Consider a central potential \(V(r)\) only, with no angular dependence. \[\begin{aligned} & \psi_{k}(\mathbf{r})=\psi_{0 k}(\mathbf{r})+\int d^{3} r^{\prime} V\left(r^{\prime}\right) G_{0+}\left(\mathbf{r}, \mathbf{r}^{\prime} ; E\right) \psi_{\mathbf{k}}\left(\mathbf{r}^{\prime}\right) \\ & G_{0+}\left(\mathbf{r}^{\prime} , \mathbf{r}^{\prime} ; E\right)=-\frac{m}{2 \pi \hbar^{2}} \frac{e^{i k\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=\langle r| \frac{1}{E-\hat{H}_{0}+i \delta}\left|\mathbf{r}^{\prime}\right\rangle \end{aligned}\] define the \(z\) direction to lie along \(\mathbf{k}\). Then \[\begin{aligned} \psi_{0 \mathbf{k}} (\mathbf{r}) & =\frac{1}{\sqrt{(2 \pi)^{3}}} e^{i k r \cos \theta} \quad\text{with}\quad \theta= \text { angle of } \mathbf{r} \text{ to } z \text{ axis}\\ & =\frac{1}{(2 \pi)^{3 / 2}} \sum_{l=0}^{\infty}(2 l+1) i^l j_{l}(k r) P_{l}(\cos \theta) \end{aligned}\] Expand \[\psi_{\mathbf{k}}(\mathbf{r}) =\frac{1}{(2 \pi)^{3 / 2}} \sum_{l=0}^{\infty}(2 l+1) i^{l} A_{l}(r, k) {P_l}(\cos \theta)\] with yet to be determined constants \(A_l(r,k)\). For \(G_{0+}\) insert \(\int dq |\psi_{0q}\rangle\langle\psi_{0q}|\) between \(\langle r|\) and the fraction \[\begin{aligned} G_{0+}\left(\mathbf{r}, \mathbf{r}^{\prime} ; E\right) & =-\int d^{3} q\left\langle\mathbf{r} \mid \psi_{0 q}\right\rangle\left\langle\psi_{0 q} \mid \mathbf{r}^{\prime}\right\rangle \frac{1}{E-\frac{\hbar^{2} q^{2}}{2 m}+i \delta}\\ & =\frac{1}{(2 \pi)^{3}} \int d^{3} q e^{i \mathbf{q}\cdot\left(\mathbf{r}-\mathbf{r}^{\prime}\right)} \frac{1}{E-\frac{\hbar^{2} q^{2}}{2m}+i \delta} \end{aligned}\] The challenge with evaluating this is that \(\mathbf{q}\) can point in any direction—it does not necessarily point along the \(z\)-axis.
Suppose a unit vector \(\mathbf{e}_{x}\) points in the \(\theta, \phi\) direction. We use \(Y_{l}^{m}\left(\mathbf{e}_{x}\right)=Y_{l}^{m}(\theta, \phi)\) to denote the spherical harmonic in that direction. Then one can generalize our plane wave calculation to \[\begin{aligned} & e^{i \mathbf{q} \cdot \mathbf{r}}=4 \pi \sum_{l m} i^{l} j_{l}(q r) Y_{l}^{m *}(\mathbf{e_{\mathbf{q}}} ) Y_{l}^{m}(\mathbf{e}_\mathbf{r}) \\ & e^{-i \mathbf{q}^{\prime} \mathbf{r}^{\prime}}=4 \pi \sum_{l m^{\prime}}(-i)^{l} j_{l^{\prime}}\left(q_r^{\prime}\right) Y_{l^{\prime}}^{m^{\prime}}(\mathbf{e}_\mathbf{q}) Y_{l^{\prime}}^{m^{\prime} *}\left(\mathbf{e}_{\mathbf{r}}\right) \end{aligned}\] and \[\begin{aligned} G_{0 +}\left(\mathbf{r}, \mathbf{r}^{\prime}; E\right)=& \frac{1}{(2 \pi)^{3}} \cdot(4 \pi)^{2} \sum_{l m} \sum_{l^{\prime} m^{\prime}} \int d^{3} q \ i^l(-i)^{l^{\prime} } j_{l}(q r) j_{l^{\prime}}\left(qr^{\prime}\right) \\ &\times Y_{l}^{m *}\left(\mathbf{e}_{\mathbf{q}}\right) Y_{l^{\prime}}^{m^{\prime}}\left(\mathbf{e}_{\mathbf{q}}\right) Y_{l}^{m}\left(\mathbf{e}_{\mathbf{r}}\right) Y_{l^{\prime}}^{m^{\prime *}}\left(\mathbf{e}_{\mathbf{r}^{\prime}}\right) \frac{1}{E-\frac{\hbar^{2} q^{2}}{2 m}+i \delta}\\ \end{aligned}\] But \[\begin{aligned} \int d \Omega_{q} Y_{l}^{m *}\left(\mathbf{e}_{\mathbf{q}}\right) Y_{l^{\prime}}^{m \prime}\left(\mathbf{e}_{\mathbf{q}}\right)=\delta_{m m^{\prime}} \delta_{l l^{\prime}} =\left\langle l m \mid l^{\prime} m^{\prime}\right\rangle\end{aligned}\] due to orthogonality of the spherical harmonics. So \[G_{0+}\left(\mathbf{r}, \mathbf{r}^{\prime} ; E\right)=\frac{2}{\pi} \sum_{l m} \int_{0}^{\infty} d q \quad q^{2} \frac{j_{l}(q r) j_{l}(q r^{\prime}) Y_{l}^{m}\left(\mathbf{e}_{\mathbf{r}}\right) Y_{l}^{m *}\left(\mathbf{e}_{\mathbf{r}}\right)}{E-\frac{\hbar^{2} q^{2}}{2 m}+i \delta}.\] Note that \(j_{l}\left(-q{r}\right)=(-1)^{l} j_{l}\left(q{r}\right)\), so we can write \[G_{0+}\left(\mathbf{r}, \mathbf{r}^{\prime} ; E\right)=\frac{1}{\pi}\frac{2m}{\hbar^2} \sum_{l m} Y_l^m(\mathbf{e}_\mathbf{r}) Y_l^{m*}(\mathbf{e}_\mathbf{r^\prime})\int_{0}^{\infty} d q \quad q^{2} \frac{j_{l}(q r) j_{l}(q r^{\prime}) }{k^2 - q^2+i \delta}\] Now use the fact that \(P_{l}(\cos \theta)=\sqrt{\frac{4 \pi}{2 (l+1)}} Y_{l}^{0}\left(\mathbf{e}_{\mathbf{r}}\right)\) to get \[\begin{aligned} & \sum_{l=0}^{\infty} \sqrt{2 l+1} i^{l} A_{l}(r, k) Y_{l}^{0}\left(\mathbf{e}_{\mathbf{r}}\right)=\sum_{l=0}^{\infty} \sqrt{2 l+1} i^l j_{l}(k r) Y_{l}^{0}\left(\mathbf{e}_{\mathbf{r}^{\prime}}\right) \\ & + \int d^{3} r^{\prime} V\left(r^{\prime}\right) \frac{1}{\pi} \frac{2 m}{\hbar^{2}} \sum_{l^{\prime} m^{\prime}} Y_{l^{\prime}}^{m^{\prime}}\left(\mathbf{e}_{\mathbf{r}}\right) Y_{l^{\prime }}^{m^{\prime * }}\left(\mathbf{e}_{\mathbf{r}^{\prime}}\right) \int_{-\infty}^{+\infty} d q \quad q^{2} \frac{j_{l^{\prime}}\left(q{r}\right) j_{l^\prime}\left(q r^{\prime}\right)}{k^{2}-q^{2}+i \delta} \\ & \times \sum_{l} \sqrt{2 l+1} i^{l} A_{l}\left(r^{\prime}, k\right) Y_{l}^{0}\left(\mathbf{e}_{\mathbf{r^\prime}}\right) \end{aligned}\]
The integral over \(r^{\prime}\) can be written as \(\int_{0}^{\infty} d r^{\prime} r^{\prime 2} \int d \Omega_{\mathbf{e}_{\mathbf{r^\prime}}}\), and \(\int d \Omega_{\mathbf{e}_{\mathbf{r}^{\prime}}} Y_{l^{\prime}}^{m^{\prime *}}\left(\mathbf{e}_{\mathbf{r}^{\prime}}\right) Y_{l}^{0}\left(\mathbf{e}_{\mathbf{r}^{\prime}}\right)=\delta_{m^{\prime} 0} \delta_{l l^\prime}\), so, the right hand side becomes \[\begin{aligned} = & \sum_{d=0}^{\infty} \sqrt{2 l+1} i^l {j_l}(k r) Y_{l}^{0}\left(\mathbf{e}_{\mathbf{r}}\right)+\int_{0}^{\infty} d r^{\prime} r^{\prime 2} V\left(r^{\prime}\right) \frac{2 m}{\pi \hbar^{2}} \sum_{l=0}^{\infty} Y_{l}^{0}\left(\mathbf{e}_{\mathbf{r}}\right) \\ & \times \int_{-\infty}^{+\infty} d q \frac{q^{2} j_{l}(q r) j_{l}\left(q r^{ \prime}\right)}{k^{2}-q^{2}+i \delta} \sqrt{2 l+1} ; i^{l}A_l\left(r^{\prime}, k\right), \end{aligned}\] so we find \[\boxed{A_{l}(r, k)=j_{l}(k r)+\frac{2 m}{\pi \hbar^{2}} \int_{0}^{\infty} d r^{\prime} r^{\prime 2} V\left(r^{\prime}\right) A_{l}\left(r^{\prime}, k\right) \int_{-\infty}^{+\infty} d q \frac{q^{2} j_{l}(q r) j_{l}\left(q r^{\prime}\right)}{k^{2}-q^{2}+i \delta}.}\]
This is an integral equation for the partial wave amplitudes.
Let us examine it for large \(r: \quad j_{l}(k r) \rightarrow \frac{1}{k r} \sin \left(k r-\frac{l \pi}{2}\right)\) The integral over \(q\) can be evaluated by contour integration using analytic properties of Bessel functions (done in Gottfried’s book). We obtain for \(r\to\infty\) \[\int_{-\infty}^{+\infty} d q \frac{q^{2} j_{l}(q r) j_{l}\left(q r^{\prime}\right)}{k^{2}-q^{2}+i \delta} \rightarrow \frac{-\pi k e^{i k r}}{k r} \int_{0}^{\infty} d r^{\prime} r^{\prime 2} j_{l}\left(k r^{\prime}\right)A_l (r^\prime,k).\] So we find as \(r \rightarrow \infty\), \[A_{l}(r, k) \rightarrow -\frac{1}{2 i k r}\left\{e^{-i\left(k r-\frac{l \pi}{2}\right)}-e^{i\left(k r-\frac{l \pi}{2}\right)} \left[\underbrace{1 - 2ik \int_0 ^\infty d r^\prime r^{\prime2}\frac{2m}{\hbar^2}V(r^\prime)j_l(kr^\prime)A_l(r^\prime,k)}_{\text{outgoing}}\right]\right\} .\] Let \[\boxed{e^{2 i \delta_{l}(k)}=1-2 i k \int_{0}^{\infty} d r^{\prime} r^{\prime 2}j_l\left(k r^{\prime}\right) \frac{2m}{\hbar^2}V(r')A_{l}\left(r^{\prime}, k\right),}\] then \[A_{l}(r, k) \rightarrow \frac{e^{i \delta_{l}(k)}}{k r} \sin \left(k r-\frac{l \pi}{2}+\delta_{l} (k)\right).\]
For large \(r\): \(\delta_{l}(k)=l\)th partial wave phase
shift. These are the important quantities to find.
