應數筆記（勒壤得方程式，Legendre's Equation）

 

Schrodinger’s Function

Def.$$i\hbar\frac{\partial}{\partial t}|\Psi(t)\rangle=\hat{H}|\Psi(t)\rangle$$

If we consider a steady-state result, we have 

$$i\hbar\frac{\partial}{\partial t}|\Psi(t)|=\hat{H}|\Psi(t)\rangle\Rightarrow\frac{-\hbar^{2}}{2m}\nabla^{2}\Psi(\vec{r})+V(r)\Psi(\vec{r})=E~\Psi(\vec{r})$$  where V(r) is the potential and E is the energy of the object(s).

Legendre’s Equation

Def.$$(1-x^{2})\frac{d^{2}y}{dx^{2}}-2~x\frac{dy}{dx}+l(l+1)y=0$$

Solution. We use the method of power series, we have $$y(x)=P_{1}(x)=\sum_{m=0}^{\lfloor\frac{l}{2}\rfloor}\frac{(-1)^{m}(2l-2m)!x^{l-2m}}{2^{l}m!(l-m)!(l-2m)!}, \quad x\in[-1,1]$$

 (Proof. See Advanced Engineering Mathematics, 10ed, Sect 5.2)

Associated Legendre’s Equation

Def.$$(1-x^{2})\frac{d^{2}y}{dx^{2}}-2~x\frac{dy}{dx}+\left(l(l+1)-\frac{k^{2}}{(1-x^{2})}\right)y=0, \quad x\in(-1,1)$$

Solution.$$y(x)=(1-x^{2})^{\frac{1}{k}}\frac{d^{k}P_{i}(x)}{dx^{k}}$$

(We skip the proof.)

Example. Solve $$\frac{-\hbar^{2}}{2m}\nabla^{2}\Psi(\vec{r})+V(r)\Psi(\vec{r})=E~\Psi(\vec{r})$$

<sol.>

Set $$\Psi(\vec{r})=R(r)\Phi(\phi)\Theta(\theta)$$ and we can expect $$\Phi(\phi)=\exp im\phi$$ because $$0\le\phi\le\pi$$. And we use spherical coordinates to solve this problem.

So, we have $$\frac{1}{r^{2}}\frac{\partial}{\partial r^{2}}r^{2}\frac{\partial}{\partial r}\Psi+\frac{1}{r^{2}}\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}\Psi+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\psi^{2}}\Psi\right)+V(r)\Psi=E\Psi$$ $$\frac{\Phi(\phi)\Theta(\theta)}{r^{2}}\frac{d}{dr}r^{2}\frac{dR}{dr}+\frac{R(r)\Phi(\phi)}{r^{2}}\left(\frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}\Theta+\frac{(-m^{2}\Theta)}{\sin^{2}\theta}\right)+\frac{2m(E-V)R(r)}{\hbar^{2}}R(r)\Phi(\phi)\Theta(\theta)=0$$

Now, we solve $$\frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}\Theta+\frac{(-m^{2}\Theta)}{\sin^{2}\theta}$$  first. We let $$x=\cos\theta$$, so we obtain $$dx=-\sin\theta~d\theta$$  and $$\frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}\Theta+\frac{(-m^{2}\Theta)}{\sin^{2}\theta} = \frac{1}{\sin\theta}\frac{d}{d\theta}\sin^{2}\theta\frac{d\Theta}{\sin\theta d\theta}-\frac{m^{2}\Theta}{\sin^{2}\theta}$$ $$\frac{1}{\sin\theta}\frac{d}{d\theta}\sin^{2}\theta\frac{d\Theta}{\sin\theta d\theta}-\frac{m^{2}\Theta}{\sin^{2}\theta} = \frac{d}{\sin\theta d\theta}(x^{2}-1)\frac{d\Theta}{dx}-\frac{m^{2}\Theta(x)}{1-x^{2}}$$ $$\frac{d}{\sin\theta d\theta}(x^{2}-1)\frac{d\Theta}{dx}-\frac{m^{2}\Theta(x)}{1-x^{2}} = \frac{-d}{dx}(x^{2}-1)\frac{d\Theta}{dx}-\frac{m^{2}\Theta(x)}{1-x^{2}}$$

We observe the above equation, we find it like a kind of special functions. Yes, it likes Associated Legendre’s Equation, so we can let $$(1-x^{2})\frac{d^{2}\Theta}{dx^{2}}-2x\frac{d\Theta}{dx}-\frac{m^{2}\Theta(x)}{1-x^{2}}=-l(l+1)\Theta$$. Actually, we can think it is an eigenvalue problem.

After the calculation, we can rewrite the equation below: $$\frac{\Phi(\phi)\Theta(\theta)}{r^{2}}\frac{d}{dr}r^{2}\frac{dR}{dr}+\frac{R(r)\Phi(\phi)}{r^{2}}\left(\frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}\Theta+\frac{(-m^{2}\Theta)}{\sin^{2}\theta}\right)+\frac{2m(E-V)R(r)}{\hbar^{2}}R(r)\Phi(\phi)\Theta(\theta)$$ $$= \frac{\Phi(\phi)\Theta(\theta)}{r^{2}}\frac{d}{dr}r^{2}\frac{dR}{dr}-\frac{R(r)\Phi(\phi)}{r^{2}}l(l+1)\Theta+\frac{2m(E-V)}{\hbar^{2}}R(r)\Phi(\phi)\Theta(\theta)=0$$$$\Rightarrow\frac{1}{r^{2}}\frac{d}{dr}r^{2}\frac{dR}{dr}-\frac{1}{r^{2}}l(l+1)R+\frac{2m(E-V)}{\hbar^{2}}R(r)=\frac{d^{2}R}{dr^{2}}+\frac{2}{r}\frac{dR}{dr}+\left(\frac{2m(E-V)}{\hbar^{2}}-\frac{l(l+1)}{r^{2}}\right)R=0$$ 

Note. $$\frac{d^{2}R}{dr^{2}}+\frac{2}{r}\frac{dR}{dr}+\left(\frac{2m(E-V)}{\hbar^{2}}-\frac{l(l+1)}{r^{2}}\right)R=0$$ is only for $$\Phi(\phi)=\exp im\phi$$ and$$\Theta=P_{1}(x=\cos\theta)=\sum_{m=0}^{\lfloor\frac{l}{2}\rfloor}\frac{(-1)^{m}(2l-2m)!(\cos\theta)^{l-2m}}{2^{l}m!(l-m)!(l-2m)!}, \quad \theta\in[0,\pi]$$.