Can one explicitly solve the shifted harmonic oscillator

Question

The quantum harmonic oscillator is described by a Hamiltonian

$$H=-\Delta + \left\lvert x \right\rvert^2.$$

By decomposing the eigenvalue problem $H\psi= E\psi$ into its angular and radial part one obtains essentially the ODE

$$-\varphi''(r) + \frac{2}{r} \varphi'(r) + \frac{l(l+1)}{r^2} \varphi(r)+r^2\varphi(r)=E\varphi(r).$$

Please let me know if you have any questions.

This ODE can be solved explicitly, i.e. one obtains the usual eigenfunctions and eigenvalues

I wonder whether one can also solve the shifted version

$$-\varphi''(r) + \frac{2}{r} \varphi'(r) + \frac{l(l+1)}{r^2} \varphi(r)+(r-r_0)^2\varphi(r)=E\varphi(r)$$

for some $r_0>0.$

It does not seem possible to map this one immediately back to the original one that's why I am asking.

Please let me know if you have questions or remarks.

Answer 1

Hint:

$-\varphi''(r)+\dfrac{2}{r}\varphi'(r)+\dfrac{l(l+1)}{r^2}\varphi(r)+(r-r_0)^2\varphi(r)=E\varphi(r)$

$\dfrac{d^2\varphi}{dr^2}-\dfrac{2}{r}\dfrac{d\varphi}{dr}-\left(r^2-2r_0r+r_0^2-E+\dfrac{l(l+1)}{r^2}\right)\varphi=0$

Let $\varphi=e^{\pm\frac{r^2}{2}}u$ ,

Then $\dfrac{d\varphi}{dr}=e^{\pm\frac{r^2}{2}}\dfrac{du}{dr}\pm re^{\pm\frac{r^2}{2}}u$

$\dfrac{d^2\varphi}{dr^2}=e^{\pm\frac{r^2}{2}}\dfrac{d^2u}{dr^2}\pm re^{\pm\frac{r^2}{2}}\dfrac{du}{dr}\pm re^{\pm\frac{r^2}{2}}\dfrac{du}{dr}+(r^2\pm1)e^{\pm\frac{r^2}{2}}u=e^{\pm\frac{r^2}{2}}\dfrac{d^2u}{dr^2}\pm2re^{\pm\frac{r^2}{2}}\dfrac{du}{dr}+(r^2\pm1)e^{\pm\frac{r^2}{2}}u$

$\therefore e^{\pm\frac{r^2}{2}}\dfrac{d^2u}{dr^2}\pm2re^{\pm\frac{r^2}{2}}\dfrac{du}{dr}+(r^2\pm1)e^{\pm\frac{r^2}{2}}u-\dfrac{2}{r}\left(e^{\pm\frac{r^2}{2}}\dfrac{du}{dr}\pm re^{\pm\frac{r^2}{2}}u\right)-\left(r^2-2r_0r+r_0^2-E+\dfrac{l(l+1)}{r^2}\right)e^{\pm\frac{r^2}{2}}u=0$

$\dfrac{d^2u}{dr^2}+2\left(\pm r-\dfrac{1}{r}\right)\dfrac{du}{dr}+\left(2r_0r-r_0^2+E\mp1-\dfrac{l(l+1)}{r^2}\right)u=0$

Let $u=e^{\mp r_0r}v$ ,

Then $\dfrac{du}{dr}=e^{\mp r_0r}\dfrac{dv}{dr}\mp r_0e^{\mp r_0r}v$

$\dfrac{d^2u}{dr^2}=e^{\mp r_0r}\dfrac{d^2v}{dr^2}\mp r_0e^{\mp r_0r}\dfrac{dv}{dr}\mp r_0e^{\mp r_0r}\dfrac{dv}{dr}+r_0^2e^{\mp r_0r}v=e^{\mp r_0r}\dfrac{d^2v}{dr^2}\mp 2r_0e^{\mp r_0r}\dfrac{dv}{dr}+r_0^2e^{\mp r_0r}v$

$\therefore e^{\mp r_0r}\dfrac{d^2v}{dr^2}\mp 2r_0e^{\mp r_0r}\dfrac{dv}{dr}+r_0^2e^{\mp r_0r}v+2\left(\pm r-\dfrac{1}{r}\right)\left(e^{\mp r_0r}\dfrac{dv}{dr}\mp r_0e^{\mp r_0r}v\right)+\left(2r_0r-r_0^2+E\mp1-\dfrac{l(l+1)}{r^2}\right)e^{\mp r_0r}v=0$

$\dfrac{d^2v}{dr^2}\pm2\left(r-r_0\mp\dfrac{1}{r}\right)\dfrac{dv}{dr}+\left(E\mp1\pm\dfrac{2r_0}{r}-\dfrac{l(l+1)}{r^2}\right)v=0$

Let $v=r^kw$ ,

Then $\dfrac{dv}{dr}=r^k\dfrac{dw}{dr}+kr^{k-1}w$

$\dfrac{d^2v}{dr^2}=r^k\dfrac{d^2w}{dr^2}+kr^{k-1}\dfrac{dw}{dr}+kr^{k-1}\dfrac{dw}{dr}+k(k-1)r^{k-2}w=r^k\dfrac{d^2w}{dr^2}+2kr^{k-1}\dfrac{dw}{dr}+k(k-1)r^{k-2}w$

$\therefore r^k\dfrac{d^2w}{dr^2}+2kr^{k-1}\dfrac{dw}{dr}+k(k-1)r^{k-2}w\pm2\left(r-r_0\mp\dfrac{1}{r}\right)\left(r^k\dfrac{dw}{dr}+kr^{k-1}w\right)+\left(E\mp1\pm\dfrac{2r_0}{r}-\dfrac{l(l+1)}{r^2}\right)r^kw=0$

$\dfrac{d^2w}{dr^2}\pm2\left(r-r_0\mp\dfrac{k-1}{r}\right)\dfrac{dw}{dr}+\left(E\pm2k\mp1\mp\dfrac{2r_0(k-1)}{r}+\dfrac{k(k-3)-l(l+1)}{r^2}\right)w=0$

Which can converts to Heun's Biconfluent Equation similar to Hunt for exact solutions of second order ordinary differential equations with varying coefficients.