Write \[f\left(\mathbf{k}^{\prime}, \mathbf{k}\right)=f(k, \theta) \quad \text{with}\quad\theta= \text{ angle } \mathbf{k}^{\prime} \text{ makes to } \mathbf{k}.\] Then \[\begin{aligned} f(k, \theta)&=(2 \pi)^{3 / 2} \lim _{r \rightarrow \infty}\left[\left(\psi_{k}(r)-\psi_{o k}(r)\right) r e^{-i k r}\right] \\ &=\sum_{l=0}^{\infty} i^{l}(2 l+1) P_{\ell}(\cos \theta) \lim _{r \rightarrow \infty}\left[A_{l}(r, k)-j_{l}(k r^\prime) e^{-i k r}\right] \\ &=\sum_{l=0}^{\infty} i l(2 l-1) P_{l}(\cos \theta) \frac{e^{2 i \delta_{l}(k)}-1}{2 i k i ^l} \end{aligned}\] Hence, we have \[\boxed{ f(k, \theta)=\sum_{l=0}^{\infty}(2 l+1) P_{l}(\cos \theta) \frac{e^{i \delta_{l}(k)} \sin \delta_{l}(k)}{k}}\] We often rewrite this in terms of the shifts for each \(l\) via defining \[f_{l}(k, \theta)=(2 l+1) P_{l}(\cos \theta)\frac{e^{i \delta_l (k) }\sin(\delta_l (k))}{k}\] In this case, we have defined \[f_{l}(k)=\frac{e^{i \delta_{l}(k)} \sin \delta_{l}(k)}{k}=\text { partial wave scattering amplitude }\] which can be rewritten as \[f_l(k) = -\int_0 ^\infty dr r^2j_l(kr)\frac{2m}{\hbar^2}V(r)A_l (r,k)\\ \boxed{\operatorname{Im}f_l (k) = \frac{1}{k}\sin ^2 \delta_l (k) = k|f_l (k)|^2}\] This is like an optical theorem for each partial wave, which yields \[\boxed{\operatorname{Im}(f_l (k)^{-1})=-k.}\]
The scattering cross section is given by \[\begin{aligned} & \sigma(k)=\int d \Omega_{k}|f(k, \cos \theta)|^{2} \end{aligned}\] Plugging in the definitions gives \[\begin{aligned} \sigma(k)&=\sum_{l l^\prime} (2l+1)(2l^\prime +1)\int d \Omega_kP_l(\cos \theta)P_{l^\prime}(\cos \theta) e^{i(\delta_l - \delta_{l^\prime})\frac{\sin \delta_l \sin \delta_{l^\prime}}{k^2}}\\ & =\sum_{l} 2(2 l+1) \frac{\sin ^{2} \delta_{l}(k)}{k^{2}} \cdot 2 \pi \end{aligned}\] So, we have that \[\boxed{\sigma(k)=4 \pi \sum_{l=0}^{\infty}(2 l+1) \frac{\sin ^{2} \delta_{l}(k)}{k^{2}}}\] which is determined entirely by the phase shifts.
