應數筆記（勒壤得方程式，Legendre's Equation）

06 Jun, 2026

Schrodinger’s Function

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

If we consider a steady-state result, we have

Legendre’s Equation

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

Solution. We use the method of power series, we have

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

Associated Legendre’s Equation

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

Solution. $$y(x)=(1-x^2{)}^{\frac{2}{k}}\frac{d^k{P}_i(x)}{d{x}^k}$$

(We skip the proof.)

Example.

Solve $\frac{-{\hslash }^2}{2m}{\nabla }^2\Psi (\overrightarrow{r})+V(r)\Psi (\overrightarrow{r})=E\Psi (\overrightarrow{r})$

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

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

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 $(1-x^2)\frac{d^2\Theta }{d{x}^2}-2x\frac{d\Theta }{dx}-\frac{m^2\Theta (x)}{1-x^2}=-l(l+1)\Theta$. Actually, we can think it is an eigenvalue problem.

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

Note. $\frac{d^{2}R}{d{r}^2}+\frac{2}{r}\frac{dR}{dr}+(\frac{2m(E-V)}{{\hslash }^2}-\frac{l(l+1)}{r^2})R=0$ is only for $\Phi (\varphi )=\exp im\varphi$ and$\Theta =P\_l(x=\cos \theta )=\sum \_{m=0}^{\lfloor \frac{l}{2}\rfloor }\frac{(-1{)}^m(2l-2m)!(\cos \theta {)}^{l-2m}}{2^{l}m!(l-m)!(l-2m)!},\theta \in [0,\pi ]$.

#物理