# Solving schrodinger, reflection coefficient

by SoggyBottoms
Tags: coefficient, reflection, schrodinger, solving
 P: 61 Consider the potential $$V(x) = \begin{cases} 0, & x < -a & (I) \\ +W, & -a < x < a & (II) \\ 0, & x > a & (III) \end{cases}$$ for a particle coming in from the left ($-\infty$) with energy E (0 < E < W). Give the solution to the Schrodinger equation for I, II and III and use these to calculate the reflection coefficient. I have the answer to this problem in front of me, but I don't understand. First they calculate the solution to the Schrodinger equation for I, II and III: $\psi_I(x) = Ae^{ikx} + Be^{-ikx}, \ \mbox{with} \ k = \frac{\sqrt{2mE}}{\hbar}$ $\psi_{II}(x) = Ce^{\kappa x} + De^{-\kappa x}, \ \mbox{with} \ \kappa = \frac{\sqrt{2m(E - W)}}{\hbar}$ $\psi_{III}(x) = Fe^{i k x}, \ \mbox{with} \ k = \frac{\sqrt{2mE}}{\hbar}$ I understand $\psi_I$, but not $\psi_{II}$ and $\psi_{III}$. Why is there no i in $\psi_{II}$? And why is $\psi_{III}$ only a single term? I imagine it has something to do with the particle coming from the left?
P: 2,193
 Quote by SoggyBottoms Why is there no i in $\psi_{II}$?
Because these are classically forbidden solutions, and the probability of the particle being found in such a region exponentially decays.

 And why is $\psi_{III}$ only a single term? I imagine it has something to do with the particle coming from the left?
Correct, the other term would represent some particle coming in from the right (which is usually not assumed).
P: 61
 Quote by Nabeshin Because these are classically forbidden solutions, and the probability of the particle being found in such a region exponentially decays.
If I were to solve the equation for II, then $Ce^{i \kappa x} + De^{-i \kappa x}$ is mathematically still a valid solution right? I don't understand why it is not allowed.

 Quote by Nabeshin Correct, the other term would represent some particle coming in from the right (which is usually not assumed).
But why do I and II have two terms if the second term represents a particle coming from the right? Shouldn't they have only one term too then?

 Quote by SoggyBottoms If I were to solve the equation for II, then $Ce^{i \kappa x} + De^{-i \kappa x}$ is mathematically still a valid solution right? I don't understand why it is not allowed.
Nope. You should go through and solve why these are the solutions here, but essentially you have something like (don't quote me exactly on this) $\kappa \sim \sqrt{E-V}$ So in the regions I and III, V=0 and this is real and fine. But in the region where V=V0>E, this is imaginary, which is why the thing you would normally write, e^(i*k*x), goes to e^(-kx) (Where now k in this expression is understood to be the real part). [I think there is a typo in what you wrote somewhere, and the sign inside your definition of kappa should be switched so that it is by definition real]