Scattering from a 1D Potential Well *

$$\bgroup\color{black} V(x)=\left\{ \matrix{0 & x<-a \cr -V_0 & -a<x<a \cr 0& x>a\cr} \right. \egroup$$

Numbering the three regions from left to right,

$$\begin{eqnarray*} u_1(x)=e^{ikx}+Re^{-ikx} \\ u_2(x)=Ae^{ik'x}+Be^{-ik'x} \\ u_3(x)=Te^{ikx} \\ \end{eqnarray*}$$

Again we have assumed no wave incident from the right (but we could add that solution if we wanted).

We now match the wave function and its first derivative at the two boundaries yielding 4 equations. That's good since we have 4 constants to determine. At $x=a$ we have 2 equations which we can use to eliminate $A$ and $B$.

$$\begin{eqnarray*} Te^{ika}=Ae^{ik'a}+Be^{-ik'a} \\ ikTe^{ika}=ik'Ae^{ik'a}-ik'B... ...t)\\ Be^{-ik'a}={1\over 2}Te^{ika}\left(1-{k\over k'}\right)\\ \end{eqnarray*}$$

At $x=-a$ we have 2 equations which can now be written in terms of $R$ and $T$ by using the above.

$$\begin{eqnarray*} e^{-ika}+Re^{ika}=Ae^{-ik'a}+Be^{ik'a} \\ ike^{-ika}-ikRe^{ik... ...\right)e^{-2ik'a}-\left({k'\over k}-1\right)e^{2ik'a}\right] \\ \end{eqnarray*}$$

We can add equations to eliminate $R$.

$$\begin{eqnarray*} e^{-ika}+Re^{ika}={1\over 2}Te^{ika}\left[\left(1+{k\over k'}\... ...-2ika}\over 2kk'\cos(2k'a)-i\left(k^2+k'^2\right)\sin(2k'a)} \\ \end{eqnarray*}$$

We can subtract the same equations to most easily solve for $R$.

$$\begin{eqnarray*} 2Re^{ika}={1\over 2}Te^{ika}\left[\left({k\over k'}-{k'\over k... ...-2ika}\over 2kk'\cos(2k'a)-i\left(k^2+k'^2\right)\sin(2k'a)} \\ \end{eqnarray*}$$

We have solved the boundary condition equations to find the reflection and transmission amplitudes

$$\begin{eqnarray*} R&=&ie^{-2ika}{(k'^2-k^2)\sin(2k'a)\over 2kk'\cos(2k'a)-i(k'^2... ...=&e^{-2ika}{2kk'\over 2kk'\cos(2k'a)-i(k'^2+k^2)\sin(2k'a)} .\\ \end{eqnarray*}$$

The squares of these give the reflection and transmission probability, since the potential is the same in the two regions.