# 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 0a. \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 $$V=0$$?

• Note that, in either case, the coefficient can be imaginary and so one could, for example, stipulate that $\alpha = ik_1$ and then write $\psi_C(x) = B_re^{\alpha x}+B_le^{-\alpha x}$. Nov 20, 2017 at 15:51
• The Ansatz is $\psi(x) = A~e^{\alpha x}$ for general (non-zero) complex A and $\alpha$. The exact form of $\alpha$ and A are determined from "boundary conditions". Nov 20, 2017 at 16:04
• hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.html
– Gert
Nov 20, 2017 at 17:08
• Just remember the Euler's formula Nov 23, 2017 at 20:42

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

• Thanks, but I have further questions about why the different results Schrodinger equation produce for the region V=0 for finite potential well and potential barrier. If you know about it, please consider answering, it would be very useful. Thanks very much. ;) Nov 23, 2017 at 15:25
• I updated the answer. Nov 23, 2017 at 20:23