spherical harmonics

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Separation of variables

In solving the time-independent Schrödinger equation for a particle moving under the influence of a potential $V(\mathbf{r})\,$ ,

$\left( -\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf{r}) \right)\psi(\mathbf{r}) = E\, \psi(\mathbf{r})\,$ ,

a great simplification occurs when the potential only depends on the magnitude of $\mathbf{r}\,$ , and can be written as $V(\mathbf{r}) = V(r)\,$ . The most important example of such a system is the hydrogen atom, and so we will investigate this system in great detail. We may express the wavefunction in spherical coordinates $r\,$ , $\theta\,$ and $\varphi\,$ . Recall that the Laplacian takes the form

$\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial }{\partial r} \right) + \frac{1}{r^2\sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial }{\partial \theta} \right) + \frac{1}{r^2\sin^2\theta} \frac{\partial^2 }{\partial \varphi^2}\,$ ,

and since

$L^2 = -\frac{\hbar^2}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta}\right) - \frac{\hbar^2}{\sin^2\theta}\frac{\partial^2}{\partial \varphi^2}$ ,

we may write the Schrödinger equation as

$\left[ -\frac{\hbar^2}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r} \right) + \frac{L^2}{r^2} + V(r) \right]\psi(r,\theta,\varphi) = E \,\psi(r,\theta,\varphi)\,$ .

It turns out that this equation can be solved by the method of separation of variables, letting $\psi(r,\theta,\varphi) = R(r) F(\theta,\varphi)\,$ . Then

$\left[ -\frac{\hbar^2}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R(r)}{\partial r} \right) + V(r) R(r) \right] F(\theta,\varphi) + \frac{R(r)}{r^2} L^2 F(\theta,\varphi)\,$

$= E \,R(r) F(\theta,\varphi)\,$ ,

and, dividing by $R(r) F(\theta,\varphi)\,$ ,

$\frac{1}{R(r)}\left[ -\frac{\hbar^2}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R(r)}{\partial r} \right) + V(r) R(r) \right] + \frac{1}{r^2 F(\theta,\varphi)} L^2 F(\theta,\varphi)\,$

$= E\,$ .

The first term depends only on $r\,$ , while the right hand side is simply a constant. It follows that the second term cannot depend on $\theta\,$ or $\varphi\,$ , and therefore

$L^2 F(\theta,\varphi) = \lambda F(\theta,\varphi)\,$

for some constant $\lambda\,$ . But this is precisely the statement that $F(\theta,\varphi)\,$ be an eigenfunction of $L^2\,$ . We have already determined the eigenvalues of $L^2\,$ to be $\ell(\ell+1)\,$ where $\ell = 0, 1, 2, 3...\,$ , so let us write the remaining radial equation as

$\left[ -\frac{\hbar^2}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r} \right) + V(r) + \frac{\ell(\ell+1)}{r^2} \right]R(r) = E\, R(r)\,$ .

Eigenfunctions of $L^2\,$ and $L_z\,$

What remains then is to find the eigenfunctions of the operator $L^2\,$ . From the discussion of angular momentum, we expect to be able to find functions that are eigenfunctions of both $L^2\,$ and $L_z\,$ , and this is indeed the case. The equation

$L^2\, Y_\ell^m(\theta,\varphi) = \ell(\ell+1)\, Y_\ell^m(\theta,\varphi)\,$

can further be solved using method of separation of variables. The only square-integrable solutions occur when $\ell\,$ is a non-negative integer, and these solutions are known as spherical harmonics, given by