Generalized Legendre differential equation | Quantum Science Philippines

## Generalized Legendre differential equation

Vanessa V. Destura, MSU-IIT

Starting with an eigenvalue equation, $L^2 \left|l,m\right> = \lambda {\hbar}^2 \left|l,m\right>$

where $\left|l,m\right> \equiv Y(\theta,\phi)$ $L^2 = -{\hbar}^2 \left[ \dfrac{1}{\sin^2\theta} \dfrac{\partial^2}{\partial\phi^2} + \dfrac{1}{\sin\theta} \dfrac{\partial}{\partial\theta} \left(\sin\theta \dfrac{\partial}{\partial\theta} \right) \right]$

Also, we set $Y(\theta, \phi) = \Theta(\theta) \Phi(\phi) = \Theta\Phi$

So, substituting the given values, we find $\left[ \dfrac{1}{\sin^2\theta} \dfrac{\partial^2}{\partial\phi^2} + \dfrac{1}{\sin\theta} \dfrac{\partial}{\partial\theta} \left(\sin\theta \dfrac{\partial}{\partial\theta} \right) \right] Y(\theta, \phi) = -\lambda Y(\theta, \phi)$

Now, by separation of variables, the equation yields to $\dfrac{1}{\Theta\Phi} \left[ \dfrac{\partial^2}{\partial\phi^2} (\Theta\Phi) + \sin\theta \dfrac{\partial}{\partial\theta} \left( \sin\theta \dfrac{\partial}{\partial\theta}\right) (\Theta\Phi) = -\lambda \sin^2\theta (\Theta\Phi) \right] \dfrac{1}{\Theta\Phi}$ $\underbrace{\dfrac{1}{\Phi} \dfrac{\mathrm{d}^2\Phi}{\mathrm{d}\phi^2}}_{-\mathrm{m}^2} + \dfrac{\sin\theta}{\Theta} \dfrac{\mathrm{d}}{\mathrm{d}\theta} \left( \sin\theta \dfrac{\mathrm{d}\Theta}{\mathrm{d} \theta} \right) = -\lambda \sin^2\theta$ $\left[ \dfrac{\sin\theta}{\Theta} \dfrac{\mathrm{d}}{\mathrm{d}\theta} \left( \sin\theta \dfrac{\mathrm{d}\Theta}{\mathrm{d}\theta} \right) = -\lambda \sin^2\theta + \mathrm{m}^2 \right] \dfrac{\Theta}{\sin^2\theta}$ $\dfrac{1}{\sin\theta} \dfrac{\mathrm{d}}{\mathrm{d}\theta} \left( \sin\theta \dfrac{\mathrm{d}\Theta}{\mathrm{d}\theta} \right) = \left( -\lambda + \dfrac{\mathrm{m}^2}{\sin^2\theta} \right) \Theta$ $\dfrac{1}{\sin\theta} \dfrac{\mathrm{d}}{\mathrm{d}\theta} \left(\sin\theta \dfrac{\mathrm{d}\Theta}{\mathrm{d}\theta} \right) + \left( \lambda - \dfrac{\mathrm{m^2}}{\sin^2\theta} \right) \Theta = 0$

Let $\mathrm{x} = \cos\theta$. So, $\mathrm{dx} = -\sin\theta \mathrm{d}\theta$ $\dfrac{\mathrm{dx}}{\mathrm{d}\theta} = -\sin\theta$

Note that $\begin{array}{rcl} \sin^2\theta + \cos^2\theta &=& 1 \\ \sin^2\theta = 1 - \cos^2\theta \\ \sin\theta = \sqrt{1-\mathrm{x}^2} \end{array}$

So, $\dfrac{\mathrm{d}}{\mathrm{d}\theta} = \dfrac{\mathrm{d}}{\mathrm{dx}}\left(\dfrac{\mathrm{dx}}{\mathrm{d}\theta}\right) = -\dfrac{\mathrm{d}}{\mathrm{dx}} \left(\sin\theta \right) = -\dfrac{\mathrm{d}}{\mathrm{dx}} \left( \sqrt{1-\mathrm{x}^2} \right)$

