分波法

中心势场具有球对称性，所以采用球谐函数为基也许会比较简单，即将本征波函数写为

$$\psi_{klm}(\vec x)=R_{kl}(r)Y_{lm}(\theta,\phi)\tag1$$

而散射的情形要求

$$\psi(\vec x)=e^{i\vec k\cdot\vec x}+\frac{e^{ikr}}{r}f(\theta,\phi)\tag2$$

所以分波法的思想为把(2)用(1)展开

形式展开

(2)两部分分别写为(1)的形式，第一部分是固定的

$$e^{i\vec k\cdot\vec x}=\sum_{lm}4\pi i^l j_l(kr)Y_{lm}^*(\hat k)Y_{lm}(\hat r)\tag{1.1}$$

第二部分只能形式地写

$$\frac{e^{ikr}}rf(\theta,\phi)=\sum_{lm}\frac{e^{ikr}}rf_{lm}Y_{lm}(\hat r)\tag{1.2}$$

其中

$$f(\theta,\phi)=\sum_{lm}f_{lm}Y_{lm}(\theta,\phi)\Leftrightarrow f_{lm}=\int\mathrm d\Omega\ f(\theta,\phi)Y^*_{lm}(\theta,\phi)\tag{1.3}$$

假设(2)用(1)展开的展开系数为$A_{lm}$，则

$$A_{lm}R_{kl}(r)=4\pi i^l j_l(kr)Y_{lm}^*(\hat k)+\frac{e^{ikr}}rf_{lm}\tag{1.4}$$

求解自由情形径向方程

既然(1)是本征函数，那么应该满足薛定谔方程

$$\left[-\frac{\hbar^2}{2m}\left(\frac1{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac1{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac1{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right)+V(r)\right]\psi_{klm}(\vec x)=\frac{\hbar^2k^2}{2m}\psi_{klm}(\vec x)\tag{2.1}$$

利用球谐函数可知上式可化为

$$\left[-\frac{\hbar^2}{2m}\left(\frac1{r^2}\frac{\mathrm d}{\mathrm d r}\left(r^2\frac{\mathrm d}{\mathrm d r}\right)-\frac{l(l+1)}{r^2}\right)+V(r)\right]R_{kl}(r)=\frac{\hbar^2k^2}{2m}R_{kl}(r)\tag{2.2}$$

自由情形$V(r)\equiv0$，所以

$$\left[\frac1{r^2}\frac{\mathrm d}{\mathrm d r}\left(r^2\frac{\mathrm d}{\mathrm d r}\right)-\frac{l(l+1)}{r^2}\right]R_{kl}(r)=-k^2R_{kl}(r)\tag{2.2}$$

两个线性无关解为

$$\left\{\begin{aligned} &R_{kl}(r)=\sum_{n=0}^\infty \frac{(-1)^n}{2^nn!(2l+2n+1)!!}(kr)^{l+2n}\quad\rightarrow j_l(kr)\\ &R_{kl}(r)=\sum_{n=0}^\infty\frac{(-1)^n}{2^nn!(-2l+2n-1)!!} (kr)^{-l-1+2n}\quad\rightarrow y_l(kr)\\ \end{aligned}\right.\tag{2.3}$$

当半径很大时，渐近形式为

$$\left\{\begin{aligned} &j_l(kr)\sim\frac{\sin(kr-l\pi/2)}{kr}\\ &y_l(kr)\sim-\frac{\cos(kr-l\pi/2)}{kr}\\ \end{aligned}\right.\tag{2.4}$$

所以方便（球面波形式）起见，采用其线性叠加，即所谓的汉克尔函数

$$\left\{\begin{aligned} &h^{(1)}_l(kr)=j_l(kr)+iy_l(kr)\sim \frac{e^{ikr-i(l+1)\pi/2}}{kr}\\ &h^{(2)}_l(kr)=j_l(kr)-iy_l(kr)\sim \frac{e^{-ikr+i(l+1)\pi/2}}{kr}\\ \end{aligned}\right.\tag{2.5}$$

分波法结论

取$R_{kl}(r)=e^{i\delta_l}h_l^{(1)}(kr)+e^{-i\delta_l}h_l^{(2)}(kr)$，代入(1.4)

$$A_{lm}\left[e^{i\delta_l}h_l^{(1)}(kr)+e^{-i\delta_l}h_l^{(2)}(kr)\right]=4\pi i^l j_l(kr)Y_{lm}^*(\hat k)+\frac{e^{ikr}}rf_{lm}\tag{3.1}$$

