Complex exponential eigenfunction blow-up

Question

Suppose $\phi_{p}$ is a square integrable eigenfunction such that $\phi_{p}=e^{ipx/\bar{h}} $where p is real.

If p is complex, there is a blow up as x tends to $\pm \infty$.

I tried to show this but was unable to.

Attempt:

Suppose p=a+ib $\forall a,b \in \mathbb{R}$.

Then, $\phi_{p}=e^{a+ix/\bar{h}}\cdot e^{-bx/\bar{h}}$

The second term decays exponentially. The first term oscillates between -1 and 1. What am I missing?

Any help is appreciated.

Answer 1

First of all, you didn't do the multiplication correctly.

$\phi_p=e^{(a+ib)ix/\bar{h}}=e^{aix/\bar{h}}\cdot e^{-bx/\bar{h}}$. As $x\rightarrow \infty$, $\phi_p$ decays but as $x\rightarrow-\infty$, $\phi_p$ blows up. That is assuming that $b>0$. However, we could have $b<0$ which would mean that $\phi_p$ blows up as $x\rightarrow \infty$ and $\phi_p$ decays as $x\rightarrow -\infty$.