The Born approximation for the partial wave replaces \(A_{l}(r,k)\) by \(j_{l}(k r)\), so \[\begin{aligned} e^{2 i \delta_{l}^{B}(k)}-1 &\cong-2 i k \int_{0}^{\infty} d r \ r^{2} j_{l}^{2}(k r) \frac{2 m}{\hbar^{2}} V(r)\\ &\cong 2 i \delta_{l}^{B}(k) \ \text{ since the phase shift is small when V is small} \end{aligned}\] Hence, \[\boxed{ \delta_{d}^{\beta}(h) \cong-k \frac{2 m}{\hbar^{2}} \int_{0}^{\infty} d r \ r^{2} j_{l}^{2}(k r) V(r).}\]
Now, let us try an example: Attractive delta shell potential \[\begin{aligned} & V(r)=-\frac{\hbar^{2}}{2 m} \lambda \delta(r-a) \qquad a=\text { range } \\ & \frac{1}{\pi} \int_{-\infty}^{+\infty} d q \ \frac{q^{2} j_{l}\left(q{r}\right) j_{l}\left(q r^{\prime}\right)}{k^{2}-q^{2}+i \delta}=-i k\left\{\begin{array}{r} j_{l}(k r)\left(j_{l}\left(k r^{\prime}\right)+i \eta_{l}(kr^\prime)\right) \\ r<r^{\prime} \\ \left(j_{l}(k r)+i \eta_{l}(k r)\right) j_l\left(kr^{\prime}\right)\\ r^{\prime} < r \end{array}\right. \end{aligned}\] So \[\begin{aligned} &A_l(r,k) - j_l(kr) = \frac{2m}{\hbar^2} \int _0 ^ \infty dr^\prime \ r^{\prime 2} V(r^\prime)A_l(r^\prime,k)(-ik) \left\{\begin{array}{r} j_{l}(k r)\left(j_{l}\left(k r^{\prime}\right)+i \eta_{l}(kr^\prime)\right) \\ r<r^{\prime} \\ \left(j_{l}(k r)+i \eta_{l}(k r)\right) j_l\left(kr^{\prime}\right)\\ r^{\prime} < r \end{array}\right. \\ & \begin{aligned} = & +\lambda a^{2} A_{l}(a, k) i k\left\{\begin{array}{l} j_{l}(k r)\left(j_{l}(k a)+i \eta_{l}(k a)\right) \quad r<a \\ \left(j_{l}(k r)+i \eta_{l}(k r)\right) j_{l}(k a) \quad r>a \end{array}\right. \\ &~\\ A_{l}(a, k) & =\frac{j_{l}(k a)}{1-i k \lambda a^{2} j_{l}(k a)\left(j_{l}(k a)+i \eta_{l}(k a)\right)}\\ &~\\ \text{So } A_l(r,k) & = j_l(kr) + \frac{ik \lambda a^2 j_l (ka) \times \left\{\begin{array}{r} j_{l}(k r)\left(j_{l}\left(k r^{\prime}\right)+i \eta_{l}(kr^\prime)\right) \\ r<a \\ \left(j_{l}(k r)+i \eta_{l}(k r)\right) j_l\left(kr^{\prime}\right)\\ r>a \end{array}\right.}{1 - ik \lambda a^2 j_l (ka)(j_l(ka) + i \eta_l (ka))}\\ \end{aligned} \end{aligned}\] This means that \[\begin{aligned} f_{l}(k) & =-\int_{0}^{\infty} d r \ r^{2} j_{l}(k r) \frac{2 m}{\hbar^{2}} V(r) A_{l}(r, k) \\ & =\lambda a^{2} j_{l}(k a) A_{l}(a, k) \\ &\boxed{ f_{l}(k) =\frac{\lambda a^{2} j_{l}^{2}(k a)}{1-i k \lambda a^{2} j_{l}(k a)\left(j_{l}(k a)+i \eta_{l}(k a)\right)}} \end{aligned}\] It is common to also note that \[\begin{aligned} & \tan \delta_{l}(k)=\frac{\operatorname{Im} f_{l}(k)}{\operatorname{Re} f_{l}(k)}, \quad \text { but } f=\frac{\alpha}{\beta-i \gamma}=\frac{\alpha(\beta+i \gamma)}{\beta^{2}+\gamma^{2}} \\ \text{so}~~& \tan \delta=\frac{\gamma}{\beta} \\ & \boxed{-\tan \delta_{l}(k)=\frac{k \lambda a^{2} j_{l}^{2}\left(k{a}\right)}{1+k \lambda a^{2} j_{l}\left(k{a}\right) \eta_{l}(k a)}} \end{aligned}\] Let us examine in some limits: \[\begin{aligned} \text{high energy } \qquad & j_{l}(k a) \rightarrow \frac{1}{k{a}} \sin \left(k a-\frac{l \pi}{2}\right) \\ &\eta_l (ka ) \rightarrow -\frac{1}{ka}\cos \left( ka - \frac{l \pi}{2}\right) \end{aligned}\]
