= -10000,C=5\) and \(\alpha =1/\sqrt{2}\) as given in tables 2 and 3. The results presented in table 2 using the two methods agree with high accuracy up to 12 decimal digits.

4.2 Energy spectrum of the 3D potential

The computations obtained for the 3D potential are divided into two tables for the angular momentum values \(\ell =0\) and 5, respectively, with the potential parameters taken as \(A=10\), \(B=-200\), \(C=10\) and \(\alpha =1/\sqrt{2}\). In table 4, the energies obtained using the AIM agree to a high degree of accuracy with those calculated using the Hamiltonian diagonalisation method (HDM) for the zero angular momentum quantum number. In addition, for \(\ell =5\), our results also agree with those obtained using the HDM in [17] as shown in table 5. Note that in both cases, and for numerical purposes, we have taken \(a=b=0\).

Table 3 Lowest 10 eigenvalues of the 1D eq. (2.1) for \(B = -10000, C=5\) and \(\alpha =1/\sqrt{2}\) associated with the odd states obtained in this work using the AIM.
Full size table
Table 4 Lowest 10 energies of the 3D potential for the s-wave case obtained using the AIM against those obtained by the HDM for the choice of parameters \(A=10, B=-200, C=10\) and \(\alpha = 1/\sqrt{2}\).
Full size table
Table 5 Comparison between the lowest energies obtained using the AIM and the HDM for \(\ell = 5\) the state with the parameters \(A=10, B=-200, C=10\) and \(\alpha = 1/\sqrt{2}\).
Full size table

5 Conclusion

In this work, we have applied the AIM for two new confining potentials that were introduced in [17]. The numerical computation of the energy spectrum agrees to a high degree of accuracy with those obtained using the HDM in [17]. This reflects the power of the AIM for solving a wider class of eigenvalue problems. Those potentials are new, and we are not aware of any direct physical application that might be suitably modelled using these interactions.