91Ó°ÊÓ

For the spherical wave, the surface probability density decreases based on the inverse square law. Which means the wave function must decrease as . The probability of scattering depends on both the polar and azimuthally angles and , so the outgoing wave function is:

$Af\left(\mathrm{θ},\mathrm{ϕ}\right)e^{ikr/r}$

Where, $f$is the scattering amplitude. The complete wave function is therefore:

$\mathrm{ψ}\left(r,\mathrm{θ},\mathrm{ϕ}\right)=A\left(e^{ikz}+\left(\mathrm{θ},\mathrm{ϕ}\right)\frac{e^{ikr}}{r}\right)$

If the potential of the scattering target is independent of $\mathrm{Ï•}$, then the probability cannot depend on $\mathrm{Ï•}$, but it still depends on the polar angle, since the direction of the incoming particles violate the symmetry of the polar angle. In this case, the complete wave function is for the two dimensional. For two dimensions, the outgoing wave function is a circular wave and the probability density falls off according to $1/r$so the outgoing wave function must fall off according to $1/\sqrt{r}$, so we have:

$\mathrm{ψ}\left(z,\mathrm{θ}\right)=A\left(e^{ikz}+f\left(\mathrm{θ}\right)\frac{e^{ikr}}{\sqrt{r}}\right)$

In one-dimensional scattering, the scattered particle can either continue on in the same direction or scatter directly backwards, so the wave function will be:

$\mathrm{ψ}\left(z\right)=A\left(e^{ikz}+f\left(\frac{z}{\left|z\right|}\right)e^{ikz}\right)$

Therefore, the graph was plotted.