11.3: Scattering in one dimension- Step function
( \newcommand{\kernel}{\mathrm{null}\,}\)
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=V0.
Assuming the particle energy E>V0, this is simply the free particle problem, the spatial solution to which is:
Φ=A exp(ikx)+B exp(−ikx)(x<0);Φ=C exp(ik′x)+D exp(−ik′x)(x>0)
where k=√2mE/ℏ and k′=√2m(E−V0)/ℏ
Figure 11.3.1: Scattering at a step function.
From the boundary condition that all particles start from x=−∞, we can immediately set D=0.
From the condition of continuity of Φ and dΦ/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
ℏkmA2(k−k′k+k′)2
and the transmitted flux is
ℏk′mA2(2kk+k′)2
Note that A2≠B2+C2. 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 V0 is negative, the transmitted flux gets smaller as |V0| 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<V0. Now the square root is imaginary and Φ(x>0)=Ce−κ′x where we define a real quantity κ′=ik′=√2m(V0−E)/ℏ. The boundary conditions are then A+B=C and ik(A−B)=κ′C, which gives the reflected amplitude B/A=(ik−κ′)/(ik+κ′) and the transmitted amplitude Cκ′/Ak=2ik/(ik+κ′).
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.