# Finite square wall with $E > V_0$

I'm working through a problem for homework and feel as if there is a typo or I am confused. The problem is with a one sided finite square wall such as this: So the energy is more than $V_0$. I'm trying to show that the wave function for x > 0 is equal to $Ae^{ikx}$ but I feel like that is a typo. I got to the solution being $Ae^{ikx} + Be^{-ikx}$ from solving Schrödinger's equation but I'm not sure how to remove the second part of this.

The problem is considering an incoming right-mover for $x<0$ and asks how it scatters off a step potential into a reflected outgoing left-mover for $x<0$ and a transmitted outgoing right-mover for $x>0$.

The last possibility -- an incoming left-mover for $x>0$ -- is not present in this scattering experiment. That's the answer to OP's question.

By the way, if it sounds strange that we can identify solutions of the time-independent Schrödinger equation (TISE) as incoming and outgoing movers which move in time, check out this Phys.SE post.

• This makes sense for the most part. But what do you mean by right/left-mover? Does that just mean the particle being measured is coming from the left to the right (and vice versa)? – TheStrangeQuark Oct 7 '15 at 19:15
• $\uparrow$ Yes. – Qmechanic Oct 7 '15 at 19:17

Note as x approaches +∞ the wavefunction vanishes. Apply $x=+∞$ boundary condition.

$Ae^{-ik(+∞)}$+$Be^{ik(+∞)}$= 0 This will apply only and only if B=0!!!

Please check on how to apply boundary conditions and understand the implied meaning

• Don't use capital letters; rather it spoils the readibility. Also, this is a mathjax-enabled site; take the advantage of it. – user36790 Oct 7 '15 at 7:48
• The wavefunction does not vanish with $\lim\limits_{x\rightarrow\infty}$. This is only true if $E<V_0$ – Jannick Oct 7 '15 at 9:29

This is a scattering problem and is therefore somewhat different from the bounded solution problems (e.g. potential well, harmonic oscillator) you may have encountered. In this case we want to model a wave function that travels from the left to the potential barrier. Part of it is scattered back towards the left and part of it is transmitted through the barrier and travels towards the right. These are our boundery conditions, which translate to the ansatz $$\psi_l = A_l \exp(-i k_l x) + B_l \exp(i k_l x)$$ $$\psi_r = A_r \exp(-i k_r x)$$ So the answer to your question is that one just sets the left-moving part on the right side to zero. You can solve this by using that the wavefunction and its derivative are continuous at the potential well boundary ($x=0$)