Lets consider the potential: $$V(x) = \begin{cases} V_0&x>0 \\ 0 & x< 0 \end{cases}$$ and $E > V_0$ and at $x = 0$ the wave gets dispersed/scattered.

My deductions:

The solution to the schrödinger equation are should have the form: $\psi(x) = Ce^{kx}$, with $k_{1,2} =\sqrt{\frac{2m(E-V_0)}{\hbar^2}}, \sqrt{\frac{2mE}{\hbar^2}}$.

The boundary condition should look like this: $\psi_1(x) = \psi_2(x), \psi_1'(x) = \psi_2'(x)$ where $\psi_1(x)$ is the wavefunction for $x>0$ and $\psi_2(x)$ is the wavefunction for $x<0$.


Does the direction of the particle matter?

How do we decide which solution has a positive and a negative part or just one of the two?

How do we decide whether the solution should me made into a complex one (so which one "swings") ?

  • $\begingroup$ You have to write the wave function solutions of the Schrödinger equation not two square roots. $\endgroup$
    – freecharly
    Jan 22, 2018 at 2:24
  • $\begingroup$ @freecharly I worded it very badly, I'm sorry I'll edit it rn $\endgroup$
    – Leroy
    Jan 22, 2018 at 2:25

1 Answer 1


The square roots for the $k_{1,2}$ are $+$ or $-$. Therefore you have to take both waves (incoming and reflected) into account on the side of the incoming particle (usually from $x=-\infty$) and one transmitted wave on the other side. For $x \gt0$ when $E$ is larger than $V_0$ you can also have an incoming oscillatory wave from $x=+\infty$ with reflection and transmission.


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