In 3D, the Lipmann-Schwinger equation still holds \[\left|\psi_k\right\rangle=\left|\psi_{0,k}\right\rangle+G_{0,+}(E)\hat{V}\left|\psi_k\right\rangle\] with \(E=\frac{\hbar^2k^2}{2m}\) and \[\hat{G}_{0,+}(E)=\frac{1}{E_0-\hat{H}_0+i\delta}.\] If we evaluate in the coordinate representation, we have \[\left\langle\mathbf{r}\middle|\hat{G}_{0,+}(E)\middle|\mathbf{r}'\right\rangle=-\frac{m}{2\pi\hbar^2}\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}.\] This derivation required contour integration and was skipped, although the derivation is fairly straightforward to complete. So, \[\psi_k(\mathbf{r})=\psi_{0,k}(\mathbf{r})-\frac{1}{4\pi}\int d^3r' \ \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}\frac{2m}{\hbar^2}V(\mathbf{r}')\psi_k(\mathbf{r}')\] for three-dimensional scattering.
Now focus on the behavior for large \(r\). If \(V(r')\) is nonzero only for small \(|\mathbf{r}'|\) and decays fast for large \(|\mathbf{r}'|\), we can expand \[\begin{aligned} |\mathbf{r}-\mathbf{r}'|\approx& \ r\left |\frac{\mathbf{r}'}{r}-\frac{\mathbf{r}'}{r}\right |=r\sqrt{1-\frac{2\mathbf{r}\cdot\mathbf{r}'}{r^2}+\frac{r'^2}{r^2}} \\ =& \ r\left(1-\frac{\mathbf{r}\cdot\mathbf{r}'}{r^2}-\frac{1}{2}\frac{(\mathbf{r}\cdot\mathbf{r}')^2}{r^4}+\frac{1}{2}\frac{r'^2}{r^2}+\cdots\right) \\ \approx& \ r-\mathbf{e}_r\cdot\mathbf{r}', \end{aligned}\] to lowest order. So, \[\begin{aligned} \frac{1}{|\mathbf{r}-\mathbf{r}'|}\approx& \ \frac{1}{r}\frac{1}{(1-\mathbf{e}_r\cdot\mathbf{e}_{r'}\frac{r'}{r})} \\ \approx& \ \frac{1}{r}\left(1+\mathbf{e}_r\cdot\mathbf{e}_{r'}\frac{r'}{r}\right) \\ =& \ \frac{1}{r}+\frac{\mathbf{e}_r\cdot\mathbf{e}_{r'}r'}{r^2} \end{aligned}\] and \[k|\mathbf{r}-\mathbf{r}'|\approx kr\left(1-\mathbf{e}_r\cdot\mathbf{e}_{r'}\frac{r'}{r}+\cdots\right).\] Define \(\mathbf{k}'=\mathbf{e}_r k\) so that \(|\mathbf{k}'|=|\mathbf{k}|=k\). Then, \[k|\mathbf{r}-\mathbf{r}'|=kr-\mathbf{k}'\cdot\mathbf{r}'+\cdots\] and \[\psi_{\mathbf{k}}(\mathbf{r})=\psi_{0,k}(r)-\frac{1}{4\pi}\int d^3r'\frac{e^{ikr-i\mathbf{k}'\cdot\mathbf{r}'}}{r}\left (1+\mathbf{e}_r\cdot\mathbf{e}_{r'}\frac{r'}{r}\right )\frac{2m}{\hbar^2}V(\mathbf{r}')\psi_{\mathbf{k}}(\mathbf{r}).\] But \(\psi_{0,k}(r)=\frac{1}{(2\pi)^{3/2}}e^{i\mathbf{k}\cdot\mathbf{r}}\), so \[\psi_\mathbf{k}(\mathbf{r})=\frac{1}{(2\pi)^{3/2}}\left(e^{i\mathbf{k}\cdot\mathbf{r}}+\frac{e^{ikr}}{r}f(\mathbf{k}',\mathbf{k})\right)+O\left (\frac{r'}{r}\right ),\] where \(f(\mathbf{k}',\mathbf{k})\) is called the scattering amplitude and is defined by \[f(\mathbf{k}',\mathbf{k})=-\frac{4\pi^2 m}{\hbar^2}\left\langle\psi_{0,\mathbf{k'}}\middle|\hat{V}\middle|\psi_{k}\right\rangle,\] which has units of length. It can be thought of as the amplitude of an outgoing spherical wave in the direction \(\mathbf{k}'\). Then, \(|f(\mathbf{k}',\mathbf{k})|^2\) is the probability to observe some particle with momentum \(\hbar\mathbf{k}'\) after scattering (where \(\hbar\mathbf{k}\) was the incident momentum).
