91Ó°ÊÓ

We want to find Green's function for the one-dimensional Schrodinger equation. Equation 11.52 is given by (Helmholtz equation):

$\left(\frac{d^2}{d{x}^2}+K^2\right)G\left(x\right)=\mathrm{δ}\left(x\right)$…â¶Ä¦.(1)

The equation 11.54 is given by:

$G\left(x\right)=\frac{1}{\sqrt{2\mathrm{Ï€}}}∫e^{isk}g\left(s\right)ds$…â¶Ä¦â¶Ä¦(2)

We want to substitute with this equation into the first one, but that require us to find,

$\frac{d^2}{d{x}^2}G\left(x\right)=\frac{1}{\sqrt{2\mathrm{π}}}∫-s^2{e}^{isk}g\left(s\right)ds$

Now substitute into the first one we get:

$$\begin{gathered} \left(\frac{d^2}{d{x}^2}+k^2\right)G=\frac{1}{\sqrt{2\mathrm{π}}}∫\left(-s^2+k^2\right)g\left(s\right)e^{isk}ds \\ \left(\frac{d^2}{d{x}^2}+k^2\right)G=\mathrm{δ}\left(x\right) \end{gathered}$$

But,

$\mathrm{δ}\left(x\right)=\frac{1}{2\mathrm{π}}∫e^{is-r}ds$

Thus,

$$\begin{gathered} \frac{1}{\sqrt{2\mathrm{π}}}∫\left(-s^2+k^2\right)g\left(s\right)e^{isx}ds=\frac{1}{2\mathrm{π}}∫e^{isx}ds \\ \frac{1}{\sqrt{2\mathrm{π}}}∫\left(-s^2+k^2\right)g\left(s\right)e^{isx}ds\frac{1}{2\mathrm{π}}∫\frac{1}{2\mathrm{π}}{e}^{isx}ds \end{gathered}$$

From this equation we can see that:

$$\begin{gathered} \left(-s^2+k^2\right)g\left(s\right)=\frac{1}{\sqrt{2\mathrm{Ï€}}} \\ g\left(s\right)=\frac{1}{\sqrt{2\mathrm{Ï€}}\left(k^2-s^2\right)} \end{gathered}$$…â¶Ä¦(3)

Substitute from (3) into (2) we get:

$G\left(x\right)\frac{1}{2\mathrm{π}}{∫}_{-∞}^{∞}\frac{e^{isk}}{k^2-s^2}ds$

Where the two integrals on the right are integrals around the little semicircles at $s=Â\pm k$. Note that the semicircle at $-k$is traversed in a clockwise direction, while that at $+k$is counterclockwise as shown in the following figure. For $x>0$, close above:

$$\begin{gathered} G\left(x\right)=-\frac{1}{2\mathrm{Ï€}}âˆ\circledR \left(\frac{e^{isx}}{s+k}\right)\frac{1}{s-k}ds \\ =-\frac{1}{2\mathrm{Ï€}}2\mathrm{Ï€}i{\overline{)\left(\frac{e^{isx}}{s+k}\right)}}_{8-k} \\ =i\frac{c^{ikx}}{2k} \end{gathered}$$

And for $x<0$, close below:

$$\begin{gathered} G\left(x\right)=+\frac{1}{2\mathrm{Ï€}}âˆ\circledR \left(\frac{e^{isx}}{s-k}\right)\frac{1}{s+k}ds \\ =-\frac{1}{2\mathrm{Ï€}}2\mathrm{Ï€}i{\overline{)\left(\frac{e^{isx}}{s-k}\right)}}_{s-k} \\ =i\frac{e^{ikx}}{2k} \end{gathered}$$

So, for both cases the results are the same, that is:

$G\left(x\right)=-\frac{i}{2k}{e}^{ik\left|x\right|}$…â¶Ä¦(4)

The integral form of the Schrodinger equation is given by:

$\mathrm{ψ}\left(x\right)=\frac{2m}{{\sout{h}}^2}∫G\left(x-x_0\right)V\left(x_0\right)\mathrm{ψ}\left(x_0\right)d{x}_0$

Substitute from (4) we get:

$\mathrm{ψ}\left(x\right)=-\frac{i}{k}\frac{m}{{\sout{h}}^2}∫e^{ik\left|x-x_0\right|}V\left(x_0\right)\mathrm{ψ}\left(x_0\right)d{x}_0$

This is the solution of the non-homogeneous Schrodinger equation, the complete solution is this solution plus any solution ${\mathrm{ψ}}_0\left(x\right)$to the homogeneous Schrodinger equation, that is:

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

So, the solution is$\mathrm{ψ}\left(x\right)={\mathrm{ψ}}_0\left(x\right)-\frac{im}{{\sout{h}}^{2}k}{∫}_{-∞}^{∞}{e}^{ik\left|x-x_0\right|}V\left(x_0\right)\mathrm{ψ}\left(x_0\right)d{x}_0$.