Solving Transcendental equation involving special functions becoming nightmare any one can help?

Question

Simply i want to solve for schrodinger equation for finite potential well problems in spherical coordinates. For case in which l=0 . It is simple but when l changes. The solution are spherical bessel and spherical hankel function. Hankel function have an imaginary argument. Using scipy python library i am not even able to define the equation properly. Is there any better way to do that ..

Answer 1

I am not sure what you are asking but I can sketch how this type of problem is usually solved.

Considering the following finite potential $$ U(r)= \begin{cases} -U_0, & r \leq R \\ 0 & r > R \end{cases} $$ The radial equation is given by \begin{equation} -\frac{\hbar^2}{2m} \frac{d^2u(r )}{dr^2}+\left[U(r)+\frac{\hbar^2 \ell(\ell+1)}{2mr^2} \right]u(r)=Eu(r). \end{equation} This is identical to the one-dimensional Schrödinger equation with an effective potential, where the centrifugal term pushes the particle outwards. You seem to be looking for the case where $\ell=0$. To solve this analytically, rewrite the equation and consider the boundary conditions. \begin{equation} \frac{d^2 u(r )}{dr^2}+\frac{M}{\hbar^2} \left[E-U(r) \right]u(r)=0, \end{equation} where I plugged in the expression for the reduced mass, $m=2M$ (two particles). For a bound state the energy eigenvalue, $E$, is negative and $E_B$ is a positive binding energy. This leads to the following expressions

\begin{equation} \label{diff1} \frac{d^2 u(r )}{dr^2}+\frac{M}{\hbar^2} \left(U_0-E_B \right)u(r)=0, \quad r \leq R, \end{equation} \begin{equation} \label{diff2} \frac{d^2 u(r )}{dr^2}-\frac{M}{\hbar^2} E_B u(r)=0, \quad r > R. \end{equation} Introduce two variables given by \begin{equation} \label{constants} k=\sqrt{\frac{M}{\hbar^2}\left(U_0-E_B\right)}, \quad \kappa=\sqrt{\frac{M E_B}{\hbar^2}}. \end{equation} Rewriting the equation and solving the differential equation yields \begin{equation} \label{solu1} \frac{d^2 u(r )}{dr^2} = -ku(r) \Rightarrow u(r) = A\sin(kr)+B\cos(kr). \end{equation} Since $R(r) = u(r)/r$ and $\cos(kr)/r$ blows up as $r \rightarrow 0 \Rightarrow B=0$ and the solution is \begin{equation} u(r)=A\sin(kr), \quad r \leq R \end{equation} Now, \begin{equation} \label{solu2} \frac{d^2 u(r )}{dr^2} = \kappa^2 u(r) \Rightarrow u(r) = Ce^{\kappa r}+De^{-\kappa r} \end{equation} Here $Ce^{\kappa r}$ blows up as $r\rightarrow \infty$. The wavefunction must be continuous and this means the two solutions must match at $r=R$. The same applies for the derivative. This leads to two equations for $r=R$. \begin{equation} \label{sin} A\sin(kR) =De^{-\kappa R} \end{equation} \begin{equation} \label{expo} Ak\cos(kR) =-D\kappa e^{-\kappa R} \end{equation} Dividing leads to \begin{equation} \label{cot} -\cot(kR) = \frac{\kappa}{k} \end{equation} This equation can be solved by requiring $kR=\pi/2$.

In relation to the Hankel functions I assume you are looking for a solution for $r>a$ with a finite $\ell$. The radial wave function with the orbital momentum $\ell$ is for $r<a$ is given by the spherical Bessel function. For a bound state, this function has to be continuously matched to the exponentially decaying Hankel function.

So to answer your question (I think) you can use the following for s-wave

\begin{equation} h_0^{(1)}(i\kappa r) = -\frac{\exp(-\kappa r)}{\kappa r} \end{equation}