The differential cross section is defined via \[\frac{d\sigma}{d\Omega_{\mathbf{k}'}}=\frac{\text{prob/time/solid angle of scattering in \textbf{k}' direction}}{\text{prob/time/area of incident flux of particles}}\] where \(\sigma\) is the cross section defined through \[\sigma=\int d\Omega \ \frac{d\sigma}{d\Omega},\] which has units of area. The cross section is a function of the incident momentum \(\hbar\mathbf{k}\) or incident energy \(E\), but does not depend on the frame of reference. Gottfried tells us how to find \[\frac{d\sigma}{d\Omega_{\mathbf{k}'}}=|f(\mathbf{k}',\mathbf{k})|^2.\]
These results are often summarized in terms of the transition matrix (or \(T\)-matrix). \[\left|\psi_k\right\rangle=\left|\psi_{0,k}\right\rangle+\hat{G}_{0,+}(E)\hat{V}\left|\psi_k\right\rangle\] \[\implies\left|\psi_k\right\rangle=(1-\hat{G}_{0,+}(E)\hat{V})^{-1}\left|\psi_{0,k}\right\rangle=\hat{\Omega}_+(E)\left|\psi_{0,k}\right\rangle\] where \(\hat{\Omega}_+(E)\) is called the Möller wave matrix. Then, \[f(\mathbf{k},\mathbf{k}')=-\frac{4\pi^2m}{\hbar^2}\left\langle\psi_{0,\mathbf{k}'}\middle|\hat{V}\middle|\psi_k\right\rangle=-\frac{4\pi^2m}{\hbar^2}\left\langle\psi_{0,\mathbf{k}'}\middle|\hat{V}\hat{\Omega}_+(E)\middle|\psi_{0,\mathbf{k}'}\right\rangle.\] Then, define the \(T\)-matrix with the following: \[\hat{T}(E)=\hat{V}\hat{\Omega}_+(E)\] so that \[f(\mathbf{k},\mathbf{k}')=-\frac{4\pi^2m}{\hbar^2}T_{\mathbf{k},\mathbf{k}'}(E).\] \(\hat{T}(E)\) also satisfies the operator equations: \[\hat{T}=\hat{V}\left[1-\hat{G}_{0,+}\hat{V}\right]^{-1}\] \[\implies\hat{T}-\hat{T}\hat{G}_{0,+}\hat{V}=\hat{V}\] or \[\boxed{\hat{T}=\hat{V}+\hat{T}\hat{G}_{0,+}\hat{V}.}\] Also, \[\hat{T}=\hat{V}+\hat{V}\hat{G}_{0,+}\hat{V}+\hat{V}\hat{G}_{0,+}\hat{V}\hat{G}_{0,+}\hat{V}+\cdots\] \[=\left[1-\hat{V}\hat{G}_{0,+}\right]^{-1}.\] Therefore, \[\hat{T}=\hat{V}+\hat{V}\hat{G}_{0,+}\hat{T}\] Since \(\hat{G}_{0,\pm}=\frac{1}{E-\hat{H}_0\pm i\delta}\), we have \(\hat{G}_{0,+}^\dagger(E)=\hat{G}_{0-}(E)\). Then, taking Hermitian conjugates yields \[\hat{T}^\dagger=\hat{V}+\hat{T}^\dagger\hat{G}_{0-}\hat{V}=\hat{V}+\hat{V}\hat{G}_{0-}\hat{T}^\dagger.