# Schrodinger equation in spherical coordinates

I read a paper on solving Schrodinger equation with central potential, and I wonder how the author get the equation(2) below. Full text.

$$-\frac{1}{2}D^2\phi+\left(V+\frac{1}{2}\frac{l(l+1)}{r^2}\right)\phi=E\phi$$

They are quite different. Can anyone explain how to deduce equation(2)?

-

The difference is due to the fact that solid harmonics are not spherical harmonic. So, equation (2) and the more conventional equation from Griffith are equations for different functions $\phi$. The Schrodinger eq. (1)

$$-\frac{1}{2r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r}\psi\right) + \frac{\hat{L}^2}{2r^2}\psi + V\psi ~=~ E\psi$$

is indeed turned by substitution

$$\psi ~=~ R(r) Y_{\ell m}(\theta,\varphi)~=~ \phi(r) r^{\ell} Y_{\ell m}(\theta,\varphi)$$ to equation (2) if you do the math correctly. Note $r^{\ell}$ here: it is what differs solid harmonics from spherical harmonics. On the other hand, Griffith's function $\phi(r)$ is defined as $rR(r)$.

-

The solid harmonics were explained by Misha, so let me fill in the rest of the details.

Laplacian operator is given by $\Delta = \partial_x^2 + \partial_y^2 + \partial_z^2$. First suppose there we are only interested in the radial part. Using chain rule (and letting $D \equiv \partial_r$, as in your references), we can write $$\partial_x = r_{,x} D, \quad \partial_x^2 = r_{,xx} D + (r_{,x} D)^2.$$ We need to compute $$r_{,x} = {\partial \sqrt{x^2 + y^2 + z^2} \over \partial x } = {x \over r}$$ and $$r_{,xx} = \partial_x {x \over r} = {1 \over r} - {x^2 \over r^3}.$$ The expressions for $\partial_y$ and $\partial_z$ are of course similar, so that $$\Delta = ({3 \over r} - {r^2 \over r^3})D + D^2 = {2 \over r}D + D^2.$$

Now, the spherical part of the Laplacian operator is given by $-{L^2 \over r^2}$ where $\mathbf L$ is the angular momentum operator. If we use spherical harmonics of level $l$ (which are eigenstates of L^2 corresponding to eigenvalue $l(l+1)$) and make a substitution $\phi \to {\phi \over r}$ we get the Griffith's result.

-