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 $V=0$?

  • 1
    $\begingroup$ Note that, in either case, the coefficient can be imaginary and so one could, for example, stipulate that $\alpha = ik_1$ and then write $\psi_C(x) = B_re^{\alpha x}+B_le^{-\alpha x}$. $\endgroup$ Nov 20, 2017 at 15:51
  • $\begingroup$ 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". $\endgroup$
    – DanielC
    Nov 20, 2017 at 16:04
  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.html $\endgroup$
    – Gert
    Nov 20, 2017 at 17:08
  • $\begingroup$ Just remember the Euler's formula $\endgroup$
    – Ruslan
    Nov 23, 2017 at 20:42

1 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.)


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


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

  • $\begingroup$ 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. ;) $\endgroup$
    – Programmer
    Nov 23, 2017 at 15:25
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Nov 23, 2017 at 20:23

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