Optical Theorem
We consider Schroedinger equation with the incident \(\vert\psi_{in}\rangle\) and emergent \(\vert\psi_{out}\rangle\) states. The scattering matrix is then defined as
\[\vert\psi_{out}\rangle=S\vert\psi_{in}\rangle\]The incident state \(\vert\psi_{in}\rangle=\vert\mathbf{k}\rangle\) is an eigenstate of the free Hamiltonian \(H_{0}=-\Delta\) of energy \(E=k_{0}^{2}\) that is
\[H_{0}\vert\psi_{in}\rangle=E\vert\psi_{in}\rangle\]The emergent state \(\vert\psi_{out}\rangle\) is an eigenstate of the total Hamiltonian \(H=H_{0}+V\) with the same energy, namely
\[H\vert\psi_{out}\rangle=E\vert\psi_{out}\rangle\]At infinity, the two states \(\vert\psi_{in}\rangle\) and \(\vert\psi_{out}\rangle\) only differ by the scattered wave, whose amplitude tends to zero. As a consequence, the two states are identical up to a phase difference. We then can obtain the relation
\[VS=(E-H_{0})(S-I)\]by the observation
\[\begin{align} H\vert\psi_{out}\rangle-H_{0}\vert\psi_{in}\rangle & =E(\vert\psi_{out}\rangle-\vert\psi_{in}\rangle)\\ HS-H_{0} & =E(S-I)\\ VS & =E(S-I)-H_{0}(S-I)=(E-H)(S-I) \end{align}\]Without interaction \(V=0\), \(S=1\). Using the resolvent operator \(G_{0}\) associated with the free problem \((E-H_{0})G=1\), we can have the Dyson equation
\[S=1+G_{0}VS\]Projecting this equation on an incident state \(\vert\mathbf{k}\rangle\) we have the relation
\[\vert\psi_{out}\rangle=\vert\mathbf{k}\rangle+G_{0}V\vert\psi_{out}\rangle\]Now we can define the scattering operator
\[T=VS\]which satisfies the Lippman-Schwinger equation
\[S=1+G_{0}T\]For the emergent state, we have by taking \(\vert\psi_{in}\rangle=\vert\mathbf{k}\rangle\)
\[\vert\psi_{out}\rangle=\vert\mathbf{k}\rangle+G_{0}T\vert\mathbf{k}\rangle\]whose projection on \(\vert\mathbf{r}\rangle\) yields
\[\langle\mathbf{r}\vert\psi_{out}\rangle=e^{i\mathbf{k}\cdot\mathbf{r}}+\int d\mathbf{r}^{\prime}\langle\mathbf{r}\vert G_{0}\vert\mathbf{r}^{\prime}\rangle\langle\mathbf{r}^{\prime}\vert T\vert\mathbf{k}\rangle\]With the asymptotic expansion,
\[\psi(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}+\frac{e^{i\mathbf{k}_{0}\mathbf{r}}}{r}f(\mathbf{k},\mathbf{k}^{\prime})\]we have
\[f(\mathbf{k},\mathbf{k}^{\prime})=-\frac{1}{4\pi}\langle\mathbf{k}^{\prime}\vert T\vert\mathbf{k}\rangle\]To derive the optical theorem, we start with the Lippman-Schwinger equation,
\[\langle\mathbf{k}\vert T^{\dagger}\vert\psi_{out}\rangle=\langle\mathbf{k}\vert T^{\dagger}\vert\mathbf{k}\rangle+\langle\mathbf{k}\vert T^{\dagger}G_{0}T\vert\mathbf{k}\rangle\] \[\mathrm{Im}\langle\mathbf{k}\vert T^{\dagger}\vert\mathbf{k}\rangle=-\mathrm{Im}\langle\mathbf{k}\vert T^{\dagger}G_{0}T\vert\mathbf{k}\rangle\]To proceed, we can (\(\epsilon(k)=k^{2}\))
\[\begin{align} \langle\mathbf{k}\vert T^{\dagger}G_{0}T\vert\mathbf{k}\rangle & =\int dk^{\prime}\langle\mathbf{k}\vert T^{\dagger}\vert k^{\prime}\rangle\langle k^{\prime}\vert G_{0}\vert k^{\prime}\rangle\langle k^{\prime}\vert T\vert\mathbf{k}\rangle\\ & =\pi\int dk^{\prime}\delta(E-H_{0}(\mathbf{k}^{\prime}))\langle\mathbf{k}\vert T^{\dagger}\vert k^{\prime}\rangle\langle k^{\prime}\vert G_{0}\vert k^{\prime}\rangle\langle k^{\prime}\vert T\vert\mathbf{k}\rangle\\ & =\frac{\pi}{2k_{0}}\int d\mathbf{k}^{\prime}\vert\langle\mathbf{k}\vert T^{\dagger}\vert\mathbf{k}^{\prime}\rangle\vert^{2}\delta(\mathbf{k}^{\prime}-\mathbf{k}) \end{align}\]by observing
\[\begin{align} \langle k^{\prime}\vert G_ {0}\vert k^{\prime}\rangle & =\langle k^{\prime}\vert\frac{1}{E-H_{0}(k^{\prime})+i0^{+}}\vert k^{\prime}\rangle\\ & =i\pi\delta(E-H_{0}(k^{\prime}) \end{align}\]In general, we have
\[\langle\mathbf{k}\vert T^{\dagger}G_{0}T\vert\mathbf{k}\rangle=\pi\left[\frac{d\epsilon(k_{0})}{dk_{0}}\right]^{-1}\int d\mathbf{k}^{\prime}\vert\langle\mathbf{k}\vert T^{\dagger}\vert\mathbf{k}^{\prime}\rangle\vert^{2}\delta(\mathbf{k}^{\prime}-\mathbf{k})\]With these basic understandings, one may think about the optical theorem in condensed matter systems. Namely, one shall consider a scattering process under the background of a periodic potential.