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!


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 $c_0 \neq 0$.


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