# Reducing angular Schrodinger equation to eigenvalue problem

• perishingtardi
In summary, the angular part of the Schrodinger equation for a positron in the field of an electric dipole moment is reduced to a matrix eigenvalue problem by considering an ansatz and using relevant equations. The final equation is almost in the required form, except for an extra term that can be moved to the LHS in a diagonal form.

## Homework Statement

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 $$\sum_{l'=|m|}^{\infty} A_{ll'}C_{l'} = \lambda C_{l}.$$

## Homework 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.
$$\cos\vartheta = \sqrt{\frac{4\pi}{3}}Y_{10}$$

## 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:
perishingtardi said:
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.
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).

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