在半径很大时，根据(2.4)(2.5)得到渐近形式

$$A_{lm}\left[(-i)^{l+1}e^{ikr+i\delta_l}+i^{l+1}e^{-ikr-i\delta_l}\right]=-2\pi i\left[e^{ikr}+(-1)^{l+1}e^{-ikr}\right]Y_{lm}^*(\hat k)+ke^{ikr}f_{lm}\tag{3.2}$$

将发散部分和汇聚部分分开

$$\left\{\begin{aligned} &i^{l+1}e^{-i\delta_l}A_{lm}=-2\pi i(-1)^{l+1}Y_{lm}^*(\hat k)\\ &(-i)^{l+1}e^{i\delta_l}A_{lm}=-2\pi iY_{lm}^*(\hat k)+kf_{lm} \end{aligned}\right.\Rightarrow\left\{\begin{aligned} &A_{lm}=2\pi i^lY_{lm}^*(\hat k)e^{i\delta_l}\\ &f_{lm}=-\frac{2\pi i}{k}Y_{lm}^*(\hat k)(e^{i2\delta_l}-1)=\frac{4\pi}kY_{lm}^*(\hat k)e^{i\delta_l}\sin\delta_l \end{aligned}\right.\tag{3.3}$$

进而根据(1.3)得到

$$f(\theta,\phi)=\frac{4\pi}k \sum_{lm}e^{i\delta_l}\sin\delta_l Y_{lm}^*(\hat k)Y_{lm}(\theta,\phi)\tag{3.4}$$

当$\hat k=\hat z$时，$Y_{lm}(\hat k)\neq0\Rightarrow m=0$，所以

$$f(\theta,\phi)=\frac{4\pi}k \sum_{l}e^{i\delta_l}\sin\delta_l\sqrt{\frac{2l+1}{4\pi}}\times\sqrt{\frac{2l+1}{4\pi}}P_l(\cos\theta) =\frac1k\sum_{l=0}^\infty(2l+1)e^{i\delta_l}\sin\delta_lP_l(\cos\theta)\tag{3.5}$$

进而微分散射截面和总截面分别为

$$\frac{\mathrm d\sigma}{\mathrm d\Omega}(\theta,\phi)=|f(\theta,\phi)|^2=\frac1{k^2}\sum_{ll'}(2l+1)(2l'+1)e^{i(\delta_l-\delta_{l'})}\sin\delta_l\sin\delta_{l'}P_l(\cos\theta)P_{l'}(\cos\theta)\tag{3.6}$$ $$\begin{aligned} \sigma_{tot}=&\frac{2\pi}{k^2}\sum_{ll'}(2l+1)(2l'+1)e^{i(\delta_l-\delta_{l'})}\sin\delta_l\sin\delta_{l'}\int_{-1}^1P_l(x)P_{l'}(x)\mathrm dx\\ =&\frac{2\pi}{k^2}\sum_{ll'}(2l+1)(2l'+1)e^{i(\delta_l-\delta_{l'})}\sin\delta_l\sin\delta_{l'}\frac{2}{2l+1}\delta_{ll'}\\ =&\frac{4\pi}{k^2}\sum_{l}(2l+1)\sin^2\delta_l \end{aligned}\tag{3.7}$$

应用：硬球势

硬球势的相移因子可根据硬球内部($r\le a$)波函数为零定出，即

$$\psi_{klm}(a,\theta,\phi)\equiv=0\Rightarrow R_{kl}(a)=e^{i\delta_l}h_l^{(1)}(ka)+e^{-i\delta_l}h_l^{(2)}(ka)=0\tag{4.1}$$ $$\Rightarrow e^{2i\delta_l}=-\frac{h_l^{(2)}(ka)}{h_l^{(1)}(ka)}=-\frac{j_l(ka)-iy_l(ka)}{j_l(ka)+iy_l(ka)}\tag{4.2}$$

从(2.3)可以看出，在$k\ll1$时，球贝塞尔函数相比于球诺伊曼函数而言是小量，所以

$$e^{2i\delta_l}=\frac{1+ij_l(ka)/y_l(ka)}{1-ij_l(ka)/y_l(ka)}\approx e^{i2j_l(ka)/y_l(ka)}\Rightarrow \delta_l\approx\frac{j_l(ka)}{y_l(ka)} \approx\frac{(ka)^l/(2l+1)!!}{-(2l-1)!!/(ka)^{l+1}}\tag{4.3}$$

因此只有$l=0,\delta_0=-ka$最为显著，即s波散射

$$\sigma_{tot}\approx\frac{4\pi}{k^2}\sin^2(-ka)\approx 4\pi a^2\tag{4.4}$$