Question regarding the solution of Schrödinger equation for finite potential well and quantum barrier When solving the Schrödinger equation for finite potential well, the solution outside of the well is
$$\psi _{1}=Fe^{{-\alpha x}}+Ge^{{\alpha x}}\,\!$$
and
$$\psi _{3}=He^{{-\alpha x}}+Ie^{{\alpha x}}\,\!. $$
However, when solving the Schrödinger equation for the quantum barrier, the solution of the regions are
\begin{align}
\psi _{L}(x)&=A_{r}e^{{ik_{0}x}}+A_{l}e^{{-ik_{0}x}}\quad x<0\\ 
\psi _{C}(x)&=B_{r}e^{{ik_{1}x}}+B_{l}e^{{-ik_{1}x}}\quad 0<x<a\\
\psi_{R}(x)&=C_{r}e^{{ik_{2}x}}+C_{l}e^{{-ik_{2}x}}\quad x>a.
\end{align}
The solutions for quantum barrier have the imaginary $i$ on the exponent of $e$ for example $A_{l}e^{{-ik_{0}x}}$.
But the solution for the finite potential well does not have the imaginary $i$ on the exponent $e$ for example $Fe^{{-\alpha x}}$.
So why is there an imaginary $i$ for quantum barrier problem while there isn't an imaginary $i$ for potential well problem when both solution of the problem is derived by solving the Schrödinger equation in the form of $(\left[{\frac {d^{2}}{dx^{2}}}\psi (x)\right]=b\psi (x))$ ?
I have a follow-up question:
Why is the region with potential $V=0$ in the finite potential well have a wavefunction of the form $$\psi _{2}=A\sin kx+B\cos kx\,\!$$
But for the quantum barrier, the regions with potential $V=0$ have a wavefunction of $$\psi _{L}(x)=A_{r}e^{{ik_{0}x}}+A_{l}e^{{-ik_{0}x}}\quad x<0\\ $$
So why does the Schrodinger equation produce different result for the region of $V=0$?
 A: It is two different situations of the TISE$^1$:

*

*A bound state has $E<0$ and the wave function
$$ \psi(x)~=~Ae^{-\kappa |x|} , \qquad \kappa~:=~\frac{\sqrt{-2mE}}{\hbar}~>~0, \tag{1}$$
decreases exponentially in the asymptotic regions $|x|\to \infty$. An exponentially decreasing wave function is the hallmark of negative kinetic energy, i.e. quantum tunneling into classically forbidden regions.


*A scattering state has $E>0$ and the wave function
$$ \psi(x)~=~A_+e^{ik x}+A_-e^{-ik x} , \qquad k~:=~\frac{\sqrt{2mE}}{\hbar}~>~0, \tag{2}$$
behaves oscillatory in the asymptotic regions $|x|\to \infty$. An oscillatory wave function is the hallmark of positive kinetic energy, i.e. classically allowed regions.
Or alternatively: Note that when the energy $E$ changes sign from negative to positive, then the square root $\kappa$ in eq. (1) becomes imaginary and can be identified with $\pm ik$ from eq. (2), cf. comments by Alfred Centauri & DanielC.
(By the way, there is another intimate relation between bound states & scattering states: If we analytically continue the real $k$ into the complex plane $\mathbb{C}$, then the scattering reflection & transmission coefficients will have poles at positions $k=i\kappa$ along the imaginary axis in the complex $k$-plane whenever $\kappa>0$ corresponds to one of the discrete bound states, cf. e.g. Ref. 1.)
References:

*

*P.G. Drazin & R.S. Johnson, Solitons: An Introduction, 2nd edition, 1989; Section 3.3.

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$^1$Short of gravity the potential function $V(x)$ is only physically relevant up to a constant. Let us here for simplicity adjust the constant, so that the potential $V(x)$ vanishes in the asymptotic regions, i.e. assume that $V(x)\to 0$ for $|x|\to \infty$.
