物理

氢原子薛定谔方程分离变量形式$

原子核带电量为 $Ze$，势能函数为 $V(r) = -\frac{Ze^2}{4\pi\varepsilon_0 r}$。

三维定态薛定谔方程形式为 $\left(-\frac{\hbar^2}{2m_\mu}\nabla^2 + V(r)\right)\psi = E\psi$，其中约化质量 $m_\mu = \frac{m'm_e}{m'+m_e}$，球

坐标尺度因子 $h_r=1, h_\theta=r, h_\varphi=r\sin\theta$。

拉普拉斯算符在球坐标系下的表达式推导如下：

$∇^2 = \frac{1}{h_r h_θ h_φ} \frac{∂}{∂ r}(\frac{h_θ h_φ}{h_r}\frac{∂}{∂ r}) + \frac{∂}{∂ θ}(\frac{h_r h_φ}{h_θ}\frac{∂}{∂ θ})$ 

$+\frac{∂}{∂ φ}(\frac{h_r h_θ}{h_φ}\frac{∂}{∂ φ} )= \frac{1}{r^2sinθ} \frac{∂}{∂ r} (r^2sinθ\frac{∂}{∂ r})$

$+ \frac{∂}{∂ θ}(sinθ\frac{∂}{∂ θ}) + \frac{∂}{∂ φ}(\frac{1}{sinθ}\frac{∂}{∂ φ})$

$= \frac{1}{r^2}\frac{∂}{∂ r}\left(r^2\frac{∂}{∂ r}\right) + \frac{1}{r^2sinθ}\frac{∂}{∂ θ}\left(sinθ\frac{∂}{∂ θ}\right)$

$+ \frac{1}{r^2sin^2θ}\frac{∂^2}{∂ φ^2}$

代入薛定谔方程得：$-\frac{\hbar^2}{2m_\mu} \frac{1}{r^2}\frac{\partial}{\partial r}( r^2\frac{\partial \psi}{\partial r})$

$+ \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial \psi}{\partial \theta}\right)$

$+ \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \psi}{\partial \varphi^2} + V(r)\psi = E\psi$

展开径向部分导数项：$-\frac{\hbar^2}{2m_\mu}\frac{\partial^2\psi}{\partial r^2} - \frac{\hbar^2}{m_\mu r}\frac{\partial \psi}{\partial r}$

$- \frac{\hbar^2}{2m_\mu r^2}\frac{\partial^2\psi}{\partial \theta^2} - \frac{\hbar^2}{2m_\mu r^2}\cot\theta\frac{\partial \psi}{\partial \theta}$

$- \frac{\hbar^2}{2m_\mu r^2\sin^2\theta}\frac{\partial^2\psi}{\partial \varphi^2} + V(r)\psi = E\psi \quad (1.1)$

引入角动量算符，动量算符 $\hat{p} = -i\hbar\nabla$，角动量分量定义为

$\hat{L}_x = \hat{y}\hat{p}_z - \hat{z}\hat{p}_y = $

$-i\hbar\left(y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}\right)$，$\hat{L}_y = \hat{z}\hat{p}_x - \hat{x}\hat{p}_z$

$= -i\hbar\left(z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z}\right)$，$\hat{L}_z = \hat{x}\hat{p}_y - \hat{y}\hat{p}_x$

$= -i\hbar\left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)$

上述形式中 $(1.1)$ 式可重写为：

$-\frac{\hbar^2}{2m_\mu}\frac{\partial^2\psi}{\partial r^2} - \frac{\hbar^2}{m_\mu r}\frac{\partial \psi}{\partial r}$

$- \frac{\hbar^2}{2m_\mu r^2} \frac{1}{\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta\frac{\partial \psi}{\partial \theta})$

$+ \frac{1}{\sin^2\theta}\frac{\partial^2 \psi}{\partial \varphi^2} + V(r)\psi = E\psi$

由角动量平方算符 $\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$

$x=r\sin\theta\cos\varphi, y=r\sin\theta\sin\varphi, z=r\cos\theta$

可得角动量分量算符的球坐标表示：

$\hat{L}_x = i\hbar\left(\sin\varphi\frac{\partial}{\partial \theta} + \cot\theta\cos\varphi\frac{\partial}{\partial \varphi}\right)$

$\hat{L}_y = -i\hbar\left(\cos\varphi\frac{\partial}{\partial \theta} - \cot\theta\sin\varphi\frac{\partial}{\partial \varphi}\right)$

$\hat{L}_z = -i\hbar\frac{\partial}{\partial \varphi}$

计算 $\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$：

$\hat{L}^2 = \left(i\hbar\left(\sin\varphi\frac{\partial}{\partial \theta} + \cot\theta\cos\varphi\frac{\partial}{\partial \varphi}\right)\right)^2$

