# Power series solution for a shifted spherical harmonic oscillator

I'm trying to solve the Schrodinger equation for a radial Harmonic oscillator whos equilibrium point has been shifted away from the origin, i.e. $$V(r) = V_0(r-1)^2$$. The standard approach is to make a substitution for the radial wavefunction $$u(r) = rR(r)$$ so that $$\frac{d^2u}{dr^2} -\left[V_0(r-1)^2+\frac{\ell(\ell+1)}{r^2} \right]u(r) = -Eu(r)$$ where I've set $$\hbar=1$$ and $$m=1/2$$ for simplicity. I then want to find the solution in terms of some nicely behaved power series. We can look at the asymptotic behaviour to make a good guess. As $$r \to \infty$$ we have $$\frac{d^2u}{dr^2} \approx V_0(r-1)^2 u$$ so that $$u \to e^{-\frac{\alpha}{2} (r-1)^2}$$ where $$\alpha := \sqrt{V_0}$$. Furthermore, as $$r\to 0$$ we have $$\frac{d^2u}{dr^2} \approx \frac{\ell(\ell+1)}{r^2}u$$ so that $$u \to r^{\ell+1}.$$ I then guess that a nice series solution which captures the desired asymptotic behaviour is $$u(r) = e^{-\frac{\alpha}{2} (r-1)^2} r^{\ell+1}\sum_k a_k r^k.$$ Plugging this into the differential equation, we get that $$0 = \sum_k a_k \left[2\alpha(\ell+1)r^k+ \left(E- \alpha(2\ell+3)\right)r^{k+1}+2\ell(\ell+1)kr^{k-1}-2\alpha(r-1)kr^k+k(k-1)r^{k-1} \right]$$ which implies $$0 = \sum_k \left[2\alpha a_k(\ell+k+1) + a_{k+1}(k+1)(2\ell(\ell+1)+k) + a_{k-1}\left(E-\alpha(2\ell+2k+3) \right) \right]r^k$$ But I am unsure of how to solve such a recursion relation where there are 3 unknown coefficients in each relation. In every case I've seen before, this reduces to equations with only two power series coefficients to the recursion relation is straightforward. Have I made a mistake in my asymptotic analysis? How should I go about solving it with this method?

There is a problem with the setup here. Note that $$r \in \mathbb{R}^+$$, since it doesn't make sense for the radius to be negative. So what your potential actually looks like is this: