Consider the Schroedinger equation in a spherically symmetric system. In the unit system under which energy is measured in Hartree and length in Bohr radius $a_0$, the schroedinger equation can be written as
$$ -{d^2 R_{l,\epsilon}(r) \over dr^2} + \left( V_d(r) + {l(l+1)\over 2 r^2} \right)R_{l, \epsilon}(r) = \epsilon R_{l, \epsilon}(r) $$
. In the above equation, $V_d(r)$ is an attractive potential that's dependent on $d$. Specifically, we will consider the following potential. $V_{0}(r) = -Z(r)/r$, where $\lim_{r \rightarrow 0}Z(r) = Z$ and $\lim_{r \rightarrow \infty} Z(r) = 1$. $V_d(r) = V_0(r) + {1 \over r} - {e^{-r/d} \over r}$. Consider the wave function $R_{l, \epsilon}^{reg}(r)$ that is regular at the origin and which is normalised to unit magnitude at infinity. In other words,
$$ R_{l, \epsilon}^{reg}(r) \rightarrow \sin(k r + \phi) $$, as $r \rightarrow \infty$. $k = \sqrt{2 \epsilon}$. My question is, for at least $\epsilon>0$, is $R_{l, \epsilon}^{reg}(r)$ a "smooth" function of $d$?
