I'm trying to solve this one dimensional time-independent Schrodinger equation:
$$ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=\frac{\hbar^2k^2}{2m}\psi $$
where
$$ V(x)=-\frac{\hbar^2b}{m}sech^2(a x) $$
,$k$ is a parameter, $a$ and $b$ are positive real numbers.
It has the general solution in this form
$$ \psi(x)=C_1*P_{\frac{\sqrt{a^2+8 b}-a}{2 a}}^{\frac{k}{a}}(\tanh (a x))+C_2* Q_{\frac{\sqrt{a^2+8 b}-a}{2 a}}^{\frac{k}{a}}(\tanh (a x)) $$
where $P^\mu_\lambda$ and $Q^\mu_\lambda$ are the associate Legendre polynomials, $C_1$ and $C_2$ are constants determined by the boundary condition.
Now I want to determine the eigenstates, which requires the wave function $\psi(x)$ is finite every where. This will give a restriction on $k$ to certain discrete values. And I have problem finding the $k's$.
Mathematically, this is equivalent to find the $k's$ that makes the associate Legendre polynomials finite on range $-1\leq x\leq1$:
$$ P_{\frac{\sqrt{a^2+8 b}-a}{2 a}}^{\frac{k}{a}}(x) $$ $$ Q_{\frac{\sqrt{a^2+8 b}-a}{2 a}}^{\frac{k}{a}}(x) $$
So what are the $k's$ in terms of $a$ and $b$ that makes the above associate Legendre polynomials finite?