How can a Step potential exist if potentials are continuous? Quantum Mechanics I was doing an example which showed step potentials, I then researched it a little and found this on wiki http://en.wikipedia.org/wiki/Solution_of_Schr%C3%B6dinger_equation_for_a_step_potential
I don't understand how a step potential can exist, I thought all fields are continuous, how could such a sharp discontinuous line exist? or am I missing something?
 A: Well, technically nothing in the Schrödinger theory exists, since the Schrödinger equation is not relativistic. But it's a useful approximation.
Similarly, there are lots of continuous functions which approximate the step function.  Consider the step function
\begin{align}
\Theta(x) &= \frac 12 \left(
1 + \tanh \frac x{x_0}
\right)
= \frac{e^{x/x_0}}{e^{x/x_0} + e^{-x/x_0}}
= \left(
1 + e^{-2x/x_0}
\right)^{-1}
%\\
%\theta(x) &= \lim_{x_0\to0} \Theta(x)
\end{align}
We can quantify the limiting behavior of this function,
\begin{align}
-x &\gg x_0:
&
\Theta(x) &\approx e^{2x/x_0}
\\
x &\gg x_0: 
&
\Theta(x) &\approx 1 - e^{-2x/x_0}
\end{align}
and define a Heaviside step function
$$
\theta(x) = \lim_{x_0\to0} \Theta(x).
$$
If you solved the Schrödinger equation for such a potential, you would end up with solutions having some momentum $p = \hbar k = h / \lambda$ corresponding to some wavelength $\lambda$.  And — here's the important bit — as long as $\lambda$ is much larger than the length scale $x_0$ over which the curvature of the potential is detectable, you'll end up with approximately the same dynamics. If the wavelength of a particle is much larger than the length scale of the potential, then the potential doesn't matter.
Furthermore, the exact Heaviside step function gives potentials that are analytically solvable.  I don't know whether that's the case for the $\tanh$ step function I gave above.  There are lots of other possible step functions, too.  Here's another one that probably doesn't admit analytic solutions:
$$
\Theta(x) = \frac12 + \frac1\pi \arctan \frac x{x_0}
$$
But because long-wavelength solutions will smooth over any details in the step, it's not necessary to go through the struggle of numerically comparing the arctangent to the hyperbolic tangent. We can skip straight to the step function and get a useful answer.