\[\begin{aligned} f_{l}(k) \rightarrow & \frac{\frac{\lambda}{k^2} \sin^2(ka - \frac{l \pi}{2})}{1 - i\frac{\lambda}{k} \sin(ka - \frac{l \pi}{2})(\sin(ka - \frac{l \pi}{2}) - i\cos(ka - \frac{l \pi}{2}))} \\ = & \frac{1}{k} \frac{\lambda \sin ^{2}\left(k a-\frac{l \pi}{2}\right)}{k-i \lambda \sin \left(k{a}-\frac{l \pi}{2}\right)\left(\sin \left(k a-\frac{l \pi}{2}\right)-i \cos \left(k a \cdot \frac{l \pi}{2}\right)\right)}\\ \rightarrow & 0 \text{ as } k \rightarrow \infty \quad \Rightarrow \quad \text{ phase shifts are small}\\ &\qquad \qquad \qquad \qquad \qquad \text{first Born is good} \end{aligned}\]
This implies that as \(k \rightarrow \infty \quad \Rightarrow\) phase shifts are small \(\Rightarrow\) first Born is good. So, we have \[\tan \delta_{l}(k) \rightarrow \frac{\lambda \sin ^{2}\left(k{a}-\frac{l \pi}{2}\right)}{k-\lambda \sin \left(k a-\frac{l \pi}{2}\right) \cos \left(h a-\frac{l \pi}{2}\right)} \rightarrow 0 \text { as } k \rightarrow \infty\]
Now for low energy \(\quad k a \ll 1\) \[\begin{aligned} & j_{l}(k a) \rightarrow \frac{(k a)^{l}}{(2 l+1)!!}\left[1-\frac{(k a)^{2}}{2(2 l+3)}+\cdots\right] \\ & \eta_{l}(k a) \rightarrow-\frac{(2 l-1)!!}{(k a)^{l+1}}\left[1+\frac{(k a)^{2}}{2(2 l-1)}+\cdots\right] \\ f_{l}(h)& \rightarrow \frac{\lambda a^{2}(k a)^{2 l}}{((2 l+1)!!)^{2}} \frac{1}{1-i k \lambda a^{2} \frac{(k a)^{l}}{(2 l-1)!!}\left(\frac{(k a)^{l}}{(2l+1)!!}-i \frac{(2 l-1)!!}{(k a)^{l+1}}\right)} \\ & =\frac{\lambda{a}^{2}\left(k{a}\right)^{2 l}}{((2 \ell+1)!!)^{2}} \frac{1}{1+\frac{k \lambda a^{2}}{k a} \frac{1}{(2l+1)}-i \lambda{a} \frac{\left(k{a}\right)^{2 l+1}}{[(2 l+1)!!]^{2}}} \\ & =a \frac{\lambda_{a}\left(k{a}\right)^{2 l}}{[(2l+1)!!]^{2}} \frac{1}{1+\frac{\lambda{a}}{2 \lambda+1}-i \lambda{a} \frac{\left(k{a}\right)^{2l + 1} }{[(2l + 1)!!]^{2}}} \\ & =O\left((k a)^{2 l}\right) \\ \tan \delta_{l}(k) & \rightarrow \frac{k{a} \lambda{a}\left(k{a}\right)^{2 l}}{[(2 l+1)!!]^{2}} \frac{1}{1+k{a} \lambda{a} \frac{(k a)^{l}}{(2 l+1)!!}\left(-\frac{(2 l-1)!!}{(k a)^{ l+1}}\right)} \\ & \rightarrow \frac{\lambda{a}\left(k{a}\right)^{2 l+1}}{[(2 l+1)!!]^{2}} \frac{1}{1-\frac{\lambda{a}}{2 l+1}} \\ \end{aligned}\] \[\boxed{ \tan \delta_{l}(k)=O\left(\left(k{a}\right)^{2l+1}\right)}\] which implies that s-wave scattering dominates at low energies.
These two results are useful rules of thumb when thinking about scattering.