$D$-dimensional Coulomb problem, is the generalization from the 3D case supposed to be simple? (Gilmore's Lie algebra, chapter 14, problem 20)

Question

I am solving problem 20 of chapter 14 of Robert Gilmore's Lie groups, physics and geometry: An Introduction for Physicists, Engineers and Chemists, which focuses on the $D$-dimensional Coulomb problem, and I'm having troubles to understand what is asked and if the problem is even correct.

Here is the problem:

In $D$-dimensional space the Schrödinger equation for the Kepler problem is, in the relativistic case (klein-Gordon Equation): \begin{equation} \label{eq: D dim KG} \left(E^2 - (mc^2)^2 + 2E\left( \frac{e^2}{r} \right) + \left( \frac{e^2}{r} \right)^2 - (-i\hbar c \nabla)^2 \right) \psi(x) = 0 \end{equation} and in the nonrelativistic case (Schrödinger): \begin{equation}\label{eq: D dim S} \left(\frac{-\hbar^2}{2m}\nabla^2 - \frac{e^2}{r} - W\right) \psi(\mathbf{x}) = 0 \end{equation} With $W$ being the nonrelativistic energy, $E = mc^2 + W$. The only difference is that the Laplacian $\nabla^2$ is on $D$ coordinates rather than three. In this case the Laplacian operator is \begin{align} \label{eq: D-dimension gratin^2} \nabla^2 &= \left( \frac{1}{r^{D-1/2}} \frac{\partial}{\partial r} r^{D-1/2}\right)^2 + \frac{\mathcal{L}^2}{r^2} \\ &= \frac{1}{r^{D-1}} \frac{\partial}{\partial r} \Bigg(r^{D-1} \frac{\partial}{\partial r} \Bigg) + \frac{\mathcal{L}^2}{r^2} \end{align} The angular part of the Laplacian operator, $\mathcal{L}$, acts on spherical harmonics on $S^{D-1}$, $\mathcal{Y}^l(S^{D-1})$. These spherical harmonics are eigenfunctions of this (Laplace–Beltrami) operator with eigenvalue $- [(l + \alpha)^2 - \alpha^2]$ , and $\alpha$ is a quantity that depends on the Lie algebra of $SO(D)$: it is half the sum over all positive roots of the algebra. For the Lie algebras of the orthogonal roots the coefficient of the sum that is important is $\alpha = D - 2$.

Since we consider half the sum, I assumed that we are supposed to take $\alpha = \frac{D-2}{2}$

In question a. we are given the following ansatz: $$ \psi(\mathbf{x}) = (1/r^{(D-1)/2} )\mathcal{Y}^l(\text{angles}) $$ And the goal is, if I understand correctly, to get the Klein-Gordon/Schrödinger equations in the form a radial equation, namely: $$ (\frac{d^2}{dr^2} + \frac{A}{r^2} + \frac{B}{r} + C) R(r) = 0$$

In question b. we are interested in how the coefficients $A, B, C$ change between the 3-dimensional case and the $D$-dimensional case. In 3D indeed, the $A$ coefficient is given by the eigenvalue of $\mathcal{L}^2$, namely $A = -l(l+1)$, and question b. asks us to show that in $D$-dimensions, only $A$ changes and the replacement is: $$ l(l+1) \rightarrow l(l+D-2)$$ Intuitively, this come as a good guess, it works if we take $D=3$.

Now here are my problems/questions:

• First, the ansatz seems off as there are no arbitrary functions of $r$ that allow to give the radial expression. I instead considered $$\psi(\mathbf{x}) = \frac{1}{r^{(D-1)/2}} R(r) \mathcal{Y}^l(\mathbf{\theta}) $$
• Now, I applied the laplacian to my ansatz which yields: $$ \nabla^2 \left( \frac{1}{r^{(D-1)/2}} R(r) \mathcal{Y}^l(\mathbf{\theta}) \right) = \frac{1}{r^{(D-1)/2}} \left[\frac{d^2}{dr^2} - \frac{l(l+D-2) + (D-1)(D-3)/4}{r^2} \right] R(r) \mathcal{Y}^l $$ Which isn't at all the simple change $l(l+1) \rightarrow l(l+D-2)$ for the $A$ coefficient! Could it be that what is asked is wrong or am I missing something?

Here are the precise questions:

a. Show that $\psi(\mathbf{x}) = (1/r^{(D-1)/2} )\mathcal{Y}^l(\text{angles})$ is a clever ansatz that reduces the Schrödinger equation in $D$ dimensions to the form of Eq.(Klein-Gordon) in the relativistic case and Eq.(Schrödinger) in the nonrelativistic case.

b. Show that the only change in Eq. (14.10) (from Gilmore's book) is the replacement \begin{equation} l(l+1) \rightarrow l(l+D-2) \end{equation}

c. Show that the relativistic and nonrelativistic energies shown in Eq. (14.12) change as follows: \begin{align*} &\text{relativistic} \qquad \quad N' \rightarrow n +\frac{1}{2} +\sqrt{ \left(l + \frac{1}{2}\right)^2 + l(D-3) - \alpha^2} \\ &\text{nonrelativistic} \quad ~ N \rightarrow n +\frac{1}{2} +\sqrt{ \left(l + \frac{1}{2}\right)^2 + l(D-3)} \end{align*}