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: The red is your actual potential, which _does not_ extend to the negatives!

The red is your actual potential, which does not extend to the negatives! This potential will be an entirely different beast from the harmonic oscillator potential that you seem to think it is. Also, trying to solve for a shifted potential doesn't make too much sense, since your choice of coordinates should always be taking full advantage of the symmetry.

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    $\begingroup$ This doesn't answer the question. I know that it's different than a harmonic oscillator, that's why I'm having trouble with the solution. Telling me not to solve for this potential doesn't answer my question. $\endgroup$ – user138458 Nov 5 '18 at 1:07

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