Monopoles - 8 - Dirac Monopoles in Quantum Mechanics - Part - 4

Let us start with the motion of an electron in the field of a magnetic monopole where the usual spherical polar coordinates are described by taking the monopole at the origin. In addition we know that Every wave function describing this system should have a singularity line starting from the origin, should pass through any closed surface.  We use the equation from previous posts (Part-1,2,3)
We consider the same wave function of the type, $$ \psi = \psi_1 e^{i\beta}$$ with corresponding definition on $\beta, \vec{k},\,etc.$
But, now it was introduced two separate things as nodal line and singular line. I couldn't understand it completely, but I just want to proceed with the next step. 
The magnetic field of the monopole is given by, $$ \vec{B} = \frac{q_m}{r^2} \hat{r} $$ and $$ \nabla\times\vec{K} = \frac{e}{\hbar{c}} \vec{B}  $$ On substitution, $$ \nabla\times \vec{K} = \frac{e}{\hbar{c}} \frac{q_m}{r^2}\hat{r} \\~\\ = \frac{e}{\hbar{c}}\frac{n\hbar{c}}{2er^2}\hat{r}\\~\\ = \frac{n}{r^2}\hat{r}$$
Thus, we get the curl of K as radial with magnitude $\frac{n}{2r^2}$ 
So, the solution of K could be worked out by expanding the curl in spherical polar coordinates as, $$ \frac{1}{r^2sin\theta} \left[ \frac{\partial(rsin\theta{k_{\phi}})}{\partial{\theta}} - \frac{\partial(rk_{\theta})}{\partial{\phi}}\right] \hat{r} + \frac{1}{rsin\theta}\left[\frac{\partial(k_r)}{\partial{\phi}} - \frac{\partial(rsin\theta{k_{\phi}})}{\partial{r}}\right] \hat{\theta}+ \\~\\\frac{1}{r}\left[ \frac{\partial(r{k_{\theta}})}{\partial{r}} - \frac{\partial(k_r)}{\partial{\theta}}\right]\hat{\phi} = \frac{n}{2r^2} \hat{r} $$
Equating the components we get a solution as, $$ k_\theta = k_r =  k_0 = 0 \\~\\ k_\phi = \frac{n}{2r} tan\frac{\theta}{2}$$
Then the Schrodinger for non-relativistic electron is given by, $$ \frac{-\hbar^2}{2m} \nabla^2\psi = E\psi$$ 
Applying $$ \psi = \psi_1 e^{i\beta}$$  we get, $$ \nabla\cdot\nabla(\psi_1e^{i\beta}) = \nabla\cdot\left[e^{i\beta}\nabla(\psi_1) + \psi_1 \nabla(e^{i\beta})\right] \\~\\= e^{i\beta} \nabla^2(\psi_1) + \nabla\psi_1\cdot\nabla(e^{i\beta}) + \nabla\cdot\left[\psi_1 \nabla (e^{i\beta})\right]$$ But $$ \nabla(e^{i\beta}) = \frac{\partial(e^{i\beta})}{\partial\vec{r}} = i e^{i\beta} \frac{\partial\beta}{\partial\vec{r}} = ie^{i\beta}\vec{k}$$ (These are 3 vectors - four vectors are separately indicated) Applying this we get, $$ \nabla^2\psi = e^{i\beta}\nabla^2\psi_1 + ie^{i\beta} \nabla\psi_1\cdot \vec{k} + \nabla(\psi_1ie^{i\beta})\cdot\vec{k} + \psi_1ie^{i\beta} \nabla \cdot\vec{k} \\~\\ = e^{i\beta}\nabla^2\psi_1 + ie^{i\beta} \vec{k}\cdot\nabla\psi_1 + ie^{i\beta}\nabla\psi_1\cdot\vec{k} + \psi_1 \nabla(ie^{i\beta})\cdot\vec{k} + \psi_1 ie^{i\beta} \nabla\cdot\vec{k} \\~\\ = e^{i\beta} \nabla^2\psi_1 + ie^{i\beta} \vec{k}\cdot\nabla\psi_1 + ie^{i\beta}\nabla\psi_1\cdot\vec{k} + ie^{i\beta}\psi_1 \nabla\cdot\vec{k} - e^{i\beta}\psi_1 \vec{k}\cdot\vec{k}$$
which finally gives, $$ \nabla^2\psi = e^{i\beta}\left[ \nabla^2\psi_1 + i\vec{k}.\nabla\psi_1 + i \left(\nabla\psi_1\cdot\vec{k} + \psi_1\nabla\cdot\vec{k}\right) - k^2\psi_1\right] $$$$ \nabla^2\psi = e^{i\beta}\left[ \nabla^2 + i\vec{k}.\nabla + i (\nabla\cdot\vec{k}) - k^2\right]\psi_1$$
Now, our initial schrodinger equation can be rewritten as,
$$ \frac{-\hbar^2}{2m}\left[\nabla^2 + i\vec{k}.\nabla + i (\nabla\cdot\vec{k}) - k^2\right]\psi_1 = E\psi_1$$ 
Substituting for the values of k, we will get, 
$$\vec{k^2} = k_\phi^2 = \frac{n^2}{4r^2}tan^2{\theta/2} $$ and 
$$ \vec{k}\cdot\nabla = (\nabla\cdot\vec{k}) = \frac{k_\phi}{rsin\theta}\frac{\partial}{\partial\phi} = \frac{n\,tan\frac{\theta}{2}}{2r^2sin{\theta/2}cos{\theta/2}}\frac{\partial}{\partial\phi} = \frac{n\,sec^2{\theta/2}}{4r^2}\frac{\partial}{\partial{\phi}}$$ 
On substitution, $$ \frac{-\hbar^2}{2m}\left[ \nabla^2 + \frac{2ni}{4r^2} sec^2{\theta/2}\frac{\partial}{\partial{\phi}} - \frac{n^2tan^2{\theta/2}}{4r^2}\right]\psi_1 = E \psi_1 $$
Applying for Laplace operator in polar coordinates and using the regular separation of variables method we finally get (it is just manipulation), for Radial part, $$ \left[ \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} - \frac{\lambda}{r^2}\right]R(r) = \frac{-2mE}{\hbar^2} R(r)$$
and angular part, $$ \left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{sin^2\theta}\frac{\partial^2}{\partial\phi^2} +\frac{ni}{2} sec^2{\theta/2}\frac{\partial}{\partial\phi} - \frac{n^2}{4}tan^2{\theta/2}\right]Y(\theta,\phi) = -\lambda{Y(\theta,\phi)}$$

