Form of scattering solutions for a square potential barrier

Question

Consider an incoming beam of particles with $E>U$ scattering from the potential

$V(x) = \begin{cases} 0 & \text{ for } x<-a\\ U & \text{ for } |x|<a \\ 0 & \text{ for } x>a \end{cases}$

So Schrodinger's (time-independent) equation reads

$\psi'' = \begin{cases} -\frac{2mE}{\hbar^2}\psi & \text{ for } |x|>a\\ -\frac{2m}{\hbar^2}(E-U)\psi & \text{ for } |x|<a \end{cases}$

Then the scattering states are given by

$\psi(x) = \begin{cases} \exp(ikx) + R\exp(-ikx) & \text{ for } x<-a\\ A\exp(-lx) + B\exp(lx) & \text{ for } |x|<a \\ T\exp(ikx) & \text{ for } x>a \end{cases}$

where $k = \sqrt{\frac{2mE}{\hbar^2}}$, $l = \sqrt{-\frac{2m}{\hbar^2}(E-U)}$.

[The above is from my lecture notes].

Now my questions are:

1. From the form of the solution in $|x|<a$ it looks like we are assuming that $l\in\mathbb{R}$, i.e. that $E<U$. Why is that? Why can't we have a combination of sines and cosines?

2. There's a theorem that says that

Any solution of the time-independent Schrodinger equation with a symmetric potential is a linear combination of solutions of definite parity.

We clearly have a symmetric potential here, and for real $l$, $\exp(\pm lx)$ doesn't have definite parity (and as far as I know can't be written as a (finite) linear combination of functions of definite parity). So given this theorem, the 'assumption' that $l\in \mathbb{R}$ seems even stranger/less justified. Do we just throw this theorem away because we are dealing with scattering solutions?

What's going on? Someone please clarify:^)

Answer 1

1. You assume in your opening statement that $E>U$, then you SHOULD actually have a combination of sines and cosines for the middle part. It's only when $E<U$ is assumed that you have exponentials.
2. It's true that the exponentials on their own do not have a definite parity, however, you can rewrite them as linear combinations of sinh and cosh that do have definite parity.