# 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?

• In first it's a theoretical exercise and then some times an approximation. But if you read the wikipedia article it notices some applications. It doesn't have to be a perfect description of the reality to use it if it can give you good results. For example, if I'm not mistaken, one of the first potentials used to describe the nuclei was the well potential. Commented May 11, 2015 at 13:50
• so in reality, would it be slightly curved? I mean in physical applications? Commented May 11, 2015 at 13:59
• Related: physics.stackexchange.com/q/1324/2451 , physics.stackexchange.com/q/105318/2451 and links therein. Commented May 11, 2015 at 14:52

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.

• Very nice explanation! Commented May 11, 2015 at 18:21