$+ \left(-i\hbar\left(\cos\varphi\frac{\partial}{\partial \theta} - \cot\theta\sin\varphi\frac{\partial}{\partial \varphi}\right)\right)^2$

$+ \left(-i\hbar\frac{\partial}{\partial \varphi}\right)^2$

$= -\hbar^2 \frac{1}{\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta\frac{\partial}{\partial \theta})$

$+ \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \varphi^2} $

将 $\hat{L}^2$ 代入 $(1.1)$ 式中，方程化为：

$-\frac{\hbar^2}{2m_\mu}\frac{\partial^2\psi}{\partial r^2} - \frac{\hbar^2}{m_\mu r}\frac{\partial \psi}{\partial r}$

$+ \frac{\hat{L}^2}{2m_\mu r^2}\psi + V(r)\psi = E\psi$

分离变量，设 $\psi(r,\theta,\varphi) = R(r)Y(\theta,\varphi)$，整理得：

$\frac{\hbar^2}{2m_υ}\frac{∂^2\psi}{∂ r^2} + \frac{\hbar^2}{m_\mu r}\frac{\partial \psi}{\partial r}+ (E-V(r))\psi = \frac{1}{2m_\mu r^2}\hat{L}^2\psi$

两边同乘 $-\frac{2m_\mu r^2}{\hbar^2 \psi}$ 并分离变量：

$r^2\frac{\partial^2 R}{\partial r^2} + 2r\frac{\partial R}{\partial r} + \frac{2m_\mu r^2}{\hbar^2}(E-V(r))R = \frac{1}{\hbar^2 Y}\hat{L}^2 Y $(1.2)

$(1.2)$ 式左边仅为 $r$ 的函数，右边仅为角度变量的函数，故等于分离常数。设 $\hat{L}^2 Y = \alpha \hbar^2 Y$，则径向方程为：

$r^2\frac{d^2 R}{dr^2} + 2r\frac{d R}{dr} + \frac{2m_\mu r^2}{\hbar^2}(E-V(r))R = \alpha R$。

角度部分 $Y(\theta,\varphi) = \Theta(\theta)\Phi(\varphi)$，代入角动量本征方程：

$ \frac{1}{sinθ}\frac{∂}{∂ θ} sinθ\frac{∂}{∂ θ} + \frac{1}{sin^2θ}\frac{∂^2}{∂ φ^2}Θ Φ = -α Θ Φ$

除以 $\Theta\Phi$ 并乘以 $\sin^2\theta$ 分离变量：$\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)$

$+ \alpha\sin^2\theta = -\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2} = m^2$

$\varphi$ 方向方程为 $\frac{d^2\Phi}{d\varphi^2} + m^2\Phi = 0$，解为 $\Phi = Ae^{im\varphi} + Be^{-im\varphi} \ (m \neq 0)$ 或

$\Phi = C + D\varphi \ (\text{特解 } m=0)$。由波函数单值性条件 $\Phi(\varphi+2\pi)=\Phi(\varphi)$ 可知 $m$ 必须为整数。

归一化条件 $\int_{-\infty}^{\infty} \psi^*(x)\psi(x)dx = 1$。Fourier变换 $\tilde{\phi}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \psi(x)e^{-ikx}dx$。

当 $\psi(x) = \frac{1}{\sqrt{2\pi}}e^{-ikx}$ 时满足波函数归一化条件 $\int_{-\infty}^{\infty} \psi(x)\psi^*(x)dx = 1$。

$\theta$ 方向方程为

$\frac{1}{\Theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \left(\alpha - \frac{m^2}{\sin^2\theta}\right)\Theta = 0$令 

$u=\cos\theta$，$\Theta(\theta)=P(u)$

得到关联勒让德方程：

$\frac{d}{du}\left[(1-u^2)\frac{dP}{du}\right] + \left(\alpha - \frac{m^2}{1-u^2}\right)P = 0$。

为使解在区间 $[-1, 1]$ 上有界，参数 $\alpha$ 必须取值为 $l(l+1)$，其中 $l=0,1,2,\dots$ 且 $|m| \le l$

此时解为关联勒让德多项式 $P(u) = P_l^{|m|}(u)$。

球谐函数定义为 $Y_{lm}(\theta,\varphi) = N_{lm} P_l^{|m|}(\cos\theta) e^{im\varphi}$，其中归一化系数

$N_{lm} = \sqrt{\frac{(l-|m|)!}{(l+|m|)!} \frac{2l+1}{4\pi}}$。