# Reducing angular Schrodinger equation to eigenvalue problem

1. Oct 8, 2013

### perishingtardi

1. The problem statement, all variables and given/known data
The angular part of the Schrodinger equation for a positron in the field of an electric dipole moment ${\bf d}=d{\bf \hat{k}}$ is, in spherical polar coordinates $(r,\vartheta,\varphi)$,
$$\frac{1}{\sin\vartheta}\frac{\partial}{\partial\vartheta} \left( \sin\vartheta\frac{\partial Y}{\partial\vartheta} \right) + \frac{1}{\sin^2 \vartheta}\frac{\partial^2 Y}{\partial\varphi^2} - 2dY\cos\vartheta + \lambda Y=0.$$
By considering the ansatz $$Y=Y_m (\vartheta,\varphi) = \sum_{l'=|m|}^{\infty} C_{l'} Y_{l'm}(\vartheta,\varphi),$$
where $Y_{l'm}$ are spherical harmonics, show that the problem of finding $\lambda$ reduces to a matrix eigenvalue problem of the form $$\sum_{l'=|m|}^{\infty} A_{ll'}C_{l'} = \lambda C_{l}.$$

2. Relevant equations
$$\frac{1}{\sin\vartheta}\frac{\partial}{\partial\vartheta} \left( \sin\vartheta\frac{\partial Y_{l'm}}{\partial\vartheta} \right) + \frac{1}{\sin^2 \vartheta}\frac{\partial^2 Y_{l'm}}{\partial\varphi^2} = -l'(l'+1)Y_{l'm}$$
$$\int_0^{2\pi}\int_0^\pi Y_{lm}^* Y_{l'm} \sin\vartheta\,d\vartheta\,d\varphi = \delta_{ll'}$$
$$Y_{lm}^* = (-1)^m Y_{l,-m}$$
$$\int_0^{2\pi}\int_0^\pi Y_{l_1 m_1}Y_{l_2 m_2}Y_{l_3 m_3} \sin\vartheta \, d\vartheta \, d\varphi = \sqrt{\frac{(2l_1 +1)(2l_2 + 1)(2l_3 + 1)}{4\pi}}\begin{pmatrix}l_1&l_2&l_3\\ 0&0&0\end{pmatrix}\begin{pmatrix}l_1&l_2&l_3\\ m_1&m_2&m_3\end{pmatrix}$$ where the arrays in parentheses are Wigner 3jm symbols.
$$\cos\vartheta = \sqrt{\frac{4\pi}{3}}Y_{10}$$

3. The attempt at a solution

By inserting the ansatz into the angular Schrodinger equation and using the first of the 'relevant equations' I got
$$\sum_{l'} C_{l'} \{ -l'(l'+1)Y_{l'm} - 2d\cos\vartheta\,Y_{l'm} + \lambda Y_{l'm} \}=0.$$
Then by multiplying through by $Y_{lm}^*$ and integrating over the unit sphere and using the other equations given, I got (after some manipulation)
$$2d(-1)^m \sum_{l'=|m|}^{\infty} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix}1&l&l'\\0&0&0\end{pmatrix}\begin{pmatrix}1&l&l'\\0&-m&m\end{pmatrix}C_{l'} = [\lambda - l(l+1)]C_l.$$
This is almost in the form required, I think, except that there is an extra term of $-l(l+1)$ on the RHS, which I don't know how to get rid of. Any help would be very much appreciated.

Last edited: Oct 8, 2013
2. Oct 23, 2013

### Staff: Mentor

Just move it to the LHS in the form of a diagonal term (you can use $\delta_{l,l'}$ if you want to put it in the sum).