Consider a step potential $V(x)$ where \begin{align} V(x) & = 0; \quad x\leq 0 \\ V(x) & = V_0; \quad x> 0 \end{align} Now consider the case where $E_0<V_0$. The solutions of the time independent Schrodinger equation are: \begin{align} \psi_L(x) & = A e^{i\theta x} + B e^{-i\theta x} \\ \psi_R(x) & = C e^{K x} + D e^{-K x} \end{align}

I understand that for $x > 0$, the amplitude $C =0$ as when $x \to\infty$ then the wavefunction goes to zero. However for $x \leq 0$, the wavefunction as I understand it is made up of an incident and reflected wave. I am confused how to understand which one is the incident and which is reflected.

Any help would be appreciated.

EDIT: enter image description here

Here is an extract from my notes which clarify that $$ A e^{i\theta x}$$ term is the one which is the incoming wave but i am still unsure why this is the case


1 Answer 1


It is not possible to tell in which direction the waves are moving without looking at the time dependence. According to the Schrodinger equation the time dependence of $\Psi_L$ is given by:

$$\Psi_L= A e^{i(\theta x - \omega t)} +B e^{i(-\theta x - \omega t)},$$

with $\omega = \frac{\hbar \theta^2}{2m}$. To see how the first term behaves, set B to $0$. We are left with a function which at $t=0$ has a constant modulus and a phase that varies in space. Pick your favorite point on this curve, e.g., $\Psi_L=A$ and analyze how the $x$ value for which $\Psi_L=A$ moves as we increase t.

For $t=0$ we obtain $\Psi_L=A$ when $x=0$ (there are many other solutions which we ignore). As $t$ increases, the $x$ value for which $\Psi_L=1$ increases:

$$\Psi_{L,B=0}=A$$ $$ A e^{i(\theta x - \omega t)} =A $$ $$ \theta x - \omega t = 0 $$ $$ x_{\Psi_L=A} = \frac{\omega}{\theta} t$$

We see that the point for which $\Psi_L=A$ moves to the right as time progresses. It does so with a velocity $v=\frac{\omega}{\theta}$ We might have chosen any other point and would have obtained the same velocity. This is why we say the wave travels to the right.

For the second term of $\Psi_L$, a similar analysis will show a left moving wave.

In general any function $f(ax+bt)$, with $a$ and $b$ real and positive, describes a left moving function, while any function $f(ax-bt)$ describes a right moving function.

  • $\begingroup$ could you please elaborate $\endgroup$
    – DJA
    Apr 16, 2019 at 15:12
  • $\begingroup$ @DavidAbraham I updated the answer $\endgroup$
    – Crimson
    Apr 16, 2019 at 18:58
  • $\begingroup$ I am unsure if that is correct since i know that the A e^i(θx+ωt) is the term that is moving to the right (i.e incident upon the barrier) $\endgroup$
    – DJA
    Apr 16, 2019 at 23:13
  • $\begingroup$ Please see the extract from my notes which clarify what i am saying. $\endgroup$
    – DJA
    Apr 16, 2019 at 23:38
  • 1
    $\begingroup$ @DavidAbraham You should look at the time-dependent schrodinger equation. There is only one sign of $\omega$ for which the time-dependent schrodinger equation is obeyed. $\endgroup$
    – Crimson
    Apr 18, 2019 at 13:31

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