# What does it mean when, in an infinite step, E=V?

Let's say that you have an step with infinite length, with potential $V_0$ at the step, and the step occurring at $z=0$.

$$V(z) = \begin{cases} 0, & z < 0 \\ V_0, & z \geq 0 \\ \end{cases}$$

We know what happens when $E>V_0$ and when $E<V_0$ at $z>0$, but what happens when $E=V_0$?

The wave function $\psi$ turns out to be linear when $z>0$

\begin{align} -\frac{\hbar ^2}{2m}\frac{\mathrm{d}^2 \psi}{dz^2} + V_0\psi &= E\psi \\ -\frac{\hbar ^2}{2m}\frac{d^2 \psi}{dz^2} &= (E-V_0)\psi \\ -\frac{\hbar ^2}{2m}\frac{d^2 \psi}{dz^2} &= 0 \\ \frac{d^2 \psi}{dz^2} &= 0 \\ \psi(z) &= Az+B \end{align}

What does this linearity of the wave function imply about what is happening to the particle when $z>0$? Is it possible to find transmission or reflection coefficients when $z>0$ and $E=V_0$?

What about if the step wasn't infinite and had length $L$? Could you get reflection and transmission coefficients out of it? $$V(z)=\begin{cases} 0, & z < 0 \\ V_0, & 0 \leq z \leq L \\ 0, & z > L \\ \end{cases}$$

If you think about it a little, you should be able to argue that $A=0$, since otherwise $|\psi|^2$ blows up as $z\rightarrow\infty$, which makes no physical sense.
It then should be fairly easy to show that the only way to satisfy the continuity equations at the boundary is for $B=0$ as well. So the transmission coefficient is zero.
This is the same as you get if you assume $E>V_0$ or $E <V_0$ and take the limit as the difference goes to zero.
With a finite barrier, there's not really anything special about the $E=V_0$ case. The wavefunction is linear instead of exponential or sinusoidal, but there's no special interpretation of that. It does make the math a little easier.
You can get reflection and transmission coefficients out of it the same way you would if $E\ne V_0$, although the math is a little tedious.