Question regarding the solution of Schrödinger equation for finite potential well and quantum barrier

Question

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 $V=0$?

Answer 1

It is two different situations of the TISE$^1$:

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.

2. 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:

1. 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 $|x|\to \infty$.