From here, 
we need to solve two differential equations for which I searched for the solution in various places. I couldn't find the complete solution but just the preview of the starting of the solution by I.Tamm. Only first two pages are free to see and the complete paper costs more money. So, I tried my own to convert the above Angular equation into the usual simple equation of Spherical Harmonics.  

We will start with the Angular part by assuming the solution of type, $$ Y(\theta,\phi) = L(\theta) e^{im\phi} $$ Upon substitution, $$e^{im\phi}\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) - \frac{m^2}{sin^2\theta}-\frac{mn}{2} sec^2{\theta/2} - \frac{n^2}{4}tan^2{\theta/2}\right]L(\theta) = -\lambda{e^{im\phi}}{L(\theta)} $$ With slight alterations, we can again rewrite the above as, $$\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) - \frac{m^2}{sin^2\theta}- \frac{mn}{1+ cos{\theta}} - \frac{n^2}{4}\frac{1-cos\theta}{1+cos\theta}\right]L(\theta) = -\lambda{L(\theta)} $$ and $$\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right)+\lambda - \frac{m^2}{sin^2\theta}- \frac{mn}{1+ cos{\theta}} - \frac{n^2}{4}\frac{1-cos\theta}{1+cos\theta}\right]L(\theta) = 0 $$Now, we will try to convert this into usual equation by replacing with suitable new variable $$ z = 1+cos\theta$$,
So that, $$ \frac{dL}{d\theta} = \frac{dL}{dz} \frac{dz}{d\theta}$$ and the first term becomes $$ \frac{dz}{d\theta} = -sin\theta$$ and $$ \frac{1}{sin\theta}\frac{d}{d\theta}\left(sin\theta\left(-sin\theta\frac{dL}{dz}\right)\right) = \frac{1}{sin\theta}\left[\frac{d}{dz}\left(-sin^2\theta\frac{dL}{dz}\right)\right]\frac{dz}{d\theta}\\~\\ = \frac{d}{dz}\left(sin^2\theta\frac{dL}{dz}\right) = \frac{d}{dz}\left(\left(2z-z^2\right)\frac{dL}{dz}\right)$$ where we used the fact that, $$ cos\theta = z - 1\\~\\ sin^2\theta = 2z-z^2$$ and the second term becomes, $$ \left[\lambda - \frac{m^2}{sin^2\theta} -\frac{m}{1+cos\theta} - \frac{n^2}{4} \frac{(1-cos\theta)}{(1+cos\theta)}\right] = \left[\lambda - \frac{m^2}{2z-z^2}-\frac{m}{z}-\frac{n^2}{4}\frac{2-z}{z}\right]\\~\\ = \lambda - \left[\frac{m^2+mn(2-z)+\frac{n^2}{4}(2-z)^2}{z(2-z)}\right]\\~\\ = \lambda - \left[\frac{\left(m+\frac{n}{2}(2-z)\right)^2}{z(2-z)}\right]$$
Finally we get our differential equation as, $$ \frac{d}{dz}\left(\left(2z-z^2\right)\frac{dL(z)}{dz}\right) + \left[\lambda - \frac{\left(m+\frac{n}{2}(2-z)\right)^2}{2z-z^2}\right]L(z) = 0$$