QM: Ground State Wave Function of Infinite Square Well

[zheng89120]

Homework Statement

I was wondering why is the ground state wave function of a particle in an infinite square well allowed? If you drawn it out on a graph, it is one-half of a full sine wave.

But the conditions for an acceptable wave equation is one that is continuous (yes) and "smooth"(no!). How is the ground state allowed, when it "breaks" at the wells?

Homework Equations

The Attempt at a Solution

 

[diazona]

The wavefunction only has to be smooth when the potential is finite.

For a non-rigorous justification, take a look at the time-independent Schrödinger equation:
$$\frac{1}{2m}\nabla^2\psi = (E - V)\psi$$
If the potential V is infinite, then the second derivative of the wavefunction can also be infinite, which corresponds to a discontinuous first derivative, a.k.a. a non-smooth function.

 

[Mindscrape]

I wouldn't worry about it too much. We know that in the real world that infinite isn't ever applicable, it's really only used for approximations and teaching. The finite square well is nicely smooth and continuous (though the transcendental equations make for a non-analytical solution that wouldn't be such a good introduction to quantum).