Step Down Potential (Griffiths)

Find the reflection coefficient of a plane wave with energy $E>V$ on a “step potential” initially 0 that drops down to $-V_0$. How is this similar and different to a classical particle? 

Step Down Potential (Griffiths)

Consider a plane wave coming from the left towards the step. Since $E>0$ it is a wave that has two components: incoming component (particle is traveling to the right) and reflected component (particle is traveling to the left).  In this region $U =0$.

$ψ_1 (x)=A_{forward}*exp⁡(i k_1 x))+A_{backward}*exp⁡(-i k_1 x)$  with $k_1=sqrt{2mE}/hbar$

After the step, the wave function will have a different k, and also two components, one traveling forward (transmitted) and one traveling backward (reflected). In this region $U=-V0$

$ψ_2 (x)=B_{forward}*exp⁡(i_(k_2 ) x)+B_{back}*exp⁡(-i k_2 x)$ with $k_2=sqrt{(2m(E+V_0))}/hbar$

At the step (that happens let’s say at $x=0$) the continuity conditions on wave function need to be imposed. That is

$ψ_1 (0)=ψ_2 (0)$    and $(dψ_1 (0))/d x=(dψ_2 (0))/d x$

and $k_1 A_{forward}-k_1 A_{back}=k_2 B_{forward}-k_2 B_{back}$


-consider no particle is coming from the right side: $B_{back}=0$

-take amplitude of the incoming wave =1:  $A_{forward}=1$

The reflection and transmission coefficients are

$R=|ψ_{1back}/ψ_{1forw} |^2=|A_{back}/1|^2$   
and $T=|ψ_{2forw}/ψ_{1forw} |^2=|B_{forward}/1|^2$

Rewrite the above two continuity conditions as:

$1+√R=√(T )$   and $k_1 (1-√R)=k_2 √T$

So that

$√R=(k_1-k_2)/(k_1+k_2 )=(√E-√(E+V_0 ))/(√E+√(E+V_0 ))$     and $R=((√E-√(E+V_0 ))/(√E+√(E+V_0 )))^2$

Consider one has a classical particle with positive energy coming from the left towards right. Since $E>0$ and $V=0$, the kinetic energy will be positive $E k=E >0$.
Differences: the particle will always fall towards the right region and there will be no reflection of it (like in the quantum case).

Similarities: from the conservation of energy, it follows that after falling towards right (where $U=-V0$) the particle will have a bigger kinetic energy ($E k=E+V0$). This is similar with the quantum case when the wave number $k2$ in the right side is bigger than the wave number $k1$ in the left side. (The wave number k is proportional to the linear momentum of the particle).