Examples of Basic Potential functions

[DaTario]

Hi All,

In teaching the basics of quantum mechanics, one has often to introduce potential functions such as the step, the barrier and the well. When I try to get some example of the physical environment of a particle that could correspond to a step function, for instance, what comes out is always the analogy with the geometric profile of a real ladder´s step. The energy is put in the form of potential gravitational energy and the analogy seems to produce some good effect in the class. But if a particle with a velocity v bumps on a vertical wall of a step, even if it has enough energy to climb up a corresponding inclined profile, it will simply bounce back.

Why this discontinous version of the step function, when treated as a potential function, produces such difficulties and what could be a better example of a real situation for a particle to experience a step potential?

Best wishes,

DaTario

 

[vanhees71]

Well, I'm also a bit sceptical about these very unrealistic exercises in solving the time-independent Schrödinger equation. They are nice examples to teach the calculational techniques and the meaning of scattering and bound-state solutions, but they are also somewhat misleading.

This is most easily revealed by asking the students to first treat the problem within classical mechanics. Then you see that it is pretty artificial. Take a potential step
$$V(x)=\begin{cases} 0 &\text{for} \quad x<0 \\
V_0 & \text{for} \quad x>0. \end{cases}$$
If you'd try to solve Newton's equation of motion, you get a pretty complicated problem involving a $\delta$ distribution,
$$m \ddot{x}=-V'(x)=-V_0 \delta(x).$$
You can solve it, however, by starting from the energy-conservation law, i.e., using a first integral
$$\frac{m}{2} \dot{x}^2+V(x)=E=\text{const}.$$
Then you see that, if the particle starts somewhere at $x<0$, it has a velocity $$v_0=\pm \sqrt{2m E}$$ and for $v_0<0$ runs with this velocity to the left forever. For $v_0>0$ you must disinguish the cases $E<V_0$ (the particle is reflected on the potential and running to the left with the velocity $-v_0$ from this moment on) or $E>V_0$ (then the particle is running through and then runs with the lower velocity $v_1 =\sqrt{2m(E-V_0)}$ to the right.

Now you can solve the time-independent Schrödinger equation and demonstrate some general features of quantum theory like the tunnel effect, how to built wave packets out of the energy eigenfunctions, and so on.

There are some better examples of exactly solvable problems that are only slightly more complicated but more realistic. The only problem is that you usually need special functions:

https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions

 

[DaTario]

Thank you.

I have tried to suggest a fast interaction capable of subtracting a definite amount of velocity from the particle. But to be honest, these system would have to store the energy that have been stolen from the particle, for if the particle is in a barrier, for instance, this energy will have to be returned.

The part involving the delta I also used. It is more or less the same as saying that we have an instantaneous interaction as above described.
It seems that the analogy with the incline is the best one. I only have to turn the "vertical wall" of the step function smoother, something like a sigmoidal curve.

 

[jtbell]

When I try to get some example of the physical environment of a particle that could correspond to a step function, for instance, what comes out is always the analogy with the geometric profile of a real ladder´s step. The energy is put in the form of potential gravitational energy and the analogy seems to produce some good effect in the class. But if a particle with a velocity v bumps on a vertical wall of a step, even if it has enough energy to climb up a corresponding inclined profile, it will simply bounce back.

Think of the step function as an idealization of a steep slope which rises during a short horizontal distance. Going further, round off the "corners" slightly so dV/dx is continuous everywhere. Similarly for the barrier and the well.