# Direction of propagating wave in quantum barrier

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. 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:

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

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.

• @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. Apr 18, 2019 at 13:31