11.3: Scattering in one dimension- Step function

Firstly, we review the problem of scattering by a step function in one dimension. Consider a particle moving from a region $(x < 0)$ where the potential is $V = 0$ to a region $(x > 0)$ where the potential is $V = V_0$.

Assuming the particle energy $E > V_0$, this is simply the free particle problem, the spatial solution to which is:

$$\Phi = A \text{ exp}(ikx) + B \text{ exp}(−ikx) \quad (x < 0); \quad \Phi = C \text{ exp}(ik'x) + D \text{ exp}(−ik'x) \quad (x > 0) \nonumber$$

where $k = \sqrt{2mE/}\hbar$ and $k' = \sqrt{2m(E − V_0)}/\hbar$

Figure $\PageIndex{1}$: Scattering at a step function.

From the boundary condition that all particles start from $x = −\infty$, we can immediately set $D=0$.

From the condition of continuity of $\Phi$ and $d\Phi /dx$ at $x = 0$ we also require $A + B = C$ and $k(A − B) = k' C$

This gives the reflected amplitude $B/A = (k − k')/(k + k')$ and the transmitted amplitude $C/A = 2k/(k + k' )$

The reflected flux is thus

$$\frac{\hbar k}{m} A^2 \left( \frac{k − k'}{k + k'}\right)^2 \nonumber$$

and the transmitted flux is

$$\frac{\hbar k'}{m} A^2 \left(\frac{2k}{k + k'}\right)^2 \nonumber$$

Note that $A^2 \neq B^2 + C^2$. The conserved quantity is the flux of particles, not the probability density. In this case the transmitted particles are moving more slowly than the incident ones.

Notice that if $V_0$ is negative, the transmitted flux gets smaller as $|V_0|$ gets larger: it is difficult to fall off a big cliff! This anomaly is due to the unphysical potential - the discontinuous first derivative at $x = 0$.

We have not considered the case of $E < V_0$. Now the square root is imaginary and $\Phi (x > 0) = Ce^{-\kappa' x}$ where we define a real quantity $\kappa' = ik' = \sqrt{2m(V_0 − E)}/\hbar$. The boundary conditions are then $A+B = C$ and $ik(A−B) = \kappa' C$, which gives the reflected amplitude $B/A = (ik−\kappa' )/(ik+\kappa')$ and the transmitted amplitude $C\kappa' /Ak = 2ik/(ik + \kappa' )$.

Now the reflected flux is equal to the incident flux, and although the wavefunction penetrates the region $x > 0$, it decays exponentially and there is no propagating wave.