Given the step potential

$$V(x)=\begin{cases} 0~~~~~~~~\text{if }~~x \leq 0 \\ V_0~~~~~~\text{if }~~x > 0 \end{cases}$$

Consider the case where $E < V_0$. In this region $x \leq 0$ we have the time-independent Schrodinger equation given as $$\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E \psi \implies \psi(x) = Ae^{ikx} + Be^{-ikx}~~~\text{where }~~k = \frac{\sqrt{2mE}}{\hbar}.$$ For the case where $x > 0$ we have $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V_0\psi = E \psi$$ and the general solution is $$\psi(x) = Fe^{-lx} + Ge^{lx}$$ where $l \equiv \frac{\sqrt{2m(V_0-E)}}{\hbar}$. At this point I want to confirm that we can take $G=0$ since when we restrict this scaterring problem to the region $x > 0$, we actually have a bounded state problem because in this region $E < V_0$ and in a bounded state $\psi$ must be bounded. Hence in the region $x > 0$ we have $$\psi(x) = Fe^{-lx}.$$

Is this reasoning fine?

  • $\begingroup$ i think it is due to physical nature of the wave function in the region -its modulus square represents the transmitted probability of finding the particle and wave function must go to zero as x goes to infinity- all physical solutions should vanish at infinite distances-otherwise one can have sources or sinks of particles and total probability will not be finite. $\endgroup$
    – drvrm
    Mar 10, 2016 at 13:51
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Gert
    Mar 10, 2016 at 16:47
  • $\begingroup$ @Gert This link does not really answer my question. It just states that there is exponential decay, which I already have. But what I want to know is are we taking $G=0$ for the reason I gave above? Which is, for the case $E < V_o$ if we consider only $x > 0$ region, then we can treat it as a bounded state and hence we require that $\psi$ is bounded and so $G = 0$ for $\psi$ to be bounded. $\endgroup$
    – user100411
    Mar 10, 2016 at 21:54

2 Answers 2


Yes you are long as the incident beam of electrons is not incoming from the right

Because the energy is greater than the potential we do not expect the particle to penetrate inside the x<0 region. This means the particle cannot exist on the right because the kinetic energy is not enough to overcome the potential.

When you get the solution

$\psi(x) = Fe^{-lx} + Ge^{lx}$

its a basic postulate of quantum mechanics that bound states must approach 0 as x goes to infinite and this can be proved mathematically. Remember that any smooth function that fails to approach zero at infinity will obviously have an infinite integral and the wavefunction must have an integral of 1.

so G must be 0 so that $Fe^{-lx} + Ge^{lx}$ does not go to infinity as x approaches infinite


This is a potential barrier, so It doesn't support bound states. $\textbf{We can forget the bound state analysis for this potential.}$

Only scattering states are possible here.

$\textbf{Scattering states:}$

On consider the scattering from left, wavefunction is given as,

$$ \psi = \begin{cases} A e^{ik x} + Be^{-ikx} \text{ for } x \leq 0 \\ Fe^{i\tilde{l}x} + Ge^{-i\tilde{l}x} \text{ for } x\gt0 \end{cases} $$ where $\tilde{l}=\sqrt{2m(E-V_0)/\hbar}$.

There is no obstacle to reflect the particles at right side ($x>0$), Hence Reflection coefficient $G$ should be zero.

$\textbf {A particular case of scattering}$, i.e. $E<V_0$,

For this particular case ($E<V_0$) of scattering, the scattering wavefunction in second region ($\psi=F e^{i\tilde{l}x}$) becomes a tunneling wave function as $\psi=F e^{-lx}$, where $\tilde{l}=il$.


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