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

In Griffiths's book, it reads

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

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

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2 Answers 2

up vote 8 down vote accepted

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 $rR(r)$.

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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.

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