In a recent paper in this journal, Ahmed [1] has pointed out an incorrect expression in [2]. The author has solved the Klein–Gordon equation without interaction in the Som–Raychaudhuri space-time with the cosmic string using the Nikiforov–Uvarov method and obtained the energy eigenvalues and the corresponding eigenfunctions of the system. The problem solved in [1] is a special case associated to \(k_{L}=0\) in Ref. [2]. Ahmed has claimed that to obtain the correct result, one should start from Eq. (4) but not from the condition (12), which is obtained from the biconfluent Heun equation (10). The purpose of this comment is to show that following an appropriate manipulation of the biconfluent Heun equation, the correct energy eigenvalues for \(k_{L}=0\) can be obtained as a particular case of [2], as opposed to what was advertised in Ref. [1].

The Klein–Gordon equation in the Som–Raychaudhuri space-time with a linear scalar potential is given by

$$\begin{aligned}&-\frac{\partial ^{2}\varPsi }{\partial t^{2}}+\frac{1}{r}\frac{\partial }{\partial r}\left( r\frac{\partial \varPsi }{\partial r}\right) +\left( \frac{1}{\alpha r}\frac{\partial }{\partial \phi }-\varOmega r\frac{\partial }{\partial t} \right) ^{2}\varPsi \nonumber \\&\qquad +\frac{\partial ^{2}\varPsi }{\partial z^{2}}=\left( M+k_{L}r \right) ^{2}\varPsi . \end{aligned}$$
(1)

Considering the solution in the form

$$\begin{aligned} \varPsi (t,r,\phi ,z)=\mathrm {e}^{i(-Et+l\phi +kz)}\psi (r), \end{aligned}$$
(2)

and by introducing the new variable and parameters

$$\begin{aligned} x= & {} \sqrt{\omega }r, \end{aligned}$$
(3)
$$\begin{aligned} \omega= & {} \sqrt{\varOmega ^{2}E^{2}+k_{L}^{2}}, \end{aligned}$$
(4)
$$\begin{aligned} \lambda= & {} E^{2}-M^{2}-k^{2}-\frac{2\varOmega El}{\alpha } \end{aligned}$$
(5)
$$\begin{aligned} \theta= & {} \frac{2Mk_{L}}{\omega ^{3/2}}, \end{aligned}$$
(6)
$$\begin{aligned} \beta= & {} \frac{\lambda }{\omega }, \end{aligned}$$
(7)

one finds that Eq. (1) becomes

$$\begin{aligned} \psi ^{\prime \prime }+\frac{1}{x}\psi ^{\prime }+\left( \beta -x^{2}-\frac{l^{2}}{\alpha ^{2}x^{2}}-\theta x \right) \psi =0. \end{aligned}$$
(8)

The solution for (8) can be expressed as

$$\begin{aligned} \psi (x)=x^{\frac{|l|}{\alpha }}\mathrm {e}^{-\frac{x^{2}}{2}} \mathrm {e}^{-\frac{\theta }{2}x}H(x), \end{aligned}$$
(9)

where H(x) can be expressed as a solution of the biconfluent Heun differential equation [3,4,5,6]

$$\begin{aligned}&H^{\prime \prime }+\left( \frac{1+\frac{2|l|}{\alpha }}{x}-\theta -2x \right) H^{\prime }\nonumber \\&\qquad +\left( \tau -2-\frac{2|l|}{\alpha }- \frac{\sigma }{x}\right) H=0, \end{aligned}$$
(10)

with

$$\begin{aligned} \tau =\beta +\frac{\theta ^{2}}{4} \end{aligned}$$
(11)

and

$$\begin{aligned} \sigma =\frac{\theta }{2}\left( 1+\frac{2|l|}{\alpha } \right) . \end{aligned}$$
(12)

The differential equation (10) has a regular singularity at \(x=0\) and an irregular singularity at \(x=\infty \). The regular solution at the origin is given by

$$\begin{aligned} H(2|l|/\alpha ,\theta ,\tau ,0;x)=\sum _{j=0}^{\infty }\frac{\varGamma (2|l|/\alpha +1)}{\varGamma (2|l|/\alpha +1+j)} \frac{A_{j}}{j!}x^{j}\,, \end{aligned}$$
(13)

where \(\varGamma (z)\) is the gamma function, \(A_{0}=1\), \(A_{1}=\sigma \) and the remaining coefficients for \(\theta \ne 0\) satisfy the recurrence relation,

$$\begin{aligned} \begin{aligned} A_{j+2}&= \frac{\theta }{2}\left( 2j+3+\frac{2|l|}{\alpha } \right) A_{j+1}\\&\quad -\,\left( j+1\right) \left( j+\frac{2|l|}{\alpha }+1 \right) \left( \varDelta -2j\right) A_{j}, \end{aligned} \end{aligned}$$
(14)

where \(\varDelta =\tau -\frac{2|l|}{\alpha }-2\). From the recurrence (14), the solution H becomes a polynomial of degree n if and only if \(\varDelta =2n\) (\(n=0,1,\ldots \)) and \(A_{n+1}=0\).

On the other hand, if \(\theta =0\), the solution H becomes a polynomial of degree n if and only if

$$\begin{aligned} \varDelta =4n \quad (n=0,1,\ldots ). \end{aligned}$$
(15)

In fact, when \(\varDelta =4n\), one has [4]

$$\begin{aligned} H(2|l|/\alpha ,0,2|l|/\alpha +2(1+2n),0;x)\propto L^{(|l|/\alpha )}_{n}(x^{2}), \end{aligned}$$
(16)

where \(L^{(\delta )}_{n}(x)\) denotes the generalized Laguerre polynomial, a polynomial of degree n with n distinct positive zeros in the range \([0,\infty )\). This result is very important for obtaining the correct energy eigenvalues for the particular case \(k_{L}=0\) (\(\theta =0\)).

The especial case \(k_{L}=0\) (\(\theta =0\)) was studied in [1]. The author of [1] concluded that the correct expression of the energy cannot be obtained as a particular case from the biconfluent Heun differential equation, but this statement is false. Considering \(k_{L}=0\) (\(\theta =0\)) and using the correct condition (15), we obtain

$$\begin{aligned} \beta -\frac{2|l|}{\alpha }-2=4n. \end{aligned}$$
(17)

Substituting (4), (5) and (7) into (17), we obtain the spectrum for \(k_{L}=0\) as (for \(E\varOmega >0\))

$$\begin{aligned} \begin{aligned} E=&\left( 2n+1+\frac{|l|}{\alpha }+\frac{l}{\alpha }\right) \varOmega \\&+\sqrt{\left( 2n+1+\frac{|l|}{\alpha }+\frac{l}{\alpha } \right) ^{2}\varOmega ^{2}+M^{2}+k^{2}}, \end{aligned} \end{aligned}$$
(18)

and the solution becomes

$$\begin{aligned} \psi (x)=N_{n}x^{\frac{|l|}{\alpha }}\mathrm {e}^{-\frac{x^{2}}{2}} L_{n}^{(|l|/\alpha )}(x^{2}), \end{aligned}$$
(19)

where \(N_{n}\) is a normalization constant. Equation (18) is the same as the result obtained in Ref. [7].

In summary, we showed that the correct expression for the energy eigenvalues for the case \(k_{L}=0\) of Ref. [2] can be obtained from an appropriate manipulation of the biconfluent Heun differential equation, in opposition to what was concluded in [1]. The results obtained in this comment are consistent with those found in [7].