Thus, the associated Legendre differential equation is $\dfrac{1}{\sin\theta} \dfrac{\mathrm{d}}{\mathrm{d}\theta} \left(\sin\theta \dfrac{\mathrm{d}\Theta}{\mathrm{d}\theta} \right) + \left( \lambda - \dfrac{\mathrm{m}^2}{\sin^2\theta}\right) \Theta = 0$ $\dfrac{\mathrm{d}}{\mathrm{dx}} \left[ \sqrt{1-\mathrm{x}^2} \dfrac{\mathrm{d}}{\mathrm{dx}}\left( \sqrt{1-x^2} \right) \Theta \right] + \left( \lambda - \dfrac{\mathrm{m^2}}{(1-\mathrm{x}^2)} \right) \Theta = 0$ $\dfrac{\mathrm{d}}{\mathrm{dx}} \left[ \sqrt{1-\mathrm{x}^2} \dfrac{\mathrm{d}\Theta}{\mathrm{dx}} \right] + \left[ \lambda - \dfrac{\mathrm{m}^2}{1-\mathrm{x}^2} \right] \Theta = 0$

Since the solutions are symmetric in $\mathrm{x}$, assume $\mathrm{m} = 0$. So, $\dfrac{\mathrm{d}}{\mathrm{dx}} \left[ (1-\mathrm{x}^2) \dfrac{\mathrm{d}\Theta}{\mathrm{dx}} \right ] + \lambda \Theta = 0$ $(1-\mathrm{x}^2) \dfrac{\mathrm{d}^2\Theta}{\mathrm{dx}^2} + (-2\mathrm{x}) \dfrac{\mathrm{d}\Theta}{\mathrm{dx}} + \lambda\Theta = 0$ $\begin{array}{rclr} \dfrac{\mathrm{d}^2\Theta}{\mathrm{dx}^2} - \mathrm{x}^2 \dfrac{\mathrm{d}^2\Theta}{\mathrm{dx}^2} - 2\mathrm{x} \dfrac{\mathrm{d}\Theta}{\mathrm{dx}} + \lambda\Theta &=& 0 & \textbf{(1)} \end{array}$

Using Frobenius method (power series method), we let $\Theta(\mathrm{x}) = \sum\limits_{n=0}^{\infty} \mathrm{a}_n \mathrm{x}^n$

Now, we take the first and second total derivative of $\Theta(\mathrm{x})$. So, $\dfrac{\mathrm{d}\Theta}{\mathrm{dx}} = \sum\limits_{n=0}^{\infty} n \mathrm{a}_n \mathrm{x}^{n-1}$ $\mathrm{x} \dfrac{\mathrm{d}\Theta}{\mathrm{dx}} = \sum\limits_{n=0}^{\infty} n \mathrm{a}_n \mathrm{x}^{n}$ $\dfrac{\mathrm{d}^2\Theta}{\mathrm{dx}^2} = \sum\limits_{n=0}^{\infty} n (n-1) \mathrm{a}_n \mathrm{x}^{n-2}$ $\mathrm{x}^2 \dfrac{\mathrm{d}^2\Theta}{\mathrm{dx}^2} = \sum\limits_{n=0}^{\infty} n(n-1) \mathrm{a}_n \mathrm{x}^{n}$

Substituting these terms to equation $\textbf{(1)}$ yields to $\sum\limits_{n=0}^{\infty} n(n-1) \mathrm{a}_n \mathrm{x}^{n-2} - \sum\limits_{n=0}^{\infty} n(n-1) \mathrm{a}_n \mathrm{x}^n - \sum\limits_{n=0}^{\infty} 2n \mathrm{a}_n \mathrm{x}^n + \sum\limits_{n=0}^{\infty} \lambda \mathrm{a}_n \mathrm{x}^n = 0$ $\sum\limits_{n=0}^{\infty} \underbrace{ \left[ (n+2) (n+1) \mathrm{a}_{n+2} - n(n-1) \mathrm{a}_n - 2n \mathrm{a}_n + \lambda \mathrm{a}_n \right]}_{=0} \mathrm{x}^n = 0$ $(n+2) (n+1) \mathrm{a}_{n+2} - n(n-1) \mathrm{a}_n - 2n \mathrm{a}_n + \lambda \mathrm{a}_n = 0$ $(n+2) (n+1) \mathrm{a}_{n+2} - \left[ n(n-1) + 2n - \lambda \right] \mathrm{a}_n = 0$ $(n+2) (n+1) \mathrm{a}_{n+2} = \left[ n(n-1) + 2n - \lambda \right] \mathrm{a}_n$

Hence, the recurrence relation for the coefficients $\mathrm{a}_n$ is given by $\mathrm{a}_{n+2} = \dfrac{n(n-1) + 2n - \lambda}{(n+2) (n+1)} \mathrm{a}_n$