Interaction potential in radial equation (central potentials)

Question

I'm studying QM from Bransden & Joachain (section 7.2). I'm at the part where one separates the Schrödinger equation in spherical polar coordinates, and looks for solutions of the form $ \psi_{Elm} (\mathbf{r}) = R_{Elm}(r) Y_{lm}(\theta, \phi)$, where $R$ is a radial function, and $Y$ are the spherical harmonics.

After introducing the new variable $u_{El}(r) = r R_{El}(r)$ we obtain the radial equation $$ - \frac{ \hbar^2}{2 \mu} \frac{ d^2 u_{El}(r)}{dr^2} + V_{eff} (r) u_{El} (r) = E u_{El}(r) \qquad (*) $$ with the boundary condition $u_{El}(0) = 0$ so that $R$ doesn't blow up. Here $V_{ef} = V(r) + \frac{l(l+1) \hbar^2}{2 \mu r^2}. $

Now, the author wants to examine more closely the behavior of the function $u_{El}(r)$ near the origin. He writes:

We shall first assume that in the vicinity of $r = 0$ the interaction potential $V(r)$ has the form $$ V(r) = r^p (b_0 + b_1 r + \ldots), \qquad b_0 \neq 0 $$ where $p$ is an integer such that $p \geq -1$. In other words, the potential cannot be more singular than $r^{-1}$ at the origin, which is the case for nearly all interactions of physical interest. Since $r = 0$ is a regular singular point of the differential equation $(*)$, we can expand the solution $u_{El}(r)$ in the vicinity of the origin as $$ u_{El} (r) = r^s \sum_{k = 0}^{\infty} c_k r^k, \qquad c_0 \neq 0 $$ Substituting this expansion in (*), we find by looking at the coefficient of lowest power of r (i.e. $r^{s-2}$) that the quantity $s$ must satisfy the indicial equation $$ s(s-1) - l(l+1) = 0 $$ so that $s = l+1$ or $s = -l$. The choice $s = -l$ corresponds to irregular solutions which do not satisfy the condition $u_{El}(0) = 0$. The other choice $s = l+1$ corresponds to regular solutions which are physically allowed, and are such that $$\lim_{r \to 0} u_{El}(r) \rightarrow r^{l+1}. $$

My questions are:

1) At the beginning, why does he expand the potential like that, and how can he even do that? Doesn't the potential just go as $1/r$ ? Also, I don't understand his remark that the potential cannot be 'more singular' than $r^{-1}$ at the origin.

2) How does he obtain that indicial equation? Does he just set $k = 0$? When I differentiate that Frobenius series twice, and look at the coefficient of lowest power, I still have the terms $(k+s)(k+s-1)$ and $l(l+1)$.

3) Also, I don't understand how $s = -l$ leads to a irregular solution that must be discarded.

Thanks in advance for any clarifications!

Answer 1

1) He explicitly assumes that he can expand the potential in that way. But it is not an outlandish assumption: It just says that, if we factor out a factor of $r^p$, that the rest can then be expanded in a Taylor series.

In other words, we expand $V$ in a Laurent series $V(r) = \sum_{k=-\infty}^\infty \tilde b_k r^k$, then all $\tilde b_k$ must be zero for $k < -1$. Those terms $r^k$ with $k < -1$ would otherwise be "more singular" than $r^{-1}$.

2) In the terms you got, there still is a $k$. But, you wanted to set $k$ to zero in order to obtain only the coefficients of lowest order. If you do that, you get the $s(s-1)$.

3) If $s$ is negative, then $u(r) = r^s \sum_{k \geq 0} c_k r^k$ has a singularity for $r = 0$. The reason is that $u(0) = 0^s \left( c_0 \cdot 1 + 0 \right)$ and $c_0 \neq 0$.