\] Recall Dirac’s identity: \[\frac{1}{x\pm i\delta}=\frac{P}{x}\mp i\pi\delta(x),\] where \(P\) denotes a principal value. So, \[\hat{T}-\hat{T}^\dagger=\hat{V}+\hat{V}\hat{G}_{0,+}\hat{T}-\hat{V}-\hat{T}^\dagger\hat{G}_{0,-}\hat{V}\] \[=\hat{T}\hat{G}_{0,+}\hat{T}-\hat{T}^\dagger\hat{G}_{0,-}\hat{V}\hat{G}_{0,+}\hat{T}-\hat{T}^\dagger\hat{G}_{0,-}\hat{T}+\hat{T}^\dagger\hat{G}_{0,-}\hat{V}\hat{G}_{0,+}\hat{T}\] \[=\hat{T}^\dagger(\hat{G}_{0,+}-\hat{G}_{0,-})\hat{T}.\] But, \[\hat{G}_{0,+}-\hat{G}_{0,-}=-2\pi i\delta(E-\hat{H}_0)\] \[\implies\boxed{\hat{T}-\hat{T}^\dagger=-2\pi i\hat{T}^\dagger\delta(E-\hat{H}_0)\hat{T}}\] This is known as the generalized optical theorem. Then, let’s take matrix elements and introduce complete sets of states on both sides of the delta function. This gives us \[T_{\mathbf{k},\mathbf{k}'}(E)-T_{\mathbf{k},\mathbf{k}'}^*(E)=-2\pi i\int d^3k''\int d^3k''' \ T_{\mathbf{k}'',\mathbf{k}'}^*(E)\left\langle\mathbf{k}''\middle|\delta(E-\hat{H}_0)\middle|\mathbf{k}'''\right\rangle T_{\mathbf{k}''',\mathbf{k}}(E)\] But \[\left\langle\mathbf{k}''\middle|\delta(E-\hat{H}_0)\middle|\mathbf{k}'''\right\rangle=\delta\left (E-\frac{\hbar^2\mathbf{k}''}{2m}\right )\left\langle\mathbf{k}''\middle|\mathbf{k}'''\right\rangle=\delta\left (E-\frac{\hbar^2\mathbf{k}''^2}{2m}\right )\delta^3(\mathbf{k}''-\mathbf{k}'''),\] so the integral can be done over \(\mathbf{k}'''\). Furthermore, \[\begin{aligned} \int d^3k''\delta\left (E-\frac{\hbar^2\mathbf{k}''^2}{2m}\right )=& \ \int_0^\infty dk'' \ k''^2\int d\Omega_{\mathbf{k}''}\delta\left (E-\frac{\hbar^2\mathbf{k}''^2}{2m}\right ) \\ =& \ \frac{mk}{\hbar^2}\int d\Omega_{\mathbf{k}''}. \end{aligned}\] So, \[T_{\mathbf{k},\mathbf{k}'}(E)-T_{\mathbf{k},\mathbf{k}'}^*(E)=-\frac{i\pi mk}{\hbar^2}\int d\Omega_{\mathbf{k}''}T_{\mathbf{k}'',\mathbf{k}'}^*(E)T_{\mathbf{k}''',\mathbf{k}}(E).