Chapter 10, Problem 10.16P

To determine

The Green’s function for the one-dimensional Schrodinger equation and the integral form.

Answer to Problem 10.16P

The Green’s function for the one-dimensional Schrodinger equation is $G\left(x\right)=-\frac{-i}{2k}{e}^{ik\left|x\right|}$ and the integral form is  $\psi \left(x\right)={\psi }_0\left(x\right)-\frac{im}{{\hslash }^{2}k}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{ik\left|x-x_0\right|}V\left(x_0\right)\psi \left(x_0\right)d{x}_0}$.

Explanation of Solution

Write the equation analogous to equation 10.52.

  $\left(\frac{d^2}{d{x}^2}+k^2\right)G\left(x\right)=\delta \left(x\right)$        (I)

Write the equation analogous to equation 10.54.

  $G\left(x\right)=\frac{1}{\sqrt{2\pi }}{\displaystyle \int e^{isx}g\left(s\right)ds}$        (II)

Put equation (II) in equation (I).

  $\frac{1}{\sqrt{2\pi }}{\displaystyle \int \left(-s^2+k^2\right)g\left(s\right)e^{isx}ds}=\delta \left(x\right)$        (III)

Write the equation for $\delta \left(x\right)$ .

  $\delta \left(x\right)=\frac{1}{2\pi }{\displaystyle \int e^{isx}ds}$

Put the above equation in equation (III).

  $\begin{array}{c}\frac{1}{\sqrt{2\pi }}{\displaystyle \int \left(-s^2+k^2\right)g\left(s\right)e^{isx}ds}=\frac{1}{2\pi }{\displaystyle \int e^{isx}ds}\\ g\left(s\right)=\frac{1}{\sqrt{2\pi }\left(k^2-s^2\right)}\end{array}$        (IV)

Write the equation for $G\left(x\right)$ ,for $x>0$ , close above.

  $\begin{array}{c}G\left(x\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{e^{isx}}{k^2-s^2}ds}\\ =-\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{e^{isx}}{\left(s+k\right)}\frac{1}{s-k}ds}\\ =-\frac{1}{2\pi }2\pi i{\frac{e^{isx}}{\left(s+k\right)}|}_{s=k}\\ =-i\frac{e^{ikx}}{2k}\end{array}$

Write the equation for $G\left(x\right)$ ,for $x<0$ , close below.

  $\begin{array}{c}G\left(x\right)=\frac{1}{2\pi }{\displaystyle \oint \frac{e^{isx}}{\left(s+k\right)}\frac{1}{s-k}ds}\\ =\frac{1}{2\pi }2\pi i{\frac{e^{isx}}{\left(s-k\right)}|}_{s=-k}\\ =-i\frac{e^{-ikx}}{2k}\end{array}$

Write the equation for $G\left(x\right)$ .

  $G\left(x\right)=\frac{-i}{2k}{e}^{ik\left|x\right|}$        (V)

Write the solution to the Schrodinger equation.

  $\psi \left(x\right)={\displaystyle \int G\left(x-x_0\right)\frac{2m}{{\hslash }^2}V\left(x_0\right)\psi \left(x_0\right)d{x}_0}$

Put equation (V) in the above equation.

  $\psi \left(x\right)=\frac{-i}{2k}\frac{2m}{{\hslash }^2}{\displaystyle \int e^{ik\left|x-x_0\right|}V\left(x_0\right)\psi \left(x_0\right)d{x}_0}$        (VI)

Write the value of $\left(\frac{d^2}{d{x}^2}+k^2\right){\psi }_0\left(x\right)$ for any solution ${\psi }_0\left(x\right)$ to the homogeneous Schrodinger equation.

  $\left(\frac{d^2}{d{x}^2}+k^2\right){\psi }_0\left(x\right)=0$

This implies ${\psi }_0\left(x\right)$ can be added to equation (VI).

Rewrite equation (VI) by adding ${\psi }_0\left(x\right)$ .

  $\psi \left(x\right)={\psi }_0\left(x\right)-\frac{-i}{2k}\frac{2m}{{\hslash }^2}{\displaystyle \int e^{ik\left|x-x_0\right|}V\left(x_0\right)\psi \left(x_0\right)d{x}_0}\frac{-i}{2k}\frac{2m}{{\hslash }^2}{\displaystyle \int e^{ik\left|x-x_0\right|}V\left(x_0\right)\psi \left(x_0\right)d{x}_0}$

Conclusion:

Therefore, the Green’s function for the one-dimensional Schrodinger equation is $G\left(x\right)=-\frac{-i}{2k}{e}^{ik\left|x\right|}$ and the integral form is  $\psi \left(x\right)={\psi }_0\left(x\right)-\frac{im}{{\hslash }^{2}k}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{ik\left|x-x_0\right|}V\left(x_0\right)\psi \left(x_0\right)d{x}_0}$.