[NOTES/QM-18004] Integral Equation for Scattering

Contents

1. Green function for free particle Schrodinger Equation
2. Set up integral equation .
3. Verify boundary condition.
4. Scattering amplitude.       

Green function for free particle Schrodinger Equation

In order to convert the Schrodinger equation \begin{equation} \left\{\frac{-\hbar^2}{2\mu}\nabla^2 +v(r)\right\}\psi=E\psi \label{E1} \end{equation} into an integral equation we first rewrite it as \begin{equation} \left(\nabla^2 + k^2\right)\psi=u(\vec{r})\psi(\vec{r}) \label{E2} \end{equation} where $k^2=\dfrac{2\mu E}{\hbar^2}$, $U(r)=\dfrac{2\mu }{\hbar^2}V(r)$ and we have defined Green function $G(\vec{r})$ as a solution of \begin{equation} \left(\nabla^2 +k^2\right)G(\vec{r})=-\delta(\vec{r}) . \label{E3} \end{equation} Two possible solutions for Green function are $G(\vec{r})$ are \begin{eqnarray} G_\pm(\vec{r})= \frac{e^{\pm ikr}}{4\pi r} \label{E4}. \end{eqnarray} Using the Green function we can now write down a "formal" solution of \eqref{E2} as a solution of integral equation. 

Set up integral equation .

Using a Green function which is a solution of \eqref{E3} a formal solution for \eqref{E2} can be written as \begin{equation} \psi(\vec{r})=\phi(\vec{r})-\int G(|\vec{r}-\vec{r}^{\,{'}}|)U(|\vec{r}^{\,{'}}|) \psi(\vec{r}^{\,{'}}) d^3r^{'} \label{E7} \end{equation} where $\phi(\vec{r})$ is a solution of the equation \begin{equation} \left(\nabla^2+k^2\right)\phi(\vec{r})=0 . \label{E8} \end{equation} It turns out that the choice \begin{equation} G(\vec{r})=\frac{e^{ikr} }{4\pi r} \label{E6} \end{equation} for the Green function leads to the correct scattering boundary condition on the wave function.

Verify boundary condition.

For the scattering problem we must select the function \(\phi(\vec{r})\) to be \begin{equation} \phi(\vec{r})=\exp(i\vec{k_i}.\vec{r}) \label{E9} \end{equation} where $\vec{k_i}$ is the momentum of the incident particles. Substituting \eqref{E6} \eqref{E9} in \eqref{E7} we get the integral equation for the scattering to be \begin{equation} \psi(\vec{r})=\exp(i\vec{k_i}.\vec{r})-\frac{1}{4\pi}\int\frac{e^{ik|\vec{r} -\vec{r^{'}}|}}{|\vec{r}-\vec{r^{'}}|} U(\vec{r^{'}})\psi (\vec{r^{'}}) d^3 r{'} . \label{E10} \end{equation} To verify that $\psi(\vec{r})$ given by \eqref{E10} does indeed have correct asymptotic property, we need to expand $|\vec{r}-\vec{r^{'}}|$ in powers of $\dfrac{\vec{r}}{\vec{r^{'}}}$ we shall assume that the potential is short range potential so that the contribution to integral over $\vec{r^{'}}$ comes from small value of $r^{'}$. \subsubsection*{Some algebraic manipulations} Expand $|\vec{r}-\vec{r^{'}}| $ in powers of $r^{'}$ \begin{eqnarray} |\vec{r}-\vec{r^{'}}| &=& \sqrt{r^2+r^{{'} 2}-2\vec{r}\cdot\vec{r}^{'} } \label{E11}\\ &=& r \left(1-2\frac{\vec{r}\cdot\vec{r}^{'}}{r^2}+\frac{r^{{'} 2}}{r^2}\right)^{1/2} . \label{E12} \end{eqnarray} Using binomial expansion we get \begin{equation} |\vec{r}-\vec{r^{'}}|=r \left(1- \frac{\vec{r}\cdot\vec{r}^{'}}{r^2} + O\left(\frac{r^{{'} 2}}{r^2}\right)^2 \right) . \label{E13} \end{equation} \subsubsection*{Large $r$ expansion of the formal solution \eqref{E10}} We substitute \eqref{E13} in the exponential and in the factor $\dfrac{1}{|\vec{r}-\vec{r^{'}}|}$ in \eqref{E10} and write $\dfrac{1}{|\vec{r}-\vec{r^{'}}|} \approx \dfrac{1}{r}$ to get \begin{eqnarray} \psi(\vec{r}) &\longrightarrow &\exp(i\vec{k}_i\cdot\vec{r}) - \frac{1}{4\pi r}\int \exp\left(ikr - ik \frac{\vec{r}\cdot\vec{r}^{'}}{r^2} \right) U(\vec{r}^{'}) \psi(\vec{r}^{'}) d^3r^{'} \label{E14}\\ &=& \exp(i\vec{k}_i\cdot\vec{r}) -\frac{e^{ikr}}{4\pi r} \int \exp(-ik \hat{n}\cdot\vec{r}^{'})U(r^{'})\psi(r^{'}) d^3r^{'} .\label{E15} \end{eqnarray} In the last step we have introduced a unit vector $\hat{n}=\vec{r}/r.$ The \eqref{E15} gives the probability amplitude ( wave function ) at $\vec{r}.$ If the particles are to reach at a detector at $\vec{r}$, the vector $\hat{n}$ must be in the direction of the final momentum and parallel to $k_f$. Note that \begin{equation} |\vec{k}_i|=|\vec{k}_f|=k \label{E16} \end{equation} holds as a consequence of energy conservation and hence \begin{equation} k(\hat{n}\cdot\vec{r}^{'}) = \vec{k}_f\cdot\vec{r} . \label{E17} \end{equation} Thus \eqref{E15} takes the form \begin{equation} \psi(\vec{r}) \approx \exp(i\vec{k}_i\cdot\vec{r}) -\frac{e^{ikr}}{4\pi r} \int \exp(-i\vec{k}\cdot\vec{r}^{'})U(r^{'})\psi(r^{'}) d^3r^{'} . \label{E18} \end{equation} This asymptotic behaviour is of the form expected for large $r$ \begin{equation} \psi(\vec{r}) \approx \exp(i\vec{k}_i\cdot\vec{r}) - \frac{e^{ikr}}{r} f(\theta,\phi) \label{E19}. \end{equation}

Scattering amplitude. 

Comparing \eqref{E18} with \eqref{E19} we see that the scattering amplitude is given by \begin{eqnarray} f(\theta,\phi) &=& -\frac{1}{4\pi} \int \exp(-i\vec{k}_f\cdot\vec{r}^{'})U(r^{'}) \psi(r^{'}) d^3r^{'} \label{E20} \\ &=& -\left(\frac{\mu}{2\pi \hbar^2}\right)\int \exp(-i\vec{k}_f\cdot\vec{r}^{'})V(r^{'}) \psi(r^{'}) d^3r^{'} \label{E21}. \end{eqnarray} It must be noted that the integral equation \eqref{E10} and the expression for the scattering amplitude in \eqref{E21} are {\bf exact} results.