\] Multiply across by \(-\frac{4\pi^2m}{\hbar^2}\) to get \[\boxed{f(\mathbf{k}',\mathbf{k})-f^*(\mathbf{k}',\mathbf{k})=\frac{ik}{2\pi}\int d\Omega_{\mathbf{k}''}f(\mathbf{k}'',\mathbf{k})f^*(\mathbf{k}'',\mathbf{k}'),}\] which is the generalized optical theorem for scattering amplitudes. For forward scattering, \(\mathbf{k}=\mathbf{k}'\) and \(\theta=0\). \[\text{Im}{f(\mathbf{k}',\mathbf{k})}=\frac{k}{2\pi}\int d\Omega_{\mathbf{k}''}|f(\mathbf{k}'',\mathbf{k})|^2\] \[=\frac{k}{2\pi}\int d\Omega_{\mathbf{k}''}\frac{d\sigma(k)}{d\Omega_{\mathbf{k}''}}=\frac{k}{2\pi}\sigma(k)\] So, \[\boxed{\sigma(k)=\frac{4\pi}{k}\text{Im}{f(\mathbf{k}',\mathbf{k})}.}\]
The scattering amplitude satisfies \[f(\mathbf{k}',\mathbf{k})=-\frac{4\pi^2m}{\hbar^2}\left\langle\psi_{0,\mathbf{k}'}\middle|\left (1-\hat{V}\hat{G}_{0,+}(E)\right )^{-1}\hat{V}\middle|\psi_{0,\mathbf{k}'}\right\rangle=\sum_{n=1}^\infty f_n(\mathbf{k}',\mathbf{k}),\] where \(n\) counts powers of \(\hat{V}\). Define the \(N\)th Born approximation by truncating the sum at \(N\). \[f^{(N)}(\mathbf{k}',\mathbf{k})=\sum_{n=1}^Nf_n(\mathbf{k}',\mathbf{k}).\] Therefore, \[\begin{aligned} f^{(1)}(\mathbf{k}',\mathbf{k})=&-\frac{4\pi^2m}{\hbar^2}\left\langle\psi_{0,\mathbf{k}'}\middle|\hat{V}\middle|\psi_{0,\mathbf{k}}\right\rangle \\ =& \ -\frac{m}{2\pi\hbar^2}\int d^3r \ e^{-i(\mathbf{k}'-\mathbf{k})\cdot\mathbf{r}}V(\mathbf{r}) \\ =& \ \tilde{V}(\mathbf{k}'-\mathbf{k}), \end{aligned}\] which is simply the Fourier transform of \(V(\mathbf{r})\). Also note that \(\left\langle\psi_{0,\mathbf{k}'}\middle|\hat{V}\middle|\psi_{0,\mathbf{k}}\right\rangle=\frac{1}{(2\pi)^3}\Tilde{V}(\mathbf{k}'-\mathbf{k})\), so \[f^{(2)}(\mathbf{k}',\mathbf{k})=-\frac{m}{2\pi\hbar^2}\left(\tilde{V}(\mathbf{k}'-\mathbf{k})+(2\pi)^3\int d^3k''\left\langle\psi_{0,\mathbf{k}'}\middle|\hat{V}\middle|\psi_{0,\mathbf{k}''}\right\rangle\left\langle\psi_{0,\mathbf{k}''}\middle|\hat{G}_{0,+}(E)\hat{V}\middle|\psi_{0,\mathbf{k}}\right\rangle\right).\] But \[\left\langle\psi_{0,\mathbf{k}''}\right|\hat{G}_{0,+}(E)=\frac{1}{E-\frac{\hbar^2k''^2}{2m}+i\delta}\left\langle\psi_{0,\mathbf{k}''}\right|,\] so \[f^{(2)}(\mathbf{k}',\mathbf{k})=-\frac{m}{2\pi\hbar^2}\left(\tilde{V}(\mathbf{k}'-\mathbf{k})+\int\frac{d^3k''}{(2\pi)^3}\frac{\tilde{V}(\mathbf{k}'-\mathbf{k}'')\tilde{V}(\mathbf{k}''-\mathbf{k})}{E-\frac{\hbar^2k''^2}{2m}+i\delta}\right)\] and so on.